let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

Calculate P{X>Y}

On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.

Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become same color?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

The answer to the snake problem is YES! They all can become one color, to proof this it is sufficient to show that there exist one solution where all snakes are one color, e.g.. one red and one yellow snake meet, this will lead to 14 blue, 14 red, and 16 yellow snakes. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. QED

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

The answer to the snake problem is YES! They all can become one color, to proof this it is sufficient to show that there exist one solution where all snakes are one color, e.g.. one red and one yellow snake meet, this will lead to 14 blue, 14 red, and 16 yellow snakes. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. QED

Nope. One red and one yellow snake meeting will lead to 15 blue, not 14 blue. Compbanker was right earlier. You need a difference of 3 between 2 colours to make this possible.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

Calculate P{X>Y}

On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.

Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become same color?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Anyone figure out number 1 yet? I don't get how to do it and it's driving me crazy.

sonyyy:

What to have some fun? Try these two.

let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

Calculate P{X>Y}

On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.

Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become same color?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Anyone figure out number 1 yet? I don't get how to do it and it's driving me crazy.

sonyyy:

What to have some fun? Try these two.

let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

Calculate P{X>Y}

On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.

Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become same color?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Anonymous Monkey

1/13/16 - 1:10am

P(X>Y) =0 since Y>X a.s.

No it is not possible since you need total number to be even at least to have that happen.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Isn't the answer to number 2 no because there is an odd number of snakes? Unless there is a way to get it so there are 15 of each, which i can't figure out.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Probability questions for S&T interview (Originally Posted: 04/01/2008)

Hey all, I have an interview coming up for a trade support position at a small trade shop. I was told ahead of time that my interview will consist of "fit" questions along with some probability questions. Does anybody have any website recommendations to refresh my probability and stat. skills? And/or what type questions might I expect? I apologize if this has been discussed already on WSO, Thanks!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

go over expectations. I had one of the hardest interview probability questions and it was just expectation however it was near impossible to structure the answer. got it though after about 5 minutes of hard work and got the job. Richard Durret has a text called Probability: theory and example that is amazing. pick it up.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Got this two parter for Bear F.A.S.T. summer analyst spot:

a) Easy: How much would you pay to roll a single die, if you got $1 if you roll a 1, $2 for a 2, etc...? b) Harder: Okay, now how much would you pay if you were allowed to roll twice and take the higher of the two?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Ok obviously for (a) the answer is $3.50 because you just mult. each amount paid by 1/6 and add them, but for (b), if each role is valued at $3.50, one's instinct is to say $7.00 because your essentially being given two roles for the price of one, but I think that is the teaser answer. What did you say and do you know the process to take for the correct answer?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

i've heard a variation of b) from a friend who heard it from a friend at Merrill. it was the same thing except you dont get the higher of the 2, if you roll the 2nd time you get the payout of the 2nd roll, not the greater of the two, but i'll give yours a shot.

this may be the long way but i still believe its correct. draw out a decsion tree.

first node is 1st roll. consists of 6 branches, each corresponding to what you roll, each with a prob of 1/6 . at the end of each branch is another node for 2nd roll, each with 6 branches of its own, same as above.

you're gonna have 36 unique combinations. at the end of each combo, you take the meax of the first and 2nd roll. then multiply that by 1/6 twice, i.e.e 1/36.

sum all of those and theres youre total expected value and therefore how much you'd be willing to pay.

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Well I don't think for b it is meant you keep the higher of the two times two, simply you get to keep the higher of the two, so I would think the value of the 2nd roll has to be either the probability of rolling higher than expected return from first roll, or >3.50 (i.e. 4, 5, 6) and then if it rolls 1, 2, 3 value is 0. So given you have a 50% chance to roll higher, and then your return could be $1.50 on avg more (5 = avg of 4, 5, 6, 5 - 3.5 = 1.50), and then 50% to get 0, you would pay an extra $0.75 for the roll or $4.25 in total.

That is how I would break it down. Never got this or anything like it, so I am not sure at all.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

using my method its, which i am sure of, the answer is 4.47222. because i am bored and lame, i did it in excel, pm me if you want the excel file, hah...

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

NICE HAHAHA... But I just used logic, I had probability questions that I didn't get like that, but at the same time they like the overall thought process, I just made a mistake in one step.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I had an interview with a top BB and not only did they ask me what I would pay for 1 roll, they 2 rolls, then 3 rolls, then they asked what I would pay for 2 rolls should I be allowed to keep the maximum of the two rolls.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

There are 6 possible outcomes for the 1st roll (1, 2, 3, 4, 5, 6), each with a probability of 1/6. For each possible outcome, there is a corresponding expected value for the "better-of-two" rolls.

First roll outcome: 6 Conditional expected value: 6

First roll outcome: 5 Conditional expected value: (5/6)5 + (1/6)6 = 5.1667 (this is expected, i.e. >5)

First roll outcome: 4 Conditional expected value: (4/6)4 + (1/6)(5+6) = 4.5

First roll outcome: 3 Conditional expected value: (3/6)3 + (1/6)(4+5+6) = 4

First roll outcome: 2 Conditional expected value: (2/6)2 + (1/6)(3+4+5+6) = 3.6667

First roll outcome: 1 Conditional expected value: (1/6)1 + (1/6)(2+3+4+5+6) = (1/6)(1+2+3+4+5+6) = 3.5 (in other words, if you roll a 1 the 1st throw, we scrap the score and roll again as if we were only given 1 throw)

Since each 1st roll outcome has a probability of 1/6, we can get the total expected value by add up the weighted expected values for each conditional outcome.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

You are presented with a fair, 6-sided die. What is the expected number of rolls needed to roll two 6's in a row? (i.e. if you roll two 6's right off the bat, that counts as 2 rolls)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

the probability for rolling any single target number with 2 dice is 2/34 because, for example you could roll a 2 and a 5 or a 5 and a 2 ( 2 ways to roll a 7 ) this is harder to demonstrate for pairs but if you thought of it as if u had 1 die with circles and 1 with squares and u were aimin to roll a total of 12 shapes, then you could possibly roll 6 squares and then 6 circles or vice-versa, givin you two possibilities to roll the 12

so your probability of rolling 2 sixes is 2/34 or 1/17( why not 2/36? because its impossible to roll a 1 since theres 2 dice )

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

"You are presented with a fair, 6-sided die. What is the expected number of rolls needed to roll two 6's in a row? (i.e. if you roll two 6's right off the bat, that counts as 2 rolls)"

Well the probability of rolling 2 sixes in a row is 1/36.....so the number of expected rolls would be 36?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Man this shit reminds me of my industrial org class. If you struggle with bayes approach (probability tree) your going to hate game theory where you then have to use backwards deduction.

"Oh - the ladies ever tell you that you look like a fucking optical illusion?"

"Oh the ladies ever tell you that you look like a fucking optical illusion" - Frank Slaughtery 25th Hour.

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

heh the best of 2 rolls tripped me up too, I came up with 4.25 but EE and Oranhoutan are def right. Oranhoutan what is the answer to your question? I have no idea, I tried coming up with prob for rolling two 6s in a row but then I have no clue what the expected # of would be. It's def not 36, that's just avg # of rolls required to roll 2 6s in general.

I had these two questions in one of the interviews: a) you have a 10x10 in cube of ice suspended in the air, it is made up of 1x1 in smaller cubes, when the ice starts to melt, the outer layer of cubes falls away, how many 1x1 cubes are left still together? and b) you have a fair 6-sided die, if you roll a 6 you win, if you roll 1,2,3,4, or 5 you keep rolling..what is your prob of winning?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Oranhoutan what is the answer to your question? I have no idea, I tried coming up with prob for rolling two 6s in a row but then I have no clue what the expected # of would be. It's def not 36, that's just avg # of rolls required to roll 2 6s in general.

Why is it not 36? Wouldn't expected # of rolls for tow 6s in general be 12? E(1/6*12)=2. Could someone please explain..

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Let's make the two-6's-in-a-row question a little simpler, what is the expected number of rolls to get one 6?

Or maybe,

Instead of having a die, suppose you had a fair coin. What is the expected number of flips to get two heads in a row? How about a head followed by a tail?

As for the 10x10x10 cube question, if the outer layer melts away, you end up with an 8x8x8 cube = 512 cubes.

For the second question, you'll eventually win, i.e. you'll eventually roll a 6 if you are allowed to keep rolling indefinitely. Am I misreading the question?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

EE and Oranhotan are right. Normally the question is, what's the expected value of two rolls assuming you can roll again after the first roll, and only your last roll counts?

The answer to that one would be 4.25, but if it's max of both rolls, then its 4.47.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I may have forgotten how the question was phrased, and yes, that does make a big difference. In any case, it must have been just the second roll, because 4.25 was the answer.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

You can answer what the probability is that out of X rolls, you will get at least one 6 using a geometric random variable. The probability that you will get at least one 6 out of 6 rolls is about 0.668.

If you wanted to, you could multiply the #rolls times the respective probability (1.167+2.139...etc) up to an infinite number of rolls and find that the expectation converges to 1/P, where P is 1/6 in our case, so the expected number of rolls to get a six (or any other number on the die) is 6.

The easiest/quickest idea to pull out of this is that the expectation of a geometric random variable is just 1/p, where p is the probability that your event will happen in a given roll.

Finding the E of 2 sixes in a row would take longer, but it seems that it would be less than 36. If you treat each event as a pair of rolls the expectation would be 36 rolls, but you have to count rolls 2 and 3, 4 and 5, etc. as well.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

alright so with a = expected # of flips to get heads with 50% chance of heads or tails on each flip: a = 1/2(1) + 1/2(1+a) -- first part of the right side is hitting heads on first try, second part is starting back at a but now with 1 flip having taken place. solving for a we get 1/2a = 1 --> a = 2

Going further for 2 heads in a row we get: a = 1/2(1+a) + 1/4(2+a) + 1/4(2) in this one we have first part denotes hitting tails right away so back to a, second part is hitting Heads and then hitting Tails, so again back to square one, finally last part is hitting 2 heads in a row on the first two flips. Solving for a again we get a = 3/4a + 3/2 --> 1/4a = 3/2 --> a=6

Extending this to the 6 sided die with your great hint of looking at it a coin with unequal weights we get: a = 5/6(1+a) + 5/36(2+a) + 1/36(2) solving for a we get 1/36a = 7/6 --> a = 42

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

You're given a loop of copper wire. You're instructed to insert the wire loop into a machine that will make three random cuts (independently, uniformly distributed) along the wire.

What is the probability that you'll end up with a piece of wire (of the 3 pieces) with at least half the length of the original loop?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Are you expected to do this all in your head where you think outloud and explain HOW you would arrive at the answer? Or are you given a paper and pen and given a couple minutes.

What concepts should one know to answer these questions? Some i saw mentioned were expected value and game theory... any others?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

D

4/11/16 - 2:09pm

Is there a book that would cover similar questions and answers? To get my head back in the game?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

roll a die, and you get paid what the dice shows. if you want, but you don't have to, you can roll the die again and get paid what the second roll shows instead of the first. what is the expected value?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

THE MOST difficult probability question (Originally Posted: 11/17/2011)

since we're posting "hardest" probability questions, here's a problem that's actually difficult to solve.

Imagine you have a biased coin. We repeatedly toss the coin and keep a running estimate of the empirical probability of heads. We want to show that after a certain number of flips, our estimate will ALWAYS stay within a certain error bound of the true probability of heads and never deviate too much from the true value. So if x is the true probability of heads and x_n_trials is the empirical probability of heads after n flips of the coin, we want to show:

P( max | x - x_n_trials | > d ) <= g

where x is the true probability, x_n_trials = (number of heads in n trials) / n d is the parameter that determines how closely we want the error bounded g is the probability bound for the event that we are not within d of the true value

Show that this ALWAYS holds if n >= O(1/d^2 * log( 1/(g*d) ) )

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I don't know why this called "Hardest Probability Questions" .. I assume it is a kind of Joke ..

Lets Solve first Rolling Once ..\ The expected value of a random variable X is denoted by E(X). For a discrete random variable, E(X) is calculated as

Rolling a "1" has a probability of 1/6. Rolling a "2" has a probability of 1/6. Rolling a "3" has a probability of 1/6. Rolling a "4" has a probability of 1/6. Rolling a "5" has a probability of 1/6. Rolling a "6" has a probability of 1/6.

Multiplying the values with their respective probability gives: 1 * 1/6 = 1/6 2 * 1/6 = 2/6 3 * 1/6 = 3/6 4 * 1/6 = 4/6 5 * 1/6 = 5/6 6 * 1/6 = 6/6

Adding them together gives: 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5

Now solvin Rolling the Dice Twice ,,

EX. The random variable X has the following probability distribution: x pX(x)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I don't know why this called "Hardest Probability Questions" ..I assume it is a kind of Joke ..

Lets Solve first Rolling Once ..\The expected value of a random variable X is denoted by E(X). For a discrete random variable, E(X) is calculated as

Rolling a "1" has a probability of 1/6.Rolling a "2" has a probability of 1/6.Rolling a "3" has a probability of 1/6.Rolling a "4" has a probability of 1/6.Rolling a "5" has a probability of 1/6.Rolling a "6" has a probability of 1/6.

Multiplying the values with their respective probability gives:1 * 1/6 = 1/62 * 1/6 = 2/63 * 1/6 = 3/64 * 1/6 = 4/65 * 1/6 = 5/66 * 1/6 = 6/6

Adding them together gives:1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5

Now solvin Rolling the Dice Twice ,,

EX. The random variable X has the following probability distribution: x pX(x)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

assuming a strategy where you only chose to roll the die again if the outcome of the first die was below the expected value of rolling a fair die (3.5)

you get 4.25 because you still have (1/6)(4 + 5 + 6) but you also add (1/12)(1 + 2 + 3 + 4 + 5 + 6)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

basically you have to relate the confidence interval (involving standard deviation of the sample mean, which is a summation of bernoullis (i.e. binomial)) to the size of the test, and produce the minimum "n" that will guarantee that your confidence interval is correct.

certainly challenging (especially with the 15 minute time frame).

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Anonymous Monkey

1/8/16 - 9:33am

This is incorrect. The only possible results of the toss with your strategy (which is correct) is 1,2,3,4,5,6, 4,5,6. Hence the expectation value (mean payout) is 4, not 4.25.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Jesus christ. This was at an interview? I'm guessing given the difficulty it was either Jane Street or DE Shaw.

I'm guessing he did a stats heavy course at university/grad level, and the interviewer had too. Neither of those companies will ask you about concepts you haven't seen before. It's a core part of their ethos.

having had that kind of training, this question gets easier. I'm not saying it isn't hard, but this is a lot more approachable if you've done similar stuff for the past 3 years

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

No no i have the hardest problem, given that f(x) is differenciatable at x = 0 and Lim/(x?>0) (f(x)?f(kx))/x = a, where a and k are constants. Show that f'(0) = a/1-k

"...all truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident."

Schopenhauer

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

No no i have the hardest problem, given that f(x) is differenciatable at x = 0 and Lim/(x?>0) (f(x)?f(kx))/x = a, where a and k are constants. Show that f'(0) = a/1-k

lim/(x->0) (f(x)/x) - lim/(x->0) (f(kx)/x) = a take derivative of top and bottom (L'Hopital's rule) and get lim/(x->0) f'(x) - lim/(x->0) kf'(kx) = a plug in 0 f'(0) - kf'(0) = a factor f'(0) (1-k) = a f'(0) = a/(1-k)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Hmm I disagree, but I could be wrong. Just did this on scrap paper and got the same thing just (d-1/2)^2 in the denominator. Are you sure thats the correct answer?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Looks like Strong Law of Large Numbers to me. Given the hint, the P( max | x - x_n_trials | > d ) < = summation (1 to n) P( | x - x_n_trials | > d ) (Boole's inequality) since each xi is bounded by 0 and 1, the latter probability is gonna be a finite (1 to n) sum of 2exp (-2d^2*n). (Hoeffding's inequality) Such thing converges since it is a geometric series. It should give you the result.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Probability Problem has me stumped (Originally Posted: 10/12/2012)

You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?

Im thinking its symetric so 1/2^3 either that or find expected first pick then prob of getting higher than that for the second card and so on

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

For each draw {A , B , C , D} , isn't there a 1/4! chance that you draw this draw in ascending order. Therefore, isn't the EV of this game (1/4!)* 10 - ( ((4! - 1)/4!)*1).

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

for your first 2 cards, the probability of the second being > first is even, yes. After that, though, the probability of each subsequent card being > than the one before decreases.

in fact, my guess would be it goes something like this - P(2>1) = 0.5

P(3>2) = 48.5/95 (we know card 1 is one of the cards less than card 2, so the probability of card 3 being greater than card 2 is always skewed by 1 card out of 95?)

P(4>3) = 49/94 (repeat logic from above but skewed by 2 cards out of 94)

So you get (1/2)(48.5/95)(49/94) = ~0.133, slightly greater than (1/2)^3, which is intuitive because these are non-independent in a way where each previous step positively impacts the probability of the subsequent step?

Caveat: I'm not 100% confident in the answer, I'm also more confident in my reasoning than my math (embarrassingly enough)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

so the 97 cards are just labeled 1-97? i'm confused

This to all my hatin' folks seeing me getting guac right now..

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Let me expand my answer: The probability that a random draw of 4 cards will be in ascending order should be 1/4!, irrespective of the sample size. Why? Consider the case where you draw {1 , 2 , 3 , 4}. Given that you have drawn {1, 2, 3, 4}, the probability that you have drawn them in exactly that order is 1/4! or 1/24. Since the situation is symmetric with respect to each such quadruplet, the probability that your random draw is in ascending order is 1/4!.

From here we can calculate the EV of the game 1/24*10 - 23/24 = -13/24 < -0.5 .

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

GS is absolutely correct. Whether you are drawing from 97, 500, or 1,000,0000 cards, the probability of drawing 4 cards in ascending order is the same. You can draw 4 cards in 4! or 24 ways. Furthermore, only 1 of those 24 ways will yield cards in ascending order.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

This to all my hatin' folks seeing me getting guac right now..

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Isn't your code drawing the cards with replacement? Looks like it to me (I don't know how the random class works but if it's a generic random num generator, then you are drawing them with replacement)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

This to all my hatin' folks seeing me getting guac right now..

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

There. That gets me a lot closer to the -13/24 value. Good catch.

This to all my hatin' folks seeing me getting guac right now..

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I believe the answer to the snake one is no. My explanation may not make a lot of sense though. Basically, in order for you to get all the snakes a single color you need to get the other 2 sets of snakes to be equal. Okay, so you have 13, 15, and 17 snakes. Every time you make a change, 2 numbers go down by 1 and one number goes up by 2. This means that you need 1 type of snake to be exactly 3 less than another in order to equal them out.

Right now the snakes are all 2 (or 4) apart in quantity. There is no change or set of changes you can do to make 2 colors seperated by 3. This is due to each mutation either closing the gap by 3 or keeping them equal. Hence, impossible.

Not the best formatted answer but that is my answer.

CompBanker

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

PROBABILITY brainteasers (Originally Posted: 10/07/2010)

Hey evryone - as it is that time of year when the rat race becomes "racier," if you will, there are many of us interviewing with prop trading firms. any firm worth its while will be asking probability brainteasers. i know of few other ways to get a sampling of some great brainteasers than by asking the fine gentlemen on this site. so, with that being said ... what are your favorite brain teaser, and/or the ones you have personally had in interviews ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

yes one or two of the previous explanations to the snake problem are correct. The snakes cannot all become the same colour simply because of the relative difference between sankes gaiven being divisible by two. By inspection , only a difference between any two snake numbers that is divisible by 3 will work . Hence the answer is no!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Question with Card Probability brainteaser (Originally Posted: 10/05/2012)

I heard this question.

You are given 16 cards 4 are hearts, 4 are clubs, 4 are spades and 4 are diamonds.

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

I assume the answer is 25%? Out of all the different combinations you can make, I assume 1/4th will end in clubs 1/4th will end in diamonds and 1/4th will end in spades and 1/4 will end in hearts.

So is 25% correct for this? I start over thinking, that you have a higher probability to draw a non club card which will increase the chances of a club card showing at the end, but I do not think this is true.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Are the cards being replaced after each draw or not? If yes, then it is 1/4 since on the 4th draw you will have 4 clubs out of 16. But if the cards are not being replaced, then you have to sum up the 4 distinct possibilities. By the last draw, there will be 13 cards left. If you had picked clubs in each of the previous 3 draws, then your probability of drawing the last club will be 1/13. If you had drawn 2 clubs in the previous 3 draws, your probability would be 2/13 and so forth. Hence, you will need to add up 1/13+2/13+3/13+4/13 to get 10/13, which should be the final answer.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

It doesn't matter. It's 1/4. The problem is perfectly symmetrical.

If I asked you about diamonds in particular, you'd still tell me 10/13. If I asked about spades, you'd still tell me 10/13. Clearly there's a problem here. The answer is 1/4.

This is actually a very clever interviewing question. Among other things, it tests how sure of yourself you are and if you're the type to just jump into math without considering the situation first.

I had two other clever questions pop up in an interview recently, both of which made you go through long-winded calculations whose answer was most of the work of the problem, but not the brief final step from there that they wanted. Luckily I noticed the second time it happened, but watch out for that.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Im sorry could you explain the answers again? I'm not sure what you mean about wrt, and it looks like unless im obviously reading it incorrectly i two different answers.

The first question with at least one is a boy chances she has a girl... your sample size is BB GB BG how can you know the chances of her having a girl (I assumed it was the 2 scenarios that have a girl in it over all scenarios so 2/3 but if you do that for the boy you see its 3/3 so that cannot be right.

The second question where the eldest is a boy the sample set is B(younger)B(older) G(younger)B(older)... same question again how do you clearly look at knowing the sample spaces if i was thinking it incorrectly in the first one

edit: I think i figured out how I was thinking wrong... if the question was posted as at least 1 is a boy what is the probability of the other being a boy... sample space is bb gb bg so it would be 1/3 not 3/3 (the way i was erroneously thiking of 3/3 thus my thinking was wrong was posing the quesiton in my head whats the probability that at least one is a boy given that one is a boy)...derf

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Im sorry could you explain the answers again? I'm not sure what you mean about wrt, and it looks like unless im obviously reading it incorrectly i two different answers.

WRT means with respect to. So, if you have drawn n cards, and don't know what any of them is, then the situation is symmetrical with respect to each suit.

brainteaser2:

The first question with at least one is a boy chances she has a girl... your sample size is BB GB BG how can you know the chances of her having a girl (I assumed it was the 2 scenarios that have a girl in it over all scenarios so 2/3 but if you do that for the boy you see its 3/3 so that cannot be right. .

It is right. P(couple has a boy | at least one boy) = 1. P(couple has a girl | at least one boy) = 2/3

brainteaser2:

The second question where the eldest is a boy the sample set is B(younger)B(older) G(younger)B(older)... same question again how do you clearly look at knowing the sample spaces if i was thinking it incorrectly in the first one thank you

The events in the sample set represent the possible configs. If eldest comes first (B G) , (B B) are the outcomes if the eldest child is a boy. In that case, the probability that the next kid is boy/girl is 50-50 as you would expect.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

let me rewrite this question in a simpler way for you.

"What is the probability of choosing a Club when choosing a card at random?".

It doesn't matter if it's the fourth or fourteenth card, if you don't know anything about the other cards being drawn.

Here's another one for you:

You have a friend who has two children. At least one of the children is a boy, what are the odds that your friend has a girl?

What are the answers to these? Why are they different?

In the first case, you have a symmetrical sample set wrt to each suit, therefore the probability is 1/4.

In the second case, your sampling space is [(B , B) (B, G) (G,B)] { (G,G ) is omitted }. Since each event in this space is equally likely, the prob that your friend has a girl is 2/3.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

let me rewrite this question in a simpler way for you.

"What is the probability of choosing a Club when choosing a card at random?".

It doesn't matter if it's the fourth or fourteenth card, if you don't know anything about the other cards being drawn.

Here's another one for you:

You have a friend who has two children. At least one of the children is a boy, what are the odds that your friend has a girl?

What are the answers to these? Why are they different?

In the first case, you have a symmetrical sample set wrt to each suit, therefore the probability is 1/4.

In the second case, your sampling space is [(B , B) (B, G) (G,B)] { (G,G ) is omitted }. Since each event in this space is equally likely, the prob that your friend has a girl is 2/3.

Oh I see. And the case where the boy is eldest makes the sample space [(B,B), (B,G)] so the probability is now 1/2. Thanks for clarifying

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I know I've already posted in your other thread but..

Think of it as a permutation of the 16 cards. The probability of any one card out of the 16 being the 4th card is 15!/16! = 1/16. And since there're 4 clubs, the probability of the 4th card being a club is 4 x 1/16 = 1/4.

And also, as one of the above posters mentioned, since the condition for each of the 4 suits is identical, the probability of the 4th card being each suit has to be equal, which is a much quicker way to get the answer.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

There are four possible outcomes, each equally likely:

Boy-Boy, Boy-Girl, Girl-Boy and Girl-Girl

Probability that couple has "a girl" is 3/4. The probability that the couple has "a boy" is also 3/4. They're not mutually exclusive of course. The probability that a couple has two boys is 1/4, two girls is also 1/4 etc.

"You have a friend who has two children. At least one of the children is a boy, what are the odds that your friend has a girl?"

You can exclude the Girl-Girl outcome. That leaves Boy-Boy, Boy-Girl, Girl-Boy. In two out of these three possible outcomes the friend has a girl, so the probability for that is 2/3.

"You have a friend who has two children. The eldest is a boy, what are the odds that your friend has a girl?"

In this case you can exclude two outcomes Girl-Boy and Girl-Girl because the older kid is a boy. That leaves Boy-Boy, Boy-Girl and in one of these the kid is a girl, so the probability is 1/2.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

I think the previous poster forgot one fact that changes things a bit.

The key here is that the problem says "you can roll the die again and get paid what the second roll shows INSTEAD of the first" (caps added). This means that you if you decide to re-roll, you should disregard your first roll and just keep the second one.

The first time you roll the dice, you have two options:

You keep you score

You re-roll the die

Since the expected value (EV) of one dice roll is 3.5, it makes sense to keep your score if it is higher than 3.5, and re-roll if you get a score less than 3.5.

If you roll over 3.5, then the score you got must have been either 4, 5, or 6. Of these three numbers, the EV is 5.

The other possibility (which is equally likely) is that you get 1, 2, or 3. Then, it makes sense to re-roll. Again, on your second roll, your EV is 3.5.

Since you have two options, which are equally likely, the EV is: 1/2(the chance you don't reroll) + 1/2(the chance you re-roll and ignore your first score) = 1/2(5) + 1/2(3.5) = 1/2(8.5) = 4.25

Note: If you had to keep the first roll and add it to your second score, it would always make sense to re-roll. Then, as the last poster mentioned, the EV would be 3.5+3.5 = 7.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

probability questions (Originally Posted: 08/16/2007)

Hi guys,

I will appreciate it very much if anyone can tell me whether there is any probability questions asked during the interview for structured finance entry level positions? How about interviews for rating agencies like FitchRathings?

Hang in there!

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

The problem was missing information, as the expected value depends on the strategy the user uses. To find the optimal strategy, we can safely argue that, by intuition/inspection, the strategy we want to use is that if we roll n or less, then we roll a second time, otherwise we keep our first roll. The expected value is then

(6 + n)/2 * (6 - n)/6 + 3.5 * n/6

Treating n as continuous will give us a maximum at n = 3.5. We then check n = 3 and n = 4 to see which value gives us the highest expectation, and when n = 3 we get 4.25 whereas when n = 4 we get 4, so we choose the strategy with n = 3 which will give us the max expectation.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

according to Bayes' Rule (what is the probability of getting a probability question, given that you have an interview for an entry level position in structured finance): about 13%

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

the odds that you get two straight heads on your 2nd roll is 1/4 the odds that you get two straight heads on your 3rd roll is (1/4)(1/2)... 1/2 is the odds that first roll is tails the odds that you get two straight heads on your 4rd roll is (1/4)(1/2)*(1/2)

Getting the two straight heads by roll 3 or roll 4 has the same likelihood as getting 3 straight (1/8 and (1/8)*(1/2)). The only difference between X and Y is that Y can't happen in 2 rolls, whereas X has a 1/4 chance

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Book to learn how to answer tough probability questions? (Originally Posted: 06/01/2010)

Does anyone have a suggestion for this? I took a Business stats course this year but it was a joke compared to the prob Qs/brainteasers seen on here. Anyone have any suggestions for how to learn how to do these? Or is it just innate skill? I'm looking for something textbook like, I already have the Crack book.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

the odds that you get two straight heads on your 2nd roll is 1/4 the odds that you get two straight heads on your 3rd roll is (1/4)(1/2)... 1/2 is the odds that first roll is tails the odds that you get two straight heads on your 4rd roll is (1/4)(1/2)*(1/2)

Downtown, you're overlooking the fact that there are two ways to create the event X=4

You can have the following for X=4 (where T is tails and H is heads)

T T H H H T H H

The actual probabiliy function of X=x is:

f(x) = F_(x-1) * (1/2)^x, where F_(x-1) represents the Fibonnaci sequence and x=2,3,4,....

Derivation of f(x) If you write it out, you will discover that there's

1 way to make X=3 2 ways to make X=4 3 ways to make X=5 5 ways to make X=6 8 ways to make X=7

The probability of two heads is (1/4) or equivalently, (1/2)^2 The probability of anything before that is (1/2)^(x-2).

If you check out the Wikipedia article on Fibonnaci Power Series, you can show that the sum of f(x) from x=2 to x=infinity does equal 1, and thus f(x) is a proper probability mass function.

Regarding event Y...

Y is a little bit more dfficult. Since X and Y are independent, the P(X>Y | Y=y) = P(X > y) = sum of f(x) from x=y+1 to x=infinity or equivalently, 1 - f(2) - f(3) ... f(y).

To show the P(X>Y) for all Y, we must calculate the infinite sum: P(Y=3)P(X>3)+P(Y=4)P(X>4)+P(Y=5)*P(X>5)...

The PMF of Y=y is:

g(x)=G_(x-2)*(1/2)^x for x=3,4,5,6,...

where G_x= G_(x-1)+G_(x-2)+G_(x-3) with the seed G_0=0, G_1=1, and G_2=1

If you write it out, there's

1 way to make Y=4 2 ways to make Y=5 4 ways to make Y=6 7 ways to make Y=7 13 ways to make Y=8

.... and I don't know how to handle that sequence...

So.... that one problem probably takes the cake for hardest probability problem... perhaps there's some elegant solution to the problem

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

5 / 3! = 5 / 6 = 83% 6 combinations of scenarios: FFF FFW FWF WWW WFW WFF the only scenario in which he loses more all games is FFF so that leave 5 success

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

First, there are 8 possible combinations not 6 (2^3). Second, you're ignoring the fact that the chances of F and W are not equal. All you have to do to answer the question is calculate the probability of FFF which equals .7*.7*.7 which equals 34.3%. That's the probabilty that no goals are scored in any game. All of the other scenarios meet your requirement of at least 1 goal scored. So the answer is 1 minus .343.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Since saying "the odds he will score at least one" is the same as saying "1 minus the odds that he won't score any", we use the "no goals" scenario as our basis.

30% chance of scoring means 70% chance of not scoring.

Since each game is an independent event, the odds of him not scoring at any of the three is simply their product: 70% x 70% x 70% = (.70)^3 = 0.343

So since we have a 34.3% chance that he will not score anything, we know that there is a 100% - 34.3% = 65.7% chance he will score at least one.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Any one know any good books on probability theory? I'm looking deeper into options and realize that I need a better base of prob knowledge to really go any further.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

you need to explain more specifically what you are trying to do. are you trying to get a qualitative understanding of the greeks?

are you looking for a textbook for an intro college probability course (with multivar + continuous distributions)? that will get you started in options, but to derive even a relatively simple model yourself (like black-scholes) you need to go down the rabbit hole into the world of stochastic calculus...

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sure, you solved this in grade 12! This is why it took more than 3 centuries to find a proof to this simple problem. It was only solved in 1994 by Andrew Wiles and Richard Taylor. (@ Art Vandelay: I'm not sure everyone got your post ;)

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Like structure sort of said, you will need to be very comfortable mathematically (relative to normal finance math).

That said, Shreve's Stochastic Calculus for Finance I and II (2 books) are great from what I've heard (although I've never read them past the introductions, the second book has 2 chapters introducing/reviewing probability theory through information & conditioning). Maybe see if your library has a copy before buying, though.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

shreve's is the best stuff out there - actually explains most of the math instead of merely laying out the underlying assumptions.

in general, it's not probability that gets most people hung up on options pricing - it's the differential equations. for example, a lot of pricing books (ie - Hull) will simply say 'and here we solve the black scholes PDE' - but they don't always explain how they derive or solve it, or really what it means.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Probability review? (Originally Posted: 08/09/2011)

Hi guys,

I'm a rising junior at an I-hope-it's-target-Midwest-state-school (well I'm sure the B-school is but ehh...) studying a combination of pure math and economics. I'm interested in trading (although the recruitment process for some of the prop shops looks absolutely terrifying/awesome). Looking around on the forum, I noticed there seems to be a big emphasis on probability.

I've done a good amount for the pre-reqs econometrics, but admittedly, have gotten a bit rusty (we covered all the standard stuff up to annoying multi integrals for multivariate distributions). Does anyone know how much material is important for interviews, as well as a good text to pick it up from again?

Courses in the math department aren't really that great an option since the only things that are appropriate for my major are measure theory based probability stuff and I have absolutely no idea how to apply measure theory.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

the probability in the interviews is nothing that is technically/theoretically advanced, its really the foundation concepts of probability like Bayes theorem but under a great amount of time pressure and in a brainteaser format.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

interview probability will be simple EV type stuff that just gets marginally harder to make sure you can think analytically/prove you can do some sort of math. nothing mind boggling. know dice/coinflip EV etc..

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Card Probability question. Anyone know how to do it? (Originally Posted: 10/05/2012)

I heard this question.

You are given 16 cards 4 are hearts, 4 are clubs, 4 are spades and 4 are diamonds.

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

I assume the answer is 25%? Out of all the different combinations you can make, I assume 1/4th will end in clubs 1/4th will end in diamonds and 1/4th will end in spades and 1/4 will end in hearts.

So is 25% correct for this? I start over thinking, that you have a higher probability to draw a non club card which will increase the chances of a club card showing at the end, but I do not think this is true.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

It depends if the cards are replaced or not...if not you need to make a path dependent tree.

doesn't matter. if u don't know what was picked then it's 25% whether you replace or not. Let me say it this way:

if there are 16 cards, same way as in the OP and if I picked 15 randomly and didn't replace, what are the odds that the 16th card is a club? 25%. Same with the first, second, etc.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

if not, you need to take into account the first 3 cards

Born in hell, forged from suffering, hardened by pain.

Want to Vote on this Content?! No WSO Credits ?

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Think of it as a permutation of the 16 cards. The probability of any one card out of the 16 being the 4th card is 15!/16! = 1/16. And since there're 4 clubs, the probability of the 4th card being a club is 4 x 1/16 = 1/4.

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

16! combinations of cards, lock in the fourth as a club (4 ways to do this), You end up with 15! of arranging the cards. Thus you get 4*15!/16!=1/4. Yep you guys are right

Sorry, you need to login or sign up using one of the blue buttons below in order to vote. As a new user, you get 3 WSO Credits free, so you can reward or punish any content you deem worthy right away. See you on the other side!

let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

Calculate P{X>Y}

Solution: 3/32 Problem can be thought of as "what is the probability t

## Comments (192)

What to have some fun? Try these two.

Calculate P{X>Y}

Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become

same color?

## Want to Vote on this Content?! No WSO Credits ?

The answer to the snake problem is YES! They all can become one color, to proof this it is sufficient to show that there exist one solution where all snakes are one color, e.g.. one red and one yellow snake meet, this will lead to 14 blue, 14 red, and 16 yellow snakes. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. QED

## Want to Vote on this Content?! No WSO Credits ?

Nope. One red and one yellow snake meeting will lead to 15 blue, not 14 blue. Compbanker was right earlier. You need a difference of 3 between 2 colours to make this possible.

## Want to Vote on this Content?! No WSO Credits ?

14+14+16 is not equal to 13+15+17

## Want to Vote on this Content?! No WSO Credits ?

Solution to problem 1 can be found here: http://pratikpoddarcse.blogspot.com/2009/10/lets-s...

## Want to Vote on this Content?! No WSO Credits ?

Anyone figure out number 1 yet? I don't get how to do it and it's driving me crazy.

## Want to Vote on this Content?! No WSO Credits ?

Solution to Problem 1:

http://pratikpoddarcse.blogspot.com/2009/10/lets-s...

Best of Luck!

## Want to Vote on this Content?! No WSO Credits ?

## Want to Vote on this Content?! No WSO Credits ?

Isn't the answer to number 2 no because there is an odd number of snakes? Unless there is a way to get it so there are 15 of each, which i can't figure out.

## Want to Vote on this Content?! No WSO Credits ?

Probability questions for S&T interview(Originally Posted: 04/01/2008)Hey all, I have an interview coming up for a trade support position at a small trade shop. I was told ahead of time that my interview will consist of "fit" questions along with some probability questions. Does anybody have any website recommendations to refresh my probability and stat. skills? And/or what type questions might I expect? I apologize if this has been discussed already on WSO, Thanks!

## Want to Vote on this Content?! No WSO Credits ?

Get some probability questions from the GMAT. www.gmatclub.com has a forum, and on it you can find a bunch of proba questions.

Remember, you will always be a salesman, no matter how fancy your title is.

- My ex girlfriend

## Want to Vote on this Content?! No WSO Credits ?

go over expectations. I had one of the hardest interview probability questions and it was just expectation however it was near impossible to structure the answer. got it though after about 5 minutes of hard work and got the job. Richard Durret has a text called Probability: theory and example that is amazing. pick it up.

## Want to Vote on this Content?! No WSO Credits ?

Got this two parter for Bear F.A.S.T. summer analyst spot:

a) Easy: How much would you pay to roll a single die, if you got $1 if you roll a 1, $2 for a 2, etc...?

b) Harder: Okay, now how much would you pay if you were allowed to roll twice and take the higher of the two?

Financial Modeling Training

Guide to Finance Interviews

<a href="http://www.wallstreetoasis.com/wso-finance-resume-

## Want to Vote on this Content?! No WSO Credits ?

Ok obviously for (a) the answer is $3.50 because you just mult. each amount paid by 1/6 and add them, but for (b), if each role is valued at $3.50, one's instinct is to say $7.00 because your essentially being given two roles for the price of one, but I think that is the teaser answer. What did you say and do you know the process to take for the correct answer?

## Want to Vote on this Content?! No WSO Credits ?

i've heard a variation of b) from a friend who heard it from a friend at Merrill. it was the same thing except you dont get the higher of the 2, if you roll the 2nd time you get the payout of the 2nd roll, not the greater of the two, but i'll give yours a shot.

this may be the long way but i still believe its correct. draw out a decsion tree.

first node is 1st roll. consists of 6 branches, each corresponding to what you roll, each with a prob of 1/6 . at the end of each branch is another node for 2nd roll, each with 6 branches of its own, same as above.

you're gonna have 36 unique combinations. at the end of each combo, you take the meax of the first and 2nd roll. then multiply that by 1/6 twice, i.e.e 1/36.

sum all of those and theres youre total expected value and therefore how much you'd be willing to pay.

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

## Want to Vote on this Content?! No WSO Credits ?

Well I don't think for b it is meant you keep the higher of the two times two, simply you get to keep the higher of the two, so I would think the value of the 2nd roll has to be either the probability of rolling higher than expected return from first roll, or >3.50 (i.e. 4, 5, 6) and then if it rolls 1, 2, 3 value is 0. So given you have a 50% chance to roll higher, and then your return could be $1.50 on avg more (5 = avg of 4, 5, 6, 5 - 3.5 = 1.50), and then 50% to get 0, you would pay an extra $0.75 for the roll or $4.25 in total.

That is how I would break it down. Never got this or anything like it, so I am not sure at all.

## Want to Vote on this Content?! No WSO Credits ?

using my method its, which i am sure of, the answer is 4.47222. because i am bored and lame, i did it in excel, pm me if you want the excel file, hah...

------

"its the running joke now, we now have fair trade with china so they send us poisoned sea food and we send them fraudulent securities."

------

## Want to Vote on this Content?! No WSO Credits ?

Alex Kap has the right answer. Having never taken probability, I didn't get it.

Financial Modeling Training

Guide to Finance Interviews

<a href="http://www.wallstreetoasis.com/wso-finance-resume-

## Want to Vote on this Content?! No WSO Credits ?

NICE HAHAHA... But I just used logic, I had probability questions that I didn't get like that, but at the same time they like the overall thought process, I just made a mistake in one step.

## Want to Vote on this Content?! No WSO Credits ?

I had an interview with a top BB and not only did they ask me what I would pay for 1 roll, they 2 rolls, then 3 rolls, then they asked what I would pay for 2 rolls should I be allowed to keep the maximum of the two rolls.

## Want to Vote on this Content?! No WSO Credits ?

There are 6 possible outcomes for the 1st roll (1, 2, 3, 4, 5, 6), each with a probability of 1/6. For each possible outcome, there is a corresponding expected value for the "better-of-two" rolls.

First roll outcome: 6

Conditional expected value: 6

First roll outcome: 5

Conditional expected value: (5/6)

5 + (1/6)6 = 5.1667 (this is expected, i.e. >5)First roll outcome: 4

Conditional expected value: (4/6)

4 + (1/6)(5+6) = 4.5First roll outcome: 3

Conditional expected value: (3/6)

3 + (1/6)(4+5+6) = 4First roll outcome: 2

Conditional expected value: (2/6)

2 + (1/6)(3+4+5+6) = 3.6667First roll outcome: 1

Conditional expected value: (1/6)

1 + (1/6)(2+3+4+5+6) = (1/6)(1+2+3+4+5+6) = 3.5 (in other words, if you roll a 1 the 1st throw, we scrap the score and roll again as if we were only given 1 throw)Since each 1st roll outcome has a probability of 1/6, we can get the total expected value by add up the weighted expected values for each conditional outcome.

That is,

(1/6)

6 + (1/6)5.1667 + (1/6)4.5 + (1/6)4 + (1/6)3.6667 + (1/6)3.5 =(1/6)*(6 + 5.1667 + 4.5 + 4 + 3.6667 + 3.5) =

(161/36) = 4.4722

## Want to Vote on this Content?! No WSO Credits ?

You are presented with a fair, 6-sided die. What is the expected number of rolls needed to roll two 6's in a row? (i.e. if you roll two 6's right off the bat, that counts as 2 rolls)

## Want to Vote on this Content?! No WSO Credits ?

its actually 1/17 if u thought about it...

the probability for rolling any single target number with 2 dice is 2/34 because, for example you could roll a 2 and a 5 or a 5 and a 2 ( 2 ways to roll a 7 ) this is harder to demonstrate for pairs but if you thought of it as if u had 1 die with circles and 1 with squares and u were aimin to roll a total of 12 shapes, then you could possibly roll 6 squares and then 6 circles or vice-versa, givin you two possibilities to roll the 12

so your probability of rolling 2 sixes is 2/34 or 1/17( why not 2/36? because its impossible to roll a 1 since theres 2 dice )

## Want to Vote on this Content?! No WSO Credits ?

"You are presented with a fair, 6-sided die. What is the expected number of rolls needed to roll two 6's in a row? (i.e. if you roll two 6's right off the bat, that counts as 2 rolls)"

Well the probability of rolling 2 sixes in a row is 1/36.....so the number of expected rolls would be 36?

## Want to Vote on this Content?! No WSO Credits ?

Man this shit reminds me of my industrial org class. If you struggle with bayes approach (probability tree) your going to hate game theory where you then have to use backwards deduction.

"Oh - the ladies ever tell you that you look like a fucking optical illusion?"

"Oh the ladies ever tell you that you look like a fucking optical illusion" - Frank Slaughtery 25th Hour.

## Want to Vote on this Content?! No WSO Credits ?

heh the best of 2 rolls tripped me up too, I came up with 4.25 but EE and Oranhoutan are def right.

Oranhoutan what is the answer to your question? I have no idea, I tried coming up with prob for rolling two 6s in a row but then I have no clue what the expected # of would be. It's def not 36, that's just avg # of rolls required to roll 2 6s in general.

I had these two questions in one of the interviews: a) you have a 10x10 in cube of ice suspended in the air, it is made up of 1x1 in smaller cubes, when the ice starts to melt, the outer layer of cubes falls away, how many 1x1 cubes are left still together? and b) you have a fair 6-sided die, if you roll a 6 you win, if you roll 1,2,3,4, or 5 you keep rolling..what is your prob of winning?

## Want to Vote on this Content?! No WSO Credits ?

Why is it not 36? Wouldn't expected # of rolls for tow 6s in general be 12? E(1/6*12)=2.

Could someone please explain..

## Want to Vote on this Content?! No WSO Credits ?

Let's make the two-6's-in-a-row question a little simpler, what is the expected number of rolls to get one 6?

Or maybe,

Instead of having a die, suppose you had a fair coin. What is the expected number of flips to get two heads in a row? How about a head followed by a tail?

As for the 10x10x10 cube question, if the outer layer melts away, you end up with an 8x8x8 cube = 512 cubes.

For the second question, you'll eventually win, i.e. you'll eventually roll a 6 if you are allowed to keep rolling indefinitely. Am I misreading the question?

## Want to Vote on this Content?! No WSO Credits ?

EE and Oranhotan are right. Normally the question is, what's the expected value of two rolls assuming you can roll again after the first roll, and only your last roll counts?

The answer to that one would be 4.25, but if it's max of both rolls, then its 4.47.

## Want to Vote on this Content?! No WSO Credits ?

I may have forgotten how the question was phrased, and yes, that does make a big difference. In any case, it must have been just the second roll, because 4.25 was the answer.

Financial Modeling Training

Guide to Finance Interviews

<a href="http://www.wallstreetoasis.com/wso-finance-resume-

## Want to Vote on this Content?! No WSO Credits ?

You can answer what the probability is that out of X rolls, you will get at least one 6 using a geometric random variable. The probability that you will get at least one 6 out of 6 rolls is about 0.668.

Broken down:

P(X=1) = 1/6=.167

P(X=2) = (5/6)(1/6)=.139

P(X=3) = (5/6)^2(1/6)=.116

P(X=4) = (5/6)^3(1/6)=.096

P(X=5) = (5/6)^4(1/6)=.080

P(X=6) = (5/6)^5(1/6)=.067

sums to .668

If you wanted to, you could multiply the #rolls times the respective probability (1

.167+2.139...etc) up to an infinite number of rolls and find that the expectation converges to 1/P, where P is 1/6 in our case, so the expected number of rolls to get a six (or any other number on the die) is 6.The easiest/quickest idea to pull out of this is that the expectation of a geometric random variable is just 1/p, where p is the probability that your event will happen in a given roll.

Finding the E of 2 sixes in a row would take longer, but it seems that it would be less than 36. If you treat each event as a pair of rolls the expectation would be 36 rolls, but you have to count rolls 2 and 3, 4 and 5, etc. as well.

## Want to Vote on this Content?! No WSO Credits ?

Think about it this way:

Let's denote the event of hitting a six as "6".

Let's also denote the event of hitting two 6's in a row as "6-6".

Notice that, in order to get to "6-6", you have to first get to "6".

That is, the expected hitting time to "6-6" = expected hitting time to "6" + expected hitting time to "6-6" from "6".

Hope that helps.

## Want to Vote on this Content?! No WSO Credits ?

If you can figure out what the expected hitting time to flipping two heads in a row for a coin...

i.e., suppose you found a coin that flips with 1/6th probability of heads and 5/6th probability of tails.

## Want to Vote on this Content?! No WSO Credits ?

alright so with a = expected # of flips to get heads with 50% chance of heads or tails on each flip: a = 1/2(1) + 1/2(1+a) -- first part of the right side is hitting heads on first try, second part is starting back at a but now with 1 flip having taken place.

solving for a we get 1/2a = 1 --> a = 2

Going further for 2 heads in a row we get: a = 1/2(1+a) + 1/4(2+a) + 1/4(2)

in this one we have first part denotes hitting tails right away so back to a, second part is hitting Heads and then hitting Tails, so again back to square one, finally last part is hitting 2 heads in a row on the first two flips. Solving for a again we get

a = 3/4a + 3/2 --> 1/4a = 3/2 --> a=6

Extending this to the 6 sided die with your great hint of looking at it a coin with unequal weights we get: a = 5/6(1+a) + 5/36(2+a) + 1/36(2)

solving for a we get 1/36a = 7/6 --> a = 42

This thing bothered me all day haha.

## Want to Vote on this Content?! No WSO Credits ?

42 is the correct answer.

## Want to Vote on this Content?! No WSO Credits ?

You're given a loop of copper wire. You're instructed to insert the wire loop into a machine that will make three random cuts (independently, uniformly distributed) along the wire.

What is the probability that you'll end up with a piece of wire (of the 3 pieces) with at least half the length of the original loop?

## Want to Vote on this Content?! No WSO Credits ?

What is the probability that a straight wire cut in two places (independent, uniformly random) will have its three pieces make a triangle?

## Want to Vote on this Content?! No WSO Credits ?

the above 2 questions are essentially the same.

## Want to Vote on this Content?! No WSO Credits ?

How applicable is all this to trading?

## Want to Vote on this Content?! No WSO Credits ?

Are you expected to do this all in your head where you think outloud and explain HOW you would arrive at the answer? Or are you given a paper and pen and given a couple minutes.

What concepts should one know to answer these questions? Some i saw mentioned were expected value and game theory... any others?

## Want to Vote on this Content?! No WSO Credits ?

^^^yea how the hell do you guys know the answers??? what are they teaching you in school that i never saw...

## Want to Vote on this Content?! No WSO Credits ?

Is there a book that would cover similar questions and answers? To get my head back in the game?

## Want to Vote on this Content?! No WSO Credits ?

roll a die, and you get paid what the dice shows. if you want, but you don't have to, you can roll the die again and get paid what the second roll shows instead of the first. what is the expected value?

## Want to Vote on this Content?! No WSO Credits ?

THE MOST difficult probability question(Originally Posted: 11/17/2011)since we're posting "hardest" probability questions, here's a problem that's actually difficult to solve.

Imagine you have a biased coin. We repeatedly toss the coin and keep a running estimate of the empirical probability of heads. We want to show that after a certain number of flips, our estimate will ALWAYS stay within a certain error bound of the true probability of heads and never deviate too much from the true value. So if x is the true probability of heads and x_n_trials is the empirical probability of heads after n flips of the coin, we want to show:

P( max | x - x_n_trials | > d ) <= g

where x is the true probability,

x_n_trials = (number of heads in n trials) / n

d is the parameter that determines how closely we want the error bounded

g is the probability bound for the event that we are not within d of the true value

Show that this ALWAYS holds if n >= O(1/d^2 * log( 1/(g*d) ) )

I had 15 min to solve this question.

## Want to Vote on this Content?! No WSO Credits ?

I don't know why this called "Hardest Probability Questions" ..

I assume it is a kind of Joke ..

Lets Solve first Rolling Once ..\

The expected value of a random variable X is denoted by E(X). For a discrete random variable, E(X) is calculated as

Rolling a "1" has a probability of 1/6.

Rolling a "2" has a probability of 1/6.

Rolling a "3" has a probability of 1/6.

Rolling a "4" has a probability of 1/6.

Rolling a "5" has a probability of 1/6.

Rolling a "6" has a probability of 1/6.

Multiplying the values with their respective probability gives:

1 * 1/6 = 1/6

2 * 1/6 = 2/6

3 * 1/6 = 3/6

4 * 1/6 = 4/6

5 * 1/6 = 5/6

6 * 1/6 = 6/6

Adding them together gives:

1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5

Now solvin Rolling the Dice Twice ,,

EX. The random variable X has the following probability distribution: x pX(x)

2 1 / 36

3 2 / 36

4 3 / 36

5 4 / 36

6 5 / 36

7 6 / 36

8 5 / 36

9 4 / 36

10 3 / 36

11 2 / 36

12 1 / 36

The random variable X assumes a value equal to the sum of two dice rolls. Its expected value is calculated as

= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36)

= (1/36) (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)

= (252/36) = 7

In the long run, the average value of two dice rolls using regular dice is 7.

## Want to Vote on this Content?! No WSO Credits ?

dude, just add the expected value of two independent variables and you get 7. no need for all that calculations.

## Want to Vote on this Content?! No WSO Credits ?

are you actually retarded?

## Want to Vote on this Content?! No WSO Credits ?

..........Are you the kind of person who thinks the "7 8 9" joke is funny?

## Want to Vote on this Content?! No WSO Credits ?

assuming a strategy where you only chose to roll the die again if the outcome of the first die was below the expected value of rolling a fair die (3.5)

you get 4.25 because you still have (1/6)(4 + 5 + 6) but you also add (1/12)(1 + 2 + 3 + 4 + 5 + 6)

## Want to Vote on this Content?! No WSO Credits ?

basically you have to relate the confidence interval (involving standard deviation of the sample mean, which is a summation of bernoullis (i.e. binomial)) to the size of the test, and produce the minimum "n" that will guarantee that your confidence interval is correct.

certainly challenging (especially with the 15 minute time frame).

## Want to Vote on this Content?! No WSO Credits ?

This is incorrect. The only possible results of the toss with your strategy (which is correct) is 1,2,3,4,5,6, 4,5,6. Hence the expectation value (mean payout) is 4, not 4.25.

## Want to Vote on this Content?! No WSO Credits ?

Jesus christ. This was at an interview? I'm guessing given the difficulty it was either Jane Street or DE Shaw.

## Want to Vote on this Content?! No WSO Credits ?

I'm guessing he did a stats heavy course at university/grad level, and the interviewer had too. Neither of those companies will ask you about concepts you haven't seen before. It's a core part of their ethos.

having had that kind of training, this question gets easier. I'm not saying it isn't hard, but this is a lot more approachable if you've done similar stuff for the past 3 years

## Want to Vote on this Content?! No WSO Credits ?

what is O? I assume its not zero...

## Want to Vote on this Content?! No WSO Credits ?

http://en.wikipedia.org/wiki/Big_O_notation

## Want to Vote on this Content?! No WSO Credits ?

No no i have the hardest problem, given that f(x) is differenciatable at x = 0 and Lim/(x?>0) (f(x)?f(kx))/x = a, where a and k are constants. Show that f'(0) = a/1-k

"...all truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident."

## Want to Vote on this Content?! No WSO Credits ?

lim/(x->0) (f(x)/x) - lim/(x->0) (f(kx)/x) = a

take derivative of top and bottom (L'Hopital's rule) and get

lim/(x->0) f'(x) - lim/(x->0) k

f'(kx) = af'(0) = aplug in 0

f'(0) - k

factor

f'(0) (1-k) = a

f'(0) = a/(1-k)

no fuckin clue about the probability problem tho

## Want to Vote on this Content?! No WSO Credits ?

Hmm I disagree, but I could be wrong. Just did this on scrap paper and got the same thing just (d-1/2)^2 in the denominator. Are you sure thats the correct answer?

**Ops read the question wrong. Disregard this.

## Want to Vote on this Content?! No WSO Credits ?

Looks like Strong Law of Large Numbers to me.

Given the hint, the P( max | x - x_n_trials | > d ) < = summation (1 to n) P( | x - x_n_trials | > d ) (Boole's inequality)

since each xi is bounded by 0 and 1, the latter probability is gonna be a finite (1 to n) sum of 2

exp (-2d^2*n). (Hoeffding's inequality)Such thing converges since it is a geometric series. It should give you the result.

## Want to Vote on this Content?! No WSO Credits ?

^ yes

## Want to Vote on this Content?! No WSO Credits ?

^ Almost surely

## Want to Vote on this Content?! No WSO Credits ?

Actually the hardest question for this forum would be prove pi is irrational!

## Want to Vote on this Content?! No WSO Credits ?

Have done proof for e and pi...

## Want to Vote on this Content?! No WSO Credits ?

lik omg infinite descent ~~!!!

look at my skill!!!!!!!!!!!!!!!!!

## Want to Vote on this Content?! No WSO Credits ?

i think proofs are over rate BC at the end of the day it has to be judged to actually be a proof proof :/

and proofs != end all to maths

TBH

/DEBATE

gogogo

## Want to Vote on this Content?! No WSO Credits ?

4.25

Another one: given a circle, make n random cuts, what is expected length of longest piece?

## Want to Vote on this Content?! No WSO Credits ?

Probability Problem has me stumped(Originally Posted: 10/12/2012)You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?

Im thinking its symetric so 1/2^3 either that or find expected first pick then prob of getting higher than that for the second card and so on

any ideas?

## Want to Vote on this Content?! No WSO Credits ?

For each draw {A , B , C , D} , isn't there a 1/4! chance that you draw this draw in ascending order.

Therefore, isn't the EV of this game (1/4!)* 10 - ( ((4! - 1)/4!)*1).

## Want to Vote on this Content?! No WSO Credits ?

for your first 2 cards, the probability of the second being > first is even, yes. After that, though, the probability of each subsequent card being > than the one before decreases.

in fact, my guess would be it goes something like this - P(2>1) = 0.5

P(3>2) = 48.5/95 (we know card 1 is one of the cards less than card 2, so the probability of card 3 being greater than card 2 is always skewed by 1 card out of 95?)

P(4>3) = 49/94 (repeat logic from above but skewed by 2 cards out of 94)

So you get (1/2)

(48.5/95)(49/94) = ~0.133, slightly greater than (1/2)^3, which is intuitive because these are non-independent in a way where each previous step positively impacts the probability of the subsequent step?Caveat: I'm not 100% confident in the answer, I'm also more confident in my reasoning than my math (embarrassingly enough)

## Want to Vote on this Content?! No WSO Credits ?

so the 97 cards are just labeled 1-97? i'm confused

This to all my hatin' folks seeing me getting guac right now..

## Want to Vote on this Content?! No WSO Credits ?

Let me expand my answer:

The probability that a random draw of 4 cards will be in ascending order should be 1/4!, irrespective of the sample size. Why?

Consider the case where you draw {1 , 2 , 3 , 4}. Given that you have drawn {1, 2, 3, 4}, the probability that you have drawn them in

exactly that orderis 1/4! or 1/24. Since the situation is symmetric with respect to each such quadruplet,the probability that your random draw is in ascending orderis 1/4!.From here we can calculate the EV of the game 1/24*10 - 23/24 = -13/24 < -0.5 .

## Want to Vote on this Content?! No WSO Credits ?

GS is absolutely correct. Whether you are drawing from 97, 500, or 1,000,0000 cards, the probability of drawing 4 cards in ascending order is the same. You can draw 4 cards in 4! or 24 ways. Furthermore, only 1 of those 24 ways will yield cards in ascending order.

## Want to Vote on this Content?! No WSO Credits ?

GS is right. I wrote a quick python script and got about -.57 over a million tries

http://www.codesend.com/view/738cb0bbd40e9642bb720...

This to all my hatin' folks seeing me getting guac right now..

## Want to Vote on this Content?! No WSO Credits ?

Isn't your code drawing the cards with replacement? Looks like it to me (I don't know how the random class works but if it's a generic random num generator, then you are drawing them with replacement)

## Want to Vote on this Content?! No WSO Credits ?

Yeah you're right. Fixing it now

This to all my hatin' folks seeing me getting guac right now..

## Want to Vote on this Content?! No WSO Credits ?

http://www.codesend.com/view/435936f17ff0e09a91f16...

There. That gets me a lot closer to the -13/24 value. Good catch.

This to all my hatin' folks seeing me getting guac right now..

## Want to Vote on this Content?! No WSO Credits ?

yea you are right the red herring threw me off

## Want to Vote on this Content?! No WSO Credits ?

I believe the answer to the snake one is no. My explanation may not make a lot of sense though. Basically, in order for you to get all the snakes a single color you need to get the other 2 sets of snakes to be equal. Okay, so you have 13, 15, and 17 snakes. Every time you make a change, 2 numbers go down by 1 and one number goes up by 2. This means that you need 1 type of snake to be exactly 3 less than another in order to equal them out.

Right now the snakes are all 2 (or 4) apart in quantity. There is no change or set of changes you can do to make 2 colors seperated by 3. This is due to each mutation either closing the gap by 3 or keeping them equal. Hence, impossible.

Not the best formatted answer but that is my answer.

CompBanker

## Want to Vote on this Content?! No WSO Credits ?

PROBABILITY brainteasers(Originally Posted: 10/07/2010)Hey evryone - as it is that time of year when the rat race becomes "racier," if you will, there are many of us interviewing with prop trading firms. any firm worth its while will be asking probability brainteasers. i know of few other ways to get a sampling of some great brainteasers than by asking the fine gentlemen on this site. so, with that being said ... what are your favorite brain teaser, and/or the ones you have personally had in interviews ?

## Want to Vote on this Content?! No WSO Credits ?

yes one or two of the previous explanations to the snake problem are correct. The snakes cannot all become the same colour simply because of the relative difference between sankes gaiven being divisible by two. By inspection , only a difference between any two snake numbers that is divisible by 3 will work . Hence the answer is no!

## Want to Vote on this Content?! No WSO Credits ?

There is some thread laying around on the net that focuses on quantitative questions

http://www.quantnet.com/forum/showthread.php?t=1152

Also, you probably heard about the "Heard on the Street". The new book by Mark Joshi is much better

http://www.quantnet.com/master-reading-list-for-qu...

## Want to Vote on this Content?! No WSO Credits ?

do they let you use paper or a calculator for these?

## Want to Vote on this Content?! No WSO Credits ?

Question with Card Probability brainteaser(Originally Posted: 10/05/2012)I heard this question.

You are given 16 cards 4 are hearts, 4 are clubs, 4 are spades and 4 are diamonds.

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

I assume the answer is 25%? Out of all the different combinations you can make, I assume 1/4th will end in clubs 1/4th will end in diamonds and 1/4th will end in spades and 1/4 will end in hearts.

So is 25% correct for this? I start over thinking, that you have a higher probability to draw a non club card which will increase the chances of a club card showing at the end, but I do not think this is true.

Thank you

## Want to Vote on this Content?! No WSO Credits ?

Seriously?

MY BLOG

## Want to Vote on this Content?! No WSO Credits ?

My first thought:

(4 C 1) (15 C 3) / (16 C 4)

This is wrong, though, since it's 1. I'll double-check and get back to you.

Second thought: Yep, it's 1/4.

## Want to Vote on this Content?! No WSO Credits ?

Are the cards being replaced after each draw or not? If yes, then it is 1/4 since on the 4th draw you will have 4 clubs out of 16. But if the cards are not being replaced, then you have to sum up the 4 distinct possibilities. By the last draw, there will be 13 cards left. If you had picked clubs in each of the previous 3 draws, then your probability of drawing the last club will be 1/13. If you had drawn 2 clubs in the previous 3 draws, your probability would be 2/13 and so forth. Hence, you will need to add up 1/13+2/13+3/13+4/13 to get 10/13, which should be the final answer.

## Want to Vote on this Content?! No WSO Credits ?

It doesn't matter. It's 1/4. The problem is perfectly symmetrical.

If I asked you about diamonds in particular, you'd still tell me 10/13. If I asked about spades, you'd still tell me 10/13. Clearly there's a problem here. The answer is 1/4.

This is actually a very clever interviewing question. Among other things, it tests how sure of yourself you are and if you're the type to just jump into math without considering the situation first.

I had two other clever questions pop up in an interview recently, both of which made you go through long-winded calculations whose answer was most of the work of the problem, but not the brief final step from there that they wanted. Luckily I noticed the second time it happened, but watch out for that.

## Want to Vote on this Content?! No WSO Credits ?

let me rewrite this question in a simpler way for you.

"What is the probability of choosing a Club when choosing a card at random?".

It doesn't matter if it's the fourth or fourteenth card, if you don't know anything about the other cards being drawn.

Here's another one for you:

You have a friend who has two children. At least one of the children is a boy, what are the odds that your friend has a girl?

MY BLOG

## Want to Vote on this Content?! No WSO Credits ?

And the relevant:

You have a friend who has two children. The eldest is a boy, what are the odds that your friend has a girl?

[It's important to know that the answers are not the same]

## Want to Vote on this Content?! No WSO Credits ?

Im sorry could you explain the answers again? I'm not sure what you mean about wrt, and it looks like unless im obviously reading it incorrectly i two different answers.

The first question with at least one is a boy chances she has a girl... your sample size is BB GB BG how can you know the chances of her having a girl (I assumed it was the 2 scenarios that have a girl in it over all scenarios so 2/3 but if you do that for the boy you see its 3/3 so that cannot be right.

The second question where the eldest is a boy the sample set is B(younger)B(older) G(younger)B(older)... same question again how do you clearly look at knowing the sample spaces if i was thinking it incorrectly in the first one

edit:

I think i figured out how I was thinking wrong... if the question was posted as at least 1 is a boy what is the probability of the other being a boy... sample space is bb gb bg so it would be 1/3 not 3/3 (the way i was erroneously thiking of 3/3 thus my thinking was wrong was posing the quesiton in my head whats the probability that at least one is a boy given that one is a boy)...derf

thank you

## Want to Vote on this Content?! No WSO Credits ?

WRT means with respect to. So, if you have drawn n cards, and don't know what any of them is, then the situation is symmetrical with respect to each suit.

It is right. P(couple has a boy | at least one boy) = 1. P(couple has a girl | at least one boy) = 2/3

The events in the sample set represent the possible configs. If eldest comes first (B G) , (B B) are the outcomes if the eldest child is a boy. In that case, the probability that the next kid is boy/girl is 50-50 as you would expect.

## Want to Vote on this Content?! No WSO Credits ?

What are the answers to these? Why are they different?

## Want to Vote on this Content?! No WSO Credits ?

What are the answers to these? Why are they different?

## Want to Vote on this Content?! No WSO Credits ?

What are the answers to these? Why are they different?

## Want to Vote on this Content?! No WSO Credits ?

In the first case, you have a symmetrical sample set wrt to each suit, therefore the probability is 1/4.

In the second case, your sampling space is [(B , B) (B, G) (G,B)] { (G,G ) is omitted }. Since each event in this space is equally likely, the prob that your friend has a girl is 2/3.

## Want to Vote on this Content?! No WSO Credits ?

Oh I see. And the case where the boy is eldest makes the sample space [(B,B), (B,G)] so the probability is now 1/2. Thanks for clarifying

## Want to Vote on this Content?! No WSO Credits ?

I know I've already posted in your other thread but..

Think of it as a permutation of the 16 cards. The probability of any one card out of the 16 being the 4th card is 15!/16! = 1/16. And since there're 4 clubs, the probability of the 4th card being a club is 4 x 1/16 = 1/4.

And also, as one of the above posters mentioned, since the condition for each of the 4 suits is identical, the probability of the 4th card being each suit has to be equal, which is a much quicker way to get the answer.

## Want to Vote on this Content?! No WSO Credits ?

After you have drawn 3 cards, the expected no of cards of each suit left are (4 - 3/4). Thus, prob of 4th being from any suit is 1/4.

## Want to Vote on this Content?! No WSO Credits ?

Correct

## Want to Vote on this Content?! No WSO Credits ?

There are four possible outcomes, each equally likely:

Boy-Boy, Boy-Girl, Girl-Boy and Girl-Girl

Probability that couple has "a girl" is 3/4. The probability that the couple has "a boy" is also 3/4. They're not mutually exclusive of course. The probability that a couple has two boys is 1/4, two girls is also 1/4 etc.

"You have a friend who has two children. At least one of the children is a boy, what are the odds that your friend has a girl?"

You can exclude the Girl-Girl outcome. That leaves Boy-Boy, Boy-Girl, Girl-Boy. In two out of these three possible outcomes the friend has a girl, so the probability for that is 2/3.

"You have a friend who has two children. The eldest is a boy, what are the odds that your friend has a girl?"

In this case you can exclude two outcomes Girl-Boy and Girl-Girl because the older kid is a boy. That leaves Boy-Boy, Boy-Girl and in one of these the kid is a girl, so the probability is 1/2.

## Want to Vote on this Content?! No WSO Credits ?

I think the previous poster forgot one fact that changes things a bit.

The key here is that the problem says "you can roll the die again and get paid what the second roll shows INSTEAD of the first" (caps added). This means that you if you decide to re-roll, you should disregard your first roll and just keep the second one.

The first time you roll the dice, you have two options:

Since the expected value (EV) of one dice roll is 3.5, it makes sense to keep your score if it is higher than 3.5, and re-roll if you get a score less than 3.5.

If you roll over 3.5, then the score you got must have been either 4, 5, or 6. Of these three numbers, the EV is 5.

The other possibility (which is equally likely) is that you get 1, 2, or 3. Then, it makes sense to re-roll. Again, on your second roll, your EV is 3.5.

Since you have two options, which are equally likely, the EV is:

1/2(the chance you don't reroll) + 1/2(the chance you re-roll and ignore your first score)

= 1/2(5) + 1/2(3.5)

= 1/2(8.5)

= 4.25

Note: If you had to keep the first roll and add it to your second score, it would always make sense to re-roll. Then, as the last poster mentioned, the EV would be 3.5+3.5 = 7.

## Want to Vote on this Content?! No WSO Credits ?

probability questions(Originally Posted: 08/16/2007)Hi guys,

I will appreciate it very much if anyone can tell me whether there is any probability questions asked during the interview for structured finance entry level positions? How about interviews for rating agencies like FitchRathings?

Hang in there!

## Want to Vote on this Content?! No WSO Credits ?

The problem was missing information, as the expected value depends on the strategy the user uses. To find the optimal strategy, we can safely argue that, by intuition/inspection, the strategy we want to use is that if we roll n or less, then we roll a second time, otherwise we keep our first roll. The expected value is then

(6 + n)/2 * (6 - n)/6 + 3.5 * n/6

Treating n as continuous will give us a maximum at n = 3.5. We then check n = 3 and n = 4 to see which value gives us the highest expectation, and when n = 3 we get 4.25 whereas when n = 4 we get 4, so we choose the strategy with n = 3 which will give us the max expectation.

## Want to Vote on this Content?! No WSO Credits ?

c'mon what're the odds you get probability questions?

## Want to Vote on this Content?! No WSO Credits ?

according to Bayes' Rule (what is the probability of getting a probability question, given that you have an interview for an entry level position in structured finance): about 13%

See my other WSO blog posts

## Want to Vote on this Content?! No WSO Credits ?

(interview for entry level position in structured finance | probability of getting a probability question) = 0.13

Or did I get the order wrong? Game theory was so long ago....

Financial Modeling Training

Guide to Finance Interviews

Banking Resume

## Want to Vote on this Content?! No WSO Credits ?

Well according to Murphy's law, there is a 100% chance that you will get a probability question, given how worried you seem.

## Want to Vote on this Content?! No WSO Credits ?

thx guys

Hang in there!

## Want to Vote on this Content?! No WSO Credits ?

Hahaha, good response.

Financial Modeling Training

Guide to Finance Interviews

Banking Resume

## Want to Vote on this Content?! No WSO Credits ?

I think that P(X>Y) is 1/4. Here's why:

the odds that you get two straight heads on your 2nd roll is 1/4

the odds that you get two straight heads on your 3rd roll is (1/4)

(1/2)... 1/2 is the odds that first roll is tails(1/2)*(1/2)the odds that you get two straight heads on your 4rd roll is (1/4)

Getting the two straight heads by roll 3 or roll 4 has the same likelihood as getting 3 straight (1/8 and (1/8)*(1/2)). The only difference between X and Y is that Y can't happen in 2 rolls, whereas X has a 1/4 chance

## Want to Vote on this Content?! No WSO Credits ?

Book to learn how to answer tough probability questions?(Originally Posted: 06/01/2010)Does anyone have a suggestion for this? I took a Business stats course this year but it was a joke compared to the prob Qs/brainteasers seen on here. Anyone have any suggestions for how to learn how to do these? Or is it just innate skill? I'm looking for something textbook like, I already have the Crack book.

## Want to Vote on this Content?! No WSO Credits ?

Downtown, you're overlooking the fact that there are two ways to create the event X=4

You can have the following for X=4 (where T is tails and H is heads)

T T H H

H T H H

The actual probabiliy function of X=x is:

f(x) = F_(x-1) * (1/2)^x, where F_(x-1) represents the Fibonnaci sequence and x=2,3,4,....

Derivation of f(x)If you write it out, you will discover that there's

1 way to make X=3

2 ways to make X=4

3 ways to make X=5

5 ways to make X=6

8 ways to make X=7

The probability of two heads is (1/4) or equivalently, (1/2)^2

The probability of anything before that is (1/2)^(x-2).

Therefore f(x) = F_(x-1)

(1/2)^(x-2)(1/2)^2= F_(x-1)*(1/2)^x

If you check out the Wikipedia article on Fibonnaci Power Series, you can show that the sum of f(x) from x=2 to x=infinity does equal 1, and thus f(x) is a proper probability mass function.

Regarding event Y...Y is a little bit more dfficult. Since X and Y are independent, the P(X>Y | Y=y) = P(X > y) = sum of f(x) from x=y+1 to x=infinity or equivalently, 1 - f(2) - f(3) ... f(y).

To show the P(X>Y) for all Y, we must calculate the infinite sum:

P(Y=3)

P(X>3)+P(Y=4)P(X>4)+P(Y=5)*P(X>5)...The PMF of Y=y is:

g(x)=G_(x-2)*(1/2)^x for x=3,4,5,6,...

where G_x= G_(x-1)+G_(x-2)+G_(x-3) with the seed G_0=0, G_1=1, and G_2=1

If you write it out, there's

1 way to make Y=4

2 ways to make Y=5

4 ways to make Y=6

7 ways to make Y=7

13 ways to make Y=8

.... and I don't know how to handle that sequence...

So.... that one problem probably takes the cake for hardest probability problem... perhaps there's some elegant solution to the problem

## Want to Vote on this Content?! No WSO Credits ?

Not a textbook, but covers just about every quant question you'll face in interviews:

http://www.amazon.com/Heard-Street-Quantitative-Qu...

If you want a textbook, here's an old, but well written probability text:

http://www.amazon.com/Probability-Stochastic-Proce...

## Want to Vote on this Content?! No WSO Credits ?

Many thanks Monkey!

## Want to Vote on this Content?! No WSO Credits ?

Also, anyone have a pdf of the solomon book?

## Want to Vote on this Content?! No WSO Credits ?

Very true. The expectation operator is a linear one.

## Want to Vote on this Content?! No WSO Credits ?

Probability Question(Originally Posted: 01/25/2008)Anybody?

a player has a 30% probability to score in a game

what is the probability he will score at least one goal in next 3 games?

Thanks.

Luke

## Want to Vote on this Content?! No WSO Credits ?

1-((1-.3)^3) = 65.7%

Author of www.IBankingFAQ.com

## Want to Vote on this Content?! No WSO Credits ?

i was thinking more like this:

5 / 3! = 5 / 6 = 83%

6 combinations of scenarios:

FFF

FFW

FWF

WWW

WFW

WFF

the only scenario in which he loses more all games is FFF so that leave 5 success

## Want to Vote on this Content?! No WSO Credits ?

First, there are 8 possible combinations not 6 (2^3). Second, you're ignoring the fact that the chances of F and W are not equal. All you have to do to answer the question is calculate the probability of FFF which equals .7*.7*.7 which equals 34.3%. That's the probabilty that no goals are scored in any game. All of the other scenarios meet your requirement of at least 1 goal scored. So the answer is 1 minus .343.

Author of www.IBankingFAQ.com

## Want to Vote on this Content?! No WSO Credits ?

Since saying "the odds he will score at least one" is the same as saying "1 minus the odds that he won't score any", we use the "no goals" scenario as our basis.

30% chance of scoring means 70% chance of not scoring.

Since each game is an independent event, the odds of him not scoring at any of the three is simply their product:

70% x 70% x 70% = (.70)^3 = 0.343

So since we have a 34.3% chance that he will not score anything, we know that there is a 100% - 34.3% = 65.7% chance he will score at least one.

## Want to Vote on this Content?! No WSO Credits ?

this is a simple binomial question

## Want to Vote on this Content?! No WSO Credits ?

1-0.7^3

## Want to Vote on this Content?! No WSO Credits ?

as someone said, simple binomial problem

Sales and Trading: Interview Guide written by Hedge Fund Trader

Everyday-Alpha.com: Investment Hacks

## Want to Vote on this Content?! No WSO Credits ?

Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.

## Want to Vote on this Content?! No WSO Credits ?

Probability help(Originally Posted: 09/01/2008)Any one know any good books on probability theory? I'm looking deeper into options and realize that I need a better base of prob knowledge to really go any further.

Thanks

## Want to Vote on this Content?! No WSO Credits ?

sorry, i don't think andrew wiles posts here anymore.

## Want to Vote on this Content?! No WSO Credits ?

never know <3

## Want to Vote on this Content?! No WSO Credits ?

Haha, we did that in IB HL math (Grade 12)

## Want to Vote on this Content?! No WSO Credits ?

you need to explain more specifically what you are trying to do. are you trying to get a qualitative understanding of the greeks?

are you looking for a textbook for an intro college probability course (with multivar + continuous distributions)? that will get you started in options, but to derive even a relatively simple model yourself (like black-scholes) you need to go down the rabbit hole into the world of stochastic calculus...

## Want to Vote on this Content?! No WSO Credits ?

Sure, you solved this in grade 12! This is why it took more than 3 centuries to find a proof to this simple problem. It was only solved in 1994 by Andrew Wiles and Richard Taylor. (@ Art Vandelay: I'm not sure everyone got your post ;)

Read "Fermat's last theorem" by Simon Singh. Highly recommendable book!

Or have a quick look into wikipedia: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

## Want to Vote on this Content?! No WSO Credits ?

If you are looking for something about the history of risk and probability, this was an interesting read.

## Want to Vote on this Content?! No WSO Credits ?

Like structure sort of said, you will need to be very comfortable mathematically (relative to normal finance math).

That said, Shreve's Stochastic Calculus for Finance I and II (2 books) are great from what I've heard (although I've never read them past the introductions, the second book has 2 chapters introducing/reviewing probability theory through information & conditioning). Maybe see if your library has a copy before buying, though.

## Want to Vote on this Content?! No WSO Credits ?

shreve's is the best stuff out there - actually explains most of the math instead of merely laying out the underlying assumptions.

in general, it's not probability that gets most people hung up on options pricing - it's the differential equations. for example, a lot of pricing books (ie - Hull) will simply say 'and here we solve the black scholes PDE' - but they don't always explain how they derive or solve it, or really what it means.

## Want to Vote on this Content?! No WSO Credits ?

hrmph well i guess i'll look to see if my library has shreve's text book. thanks

## Want to Vote on this Content?! No WSO Credits ?

yesman has it spot on. PDEs are very important for trading options.

## Want to Vote on this Content?! No WSO Credits ?

Grimmet and strizaker if you are serious about it

## Want to Vote on this Content?! No WSO Credits ?

and if he did, no one here would understand the proof anyway.

## Want to Vote on this Content?! No WSO Credits ?

Probability review?(Originally Posted: 08/09/2011)Hi guys,

I'm a rising junior at an I-hope-it's-target-Midwest-state-school (well I'm sure the B-school is but ehh...) studying a combination of pure math and economics. I'm interested in trading (although the recruitment process for some of the prop shops looks absolutely terrifying/awesome). Looking around on the forum, I noticed there seems to be a big emphasis on probability.

I've done a good amount for the pre-reqs econometrics, but admittedly, have gotten a bit rusty (we covered all the standard stuff up to annoying multi integrals for multivariate distributions). Does anyone know how much material is important for interviews, as well as a good text to pick it up from again?

Courses in the math department aren't really that great an option since the only things that are appropriate for my major are measure theory based probability stuff and I have absolutely no idea how to apply measure theory.

Thanks!

## Want to Vote on this Content?! No WSO Credits ?

the probability in the interviews is nothing that is technically/theoretically advanced, its really the foundation concepts of probability like Bayes theorem but under a great amount of time pressure and in a brainteaser format.

Sales and Trading: Interview Guide written by Hedge Fund Trader

Everyday-Alpha.com: Investment Hacks

## Want to Vote on this Content?! No WSO Credits ?

Are you from IU?

## Want to Vote on this Content?! No WSO Credits ?

interview probability will be simple EV type stuff that just gets marginally harder to make sure you can think analytically/prove you can do some sort of math. nothing mind boggling. know dice/coinflip EV etc..

## Want to Vote on this Content?! No WSO Credits ?

Thanks. Guess I'll just look over some of my old notes, good to know it's not going to be too bad.

## Want to Vote on this Content?! No WSO Credits ?

the answer to the snake question is YES, agree with ralph! Coin question is 4.25, and dice quesiton is 3.5

## Want to Vote on this Content?! No WSO Credits ?

Card Probability question. Anyone know how to do it?(Originally Posted: 10/05/2012)I heard this question.

You are given 16 cards 4 are hearts, 4 are clubs, 4 are spades and 4 are diamonds.

What is the probability that the fourth card drawn is a club. (you aren't told what you drew in cards 1 2 or 3).

I assume the answer is 25%? Out of all the different combinations you can make, I assume 1/4th will end in clubs 1/4th will end in diamonds and 1/4th will end in spades and 1/4 will end in hearts.

So is 25% correct for this? I start over thinking, that you have a higher probability to draw a non club card which will increase the chances of a club card showing at the end, but I do not think this is true.

Thank you

## Want to Vote on this Content?! No WSO Credits ?

It depends if the cards are replaced or not...if not you need to make a path dependent tree.

## Want to Vote on this Content?! No WSO Credits ?

doesn't matter. if u don't know what was picked then it's 25% whether you replace or not. Let me say it this way:

if there are 16 cards, same way as in the OP and if I picked 15 randomly and didn't replace, what are the odds that the 16th card is a club? 25%. Same with the first, second, etc.

//www.wallstreetoasis.com/forums/question-with-card-...brainteaser

## Want to Vote on this Content?! No WSO Credits ?

if the card is replaced, it is 25%

if not, you need to take into account the first 3 cards

Born in hell, forged from suffering, hardened by pain.

## Want to Vote on this Content?! No WSO Credits ?

Think of it as a permutation of the 16 cards. The probability of any one card out of the 16 being the 4th card is 15!/16! = 1/16. And since there're 4 clubs, the probability of the 4th card being a club is 4 x 1/16 = 1/4.

## Want to Vote on this Content?! No WSO Credits ?

Miscer the probabilities might match so you could be right, the three will spit at 1/4 as well

## Want to Vote on this Content?! No WSO Credits ?

Or you could do it like this.

16! combinations of cards, lock in the fourth as a club (4 ways to do this), You end up with 15! of arranging the cards. Thus you get 4*15!/16!=1/4. Yep you guys are right

## Want to Vote on this Content?! No WSO Credits ?

Calculate P{X>Y}

Solution: 3/32

Problem can be thought of as "what is the probability t