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Comments (195)

Feb 12, 2007 - 9:09am

What to have some fun? Try these two.

  1. 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}

  1. 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?

Aug 13, 2010 - 2:57am

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

Aug 14, 2010 - 9:29am
ralph240574:
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.

Nov 16, 2010 - 12:24pm

Solution to problem 1 can be found here: http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-…

sonyyy:
What to have some fun? Try these two.
  1. 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}

  1. 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?

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Nov 16, 2010 - 6:38pm

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.
  1. 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}

  1. 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?

Nov 23, 2010 - 3:26pm

Solution to Problem 1:

http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-…
Best of Luck!

Anal Analyst:
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.
  1. 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}

  1. 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?

Sep 26, 2020 - 9:08pm

sonyyy

What to have some fun? Try these two.

  1. 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}

  1. 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?

sonyyy

What to have some fun? Try these two.

  1. 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}

  1. 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?

Thank you for the problems. I think I got the answers right?

1- rephrased, it asks "What is the probability of getting 3 consecutive heads before 2 consecutive heads". This would be zero

2 - represent number of reds blues and yellows with 'a', 'b' & 'c' where a>b>c. Possible only when n is a whole number:

n = log ((a+b+c)/(b-c)) / log (2) 

so not possible in this case 

Apr 11, 2016 - 1:37pm

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!

Apr 11, 2016 - 1:39pm

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.

Apr 11, 2016 - 1:41pm

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?

Apr 11, 2016 - 1:42pm

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."

  • 2
Apr 11, 2016 - 1:43pm

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.

Apr 11, 2016 - 1:44pm

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."

Apr 11, 2016 - 1:47pm

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.

Apr 11, 2016 - 1:48pm

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.

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

Apr 11, 2016 - 1:49pm

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)

Apr 11, 2016 - 2:06pm

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 )

Apr 11, 2016 - 1:50pm

"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?

Apr 11, 2016 - 1:51pm

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.
Apr 11, 2016 - 1:52pm

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?

Apr 11, 2016 - 1:56pm
mykyta:

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..

Apr 11, 2016 - 1:53pm

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?

Apr 11, 2016 - 1:54pm

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.

Apr 11, 2016 - 1:57pm

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.

Apr 11, 2016 - 1:58pm

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.

Apr 11, 2016 - 2:00pm

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.

Apr 11, 2016 - 2:02pm

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?

Apr 11, 2016 - 2:07pm

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?

Feb 12, 2007 - 11:44am

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?

Apr 11, 2016 - 2:10pm

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 ) = O(1/d^2 * log( 1/(g*d) ) )

I had 15 min to solve this question.

Jul 18, 2009 - 7:24am

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.

Apr 3, 2010 - 2:34pm
IdrisAlMalki:
I don't know why this called "Hardest Probability Questions" ..
I assume it is a kind of Joke ..

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.

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

Jan 8, 2016 - 8:56pm
IdrisAlMalki:

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.

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

May 9, 2014 - 1:01am

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)

Apr 11, 2016 - 2:11pm

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).

  • Anonymous Monkey's picture
  • Anonymous Monkey
  • Rank: Chimp
Jan 8, 2016 - 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.

Apr 11, 2016 - 2:16pm
Brady4MVP:
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

Apr 11, 2016 - 2:14pm

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
Apr 11, 2016 - 2:15pm
seabird:
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) - k
f'(0) = a
factor
f'(0) (1-k) = a
f'(0) = a/(1-k)

no fuckin clue about the probability problem tho

Apr 11, 2016 - 2:19pm

Looks like Strong Law of Large Numbers to me.
Given the hint, the P( max | x - x_n_trials | > d ) 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.

Apr 11, 2016 - 2:26pm

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?

Apr 11, 2016 - 2:27pm

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).

Apr 11, 2016 - 2:31pm

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)

Apr 11, 2016 - 2:28pm

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

This to all my hatin' folks seeing me getting guac right now..
Apr 11, 2016 - 2:29pm

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

Apr 11, 2016 - 2:30pm

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.

Apr 11, 2016 - 2:34pm

Yeah you're right. Fixing it now

This to all my hatin' folks seeing me getting guac right now..
Apr 11, 2016 - 2:37pm

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 ?

Feb 12, 2007 - 2:09pm

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

  • 2
Apr 2, 2010 - 1:43pm

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!

Apr 11, 2016 - 2:39pm

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

Apr 11, 2016 - 2:41pm

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.

Apr 11, 2016 - 2:42pm

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.

Apr 11, 2016 - 2:43pm

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.

Apr 11, 2016 - 2:44pm
brainteaser:
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?

Apr 11, 2016 - 2:45pm
SirTradesaLot:
brainteaser:
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?

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]

Apr 11, 2016 - 2:48pm

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?

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]

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

Apr 11, 2016 - 2:46pm

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.

Apr 11, 2016 - 2:56pm
GS:
Correct

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.

Apr 11, 2016 - 2:57pm

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!
Best Response
Dec 9, 2009 - 2:45pm

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.

  • Claudia
Nov 5, 2013 - 12:16am

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.

Apr 11, 2016 - 2:58pm

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

  • 2
Apr 11, 2016 - 3:04pm

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.

Dec 9, 2009 - 3:34pm

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
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

Dec 9, 2009 - 5:32pm
Downtown:
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).

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

Apr 11, 2016 - 3:08pm

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

Apr 11, 2016 - 3:11pm

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
Apr 11, 2016 - 3:12pm

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.

Apr 11, 2016 - 3:16pm

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

Jun 4, 2010 - 5:53am
Art Vandelay:
balbasur:
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.

sorry, i don't think andrew wiles posts here anymore.

never know

Apr 11, 2016 - 3:17pm

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...

Aug 14, 2010 - 12:43pm
HopefulIBGuy:
balbasur:
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.

Haha, we did that in IB HL math (Grade 12)

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

Apr 11, 2016 - 3:19pm

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.

Apr 11, 2016 - 3:20pm

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.

  • 1
Apr 11, 2016 - 3:24pm

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!

Apr 11, 2016 - 3:29pm

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

Apr 11, 2016 - 3:32pm
blackthorne:
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.

//www.wallstreetoasis.com/forums/question-with-card-probability-brainteas…

  • 1
Apr 11, 2016 - 3:31pm

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.
Apr 11, 2016 - 3:35pm

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

Oct 13, 2010 - 11:45am
  1. 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 that B gets 3 consecutive heads and A doesn't get 2 consecutive heads". Doesn't matter when A gets his heads as long as B gets it first.

This covers all cases because this defines when X>Y, and it becomes an easy problem.

P{A doesn't gets 2 consecutive heads} = 1 - P(A gets 2 consecutive heads} = 1 - (1/2)(1/2) = 3/4.
P{B gets 3 consecutive heads} = (1/2)
(1/2)*(1/2) = (1/8).

P{both occuring} = (3/4) * (1/8) = 3/32.

edit nvm i think this is wrong

Apr 11, 2016 - 3:36pm

probability question (Originally Posted: 06/21/2008)

i heard this brainteaser from somewhere and it has been eating at me for a while:

There is a baseball team with a .600 winning record. Currently it's lost the last four games. What is the probability that it will win the 5th game? basically, as the team continues losing (and assuming that the .600 success rate does not change) should I bet heavier in favor for a win?

Apr 11, 2016 - 3:37pm

Do past events effect future probabilities? I am going to assume no, and say that the team still has a 60% chance of winning their next game.

Apr 11, 2016 - 3:38pm

I agree with the past results (losing streak) not affecting the next game. Not sure about the probability, because you could make an argument about the .600 might not be a good predictor if it is not many games, and it probably depends on what team they are playing. Maybe I'm reading too much into it though haha.

Jack: They’re all former investment bankers who were laid off from that economic crisis that Nancy Pelosi caused. They have zero real world skills, but God they work hard.
-30 Rock

  • 1
Apr 11, 2016 - 3:39pm

Dude, you can't figure this out? That's very basic probabilities.

Either you consider the probability of losing 4 games then winning 1, P(LLLLW)=.4^4*.6^1 assuming independance, or you consider the probability of winning one game, 60%.

PS : obviously, P(LLLLL)

Apr 11, 2016 - 3:41pm

"as the team continues losing (and assuming that the .600 success rate does not change)"

just a question, if the .600 is the win-loss RATIO (which doesn't seem the same as the probability of winning a single game), then how can that ratio NOT change if the team continues losing???

Apr 11, 2016 - 3:44pm

I mean, I can see how it could be done with a weighted coin - you can theoretically adjust the weight 60-40, so it comes out heads 60% of the time and tails 40%. In this case we can say both the long term percentage of heads AND the probability for a single event will approach 60%.

So if we make the analogy that we have a 60-40 weighted coin, and it just came up tails the last 4 times, then we'd still expect it to come up heads the 5th time because of the weighting.

But I'm not sure you can say the same with a baseball game, so this is the crux:

Is the HISTORICAL winning percentage necessarily equal to the probability they will win a certain game?

Apr 11, 2016 - 3:45pm

I think i'm understanding this better. P(LLLLW) is really only the probability of a four game losing streak with one win, which is different than the probability of one win.

Apr 11, 2016 - 3:47pm

how can I solve this card probability question? (Originally Posted: 09/22/2010)

Not sure whether I remember the question correctly, but here it goes -

Say we have a deck of card (=52). Now, I choose 13 cards out of 52 and flip one over - it turns out to be a queen (or king, or ace, basically whatever). Then, what is the probability that I have a four of a kind in queen ?

Apr 11, 2016 - 3:48pm

I completely read the problem wrong, I believe drexel is correct.

"Greed, in all of its forms; greed for life, for money, for love, for knowledge has marked the upward surge of mankind. And greed, you mark my words, will not only save Teldar Paper, but that other malfunctioning corporation called the USA."
Apr 11, 2016 - 3:49pm

^ Disagree. If I remember probability correctly, that would be the odds of four Queens in a row. Also, you should be multiplying, not adding. He's drawn 13 cards, and already knows the first is a Queen, so the question is what are the odds there are another three Queens in that. Intuitively, I think it would be:

((3/51)12)((2/50)11)((1/49)*10)

Apr 11, 2016 - 3:50pm

leaving out the first queen, you now have 51 cards to choose from to form the remaining 12 cards in your hand. There are (51 choose 12) possible hands.

Of these hands, there are (3 choose 3) * (48 choose 9) hands that have all 3 remaining queens in them

So your probability is (3 choose 3) * (48 choose 9) / (51 choose 12), which is roughly 1%

Apr 11, 2016 - 3:57pm

i'm no math major, but wouldn't it be

P(four queens)/P(at least one queen) ? (edit)

we know the denominator is true

this would be

[(4 c 4)(48 c 9)/(52 c 13)] / [1 - (48 c 13)/(52 c 13)]

if my plugging-into-calc skills are good, this equals ~0.38%

Apr 11, 2016 - 3:52pm

Not a math major either but I agree with drexel

People like Coldplay and voted for the Nazis, you can't trust people Jeremy
Apr 11, 2016 - 3:53pm
Toronto:
I have taken a probability class and I agree with drex as well

Oh this is rich.

Drexel is actually wrong, jackpot779 is correct.

Apr 11, 2016 - 3:54pm

Notice he said that you flip the first one over and it turns out to be a queen/king/ace/whatever. So the question really is what's the probability that there will be 3 more of the first card in the remaining 12 face down.

You assume that the other 3 of the same cards are somewhere within the 12 remaining cards. That leaves you 9 spots to fill, and 48 choices to fill them with (since the 4 of the same card are already there). Then count how many ways that can happen, which is (48 C 9).

Agree with jackpot's solution.

Apr 11, 2016 - 3:59pm
chewingum:
So the question really is what's the probability that there will be 3 more of the first card in the remaining 12 face down

Missing this speaks volumes of my attention to detail lol

Sorry OP, hope it wasn't that important...

People like Coldplay and voted for the Nazis, you can't trust people Jeremy
Apr 11, 2016 - 4:04pm

thanks guys for posting your solutions

Shouldn't it be then (3 choose 3) * (48 choose 8) / (51 choose 12), since we have 8(instead of 9) more cards to choose out of 48 ? There are four queens so it seems to me that we have to subtract 4 from 12, not 3.

Apr 11, 2016 - 4:05pm

jackpot is right i think

it is 9. out of the other 12 cards you have, you want to figure out how many combinations of hands with
- 3 queens and
- 9 other cards
there are

Apr 11, 2016 - 4:07pm

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Apr 11, 2016 - 4:08pm

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Apr 11, 2016 - 4:09pm

Eum quos odit aut voluptatem aut tempora consequuntur. Omnis autem aperiam cumque enim molestiae consequatur soluta voluptas. Recusandae hic inventore praesentium debitis unde tempore eius. Consequuntur rerum iure autem odit aperiam ut ratione.

Labore perferendis natus cupiditate et. Tenetur cum repellat molestiae sed exercitationem itaque. Quo earum ad ipsa eum velit omnis doloremque. Blanditiis dicta nobis blanditiis nemo animi. Laboriosam et est quis dolores voluptatum nobis. Repellendus voluptate dolor similique vel quod dignissimos. Ut voluptatem doloremque enim velit laborum quia. Nulla magnam vitae corporis et.

Quia illo quidem neque recusandae consectetur voluptatem. Repellat dolor reiciendis sed nostrum. Aut ipsa impedit doloribus et dolorem. Sed rerum minima vel.

Tempora odit repudiandae aut perspiciatis quia et perferendis. Dolor dicta recusandae molestiae porro quidem voluptatibus qui. Et corporis optio tempora sint delectus. Suscipit qui fugiat aut temporibus perferendis alias quam.

Mar 16, 2013 - 2:58am

Eaque accusantium rem quod. Cumque quidem quia nesciunt voluptas deserunt quas numquam omnis. Voluptas vero aperiam voluptas rerum. Mollitia sapiente aut cumque ipsa qui ea quidem accusantium. Ullam laudantium reiciendis eum corporis ut error totam.

Voluptate sit eveniet incidunt. Ea adipisci soluta eius ducimus et optio. Sequi consequatur vero placeat molestias maiores.

Hic atque quae qui facere similique. Culpa cupiditate assumenda odio soluta sunt quo et. Dignissimos pariatur minima aperiam molestias. Eum nobis quo iure consequuntur et qui. Aut a totam atque voluptates.

Officiis qui fugit quaerat qui atque. Aut ratione non eum et. Quam laboriosam rem et. Tempora tempora ratione perspiciatis autem modi. Delectus aliquid dignissimos modi nesciunt eos maxime sed reprehenderit.

Jul 27, 2015 - 12:11am

Laboriosam dolor officiis tempore nihil hic consequuntur aspernatur. Qui omnis in autem culpa ea. Eum dolor assumenda dicta enim aspernatur. Asperiores veritatis vel fuga dolorum animi.

Autem accusamus molestiae incidunt facilis beatae. Rerum dolorem sed ea necessitatibus quos. Sapiente quidem aut unde. Modi earum dignissimos voluptatum aliquid.

Nov 5, 2013 - 12:10pm

Ratione est fuga omnis dolore quam. Ea exercitationem ut doloremque. Debitis necessitatibus veniam nihil sequi. Modi ut et sit laborum eum repellat. Et laborum enim illum debitis corrupti accusamus excepturi blanditiis. Aut perspiciatis rerum autem odio sint aut. Officiis quidem necessitatibus quos provident repellendus aut odit.

Et quaerat expedita tenetur quia perferendis maxime et. Et consequatur id corporis est ad earum quas. In ducimus maiores beatae omnis. Consequuntur aut voluptas facilis accusamus neque.

Doloribus voluptatem enim qui voluptas iusto sapiente quidem. Qui veritatis qui nesciunt. Cumque corporis omnis possimus totam. Ipsa fuga iure adipisci corporis cum et. Minus cum fuga illum quaerat autem minima.

In vitae molestias unde et laudantium et sint. Velit beatae sit soluta sit aliquid atque et. Quia non est et expedita. Aut sunt ut eaque deleniti sed similique odio itaque.

Competition is a sin. -John D. Rockefeller
  • Anonymous Monkey's picture
  • Anonymous Monkey
  • Rank: Chimp
Dec 11, 2015 - 9:19pm

Non sint molestiae in temporibus aut in et. Blanditiis nam reiciendis dolorem voluptatem totam sint eum. Maxime inventore est eligendi non aspernatur ea maxime. Itaque incidunt corrupti ipsum qui rem ut et.

Dolorem ea perferendis corrupti placeat quod iusto. Voluptatem sunt facere error quia cumque. Cumque id est consequatur eaque est perferendis et.

Quia beatae dolor ratione qui itaque maxime. Accusamus adipisci qui aut voluptatem. Ipsam quia minima aut praesentium rem eum. Dolores voluptas quae eos nobis eaque.

Earum est aut laborum earum. Quas corrupti voluptatibus facere autem. Est non quia eum ipsum. Dolores necessitatibus voluptatibus qui quia est nam accusamus.

Aug 15, 2019 - 10:59pm

Voluptatem sit consectetur distinctio delectus. Ipsum explicabo minus aperiam quis commodi consequatur. Dolores ut eaque beatae laborum ab placeat error.

Dolor iste voluptatum iste vitae expedita. Tenetur vero occaecati animi sunt ea autem. Sed rerum optio eos. Repellendus doloribus optio aperiam expedita.

Pariatur ut culpa est et debitis a aut. Est quasi nam at omnis debitis eos esse. Ipsa aut aut alias et quo. Illum soluta ut aliquid qui neque nam. Quis et quibusdam rerum nostrum maiores.

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  • Anonymous Monkey's picture
  • Anonymous Monkey
  • Rank: Chimp
Dec 15, 2015 - 10:49pm

Et sapiente fugit sequi consectetur. Soluta eum ea consectetur. Voluptas eveniet in quia est quia ut. Sed consequatur omnis in omnis quod et iusto. Sint ut ut consequatur dicta unde illo. Et blanditiis amet commodi dicta sapiente quis occaecati officia.

Cum reprehenderit quaerat sit quae. Ut culpa autem quia quisquam officia corporis. Quas in omnis voluptate suscipit labore. Magni et dolores quis. Rerum veniam dolores quis omnis. Nam fugit id animi aspernatur autem distinctio accusamus maiores.

Aug 15, 2019 - 11:11pm

Facere a consequatur libero reprehenderit ea veniam sit in. Aut explicabo omnis voluptatem eum quaerat enim. Eveniet labore exercitationem qui. Aut minima facere qui et nulla soluta vero.

Nisi molestiae illum cum beatae in quia. Est ipsa non optio sint quos. Commodi aliquam dicta incidunt consequuntur cum quasi. Iure ut commodi quia consectetur commodi voluptate enim. Eligendi reiciendis iure itaque corporis assumenda. Et sed rerum corrupti iste tempora.

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