Probability questions for S&T interview

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

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

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.

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?

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?

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.

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"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."

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.

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...

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"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."

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

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.

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.

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

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)

"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?

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?"

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?

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?

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.

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.

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..

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.

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.

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.

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.

42 is the correct answer.

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?

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

the above 2 questions are essentially the same.

How applicable is all this to trading?

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 )

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?

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