Jane Street Capital brainteaser help

Suppose the teams play forever and the results are recorded. Then you ask about the probability that team A (or equivalently team B) wins exactly 3 times in the first 6 games. This may be computed e.g. from the binomial distribution. Answer = 5/16. Hope that helps.

Quag mire think about it like this

the sample space

(A,A,A,B,B,B) (A,A,B,A,B,B) (B,B,B,A,A,A) and so on. also think bout it when either team A or B wins it 4,5,6 games take those into effect.then obviously no game 5,6,7 is needed. hope this helps.I haven't done the problem but its fairly simple.

Kucion has it. You are calculating the likelihood that Team A wins exactly 3 of the first 6 games. Do NOT multiply this number by 2, because if A wins exactly 3 of 6 then clearly so does Team B.

this is just a coin flip problem [assuming equal probability of a win for either team]. getting to game 7 means that each team had 3 wins in the first 6 games --so if you flip 6 coins, what is the probability of getting exactly 3 heads?

i made it to the final round at JSC; i did fine on the math but failed miserably at the gambling portion. very sharp folks.

Yea I was on the verge of using binomial distribution, but was hesitant because that seemed too simple for a JSC question. Initially I had (6C3)*(0.5^3)(0.5^3), but didn't know it was right. Thanks for the confirmation. There is another brainteaser which I'm not even sure how to attempt...here it goes

You have 25 horses and a racetrack that has 5 lanes (i.e. You can simultaneously race up to 5 horses at a time). What is the minimum number of races you need to find the top 3 horses in the whole bunch?

any ideas?

Assuming that you can't record the time of each horse and then compare the results and that the relation ordering horses according to their speed remains unchanged I would say 8 races. First you create 5 disjoint groups of 5 horses each and run 5 races. What you end up with is the 5 sets of 3 horses each (the top 3 out of each race). Lets say that you denote the fastest horse in each set by A, the second fastest by B and the slowest by C. Then it is sufficient to run 3 races: one with A's, one with B's and one with C's. The winners of these races form the top 3 of 25.

Suppose the payout is $10 for a roll of a 6 on a fair die (i.e., get $10 if a 6 comes up, losing your premium paid for 1-5 sides up.) How much do you pay to play the game? (Rather, what's the FV of the game, since one would always pay below FV..) Will post the answer shortly.

Feel free to use odds/probability/expected return approaches.

Ok, I thought about it for a minute and came up with this idea: You only need seven races to find the fastes 3 horses.

-First you race all 25 horses in five seperate races. The winners are labeled A1-A5, second place finishers B1-B5, and third place finishers C1-C5.

-In the Sixth race you race all of the winners A1-A5. Lets say A1 wins, A2 finishes second and A3 finishes third. Now you know for sure that A1 is the fastest horse. But it is still possible that B1 and C1 is faster than A2 and A3 or that B2 is faster than A3. These are your possible trios at this point:

A1,A2,A3 A1,A2,B1 A1,A2,B2 A1,B1,C1

-In your seventh and last race you start A2,A3,B1,B2,C1. The top two finishers will be the second and third fastest horse.

I think that is correct. It gives you the top three horses in the correct order.

fairly certain that logic doesnt work, stk...

there are 2 ways to interpret your answer: either 1. you'll pay 3.58 to roll as many times as possible (that obviously doesn't make sense because if you have infinite rolls you know you'll get 10 eventually)

or 2. you'll pay 3.58 per roll. that doesn't make sense because then you're paying 2x the expected value of any single roll (1.67) per roll.

my guess is the right answer would be 1.67, at least hte way the question was stated here.

Nice solution of the horses brainteaser. As to the fair value question (in its current form) the xqtrack is right. If you can roll until you hit 10, the fair value is obviously 10. If you pay for one roll, then the cost of the game is its expected value, which is equal 1.67.

great posts here all. just curious, how does stuff like this come into play for daily work? don't they use much more sophisticated models for their work?

The fair value of the dice roll would be $1.67 only if you planned on playing it an infinite number of times.

Think of it this way: if I allowed you to play for $1.67 then you would have up to six rolls to hit a 6 and at least break even. Now look at the probability of hitting a 6 within the first 6 rolls:

These are the probabilities of hitting a 6 in exactly the: 1st roll - prob = .167 2nd roll - prob = 0.8330.167 = 0.138 3rd roll - prob = 0.8330.138 = 0.115 4th roll - prob = 0.8330.115 = 0.096 5th roll - prob = 0.8330.096 = 0.080 6th roll - prob = 0.833*0.080 = 0.066

Once, you hit a 6 you stop playing, take the $10 and call it a day.

Now if you add up those probabilities then you come up with 0.665, which means that you would have a 66.5% chance of at least braking even (most likely making some money). No one would give you those odds, so you can see that $1.67 is not the answer. I still stand by $3.58, but we will have to wait till Faust posts the solution.

Oops, I meant to say 1.67, not 1.34 in my above post. Regardless, I think stk is unnecessarily complicating things. It's a simple expected value question, the answer is clear given the wording of the original problem.

Thanks all, great logic and thought processes, here is the answer:

Everyone was right about P|roll 6| = 1/6, or .1667, or 16.67% Prob of Success.

Let's translate it into odds: probability of 1/6 = 5:1 odds (5/6 against rolling a 6.) I.e., one should expect to get paid $5 for every $1 bet.

So, if we're already given a known payoff of $10, and the odds are 5:1, ergo, the FV is $2.

Proof: Risk $2 to win $10, Risk/Reward ratio = 2/12 (12 is because the original bet is returned to the bettor on top of the payout.) 2/12 = .1667, (or the same 1/6) 16.67% probability of winning..

IMPLICATIONS: Using one of the given answers, if someone would be willing to pay $3.58, (R/Rw = 3.58/13.58) = .2636%, they're IMPLYING a 26.36% probability of success, which we know from historical observations (Historical Volatility, anyone?) is only 16.67%. They would have overpaid rich vol by offering too high (and over the long run would rack up net negative P&L.) Inversely, if someone could get away with offering $1.67 (the most common, albeit incorrect answer given) 1.67/11.67 = .1431, or 14.31% Prob of Winning, which would still leave a cushion of couple of volatility points to increase in one's favor to reach historical levels, giving them pricing edge in the long run.

Hence the elusive trick, correctly pricing IV% to make theo = market price, backing out from the known payoff.

Great book on the topic is Robert Ward's "Options and Options Trading: A Simplified Course that Takes You From Coin Tosses to Black-Scholes," along with anything by David Sklansky and Edward O. Thorp's "Beat the Dealer."

i kinda see your logic, and this is why i didn't become a trader (knew i couldnt handle this kind of stuff).

that being said, the EV of a die that pays $10 when it rolls a 6 is still $1.67, right (5/60 + 1/610)?

from a pure theoretical perspective (ie not using the sample calcs), can you explain to me why the fair value of the game does not equal the game's expected value?

The fair value of the game does equal the game's expected value. Faust assumes in his answer that the premium is returned in addition to the $10 for a roll of a six, which was not specified in the question. If this assumption is not made, the answer is 10/6.

Let X = fair value

Then

(10-X)(1/6)=(5/6)X

What you get if you win (10-X) * Probability of winning (1/6) = What you pay if you lose (X) * Probability of losing (5/6)

X = 10/6

thanks for that explanation...for a second i thought i was a total retard. good to know that my instincts are still somewhat intact...too often simple things are clouded by jargon

the EV question is beyond elementary and stk will be welcome at any of my poker games :-X

Faust is trying to compare apples to oranges and probably confusing you in the process. read the books but ignore his answer. why even think about options pricing when looking at a simple one time dice roll with only two possible outcomes (implied volatility? are you serious?) .

I have to say though that the generally agreed upon answer above that what you would pay to roll the dice is "somewhere near or below 1.67" isn't exactly correct. Has to do with your own unique risk profile and utility of income.

ANYONE that gambles in Vegas pays more than 1.67

Are there any good websites with a lot of similar brainteasers?

draw out a probability tree. Answer comes out to 20/50 = 40%. I'll try to explain below sans the tree:

We want to find outcomes with the following mix: AAA-BBB (each team, A and B, wins 3 games).

To find the number of possibilities we take the number of choices at each spot and multiply them together, 222222 = 64 choices (I'll get back to this point later).

We also now have to find the number of ways to arrange AAA-BBB. We can look at the following categories: AAA-BBB --> 1 choice = 1 AAB-BBA --> [AAB/ABA/BAA]-[BBA/BAB/ABB] = 33 choices = 9 ABB-BAA --> [BBA/BAB/ABB]-[AAB/ABA/BAA] = 33 choices = 9 BBB-AAA --> 1 choice = 1

Summing them up, we get 1+9+9+1 = 20 choices. Therefore the answer WOULD be 20/64; however, the denominator, 64, takes into account such permutations as AAA-ABB, which would be impossible under the rules since the series stops when a team reaches 4 wins. By drawing out the probability tree, you will see this as well, and should come up with a reduced denominator of 50. Hence the 20/50.

@zyphNOR ==> Get Timothy Falcon Crack's book, and check out techinterview.org (and similar sites via a google search).

there is...

...a 1/8 chance its over in game 4 ...a 3/8 chance its over in game 5 ...a 11/16 chance its over in game 6

or there is a 5/16 or 31.25% chance there will be a game 7.

read em and weap.

Are being good at these types of brain teasers (or anything more than basic algebra) even necessary to be a successful trader? Seems to me that most of the actual math used while trading would be computer generated anyway to make sure no errors were made.

i dunno the answer to your question b/c I have never traded for a firm...

but from my experience in education if you cannot understand the math behind the the computer calculations its unlikely you will be able to fully appreciate the results of such calculations...

The "beyond elementary" question (which out of a cross-section of surveyees even Cornell/Stanford Fin. Eng. majors had gotten intuitively wrong when asked) is meant to illustrate that for observed (known) payout of $10, the price of $1.67 or $3.58 can only be justified if the "volatility" (or Prob of Success/Winning) is 14.31% and 26.36% respectively, while realized (historical observation) of rolling a 6 is 16.67%. Of course, we're not making it as easy as "sell premium when IV > HV, long premium when IV

I think the dice rolling question is being made way too complicated, and one guy had the right answer at 2.

A way to think about it is that you pay X to play the game, so in order to make the game fair it would have to be that:

5X/6=10/6 or 10/6 - 5X/6 = 0

therefore X=2

The EV calcs of the game before saying 1.67 disregrd the fact that you lose money on 5/6 times for one roll.

Yeah, the answer is very simply 1/3. JSC asks these sort of questions precisely because their interviewees make the problem way too complicated and get caught up in a mess of math / statistical theorems. They don't give a fuck how much complex math you've learned and whether you can spew that on command at an interview. They want to see if you can logically and systematically work through a practical calculation and do it in an efficient way, you're interviewing to be a trader.

The questions are all pretty straightforward, if you've taken a basic class in stats, and are rigorous with your basic theorems. With the horse race question the answer is 7, and here too if you just draw out possibilities and work through the problem systematically, it's doable.

Does anyone have more questions from Jane Street interviews? (Assistant Trader position)

I think it is possible to find the top 3 horses in 7 races. First we separate the 25 horses into 5 groups of 5. The top 3 horses in each race are then labeled A, B, C. So the 2nd place horse in the 3rd group can be labeled as horse 3B. So now we have 1-5A, 1-5B, 1-5C, a total of 15 horses to consider. We now race all of the winners together in a 6th race. The top 3 horses in that race keep their groups in contention, while we now eliminate consideration of the groups of the bottom 2 horses. Out of those 3 groups remaining, the group with the 3rd place horse in the 6th race has only 1 horse (A) remaining in consideration. The group with the 2nd place horse has 2 horses (A and B) in consideration. The group with the 1st place horse still has all 3 horses (A, B, C) remaining for consideration. Now we race five horses: every horse except for the one that won the 6th race, because he is the top ranked horse for sure. This 7th race gives us the remaining information that we need to spot the 2nd and 3rd fastest horses; the 2nd and 3rd ranked horse are the 1st and 2nd place winners in this 7th race.

It is 7 for sure.. I was asked this question in an interview before

The 7-game solution could be wrong.

If we use the binomial distribution, we have made some wrong assumption. For instance, we implicitly included the probability of getting AAAABB for the first 6 games but we don't actually need the "BB" at the end since they are not actually played. So it could be AAAAAA or AAAAAB or AAAABA too since it's not real anyway.

My solution was to list out every possible outcome of not playing the 7th game.

AAAA

AAABA AABAA ABAAA BAAAA

AAABBA AABABA AABBAA ABABAA ABBAAA BABAAA BBAAAA

And the symmetrical ones for B

so the answer I got would be : 1 - [ (1/2)^4 + (1/2)^54 + (1/2)^67 ] * 2 = 13/32

Please correct me...

victormai, 6 choose 3 does not include AAAABB in the probability. Think of it as flipping a coin 6 times and getting 3 heads and 3 tails. That's the same as the first 6 games with A and B each winning 3 times.

Also, your solution is not the right way interviewers will want to see you do it. This problem can be done in like 1 minute, you just have to do (6c3) / (2^6) in your head. Your way will take much longer.

You got the wrong answer via your solution because you missed some outcomes, like ABAABA, BAAABA, etc. Listing outcomes is usually a very naive way to do these kinds of problems.

Agreed with above. If you want to be a trader, you should never list out the whole sample space. You have to be able to think probabilistically. Even for the probability problems regards to discrete RV, the cardinality of the sample space is infinite, most of the time.