 Jane Street

## Interview Question

Quantitative Researcher Summer Intern Interview

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# 3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn?

9

There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents.

SG on

3

The problem should be random draw card and dont put it back. Every draw you have one chance to guess. So the strategy is after first draw you random guess it's red. If correct you get one dollar, next draw you know there is less red than black. So you guess black on next draw. Else if first guess you are wrong, you guess red on next round. It's all about conditioning on the information you know from the previous drawings

xc on

5

The problem statement is not very clear. What I understand is: you take one card at a time, you can choose to guess, or you can look at it. If you guess, then if it's red, you gain \$1. And whatever the result, after the guess, game over. The answer is then \$0.5, and under whatever strategy you use. Suppose there is x red y black, if you guess, your chance of winning is x/(x+y). If you don't, and look at the card, and flip the next one, your chance of winning is x/(x+y)*(x-1)/(x+y-1) + y/(x+y)*x/(x+y-1) = x/(x+y), which is the same. A rigorous proof should obviously done by induction and start from x,y=0,1.

JY on

0

This should be similar to the brainteaser about "picking an optimal place in the queue; if you are the first person whose birthday is the same as anyone in front of you, you win a free ticket." So in this case we want to find n such that the probability P(first n cards are black)*P(n+1th card is red | first n cards are black) is maximized, and call the n+1th card?

SHF on

0

scheme: guess when the first one is black, p(guess) x p(right) x 1=1/2 x 26/51=13/51

hex on

0

Doesn't matter what strategy you use. The probability is 1/2. It's a consequence of the Optional Stopping Theorem. The percent of cards that are left in the deck at each time is a martingale. Choosing when to stop and guess red is a stopping time. The expected value of a martingale at a stopping time is equal to the initial value, which is 1/2.

Will on

0

My strategy was to always pick that colour, which has been taken less time during the previous picks. Naturally, that colour has a higher probability, because there are more still in the deck. In the model, n = the number of cards which has already been chosen, k = the number of black cards out of n, and m = min(k, n-k) i.e. the number of black cards out of n, if less black cards have been taken and the number of red cards out n if red cards have been taken less times. After n takes, we can face n+1 different situations, i.e. k = 0,1,2, ..., n. To calculate the expected value of the whole game we are interested in the probability that we face the given situation which can be computed with combination and the probability of winning the next pick. Every situation has the probability (n over m)/2^n, since every outcome can happen in (n over m) ways, and the number of all of the possible outcomes is 2^n. Then in that given situation the probability of winning is (26-m)/(52-n), because there are 26-m cards of the chosen colour in the deck which has 52-n cards in it. So combining them [(n over m)/2^n]*[(26-m)/(52-n)]. Then we have to sum over k from 0 to n, and then sum again over n from 0 to 51. (After the 52. pick we don't have to choose therefore we only sum to 51) I hope it's not that messy without proper math signs. After all, this is a bit too much of computation, so I wrote it quickly in Python and got 37.2856419726 which is a significant improvement compared to a basic strategy when you always choose the same colour.

Barna on

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dynamic programming, let E(R,B) means the expected gain for R red and B blue remain, and the strategy will be guess whichever is more in the rest. E(0,B)=B for all Bs, E(R,0)=R for all Rs. E(R,B)=[max(R,B)+R*E(R-1,B)+B*E(R,B-1)]/(R+B). I don't know how to estimate E(26,26) quickly.

Anonymous on

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The question, to me, is not clear. Perhaps on purpose. If so, the best answers would involve asking for clarification.

Stephen Jones on

2

I would start picking cards without making a decision to reduce the sample size. This is risky because I could just as easily reduce my chances of selecting red by taking more red cards to start, as I could increase my chances of selecting red by picking more black cards first. But I like my chances with 52 cards, that at some point, I will at least get back to 50% if I start off by picking red. Ultimately, I can keep picking cards until there is only 1 red left. But I obviously wouldn't want to find myself in that situation so I would do my best to avoid it, by making a decision earlier rather than later. Best case scenario, I pick more blacks out of the deck right off the bat. My strategy would be to first pick 3 cards without making a decision. If I start off by selecting more than 1 red, and thus the probability of guessing red correctly is below 50%, then I will look to make a decision once I get back to the 50% mark. (The risk here is that I never get back to 50%) However, if I pick more than 1 black card, then I will continue to pick cards without making a choice until I reach 51% - ultimately hoping that I get down to a much smaller sample size, and variance is reduced, while odds are in my favor that I choose correctly. The expected return, in my opinion, all depends on "when" you decide to guess. If you decide to guess when there is a 50% chance of selecting correctly, then your expected return is 50 cents (50% correct wins you \$1 ; 50% incorrect wins \$0 --- 0.5 + 0 = .5) If you decide to guess when there is a 51% chance of selecting red correctly, then the expected return adjusts to (0.51* \$1) + (0.49 * \$0) = 51 cents. So, in other words, your expected return would be a direct function of the percentage probability of selecting correctly. i.e. 50% = 50 cents, 51% = 51 cents, 75% equals 75 cents. Thoughts?

Stephen on

0

The answer above is not 100% correct, for second scenario, if you don't guess, and only look, the total probability of getting red is indeed the same. However, the fact that you look at the card means you know if the probability of getting red is x/(x+y)*(x-1)/(x+y-1) or y/(x+y)*x/(x+y-1). Therefore, this argument only holds if you don't get to look at the card, or have any knowledge of what card you passed

Anonymous on

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0.5, just turn the first card to see if it's red. I think it's more about trading psychology. If you don't know where the price is going, just get out of the market asap. Don't expect anything.

Yang on

2

expected earn is 25 cents. 1/2*1/2*1, prob of choosing to guess is 1/2, prob of guessing right is 1/2, and the pay is \$1

interviewee on 