7.7 Jane Street Capital Second Round Interview

HarishKumar

HarishKumar

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Hello guys, I just had a brutal second round interview for a trader summer internship at Jane Street Capital. I didn't do well but I'm curious in finding out the answer to one of their questions. Basically, this is how it goes.

You are given a 100 sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game?(There is no limit on the number of rolls.)
 
each side of landing is 1/100 probabilty.
each landing you have an equal payoff from $1 to $100.
expected value = 1/100 x sumFrom(1 to 100) = 1/100 x ((100 x (100+1)/2) = 50.50.

i could be wrong. correct me!
 
I think that is incorrect. The interviewer told me that the expected value should be above 80. He then asked me to calculate the expected value of the game if the player adopted a strategy of opting to roll again only if he/she rolls <=84 which I then worked out to be 87.5 according to the following calculation.

E(game) = (Pr(>84) x (92.5)) + (Pr(<=84) * (E(game) - 1))

Using the above recurrence relation, E(game) works out to be $87.50. So after he asked me this, I learnt that the expected value of the game is at least $87.50. However, I am confused on how to find the maximum expected value(which I think is what the interviewer wanted) systematically.
 
Keith, you need to take into account the fact that you can roll again.

If we let E be the expected value of the game, then the optimal strategy is to roll again if we roll any number less than E-1. The probability of rolling less than E-1 is approximately (E-1)/100 (need to round E down to an integer for this to be exact). If we roll greater than E-1, we stop and take the money. Our expectation, assuming we roll greater than E-1 is 0.5*(100+E-1) (again E needs to be rounded off to be exact). We have 1-(E-1)/100 probability of rolling a number greater than E-1.

Putting it all together then gives us that

E=0.01*(E-1)*(E-1) + 0.5*(100+E-1)*(1-0.01(E-1)) . Solving for E gives us E=86.85 (or E=115 which we can rule out as E needs to be between 0 and 100).

So this would give a strategy where you'd roll again any time you throw 86 or less. Given the rounding errors above, it's probably best to check if a strategy where you roll again if you throw less than 87 would be better, or a strategy where you roll again if you throw less than 85.
 
HarishKumar said:
Hello guys, I just had a brutal second round interview for a trader summer internship at Jane Street Capital. I didn't do well but I'm curious in finding out the answer to one of their questions. Basically, this is how it goes.

You are given a 100 sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game?(There is no limit on the number of rolls.)
Click to expand...

There is a very similar question in this book...the strategy is to go roll by roll and only continue if you roll a number lower than the expected winnings on each roll and find the expectation based on each roll.
 
Let the expected value of the game be E[X].
If you end up with a number <=50, you would roll again.
Pr(you would roll again) = 0.5
E[X] = 0.5(sum(51-100)/50) + 0.5(0.5(sum(51-100)/50) – 1 + 0.5(0.5(sum(51-100)/50) – 1 +….

looks like geometric series with r <1. sum = a/1-r where a is the first term.
 
Correction:
E[X] = 0.51(sum(50-100)/51) + 0.49(0.51(sum(50-100)/51) – 1 + 0.49(0.51(sum(50-100)/51) – 1 +….

It wouldn't make sense to roll again at 50 since your only 0.5 below expectation and the charge to roll again is 1.
 
Yes, agreed. That's what I am doing in the equation. Roll. scored 50 or above -> OK. Otherwise pay 1 and roll again. Since E[x]-1 = 49.5. So what's wrong ? It comes out = 75 - (49/51)= 74.04.
 
But if the expectation is 74.04, why would you accept a roll of say 51? Wouldn't it be better to pay $1 and get the expectation of 69.04? Where does your "50" come from? It is the expectation of 1 roll, but there's no need to limit yourself to one roll.
 
You would accept 51 (for that matter 50) if you hit since if you rolled again the chance you get 74 and above hasn't changed and is still 0.26. Yes, the probability that you get 74 or more over multiple tries increases considerably. But, every time you try it, its the same since each try is independent.This, in essence, is captured by the relationship between conditional and absolute probability over multiple tries.

50 comes from the fact that you wouldn't pay 1 to play a game with expectation 50.5 since that is 0.5 -EV.

The reason why the expectation is higher than numbers you'd accept is because it captures effects of scenarios where you get a string of low scores below 49.5 and eventually hit a score above 49 thereby converting probabilities associated with low payoffs in a conventional game with "at least" a respectable one in this game.

It, intuitively speaking, essentially is a premium you pay in order to ensure that the minimum amount you win from the game is going to be 50 regardless of the investment in the game (which depends on the number of tries) as opposed to a normal where you are guaranteed to win a minimum of 1 only.

Please correct me if your wrong.
 
On thinking further about this, there might be a few things I haven't captured in the equation. For instance, if I applied the equation's logic to a hypothetical where it was free to retry, I do not get 100 as the answer!
 
Advait said:
You would accept 51 (for that matter 50) if you hit since if you rolled again the chance you get 74 and above hasn't changed and is still 0.26. Yes, the probability that you get 74 or more over multiple tries increases considerably. But, every time you try it, its the same since each try is independent.This, in essence, is captured by the relationship between conditional and absolute probability over multiple tries.

50 comes from the fact that you wouldn't pay 1 to play a game with expectation 50.5 since that is 0.5 -EV.

The reason why the expectation is higher than numbers you'd accept is because it captures effects of scenarios where you get a string of low scores below 49.5 and eventually hit a score above 49 thereby converting probabilities associated with low payoffs in a conventional game with "at least" a respectable one in this game.

It, intuitively speaking, essentially is a premium you pay in order to ensure that the minimum amount you win from the game is going to be 50 regardless of the investment in the game (which depends on the number of tries) as opposed to a normal where you are guaranteed to win a minimum of 1 only.

Please correct me if your wrong.
Click to expand...

Want to play the game? :). I will pay 74
 
Want to play the game? :). I will pay 74

Well! If 74.04 were correct, I'd at least charge 75 to make a profit!
 
Hey Dieter, I guess something's wrong. I'll work it out and let you know of my price! I'm sure your not right either since as per your logic, you should take the money for anything greater than equal to E-1; and not just "greater than".
 
Hey Dieter, incorporating your logic of choosing outcomes over E only into my equation, I get this :


E=(100*101-E*(E+1))/200 + (E-1)/100((100*101-E*(E+1))/200 -1 + (E-1)/100((100*101-E*(E+1))/200 -1 +…
 
Correction (same error!):


E=(100*101-(E-1)*(E-2))/200 + (E-2)/100((100*101-(E-1)*(E-2))/200 -1 + (E-1)/100((100*101-(E-1)*(E-2))/200 -1 +…
 

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