## Questions d'entretien

Entretien pour Prop Trading Summer Intern

-New York, NY

# You are playing a game where the player gets to draw the number 1-100 out of hat, replace and redraw as many times as they want, with their final number being how many dollars they win from the game. Each "redraw" costs an extra \$1. How much would you charge someone to play this game?

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## Réponses aux questions d'entretien

10 réponse(s)

8

limitmove le

7

All of the above answers are way off. For a correct answer, see Yuval Filmus' answer at StackExchange: http://math.stackexchange.com/questions/27524/fair-value-of-a-hat-drawing-game

Utilisateur anonyme le

5

Let x be the expected value and n be the smallest number you'll stop the game at. Set up equation with x and n, get x in terms of n, take derivative to find n that maximizes x, plug in the ceiling (because n must be integer) and find maximum of x. ceiling ends up being 87, x is 87.357, so charge \$87.36 or more

Utilisateur anonyme le

0

Yuval Filmus proves that the value of the game is 1209/14=86.37 and the strategy is to stay on 87 and throw again on 86 and below..

Utilisateur anonyme le

1

I guess the question asks for the expected value of the game given an optimal strategy. I suppose the strategy is to go on the next round if the draw is 50 or less. Hence, the expected value of each round is: (1) 1/2*1/50(51 + 52 + ... + 100) (2) 1/2*1/2*1/50(51 + 52 + ... + 100) - 1/2 (3) 1/2^3 * 1/50 (51 + 52 + ... + 100) - 1/4 .... Sum all these up to infinity, you'd get 74.50.

Viet le

0

the average draw will pay out \$50.50

prm le

1

Every time when you are deciding whether to play once more, we consider the two options: stop now, then you get current number (the cost of \$1 is sunk cost); continue, then the expectation of benefit would be \$50.50-1=49.50. This means, as long as you have get more than \$50 (inclusive), then you should stop the game. Suppose the game ends after N rounds (with probability (49%)^(N-1) x 51%, and in the last round, the expected number is (50+100)/2=75, and thus the expected net benefit would be 75-N . This shows N<=74. Then we take the sum: \$Sigma_{N=1}^{74} (49%)^(N-1) x 51% x (75-N), which is 73.

Nova le

0

redraw 10 times and get the payoff around 77?

Utilisateur anonyme le

0

10?

Utilisateur anonyme le

0

Every time when you are deciding whether to play once more, we consider the two options: stop now, then you get current number (the cost of \$1 is sunk cost); continue, then the expectation of benefit would be \$50.50-1=49.50. This means, as long as you have get more than \$50 (inclusive), then you should stop the game. Suppose the game ends after N rounds (with probability (49%)^(N-1) x 51%, and in the last round, the expected number is (50+100)/2=75, and thus the expected net benefit would be 75-N . This shows N<=74. Then we take the sum: \$Sigma_{N=1}^{74} (49%)^(N-1) x 51% x (75-N), which is 73.

Someone le

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