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Quantitatiave Research Associate at Susquehanna International Group was asked...

Oct 30, 2013

There is a game where you have three coins, and your opponent has three coins. You both flip your coins at the same time. If the number of heads you have matches the number of heads he has, then you win $2. Otherwise you lose $1. Do you want to play this game?

4 Answers

No, the expected win from the game is negative

Here's my solution. We consider 2 sequences (outcomes for the players). Since these sequences do not depend on each other, the probability that they will have the same number of heads is prob(no heads for player 1) * prob(no heads for player 2) + ... + prob(3 heads for player 1) * prob(3 heads for player 2) = 2* (1/8)^2 + 2 * (3/8)^2 = 20/64 = 5/16 (since prob of having no heads = prob of having 3 heads and prob of having 1 head = prob of having 2 heads). Hence the prob that the players will have different number of heads is 11/16. So the expected value of this game is (5/16)*2 + (11/16)*(-1) = -1/16. Hence we should not play this game.

you win 2 dollars with 25% probability and lose one with 75%. Therefore you expect to lose .25 dollars any time

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