I'm giving you a choice between two games. In the first game, you roll two six-sided dice. For every number by which their sum exceeds 10, I will give you $1 and $0 if their sum is equal to or less than 10. In the second game, you still have two dice but once is already set to 5 so you can only roll one die. The same rules apply: you get a dollar for each number over 10 and get nothing if the sum is equal to or less than 10. Which game would you prefer to play, and how much would you be willing to pay me to play the preferred game over the other game?

Question

Anonymous · Accepted Answer

Expected value of the first game is (2/36)($1) + (1/36)($2) = 1/9 of $1. The expected value of the second game is (1/6)($1) = 1/6 of a dollar. Therefore, the second game is preferred and you would pay (1/6 of $1) - (1/9 of $1) = 1/18 of $1 to play the second game instead of the first game.

Anonymous · Answer

I believe the interview candidate's answer is incorrect.

The expected value of game #1 is (3/36)($1) + (33/36)($0) = 1/12 of $1.
The expected value of game #2 is (1/6)($1) = 1/6 of $1.

The second game is preferred, and you'd pay 1/12 of $1 to play the second over the first.

Anonymous · Answer

The interview candidate's answer is correct. The wording states that in the first game, you receive $1 for every number by which the sum of the dice exceeds 10. This means that rolling a sum of 12 gives you $2 and rolling a sum of 11 gives you $1, so the expected value of the first game is indeed 1/9 of $1.

Anonymous · Answer

the E(X1) = 1/12, and E(X2) = 1/6

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