Question

A pair of fair dice is rolled once. Suppose that you lose

​$11

if the dice sum to

4

and win

​$14

if the dice sum to

3

or

2

How much should you win or lose if any other number turns up in order for the game to be​ fair?

Answer 1

Number of combinations to get sum of 4 are (1, 3), (2, 2) , (3, 1)

Probability of getting a sum of 4 = 3/36 = 1/ 12

Number of combinations to get sum of 2 or 3 are (1, 1), (1, 2) , (2, 1)

Probability of getting a sum of 2 or 3 = 3/36 = 1/ 12

Let x be the amount that  you win or lose if any other number turns up.

Probability of any other outcome = 1 - (1/12 + 1/12) = 5/6

For fair game, the expected value of the game should be 0,

Expected value of the game = -11 * (1/12) + 14 * (1/12) + x * 5/6 = 0

=> 3 * (1/12) + x * 10/12 = 0

=> 3 + 10x = 0

=> x = -3/10 = -$0.3

Thus, if you lose $0.30 on any other number turns up then the game is​ fair.