A pair of fair dice is rolled once. Suppose that you lose $8 if the dice sum to 10 and win $11 if the dice sum to 11 or 12. How much should you win or lose if any other number turns up in order for the game to be fair?
Accepted Solution
A:
First find the possible dices rolls that give you 10, 11, and 12
Gives 10: 5,5 4,6 6,4
Gives 11 or 12: 5,6 6,5 6,6
The dice rolls have 6 * 6 = 36 possible combinations
The probability of getting a 10 is 3/32 The probability of getting an 11 or 12 is 3/32
multiplying those probabilities together with their reward/loss we have:
3/32 * -8 = -3/4 3/32 * 11 = 33/32
to make the game "fair" we need the probable amount of money you can win to be $0, so we can use the equation