A pair of dice is rolled once. suppose you lose $9 if the dice sum to 9 and win $11 if the dice sum to 4 or 6. how much should you win or lose if any other number turns up in order for the game to be fair?

Answer 1

There are 6 combinations for the first dice, and 6 combinations for the second dice, so the total number of possibilities is 36

There are only 4 ways to get a sum of 9: (4,5), (5,4), (3, 6), (6, 3)

There are only 3 ways to get 4: (2, 2), (1,3), (3,1)

There are only 5 ways to get 6: (1,5), (5,1), (2,4), (4,2), (3,3)

So there are 8 ways to get 4 or 6

There are 36-8-4 = 24 other possible ways to get another number

For the game to be fair, the sum of all the probabilities multiplied by the gain/loss for each of them should be 0:

-$9 *(4/36) + $11*(8/36) + x*(24/36) = 0

solve for x