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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?

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    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

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