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12 May, 18:33

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?

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  1. 12 May, 21:39
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    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

    -3/4 + 33/32 + (32/32 - 6/32) * x = 0

    then you can just solve for x
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