 Mathematics
27 January, 13:45

# Suppose that you lose \$10 if the dice sum to 7 and win \$11 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?

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Answers (1)
1. 27 January, 14:03
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A fair game is when the expected value/gain is zero, and is calculated by

E[X] = ∑ P (x) * x.

We assume we don't need to pay to play the game.

When throwing two dice, the probabilitiy of throwing a 7 is

P (7) = 6/36 (pays x=\$10)

(6 because there are six possible outcomes, { (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }.

Similarly,

P (2) = 1/36 (pays x=\$11)

P (3) = 2/36 (pays x=\$11)

We assume the payout is x=k for the other outcomes, with total probability of (36-6-1-2) / 36=27/36

Therefore the expected gain is

E[X]=P (7) * 10+P (2) * 11+P (3) * 11+P (others) * k

For the game to be fair, we have E[X]=0, or

P (7) * 10+P (2) * 11+P (3) * 11+P (others) * k=0

=>

k = - (P (7) * 10+P (2) * 11+P (3) * 11) / P (others)

= - (6*10+2*11+3*11) / (27)

= - 115/27

=-4.26 approximately.
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