You roll two fair dice. If they land with a sum of 7, you get a $10. If they land with a sum of 6 or 8, you get $5. Everything else earns $0. Which of the following would be the highest value you could set the price and the player would still win? $8
$5
$2
$10

Using the expected value, it is found that the highest value you could set the price is: $5.

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A game is fair, that is, the player would still win, if the expected value is 0.The expected value is the sum of each outcome multiplied by it's probability.A probability is the number of desired outcomes divided by the number of total outcomes.For two fair dice, there are 36 total outcomes, as .The charge is x.6 outcomes result in a sum of 7, which are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), thus, probability of getting 10.10 outcomes result in a sum of 6 or 8, which are (1,5), (2,4), (2,6), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3) and (6,2), thus, probability of getting x.36 - 14 = 22, thus probability of losing x.

Since we want the expected value to be 0:

The highest value you could set the price is: $5.

A similar problem is given at

The expression of this would be 3.75x+y= 23