Question

In a game, two fair number cubes with faces numbered 1 through 6 are rolled. If the sum of the two numbers rolled is 12, the player wins c coupons. 11, the player wins 4 coupons. less than 11, the player loses 2 coupons. What value of c makes this game fair? A.10B.14 C.58 D.64

Answer 1

the probability to get a number less than 11 is 33/36, to get an 11 is 2/36 and to get an 12 is 1/36. if we want the game to be fair, the expected value should be 0, so we get

$-2\cdot\frac{33}{36}+4\cdot\frac{2}{36}+\frac{c}{36}=0\rightarrow-66+8+c=0\rightarrow c=66-8=58$