Suppose that a game of chance is played with a pair of fair 8-sided dice (with t

Question

Suppose that a game of chance is played with a pair of fair 8-sided dice (with the sides numbered 1 to 8). In the game you can pick any number from 1 to 8 and the two dice are then "rolled" in a cage. If dollar 1 is bet and exactly one of the number that you picked is rolled you win dollar 2, and if both of the dice are the number that you picked you win dollar 10 (in each of those cases you also get your initial dollar 1 bet back). If none of your number winds up being rolled you lose your dollar 1 bet. Suppose that you play this game 10 times and pick the same number each time. What is the probability that doubles of YOUR number (both dice come up your number) does not occur in the 10 rolls? What is your total expected win or loss? Indicate in your answer both the amount (rounded to the nearest dollar 0.01 if necessary) and whether it is a win or loss.

Explanation / Answer

A) P(doubles of your number) = (1/8)x(1/8) = 1/64

P(not doubles of your number) = 1-1/64 = 63/64

P(not doubles of your number in 10 rolls) = (63/64)10

= 0.8543

B) P(winning $2) = 14/64 = 7/32

P(winning $10) = 1/64

P(losing $1) = 1-((14/64)+(1/64)) = 49/54

Expected outcome from 10 rolls

= 2x14/64 + 10x1/64 - 1x49/64

= -0.23

So, the total expected LOSS is $0.23

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