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 $1 is bet and exactly one of the numbers that you picked is rolled you win $1, and if both of the dice are the number that you picked you win $15 (in each of those cases you also get your initial $1 bet back). If none of your number winds up being rolled you lose your $1 bet. Suppose that you play this game 4 times and pick the same number each time.

a. What is the probability that doubles of YOUR number (both dice come up your number) does not occur in the 4 rolls?

b. What is your TOTAL expected win or loss? Indicate in your answer both the amount (rounded to the nearest $0.01 if necessary) and whether it is a win of a loss.

Explanation / Answer

(a)

   probability of exactly same number appearing two dice is =1/64

probability of a apperaing one number in two dice is =15/64

hence probability that doubles appear at least does not appear) =1 -4C0p0(1-p)^4 where p=1/64

                                               =1-0.938949644

                                                                                                                         =0.061050355

b)expected win/loss in one trial =win from exactly one number matches+win from exactly same number comes -loss of 1 dooar

                            =(15/64)*1+(1/64)*15-(48/64)*1

                            =   -18/54   (hence it is a loss)

and expected loss from 4 rolls =   -(18/64)*4

                                                 = -1.125

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