Consider the following game of chance. Two fair dice are rolled. If the sum of t

Question

Consider the following game of chance. Two fair dice are rolled. If the sum of the outcomes is either 7 or 11, the player wins immediately, while if the sum is either 2 or 3 or 12, the player loses immediately. if the sum is either 4 or 5 or 6 or 8 or 9 or 10, the player continues rolling the dice until either the same sum appears before a sum of 7 appears in which case he wins, or until a sum of 7 appears before the original sum appears in which case the player loses. it is assumed that the game terminates the first time the player wins or loses. what is the probability of winning?

Explanation / Answer

Wins: Die 1 Die 2 1 6 6 1 2 5 5 2 3 4 4 3 5 6 6 5 8 combinations will give a winning result Loses: Die 1 Die 2 1 1 1 2 2 1 6 6 There are 4 combinations that give a losing result 12 total combinations that will terminate the game either winning or losing Probability of winning = number of winning combinations / total combinations Prob = 8/12 = 0.667 or 66.7%

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