If the gambler rolls a \( 4,5,6,8,9 \), or 10 , that number is now considered the target. The dice are then rolled again and continued to be rerolled until either the sum the target, in which case the gambler wins, or the sum is a 7 in which case the gambler loses. So, if the initial dice roll had a sum of 9 and then the next sums were \( 4,4,11,3,9,12,7 \)... the gambler would win because another 9 was rolled before a 7. The probabilities that various numbers will be rolled before a 7 is as follows \[ \begin{aligned} P(4 \text { before } 7) & =\frac{3}{9} \\ P(5 \text { before } 7) & =\frac{4}{10} \\ P(6 \text { before } 7) & =\frac{5}{11} \\ P(8 \text { before } 7) & =\frac{5}{11} \\ P(9 \text { before } 7) & =\frac{4}{10} \\ P(10 \text { before } 7) & =\frac{3}{9}\end{aligned} \] So, if a gambler rolls a 6 first, the probability that they will win after that is \( \frac{5}{11} \).
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