2006 iTest Problems/Problem 23

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Jack and Jill are playing a chance game. They take turns alternately rolling a fair six sided die labeled with the integers 1 through 6 as usual (fair meaning the numbers appear with equal probability.) Jack wins if a prime number appears when he rolls, while Jill wins if when she rolls a number greater than 1 appears. The game terminates as soon as one of them has won. If Jack rolls first in a game, then the probability of that Jill wins the game can be expressed as $\tfrac mn$ where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.


The probability of Jack winning in one round is $\tfrac36 = \tfrac12$, while the probability of Jill winning in one round is $\tfrac56$. In order for Jill to win, Jack must not roll winning conditions, while Jill must roll a winning condition.

If Jill wins the game in round 2, the probability is $\tfrac12 \cdot \tfrac56$. If Jill wins the game in round 4, she must fail to roll winning conditions in round 2, so the probability is $\tfrac12 \cdot \tfrac16 \cdot \tfrac12 \cdot \tfrac56$. With similar reasoning, we can find the probability where Jill wins the game in a certain round. Since all cases are mutually exclusive, we can let $P$ be the probability of Jill winning and add all of the probabilities. \[P = \tfrac12 \cdot \tfrac56 + \tfrac12 \cdot \tfrac16 \cdot \tfrac12 \cdot \tfrac56 + \tfrac12 \cdot (\tfrac16 \cdot \tfrac12)^2 \cdot \tfrac56 + \cdots\] The expression is an infinite geometric series with the common ratio between 0 and 1, so we can use the infinite geometric series formula. \begin{align*} P &= (\tfrac12 \cdot \tfrac56) \div (1 - \tfrac16 \cdot \tfrac12) \\ &= \tfrac{5}{12} \div (1 - \tfrac{1}{12}) \\ &= \tfrac{5}{12} \cdot \tfrac{12}{11} \\ &= \tfrac{5}{11} \end{align*} The probability of Jill winning is $\tfrac{5}{11}$, so $m+n = \boxed{16}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 22
Followed by:
Problem 24
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