Question

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m + n.$

Answer

Answer: 41

Concepts:geometric distribution geometric sequence symmetry

Difficulty rating: 2560

Solution:

The probability that a player's first six appears on roll $k$ is $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}.$ The probability of a tie is $\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.$

The probability that Linda needs exactly one more roll than Dave is $\sum_{k=1}^{\infty} p_k p_{k+1}$ $= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2$ $= \frac{5}{66},$ and by symmetry the same holds with the players swapped.

The total probability is $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33},$ so $m + n = 8 + 33 = 41.$

2009 AIME II Problem 8

Answer: 41

Difficulty rating: 2560

Solution:

Problem 8 in Other Years