2009 AIME II Problem 8

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

Small hint:

The chance the first six appears on roll

k

is

p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}

Big hint:

Compute

\sum p_k^2

for a tie and

\sum p_k p_{k+1}

for a difference of one; both are geometric series with ratio

\frac{25}{36}

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.