Question

Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What are the last three digits of $m+n?$

Answer

Answer: 93

Concepts:basic probability recursion recursive probability

Difficulty rating: 2370

Small Hint:

Compute Alfred’s chance to win a single game when he goes first and when Bonnie goes first

Big Hint:

If $p_r$ is Alfred’s chance to win game $r$ , express $p_{r+1}$ in terms of $p_r$

Solution:

Alfred wins a game with probability $\frac{2}{3}$ when he starts and $\frac{1}{3}$ when Bonnie starts. Since the loser starts the next game, $\begin{aligned}p_{r+1}&=\frac23(1-p_r)+\frac13p_r\\&=\frac23-\frac13p_r,\qquad p_1=\frac23.\end{aligned}$ Thus $p_r-\frac12=\frac{(-\frac{1}{3})^{r-1}}{6}.$ In particular, $p_6=\frac12-\frac1{1458}=\frac{364}{729}.$ Therefore $m+n=1093,$ whose last three digits are $093.$

1993 AIME Problem 11

Answer: 93

Difficulty rating: 2370

Small Hint:

Big Hint:

Solution:

Problem 11 in Other Years