1993 AIME Problems/Problem 11

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Problem

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 $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$ ?

Solution

The probability that the $n$ th flip in each game occurs and is a head is $\frac{1}{2^n}$ . The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}$ , and the probability of the second person winning is $\frac{1}{3}$ .

Let $a_n$ be the probability that Alice wins the $n$ th game, and let $b_n$ be the probability that Bonnie wins the $n$ th game.

If Alfred wins the $n$ th game, then the probability that Alfred wins the $n+1$ th game is $\frac{1}{3}$ . If Bonnie wins the $n$ th game, then the probability that Alfred wins the $n+1$ th game is $\frac{2}{3}$ .

Thus, $a_{n+1}=\frac{1}{3}a_n+\frac{2}{3}b_n$ .

Since Alfred goes first in the $1$ st game, $(a_1,b_1)=\left(\frac{2}{3}, \frac{1}{3}\right)$ .

Using these recursive equations:

$(a_2,b_2)=\left(\frac{4}{9}, \frac{5}{9}\right)$

$(a_3,b_3)=\left(\frac{14}{27}, \frac{13}{27}\right)$

$(a_4,b_4)=\left(\frac{40}{81}, \frac{41}{81}\right)$

$(a_5,b_5)=\left(\frac{122}{243}, \frac{121}{243}\right)$

$(a_6,b_6)=\left(\frac{364}{729}, \frac{365}{729}\right)$

Since $a_6=\frac{364}{729}$ , $m+n = 1093 \equiv \boxed{093} \pmod{1000}$ .

1993 AIME (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1993 AIME Problems/Problem 11

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