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1993 AIME Problems/Problem 11: Difference between revisions

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==Solution 2==
==Solution 2==
In order to being this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a <math>\frac{1}{2}</math> chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a <math>\frac{1}{2}</math> * <math>\frac{1}{2}</math>, or <math>\frac{1}{4}</math>, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is <math>\frac{1}{4}</math> * <math>\frac{1}{2}</math>, or <math>\frac{1}{8}</math>
From this, we can see that Alfred’s (who goes first) chance of winning the first round is the geometric sequence:
<math>\frac{1}{2}</math> + <math>\frac{1}{8}</math> + <math>\frac{1}{32}</math> + /cdots = $\frac{2}{3}.


== See also ==
== See also ==
{{AIME box|year=1993|num-b=10|num-a=12}}
{{AIME box|year=1993|num-b=10|num-a=12}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 16:54, 25 February 2018

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

Similarly, $b_{n+1}=\frac{2}{3}a_n+\frac{1}{3}b_n$.

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

Solution 2

In order to being this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a $\frac{1}{2}$ chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a $\frac{1}{2}$ * $\frac{1}{2}$, or $\frac{1}{4}$, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is $\frac{1}{4}$ * $\frac{1}{2}$, or $\frac{1}{8}$

From this, we can see that Alfred’s (who goes first) chance of winning the first round is the geometric sequence: 

$\frac{1}{2}$ + $\frac{1}{8}$ + $\frac{1}{32}$ + /cdots = $\frac{2}{3}.

See also

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

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