## Tossing Die

Amy, Ben, and Charlie repeatedly take turns tossing a fair die in the given order $(A,B,C,A,B,C,A,\cdots)$. What is the probability that Charlie will be the first one
to toss a $5$?
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
The possible events that Charlie is the first one to toss a $5$ are
$XX5$
$XXXXX5$
$XXXXXXXX5$
$XXXXXXXXXXX5$
$\cdots$
The probability of tossing a $5$ is $\dfrac{1}{6}$ and the probability of tossing anything else is $\dfrac{5}{6}$.
Probability of $XX5$
$\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}=\dfrac{5^2}{6^3}$
Probability of $XXXXX5$
$\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}=\dfrac{5^5}{6^6}$
Probability of $XXXXXXXX5$
$\dfrac{5^8}{6^9}$
Probability of $XXXXXXXXXXX5$
$\dfrac{5^{11}}{6^{12}}$
Let $p$ be the probability that Charlie is the first one to toss a $5$
$p=\dfrac{5^2}{6^3}+\dfrac{5^5}{6^6}+\dfrac{5^8}{6^9}+\dfrac{5^{11}}{6^{12}}+\cdots$
The right hand side is the sum of the infinite geometric series with the first term $\dfrac{5^2}{6^3}$ and the common ratio $\dfrac{5^3}{6^3}$.
$p=\dfrac{\dfrac{5^2}{6^3}}{1-\dfrac{5^3}{6^3}}$
$=\dfrac{25}{216}\div\dfrac{91}{216}$
$=\dfrac{25}{91}$

Answer: $\dfrac{25}{91}$

Alternative solution
Let $p$ be the probability that Charlie is the first one to toss a $5$. Charlie can do this in two ways:
(1) both Amy and Ben fail to toss a $5$ and Charlie tosses a $5$
probability $=\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}$
(2) all three players fail to toss a $5$ and Charlie is the first one to toss a $5$ from that point. The probability of Charlie winning from that point is $p$, since whether from that point or from the beginning the situation is the same, i.e. Charlie is the first one to toss a 5.
probability $=\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}\times p$
Thus,
$p=\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}+\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}\times p$
$=\dfrac{25}{216}+\dfrac{125p}{216}$
$216p=25+125p$
$91p=25$
$p=\dfrac{25}{91}$