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}

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , . Bookmark the permalink.

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