# Probability of a person being first to obtain a head when three people are flipping coins

A, B and C are people. They take turns flipping a coin with A flipping first, B second, C third..The winner is the one who first obtains a head. What is A's probability of winning?

What is the procedure for solving this problem?

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You're asking for a slightly more general question than this one math.stackexchange.com/questions/229658/… Try to see if you can make something out of it. – Jean-Sébastien Nov 11 '12 at 19:56

Suppose we know that Player A won. The probability that he won on his first toss is $\frac 12$. The probability that he won on his second toss is $\frac 12 * \frac 12 * \frac 12 * \frac 12$ = $(\frac 12)^4$. (The first three $\frac 12$'s are the probability that A got tails, then B got tails, then C got tails, so that A may flip again). Using the same procedure, we see that the probability of player A winning on his third flip is $(\frac 12)^7$. We quickly see there is a pattern. Player A's total probability of winning is: $$\frac 12,(\frac 12)^4,(\frac 12)^7,...(\frac 12)^{3n-2}$$ This is a geometric progression with $a=\frac 12$ and $r = \frac 18$. That means that the sum $$S=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{8}} = \frac{4}{7}$$ represents $P(A)$, the probability of player A wining.

We can do the same thing with player B and player C. Note that since player B goes second, he must count on player A getting tails the first time, which has a $\frac 12$ probability, so all we need to do is multiply $\frac 12$ with each term in the series probided above for A. That means that $P(B)$ is simply $\frac 12$ of the sum given for A, which would make $P(B) = \frac 12 * \frac 47 = \frac 27$.

Same thing with player C. He must count on the first two players getting tails, which each have a $\frac 12$ probability, giving $P(C) = \frac 14 P(A) = \frac 14 * \frac 47 = \frac 17$.

In the end, $P(A) = \frac 47, P(B) = \frac 27, P(C) = \frac 17$. Adding up all the probabilities gives you $1$.

As Ross describes in the commnets below, we know that $P(B) = \frac 12 P(A),P(C) = \frac 14 P(A),P(A)+P(B)+P(C) = P(A) + \frac 12 P(A) + \frac 14 P(A)= 1$ and so on.

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In fact, you can argue simply that $P(B)=\frac 12 P(A)$ as if A throws tails B is in the same position. Then $P(C)=\frac 14 P(A)$. This is what you did in your check. – Ross Millikan Nov 11 '12 at 20:27
Very nice cheers. – dukenukem Nov 11 '12 at 20:34
Ross, Thank you for pointing that out. That is actually what I would have done. Noting those equalities and noting that the sum of all the probabilities equal 1 will give you each of the probabilities. – mathguy Nov 11 '12 at 21:05