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?
 A: 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.
