Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first. If the coin is fair, what is the probability that $A$ wins?
So $A$ only flips on odd tosses. So the probability of winning would be $$ P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$
Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, …., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
Best Answer
Let $p$ be the probability that A wins. This can happen in two ways: (i) A wins immediately (probability $1/2$ or (ii) A tosses a tail, but ultimately wins.
If A tossed a tail (probability $1/2$, then in effect B is now "first" so the probability she does not win is $1-p$. We conclude that $$p=\frac{1}{2}+\frac{1}{2}(1-p).$$ Solve for $p$. We get $p=2/3$.
Comment: There are nice expressions for $p$ as infinite geometric series. So we can think of the above argument as a probabilistic method for summing a very particular geometric series. By varying the probability that the coin lands heads, we can use the same idea to find the sum of any infinite geometric series, as long as the "common ratio" is positive.