<s>[INST] Two players alternately flip a coin; what is the probability of winning by getting a head? 
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 $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
 [/INST] The probability that the first head occurs on toss $n$ is $2^{-n}$, so the probability that the first head happens on an odd $n$ is
$$
\sum_{k=0}^{\infty}2^{-(2k+1)}=\frac{1}{2}\frac{1}{1-1/4}=2/3
$$
</s>