<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] \mathsrc{A_k}={A losing in his first k tosses and B losing in his first k tosses and A winning in his k+1 toss}
$$\begin{align}
P(\text{A winning}) &= P(\text{A winning in his first toss or } \mathrm{A_1} \text{ or } \mathrm{A_2} \text{ or } \ldots)\\
&= P(\text{A winning in his first toss}) + P(\mathrm{A_1}) + P(\mathrm{A_2}) + \dots\\
&= 0.5 + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5)(0.5)(0.5) + \dots \\
&=0.5 + (0.5)^3 + (0.5)^5 + .........\\
&= \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{50}{75} = \frac{2}{3}\\
\end{align}$$
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