<s>[INST] Two players playing a coin game - probability of winning Alex and Beth take turns flipping a pair of coins. The first person to flip a pair of heads wins the game. Alex flips first. Beth eventually wins. What is the probability she flipped a pair of heads on her second turn?
Can someone verify if this reasoning is correct or not:
Question can be converted to Bayes rule:
P(someone wins given that they don't flip first) = 3/7 (solved the equation p=(.75)(.25) + (.75)(.75)(p))
Desired probability: P(Beth wins on 2nd turn | she wins and Alex flips first) = (.75)(.75)(.75)(.25)/(3/7)
This equals approximately 1/4.
 [/INST] my approach
$$
\begin{cases}
A   & ... & A   & B  \\
!2H & ... & !2H & 2H \\
{3\over 4} & ... & {3\over 4} & {1\over 4}
\end{cases} 
$$
$$
X: \text{B flipped a pair of heads on her second turn} \\
Y: \text{Alex flips first. Beth eventually wins.} \\
\Pr[X|Y] = {\Pr[X\cap Y] \over \Pr[Y] } = { ({3\over 4})^3 {1\over 4} \over \Pr[Y] } \\
\Pr[Y] = \sum_{n=1}^{\infty} ({3\over 4})^{2n-1} {1\over 4} = {1\over 3} \sum_{n=1}^{\infty} ({9\over 16})^n = {1\over 3} {9\over 7} = {3\over 7}\\
\Pr[X|Y] = { ({3\over 4})^3 {1\over 4} \over {3\over 7} } = {63\over 4^4}
$$
your approach
$$
... !2H, !2H \begin{cases}
(A) & (B)  \\
!2H & !2H & {3\over 4} {3\over 4} \Pr[Y] \\
!2H & 2H &  {3\over 4} {1\over 4} \cdot 1 \\
2H &  & {1\over 4} \cdot 0 \\
\end{cases} \\
\Pr[Y] = {3\over 4} {3\over 4} \Pr[Y] + {3\over 4} {1\over 4} \cdot 1 + {1\over 4} \cdot 0 \\
\Pr[Y] = {3\over 7}
$$
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