# 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.

• Looks right to me. Jul 18, 2020 at 7:51
• Also looks right to me! Outside Baye's rule, I believe this is equivalent to ignoring Alex from this specific scenario and focusing on Beth. We could rephrase it to: "Beth flipped a coin until she eventually flipped heads. What is the probability it only took her two flips?" Which can be easily be shown to be 25%. Jul 18, 2020 at 7:52
• the answer is right. Jul 18, 2020 at 11:17

$$\begin{cases} A & ... & A & B \\ !2H & ... & !2H & 2H \\ {3\over 4} & ... & {3\over 4} & {1\over 4} \end{cases}$$
$$X: \text{B flipped a pair of heads on her second turn} \\ Y: \text{Alex flips first. Beth eventually wins.} \\ \Pr[X|Y] = {\Pr[X\cap Y] \over \Pr[Y] } = { ({3\over 4})^3 {1\over 4} \over \Pr[Y] } \\ \Pr[Y] = \sum_{n=1}^{\infty} ({3\over 4})^{2n-1} {1\over 4} = {1\over 3} \sum_{n=1}^{\infty} ({9\over 16})^n = {1\over 3} {9\over 7} = {3\over 7}\\ \Pr[X|Y] = { ({3\over 4})^3 {1\over 4} \over {3\over 7} } = {63\over 4^4}$$
$$... !2H, !2H \begin{cases} (A) & (B) \\ !2H & !2H & {3\over 4} {3\over 4} \Pr[Y] \\ !2H & 2H & {3\over 4} {1\over 4} \cdot 1 \\ 2H & & {1\over 4} \cdot 0 \\ \end{cases} \\ \Pr[Y] = {3\over 4} {3\over 4} \Pr[Y] + {3\over 4} {1\over 4} \cdot 1 + {1\over 4} \cdot 0 \\ \Pr[Y] = {3\over 7}$$