公平硬币 vs 偏硬币：先出正面者胜

Abby 在第 $k$ 次轮到她时获胜的概率为

$$p\left(\frac{1-p}{2}\right)^{k-1}.$$

因此 Abby 获胜概率为几何级数：

$$P(A)=\sum_{k=1}^\infty p\left(\frac{1-p}{2}\right)^{k-1}=\frac{p}{1-\frac{1-p}{2}}=\frac{2p}{1+p}.$$

令其等于 $1/2$ 得

$$\frac{2p}{1+p}=\frac{1}{2}\Rightarrow p=\frac{1}{3}.$$

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Original Explanation

Let $A$ be the event that Abby wins the game. Abby either gets the head on her first flip, both her and Ben get tails on the first flip and Abby gets it on her second flip, etc. The probability that Abby gets the head on the $k$th flip means that both her and Ben obtain heads for the first $k-1$ flips and then Abby flips a head on her $k$th flip. The probability of this is $p\left(\frac{1-p}{2}\right)^{k-1}$. This is because of the independence of each of their flips and taking the complement of their respective heads probabilities. We now just sum this up from $k=1$ to $\infty$ to get

$$\mathbb{P}[A] = \sum_{k=1}^{\infty} p\left(\frac{1-p}{2}\right)^{k-1} = \frac{p}{1 - \frac{1-p}{2}} = \frac{2p}{1+p}$$

We want this probability to be equal to $\frac{1}{2}$, as we want Abby and Ben to have equal winning probabilities. Therefore, we must find $p$ such that

$$\frac{2p}{1+p} = \frac{1}{2}$$

Solving this yields $p = \frac{1}{3}$.