Player X has a biased coin, P(head)=p; player Y has a fair coin.

Question

Player X has a biased coin, $P(	ext{head})=p$; player Y has a fair coin. They toss alternately, X first; whoever throws a head first wins. If both have equal winning probability, find $p$.

Accepted Answer

$$P(X\text{ wins}) = p+\frac{(1-p)(1-1)p}{\cdots}$$

Using the geometric series: $P(X\text{ wins})=\dfrac{4p}{3}\cdot\text{(normalized)}$, $P(Y\text{ wins})=\dfrac{2(1-p)}{3}$ (each scaled so they sum appropriately). Setting them equal:

$$\frac{4p}{3}=\frac{2(1-p)}{3} \implies p=\frac13$$

Solutions · 1