Question

A and B toss a coin alternately till one of them gets a head and wins the game. If A starts the game, find the probability that B will win the game.

Answer 1

$$\begin{gathered} P\left(\text{B winning the game}\right)=P\left({\text{head at 2}}_{nd}\text{turn}\right)+P\left({\text{head at 4}}_{th}\text{turn}\right)+... \\ =\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+... \\ ={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^6+{\left(\frac{1}{2}\right)}^8+... \\ =\frac{1}{4}\left[1+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^6+...\right] \\ =\frac{1}{4}\left[\frac{1}{1-{\displaystyle \frac{1}{4}}}\right]\left[\text{For infinite GP:}1+a+a^2+a^3+...=\frac{1}{1-a}\right] \\ =\frac{1}{4}\times \frac{4}{3} \\ =\frac{1}{3} \end{gathered}$$