In a game, two players $X$ and $Y$ toss a coin alternately. Whosoever gets a 'head' first wins the game and the game is terminated. Assuming that player $X$ starts the game. Find the probability of player $X$ winning the game.

In a game, two players $X$ and $Y$ toss a coin alternately. Whosoever gets a 'head' first wins the game and the game is terminated. Assuming that player $X$ starts the game. The probability of player $X$ winning the game is 

(A) $\dfrac{1}{3}$ 

(B) $\dfrac{1}{2}$ 

(C) $\dfrac{2}{3}$ 

(D) $\dfrac{2}{4}$ 

Answer: (C) $\dfrac{2}{3}$ 

Explanation: 

Here, the probability of getting a 'head' is, $p = \dfrac{1}{2}$, and 

the probability of getting a 'tail' is, $q = \dfrac{1}{2}$. 

Therefore, the required probability is 

$p + q^2p + q^4p + \cdots$ 

$= p(1 + q^2 + q^4 + \cdots)$ 

$= p \times \dfrac{1}{1 - q^2}$ 

$= \dfrac{1}{2} \times \dfrac{1}{1 - \dfrac{1}{4}}$ 

$= \dfrac{2}{3}$