In a game, two players X and Y are tossing a coin alternately. Whoever gets a ‘head’ first, wins the game and the game is terminated. Find the chance that player X will win the game if he starts? A. $$\frac{1}{3}$$ B. $$\frac{1}{2}$$ C. $$\frac{2}{3}$$ D. $$\frac{3}{4}$$

The correct answer is $\boxed{\frac{1}{2}}$.

Player X has a $\frac{1}{2}$ chance of winning on the first toss. If he does not win on the first toss, then player Y has a $\frac{1}{2}$ chance of winning on the first toss. In this case, player X has a $\frac{1}{2}$ chance of winning on the second toss. This process continues until one player wins. The probability that player X wins is the sum of the probabilities of winning on each toss, which is $\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$.

Option A is incorrect because it is the probability that player X loses the game. Option B is incorrect because it is the probability that player X wins the game if he starts and player Y tosses the coin first. Option C is incorrect because it is the probability that player X wins the game if he starts and player Y tosses the coin second. Option D is incorrect because it is the probability that player X wins the game if he starts and player Y tosses the coin third.