A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is:

Let us denote by $$P_X$$ the probability that player X finally wins the game and by $$P_Y$$ the probability that player Y finally wins the game. Because somebody must eventually show a head, we obviously have

$$P_X+P_Y=1.$$

According to the statement, the chances of the two players are the same, that is

$$P_X=P_Y.$$

Substituting $$P_Y=1-P_X$$ into the above equality we get

$$P_X=1-P_X \;\;\Longrightarrow\;\; 2P_X=1 \;\;\Longrightarrow\;\; P_X=\dfrac12.$$

So our task is to find the value of $$p$$ (the probability of head for X’s biased coin) that makes $$P_X=\dfrac12.$$

Now we analyse the game step by step. Player X starts the first toss.

Formula we use: When a process can return to its initial state, we can write a recurrence $$P_X=\text{(probability X wins immediately)}+\text{(probability game resets)}\times P_X.$$

We have:

1. X wins right away if he tosses a head on his first attempt. The probability of this event is simply $$p.$$

2. If X fails (i.e. gets a tail) and Y also fails (i.e. gets a tail), the situation is exactly the same as at the very beginning, with X to play next.    • Probability that X fails = $$1-p.$$    • Probability that Y fails = $$1-\dfrac12=\dfrac12.$$    • Hence, probability that both fail consecutively = $$(1-p)\times\dfrac12.$$    • After this double failure, the game is in its initial state, so the chance that X eventually wins from there is again $$P_X.$$

Putting these cases together, the recurrence relation becomes

$$P_X \;=\; p + \Bigl[(1-p)\times\dfrac12\Bigr]\;P_X.$$

We now solve this equation for $$P_X$$ in terms of $$p.$$

First distribute the bracket:

$$P_X = p + \dfrac{1-p}{2}\,P_X.$$

Bring the term containing $$P_X$$ on the right to the left-hand side:

$$P_X - \dfrac{1-p}{2}\,P_X = p.$$

Factor out $$P_X$$ on the left:

$$P_X\Bigl[\,1-\dfrac{1-p}{2}\Bigr]=p.$$

Simplify the expression inside the brackets:

$$1-\dfrac{1-p}{2}=\dfrac{2}{2}-\dfrac{1-p}{2}= \dfrac{2-(1-p)}{2}= \dfrac{1+p}{2}.$$

So the equation now reads

$$P_X \times \dfrac{1+p}{2}=p.$$

Isolate $$P_X$$:

$$P_X=\dfrac{2p}{1+p}.$$

But earlier we proved that $$P_X=\dfrac12.$$ Setting the two expressions equal gives

$$\dfrac{2p}{1+p}=\dfrac12.$$

Cross-multiplying, we get

$$4p = 1 + p.$$

Subtract $$p$$ from both sides:

$$4p - p = 1 \;\;\Longrightarrow\;\; 3p = 1.$$

Finally, dividing by 3 gives

$$p = \dfrac13.$$

This value exactly matches Option A.

Hence, the correct answer is Option A.