Coin Tossing Game: No Reset

Two players, A and B, toss a fair coin. A tosses the coin first, then B tosses the coin, then A, etcetera. The sequence of heads and tails is recorded. If there is a head followed by a tail (HT), the game ends and the person who tosses tail wins. What is the probability that player A wins the game?

Answer

Condition on the outcome of the first toss:

P(A) = \frac{1}{2} \cdot P(A \mid H) + \frac{1}{2} \cdot P(A \mid T).

If the first toss is T, the game essentially resets with player B now effectively in the first position and us on second; hence, our probability of winning in this position is the same as the probability of B winning (on initial conditions). Basically we swapped the roles here.

P(A \mid T) = P(B) = 1 - P(A).

Now, if the first toss is H, for A to win the game the eventual winning sequence must be one of the forms:

\text{HHT, HHHHT, HHHHHHT,} \dots

which occurs with probability

P(A \mid H) = \frac{1}{4} + \frac{1}{4^2} + \cdots = \frac{1/4}{1 - 1/4} = \frac{1}{3}.

Substituting into the total probability equation:

P(A) = \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot (1 - P(A)).

That is,

P(A) = \frac{1}{6} + \frac{1}{2} - \frac{1}{2} P(A).

Adding $\frac{1}{2}P(A)$ to both sides gives:

\frac{3}{2}P(A) = \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}.

Thus,

P(A) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}.

Notes and comments

Comment 1: Note that this question cannot be solved by a pure recursive method because there is no complete reset of the game. Even if no one wins in the first round, the last toss (for example, a trailing H in HHH) influences the next round.