Coin Toss Game

A and B alternately toss a fair coin, A first. The instant a head is immediately followed by a tail, the game stops and whoever tossed that tail wins. What is the probability that A wins?

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#The states that matter

Going into a toss, only two things matter, the last symbol and whose turn it is. The whole game lives in two states, last tail (which includes the start) and last head.

graph LR
  T["last tail or start"] -->|heads| H["last head"]
  T -->|tails| T
  H -->|heads| H
  H -->|tails| W["this tosser wins"]

#Win chances by symmetry

Write $a$ for the chance the player about to toss eventually wins from last tail, and $b$ from last head. By symmetry these do not depend on which player it is.

From last head, the mover tosses tails and wins outright with chance $\tfrac{1}{2}$ , or tosses heads and hands a last-head position to the opponent, winning with chance $1 - b$ . So

b = \tfrac{1}{2} + \tfrac{1}{2}(1 - b) = 1 - \tfrac{b}{2}, \qquad b = \tfrac{2}{3}. \tag{1}

From last tail, no toss can end the game this step, and the mover passes either a last-head or a last-tail position to the opponent,

a = \tfrac{1}{2}(1 - b) + \tfrac{1}{2}(1 - a), \qquad 3a + b = 2, \qquad a = \tfrac{4}{9}. \tag{2}

#The answer

A moves first from the start, a last-tail state, so

\PP(A \text{ wins}) = a = \frac{4}{9}. \tag{3}

B keeps the edge at $\tfrac{5}{9}$ , since A is the one who tends to lay down the head that B finishes.