The Game

  • Two players, A and B, take turns flipping a fair coin.
  • Player A flips first, then B, and so on.
  • The game ends as soon as the pattern HT (a Head immediately followed by a Tail) appears.
  • The player who flipped the Tail in the HT wins.

Question

What is the probability that Player A wins the game?

Strategy and State Analysis

Let’s define the game states based on whose turn it is and the previous flip:

  • \( S_A \): It is A’s turn, and there is no previous H (e.g., start of game, or after a T).
  • \( S_B \): It is B’s turn, and there is no previous H.
  • \( H_A \): Last flip was H by Player A (it is B’s turn).
  • \( H_B \): Last flip was H by Player B (it is A’s turn).

From \( S_A \):

  • A flips:
    • H with probability \( \frac{1}{2} \) → state \( H_A \)
    • T with probability \( \frac{1}{2} \) → state \( S_B \) (it is now B’s turn)

From \( S_B \):

  • B flips:
    • H with probability \( \frac{1}{2} \) → state \( H_B \)
    • T with probability \( \frac{1}{2} \) → state \( S_A \) (it is now A’s turn)

From \( H_A \):

  • B flips:
    • T → B wins (completes HT, A wins with prob 0)
    • H → state \( H_B \)

From \( H_B \):

  • A flips:
    • T → A wins (completes HT, A wins with prob 1)
    • H → state \( H_A \)

Recursive Probabilities

Let \( P(\text{State}) \) be the probability that Player A wins starting from that state. We want to find \( P_{S_A} \).

First, solve the subsystem for the states where an H was just flipped:

\[ P_{H_A} = \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot P_{H_B} = \frac{1}{2} P_{H_B} \]

\[ P_{H_B} = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot P_{H_A} = \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2} P_{H_B} \right) = \frac{1}{2} + \frac{1}{4} P_{H_B} \]

Solving gives \( P_{H_B} = \frac{2}{3} \) and therefore \( P_{H_A} = \frac{1}{3} \).

Next, use the initial states to find \( P_{S_A} \):

\[ P_{S_B} = \frac{1}{2} P_{H_B} + \frac{1}{2} P_{S_A} = \frac{1}{3} + \frac{1}{2} P_{S_A} \]

\[ P_{S_A} = \frac{1}{2} P_{H_A} + \frac{1}{2} P_{S_B} = \frac{1}{6} + \frac{1}{2} P_{S_B} \]

Substitute \( P_{S_B} \) into \( P_{S_A} \):

\[ P_{S_A} = \frac{1}{6} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{2} P_{S_A} \right) = \frac{1}{3} + \frac{1}{4} P_{S_A} \]

\[ \frac{3}{4} P_{S_A} = \frac{1}{3} \implies P_{S_A} = \frac{4}{9} \]

Final Answer

The probability that Player A wins is:

\[ \boxed{\frac{4}{9}} \]

Reference