A flips first, then B, alternating forever. The first time HT appears,
whoever flipped the T wins.
Exact answer
4/9 ≈ 44.44%
Player A wins with probability 4/9
The trick is that the game is not determined only by whose turn it is.
We also need to know whether the previous flip was H.
x
A's turn, previous ≠ H
x = 4/9
y
A's turn, previous = H
y = 2/3
z
B's turn, previous ≠ H
z = 5/9
w
B's turn, previous = H
w = 1/3
State transitions
x
A flips H → w A flips T → z
x = ½w + ½z
y
A flips T after H → A wins A flips H → w
y = ½(1) + ½w
z
B flips H → y B flips T → x
z = ½y + ½x
w
B flips T after H → B wins B flips H → y
w = ½y
Solve the equations
Click to reveal the algebra.
y = ½ + ½w
w = ½y
y = ½ + ¼y
¾y = ½ y = 2/3
therefore w = 1/3
z = ½y + ½x
z = 1/3 + ½x
x = ½w + ½z
x = 1/6 + ½(1/3 + ½x)
x = 1/3 + ¼x
¾x = 1/3 x = 4/9
The game starts in state x, because A goes first and there is no
previous Head.
Therefore P(A wins) = 4/9.
Interactive simulation
Play one game to watch the sequence, or simulate many games and see the
estimate converge toward 44.44%.
Latest game
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Play a game to see what happens.
Estimated P(A wins)
—
0 games
Exact value: 44.44%
Why isn't it 50–50?
Because the winner is determined by who flips the T immediately after
an H. Alternating turns interacts with the two-flip pattern HT, so the
two players are not symmetric even though the coin itself is fair.
Tracking the previous flip exposes that asymmetry.