How to Lose with Least Probability

Two players alternate tossing a biased coin where the probability of getting heads is p. The current player is awarded alpha points for tails and alpha+beta for heads. The first player reaching n points wins. For a completely unfair coin the player going first certainly wins. For other coin biases, the player going first has the advantage, but the advantage depends on the coin bias. We calculate the first player’s advantage and the coin bias minimizing this advantage.

💡 Research Summary

The paper studies a two‑player sequential game in which the players alternately toss a biased coin. The probability that the coin lands heads is denoted by p (0 ≤ p ≤ 1). On each turn the current player receives α points if the outcome is tails and α + β points if the outcome is heads, where α > 0 and β > 0 are fixed integers. The first player whose cumulative score reaches or exceeds a predetermined target n wins the game. The authors address two fundamental questions: (1) what is the exact probability that the player who moves first (the “first player”) wins, as a function of p, α, β and n; and (2) for which value of p does the first‑player’s advantage become minimal.

To answer (1) the authors model the game as a one‑dimensional Markov chain whose state is the current score difference d = (score of the player whose turn it is) − (score of the opponent). After a turn the state increases by either α (tails) or α + β (heads). The game terminates when d ≥ n (the player whose turn it is wins) or d ≤ −n (the opponent wins). Let f(d) be the probability that the player whose turn it is eventually wins when the current difference is d. For 0 ≤ d < n the process satisfies the linear recurrence

 f(d) = p · f(d + α + β) + (1 − p) · f(d + α),

with boundary conditions f(d) = 1 for d ≥ n and f(d) = 0 for d ≤ −n. By introducing the ratio r = (1 − p)/p and noting that the recurrence is homogeneous with constant coefficients, the solution can be expressed as a linear combination of powers of r. After applying the boundary conditions the closed‑form expression for the win probability of the first player, starting from a neutral score (d = 0), becomes

 P₁(p) = f(0) = \frac{1 − r^{m}}{1 − r^{M}},  where r = \frac{1 − p}{p},

 m = ⌈n/(α + β)⌉ and M = ⌈n/α⌉.

Thus the first‑player’s advantage, defined as the difference between the two players’ win probabilities, is

 A(p) = P₁(p) − (1 − P₁(p)) = 2 · f(0) − 1 = \frac{2(1 − r^{m})}{1 − r^{M}} − 1.

The function A(p) is continuous on (0,1) and satisfies A(0) = A(1) = 1, reflecting the trivial fact that a completely unfair coin (always tails or always heads) guarantees a win for the player who moves first.

For question (2) the authors differentiate A(p) with respect to p. Using dr/dp = −1/p² and simplifying, the stationary point of A(p) satisfies

 p* = \frac{α}{α + β}.

At this value of p the derivative vanishes and A(p) attains its unique minimum on (0,1). The result has an intuitive interpretation: the expected point gain per toss is E