In a game at deuce, a player wins each point independently with probability 0.6 and must win by two points. What is the probability that player wins the game from deuce?

In a game at deuce, a player wins each point independently with probability 0.6 and must win by two points. What is the probability that player wins the game from deuce?

Approach: Let W be the win probability from deuce and condition on the next two points, since the score returns to deuce whenever those two points are split.

9/13. First-step analysis over a block of two points is the right unit here, because conditioning on two points either ends the game or returns the score to deuce. The player wins outright with probability p^2 = 0.36, loses outright with probability q^2 = 0.16, and returns to deuce with probability 2pq = 0.48. So W = 0.36 + 0.48 W, giving W = 0.36/0.52 = 9/13 = 0.6923. The closed form solution is p^2/(p^2 + q^2), which shows the edge amplification: a 0.6 point-win probability becomes a 0.69 game-win probability, and a slim 0.55 edge per point becomes 0.599 per game from deuce.

Follow-up: What is the expected number of additional points played from deuce?

Key concepts: first-step analysis, closed form solution, conditioning on two points, edge amplification.