Tiebreak Mania

Solution

Answer: 90%

Let \(p = 0.75\) be the probability that you win a point, and let \(q = 0.25\) be the probability that you lose a point. At a score of \(5\text{-}5\), the tiebreaker continues until one player leads by \(2\) points. First consider the next two points. You win immediately if you win two points in a row: \[ P(\text{win }7\text{-}5) = p^2 \] You lose immediately if you lose two points in a row: \[ P(\text{lose }5\text{-}7) = q^2 \] If the players split the next two points (WL or LW), the score becomes \(6\text{-}6\), which returns the game to the same “win by 2” situation. The probability of splitting the two points is \[ P(\text{split}) = 2pq \] Let \(W\) be the probability that you eventually win the tiebreaker starting from \(5\text{-}5\). Conditioning on the outcomes of the next two points gives \[ W = p^2 + (2pq)W \] The first term represents winning immediately, while the second term represents returning to the same state and still having probability \(W\) of eventually winning. Solving for \(W\), \[ W - (2pq)W = p^2 \] \[ W = \frac{p^2}{1 - 2pq} \] Substitute \(p = 0.75\) and \(q = 0.25\). \[ p^2 = 0.75^2 = 0.5625 \] \[ 2pq = 2(0.75)(0.25) = 0.375 \] Therefore, \[ W = \frac{0.5625}{1 - 0.375} \] \[ W = 0.9 \] Thus, the probability that you win the tiebreaker is \[ 0.9 \] so you have a \(90\%\) chance of winning from \(5\text{-}5\).