Winning From Deuce With a Weaker Serve
You are playing a ping pong game, and the score is tied 10-10 (deuce). From this point, you must win by 2 clear points to take the game. Your probability of winning any individual point is $p = 0.4$.
What is the probability that you win the game?
Hints
- The game resets every time the score returns to deuce. What does that tell you about the structure of the problem?
- Set up a single unknown $P$ for the win probability from deuce. Write expressions for winning from advantage and from behind, both in terms of $P$.
- From deuce: $P = p \cdot P_{+} + q \cdot P_{-}$, where $P_{+} = p + qP$ and $P_{-} = pP$. Substitute and solve the linear equation.
Worked Solution
How to Think About It: This is a classic Markov chain / recursive probability setup. At deuce, neither player has an advantage in score, but the weaker player (you, at 40%) is at a disadvantage. The game from deuce is memoryless -- every time you return to deuce, it is as if you started over. So the whole problem reduces to: what is the probability of going on a net +2 run before a net -2 run, where each point is an independent Bernoulli trial?
Before any math, your gut should say: since you win each point less than half the time, your chance of winning must be well below 50%. And because the "win by 2" rule gives the stronger player repeated chances to reassert dominance, it should be noticeably below the naive $p^2 = 16\%$ that ignores the possibility of returning to deuce.
Quick Estimate: The two immediate paths from deuce are: you win the next two points outright (prob $0.4^2 = 0.16$), or the opponent wins both ($0.6^2 = 0.36$). The remaining $2 \times 0.4 \times 0.6 = 0.48$ of the time you split the points and return to deuce. So roughly, conditional on the game not returning to deuce, you win with probability $\frac{0.16}{0.16 + 0.36} = \frac{0.16}{0.52} \approx 0.308$. That shortcut actually gives the exact answer here because the deuce reset is memoryless.
Approach: Model the game as a three-state Markov chain (Deuce, Advantage You, Advantage Opponent) and solve the resulting linear equation for the win probability.
Formal Solution:
Let $p = 0.4$, $q = 1 - p = 0.6$, and let $P$ be your probability of winning from deuce. Define three states:
- Deuce: score is tied.
- Advantage You (A+): you lead by 1 point.
- Advantage Opponent (A-): you trail by 1 point.
From deuce, with probability $p$ you move to A+, with probability $q$ you move to A-.
From A+: with probability $p$ you win the game (score +2), with probability $q$ you return to deuce.
From A-: with probability $p$ you return to deuce, with probability $q$ you lose (score -2).
Let $P_{+}$ be the probability of winning from A+ and $P_{-}$ from A-:
$$P_{+} = p + q \cdot P$$
$$P_{-} = p \cdot P$$
The main equation:
$$P = p \cdot P_{+} + q \cdot P_{-} = p(p + qP) + q(pP)$$
$$P = p^2 + pqP + pqP = p^2 + 2pqP$$
Solving:
$$P - 2pqP = p^2$$
$$P(1 - 2pq) = p^2$$
$$P = \frac{p^2}{1 - 2pq}$$
Plugging in $p = 0.4$, $q = 0.6$:
$$P = \frac{0.16}{1 - 2(0.24)} = \frac{0.16}{1 - 0.48} = \frac{0.16}{0.52} = \frac{4}{13} \approx 0.3077$$
Answer: The probability of winning from deuce is $P = \dfrac{p^2}{1 - 2pq} = \dfrac{4}{13} \approx 30.8\%$.
Intuition
This is a random walk on a small state space with absorbing barriers. The "win by 2" rule means the game from deuce is a memoryless renewal process -- every return to deuce is a fresh start, so the win probability satisfies a single self-referencing equation. The beautiful shortcut is to notice that the game must eventually end in a two-point streak (either $pp$ or $qq$), and the probability of that streak being yours is $p^2 / (p^2 + q^2)$, which simplifies to $p^2 / (1 - 2pq)$.
This pattern shows up constantly in quant interviews and in practice: any time you have a symmetric stopping rule on a biased random walk, the weaker side's disadvantage is amplified. At 40% per point you might expect roughly a 40% chance, but the "win by 2" rule drops you to about 31%. The general formula $p^2/(p^2 + q^2)$ is worth memorizing -- it applies to tennis deuce, overtime scoring, and any repeated head-to-head contest with a "must win by 2" rule.