Tennis probability: how winning 55% of points means 62% of games

September means the US Open, and anyone who watches tennis knows the feeling: two players look evenly matched point for point, yet the final score is lopsided. That’s not an illusion. Tennis scoring works like an amplifier for small edges, and you can measure the effect with high school probability and one geometric series. All numbers here are hypothetical; the goal is to understand the mechanism, not to predict any match.

The model: every point is a biased coin

Assume player A wins each point with the same probability $p$ , independent of earlier points, and let $q = 1 - p$ be the chance of losing a point. In real matches the server has an edge and pressure matters, but this simple model already captures what counts.

The rule for a game: the first player to reach 4 points with a lead of at least 2 wins. The labels 15, 30 and 40 are just names for 1, 2 and 3 points. At 3–3 (40–40, or deuce), the game only ends once someone gets 2 points ahead.

Breaking the game into cases

Player A can win in four ways: 4–0, 4–1, 4–2, or by going through deuce. The first three are pure counting. The final point always goes to A, so only the earlier points can be shuffled.

4–0: four straight points, probability $p^4$ .
4–1: A loses exactly 1 of the first 4 points, in any order; that’s $\binom{4}{1} = 4$ orders. Probability $4p^4q$ .
4–2: A loses 2 of the first 5 points; $\binom{5}{2} = 10$ orders. Probability $10p^4q^2$ .
Deuce: 3 points each in the first 6; $\binom{6}{3} = 20$ orders. Probability of getting there: $20p^3q^3$ .

Note there is no 4–3 case: at 3–3 the game has already gone to deuce.

Deuce and the geometric series

From deuce, look at points in pairs. Each pair has three outcomes: A wins both ( $p^2$ ) and takes the game; B wins both ( $q^2$ ) and takes the game; or they split ( $2pq$ ) and it’s back to deuce. So A can win on the first pair, or after one return to deuce, or after two, and so on. Adding up all those paths gives a geometric series with ratio $2pq$ , which is less than 1, so the sum converges.

P(\text{win from deuce}) = \sum_{k=0}^{\infty} p^2(2pq)^k = \frac{p^2}{1-2pq}

Each trip back to deuce multiplies by 2pq; the infinite sum is finite.

From deuce with p = 0.6: A wins both points with 0.36, B with 0.16, and a split (0.48) returns to deuce. Hence p²/(1 − 2pq) = 0.36/0.52 ≈ 0.6923.

Putting it all together, the probability of winning the game is:

P(\text{game}) = p^4\left(1 + 4q + 10q^2\right) + 20p^3q^3\cdot\frac{p^2}{1-2pq}

Probability of winning a game when each point is won with probability p, where q = 1 − p.

Quick sanity check: with $p = 0.5$ the formula must give exactly 50%, since the players are identical. It does: the terms add up to 0.5.

Worked example: p = 0.6

Finding the chance of holding a game with 60% of the points

$\displaystyle p = 0.6,\quad q = 0.4$ Set the values. With p = 0.6, q = 0.4.
$\displaystyle p^4 = 0.1296$ Win 4–0: raise p to the fourth power.
$\displaystyle 4p^4q = 4 \cdot 0.1296 \cdot 0.4 = 0.20736$ Win 4–1: four possible orders.
$\displaystyle 10p^4q^2 = 10 \cdot 0.1296 \cdot 0.16 = 0.20736$ Win 4–2: ten possible orders.
$\displaystyle 20p^3q^3 = 20 \cdot 0.216 \cdot 0.064 = 0.27648$ Chance of reaching deuce: twenty orders with 3 points each.
$\displaystyle \frac{p^2}{1-2pq} = \frac{0.36}{1 - 0.48} = \frac{0.36}{0.52} \approx 0.6923$ Chance of winning from deuce, via the geometric series.
$\displaystyle 0.1296 + 0.20736 + 0.20736 + 0.27648 \cdot 0.6923 \approx 0.7357$ Add the direct cases and the deuce case.
Check: as an exact fraction the value is 29889/40625, which is 0.7357. A simulation of 200,000 random games with p = 0.6 also lands at about 73.6%.

How a small edge grows

Running the same formula for other values of $p$ (all hypothetical):

Chance of winning a point	Chance of winning the game
50%	50.0%
55%	62.3%
60%	73.6%
65%	83.0%

Five extra percentage points per point turn into more than 12 extra percentage points per game. The reason is that a game requires a run of successes: you have to reach 4 and still be 2 ahead. The more events a format demands, the more the underlying average beats luck, the same idea behind the law of large numbers.

A set repeats the trick one level up: you need 6 games with a 2-game margin (or a tiebreak). So an edge that was already amplified in the game gets amplified again, and a best-of-five match amplifies it a third time. Computing a set means tracking who serves each game, but the structure is the same: direct cases counted with combinations, plus a stretch that repeats.

The game's S-curve: 50% per point gives 50.0%, 55% gives 62.3%, 60% gives 73.6% and 65% gives 83.0%. The straight line is the point with no amplification.

Common mistakes

Forgetting the orders. Writing $p^4q$ for 4–1 ignores that the lost point can be any of the first 4. The binomial coefficient is required.
Letting the loss land on the last point. The final point always goes to the winner, which is why 4–2 uses $\binom{5}{2}$ , not $\binom{6}{2}$ .
Treating deuce as a single point. From deuce you need two in a row, and a split restarts the cycle.
Getting the ratio wrong. It’s $2pq$ , not $pq$ : a split pair can happen in two orders.
Mistaking the model for reality. Independent points with a fixed $p$ is a useful simplification, not an exact description of a real match.

Frequently asked questions

Why does the deuce series converge?

Because the ratio 2pq is never more than 0.5 (the maximum happens at p = 0.5). A geometric series with ratio between 0 and 1 sums to a/(1 − r), here p²/(1 − 2pq).

Does the formula work for a tiebreak?

Same idea, different coefficients: a tiebreak goes to 7 points with a 2-point margin, so you count wins from 7–0 through 7–5 and then the 6–6 tie, which also resolves with p²/(1 − 2pq).

What if a player is stronger on serve than on return?

Within a game the same player serves every point, so a fixed p is reasonable. The difference shows up in a set, where servers alternate; you compute one game probability for each case and combine them.

Can this predict real results?

It explains the structure. Real forecasts need p estimated from reliable data and have to account for points not being perfectly independent.