Charles Petzold

Many tennis fans seemed to enjoy the 5th set of the recent Wimbledon match between John Isner and Nicolas Mahut. In the absence of a tie-break for the final set, it went on for 138 games, finally ending at an unfathomable score of 70-68 with only one break of serve at the end.

To me, it was a frightening but not altogether surprising phenomenon. The men's game has increasingly emphasized power serves, and modern rackets help the players achieve serves that are simply unreturnable. The sheer number of aces in the Isner/Mahut match should be a warning that this problem will only get worse in the future. I'm not sure if we'll ever see something quite so extreme, but I suspect we'll see a match come close in the years ahead, and we'll see more matches achieve double-digit 5th sets. (And while I'm at it, I'll also predict that Wimbledon will implement a last-set tie-break sometime in the next decade.)

But I was curious about the mathematics of this 5th set. Just how unlikely was it?

Let's assume you have two equally matched players who are both powerful servers and who happen to be playing very consistently. Both players have a probability P of holding serve against the other. In other words, for each point, the server has a probability P of winning the point, and the returner has a probabiliy of (1 – P) of winning the point. Just as a sample number, let's assume P is 0.9. Each player, when serving, wins 90% of the points.

It's well known based on the Law of Large Numbers that if P is greater than 0.5, the more points that are played, the more likely it is the server will win overall. For example, for one point, the probability that the server will win is just P or (in my example) 0.9. But for "best two out of three points" the server can win in one of three ways:

Win-Win with a probability of P² or 0.81.

Lose-Win-Win with a probability of (1 – P) • P² or 0.081

Win-Lose-Win also with a probability of (1 – P) • P² or 0.081

The total is 0.972, which means that "best two out of three" increases the probability of the server to win from 90% to 97%.

What is the probability of the server winning a regular 4-point tennis game?

The server can win with 4 points in a row with a probability of P⁴ or (using P equal to 0.9) a probability of 0.6561.

The server can win with 5 total points in 4 different ways: LWWWW, WLWWW, WWLWW, and WWWLW. Each has a probability of P⁴ • (1 – P), so the total probability is 4 • P⁴ • (1 – P) or 4 • 0.06561 which is 0.26244.

The server can win with 6 total points in 10 different ways: LLWWWW, LWLWWW, LWWLWW, LWWWLW, WLLWWW, WLWLWW, WLWWLW, WWLLWW, WWLWLW, and WWWLLW. Each has a probability of P⁴ • (1 – P)², so the total probability is 10 • P⁴ • (1 – P)² or 10 • 0.006561 which is 0.06561.

Already we're up to a total probability of 0.6561 + 0.26244 + 0.06561 = 0.98415, and we haven't even gotten to the messier math of the deuce.

There are 6! / (3! • 3!) or 20 different ways to get to deuce. (OK, I'll list them: WWWLLL, WWLLLW, WWLLWL, WWLWLL, WLLLWW, WLLWLW, WLLWWL, WLWLLW, WLWLWL, WLWWLL, and now switch all the W's and L's.) The probability of winning 3 points and losing 3 points is P³ • (1 – P)³, so the total probability of getting to deuce is 20 • P³ • (1 – P)³ or 20 • 0.000729 which is 0.01458.

At deuce, you must win two games in a row. The probability of that is P² or 0.81. The probability of losing two games in a row is (1 – P)² or 0.01. There are two different ways to win one and lose one, each with a probability of P • (1 – P) or 2 • P • (1 – P) or 0.18, at which point it starts over again. (Notice how the three probabilities add to 1, which is always comforting.)

So the total probability of winning deuce is a power series:

P² + P² • [2 • P • (1 – P)] + P² • [2 • P • (1 – P)]² + P² • [2 • P • (1 – P)]³ + ...

Or:

P² • ∑ [2 • P • (1 – P)]^N

where N goes from 0 to infinity. It's not as bad as it looks. It is well known that the infinite geometric series

∑ X^N

where X is less than 1 converges to 1 / (1 – X), which means that the probability of winning at deuce is:

P² / (1 – 2 • P • (1 – P)) or (setting P to 0.9), 0.81 / 0.82 = 0.9878.

So, if the probability of holding serve is P, the probability of winning a 4-point service game is the sum of the following:

Four points in a row: P⁴ or 0.6561.

Win four, lose one: 4 • P⁴ • (1 – P) or 0.26244.

Win four, lose two: 10 • P⁴ • (1 – P)² or 0.06561.

Get to deuce and win: 20 • P³ • (1 – P)³ • P² / (1 – [2 • P • (1 – P)]) or 0.01458 • 0.9878 = 0.01440.

The total is 0.65610 + 0.26244 + 0.06561 + 0.01440 = 0.99855.

In other words, if the probability of the server winning the point is 90%, the probability of the server winning the four-point game is 99.8%.

Now for the big question. If the probability of the server winning the game is 0.99855, what is the probability of a set going on for 137 games without a break of serve? That's simply:

0.99855¹³⁷ = 0.82

That is a very high number, and surely we won't be seeing 82% of all matches last for 11 hours!

But let's try something a little more reasonable: Let's say that each of the two players consistently wins 75% of points when serving against the other player. The probability of winning a four-point game is 94.9%. The probability of 137 straight games without a loss of serve is about 0.08% or about 1 in every 1,250 matches.

Addendum, June 26, 2010

Here are results for a few other values of P (the probability of winning a point on serve). All numbers are probabilities.

Suppose players A and B are in the 5th set of a match. Player A is serving well enough against B to win, on average, 80% of all points. Player B is not quite as good but wins 70% of all points when serving against A. The probability of going two games without a break of serve is the product of the probability of winning a game for these two cases: 0.97822 × 0.90079 = 0.88928. The probability of going 136 games without a break of serve is that number to the 68th power, or .00034, or about one match in every 3,000.