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The Mathematics of Pathological Tennis

June 25, 2010
Roscoe, N.Y.

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:

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?

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:


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

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

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

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:

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.

Winning a point on serve: 0.8 0.75 0.7 0.6 0.5
Winning a game in 4 points: 0.40960   0.31641   0.24010 0.12960 0.06250
Winning a game in 5 points: 0.32768 0.31641 0.28812 0.20736 0.12500
Winning a game in 6 points: 0.16384 0.19775 0.21609 0.20736 0.15625
Getting to deuce: 0.08192 0.13183 0.18522 0.27648 0.31250
Winning at deuce: 0.94118 0.90000 0.84483 0.69231 0.50000
Winning the game: 0.97822 0.94922 0.90079 0.73572 0.50000
137 games without break of serve: 0.049 0.00079 6.07 × 10–7 5.50 × 10–19 5.74 × 10–42

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.


This does assume that both players are of equal skill. Changing it so that supposing one player wins P_1 on the serve, and the other P_2 on the serve would change things a little bit. The probability being maximized is when P_1 = P_2 of course, but I think that in general |P_1 - P_2| may not be zero for most competitors.

And this also doesn't take in to account various psychological effects, such as when you're down, you tend to not do as well.

Kevin H, Fri, 25 Jun 2010 16:29:29 -0400

Suppose two players have probability P1 and P2 of winning a point, and P1 = P2, and you calculate the probability of a pathological set (137 games) as Pset.

Now suppose P2 is higher than the original P1. The overall possibility of breaking serve is less, and Pset becomes greater. In fact, look at the extreme case: If P2 = 1 (ie, an ace every serve) then the original Pset is now associated with a set of 274 games because half the games are always won. — Charles

"The men's game has increasingly emphasized power serves"

That's just because of bias in your sample. Instead of Wimbledon if you had studied a competitor such as Lindlenugz, you would have seen that the eunuch's game increasingly emphasizes power serves. Or if you let Apaches play then all serves can be powerful.

— That happens when programmers talk about sports, Sun, 27 Jun 2010 20:28:42 -0400

In the Sunday New York Times, an article by Leonard Cassuto "Men's Tennis Finds Itself at the Point of No Returns" discusses some of these issues, partially in reference to last year's Wimbledon final between Roger Federer and Andy Roddick, which was also a "sonic-boom serving" match:

    The match ... spotlighted the way that advancing equipment technology haunts rather than enhances the men's game... [It] delivered a reminder that an avalanche of unreturnable serves remains the bane of the men's game. It also featured some dazzling shot-making, which appeared only in flares that emerged from the smoke left by cannonball serves....

Cassuto concludes his article by wondering whether a return to wooden rackets (long advocated by John McEnroe) might make the men's game more interesting with "fewer aces and service winners, fewer stab returns and more points that were fun to watch."

— Charles

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