## Thursday, June 24, 2010

### What were the odds of the 70-68 score at Wimbledon?

Two minutes after posting this, I found that Carl Bialik covered this topic much better, and answered some of my questions. You might want to check out his post instead.

---

At Wimbledon, the fifth set of the Nicolas Mahut / John Isner match took 138 games to decide. It was won by Isner, 70-68.

As I understand it, the scoring system for the fifth set at Wimbledon works like this: the first player to get to 6 games wins, except that he has to win by two. That means that the score must have been 6-6 at one point. Then, for every pair of games after that, it must have been that Mahut won one of them and Isner won one of them (in either order), until the last pair, where Isner won them both.

What are the odds of a pair of games being split? The player serving has a strong advantage, and therefore has a much better than 50% chance of winning the game. I'm not sure how much that advantage is (anyone know?), but let's start with a supposition that it's 80%. Since players alternate serves in consecutive games, that means the chance of splitting is 68%. There's a 64% chance that the server wins both games (80% times 80%), and a 4% chance that the non-server wins both games (20% times 20%).

After 6-6, there were 62 such splits, bringing the score to 68-68. The chance of 62 consecutive results with a probability of .64 is just over 1 in a trillion. If you played one set a second, it would take 32,963 years until 68-68 happened.

It seems safe to assume, then, that our estimate of an 80% chance that the server wins is a little too low. Suppose it's 90%. Then the probability of two split games is .82 instead of .68. That brings the chance of 68-68 to only about 1 in 220,000. Better.

What if we go 95%? The probability is now .905 to the 62nd power, which is 1 in 487. That seems much more reasonable -- perhaps even too likely, since 70-68 happens a lot less often than 1 in 487.

Maybe 92%? That's approximately 1 in 19,000. That seems more reasonable to me. Assuming 91% gives about 1 in 66,000. In a chart:

80% -- 1 in 1,000,000,000,000
90% -- 1 in 220,000

91% -- 1 in 66,000
92% -- 1 in 19,000
95% -- 1 in 487

These, by the way, are the maximums that occur when both players are of exactly even skill. If one player is better than the other, or even slightly better, the odds go way down. It's fair to assume, then, that these two players are fairly evenly matched. Of course, when a set is tied after 136 games, you don't need serious probability calculations to guess that.

In the fifth set, Mahut's "receiving points won" was 23% (which means he won 23% of points when his opponent served). Isner's was 17%. Those are indeed lower than in some of their other sets (which ranged from 18 to 41 percent). And a quick look at some of the other matches suggests that 30% is typical (I'm sure some reader knows where to find actual numbers). But this other close match had 28% and 24% overall, which isn't that much bigger (although small changes lead to very large changes in probabilities when you extend them to 124 games).

Anyway, my question to you tennis watchers is this: 70-68 is so unusual that the best guess is that, in this particular set, the server had a much greater advantage than normal. Is that correct? Were either or both players playing a style of tennis that somehow limited their probability of winning receiving points or breaking serve?

Maybe when you get tired after playing so many hours, your ability to return serves drops substantially? But, the match was called at 59-59 on account of darkness, so there were another 9 split pairs when the players resumed this morning, presumably refreshed.

What really happened here?

Labels: ,

At Thursday, June 24, 2010 4:03:00 PM,  Chris said...

You had two people who's chances of holding serve are high, playing on a service where service holds are more likely. It was still utterly mind-boggling. Carl's one in a million sounds close to right - this was far, far beyond anything that's ever happened in tennis, and I can't conceive of it being matched.

At Thursday, June 24, 2010 4:37:00 PM,  Devin said...

Obviously it was an extremely rare event, but I don't think the probabilities can be thought of as independent like that. I watched about 4 hours of the match yesterday plus the conclusion today, and clearly playing such a long time had taken a huge toll on their return games. Isner was barely moving on many returns, and there were tons of games held at love and 15. I think he lost something like 20 to 30 mph on his serves by the end of the match, to give you an idea of the physical toll it took on him. I think a careful look at the point percentages as the match went on would likely show a growing advantage to the server as the match went on, especially Mahut, who seemed much fresher, if still unable to break through. Fwiw, betting markets had the over under at 130.5 games after yesterday when they stopped at 118, albeit most likely in a tiny market.

At Thursday, June 24, 2010 8:01:00 PM,  Phil Birnbaum said...

If the betting markets expected between 6 and 7 extra "pairs" of unbroken serves, that means a 50% probability. That implies an 89.1% chance of a pair for 6 extra pairs, and a 90.6% chance of a pair for 7 extra pairs. Taking the midpoint gives roughly 89.8%.

The square root of .898 is just below .95. So the betting market was saying there was a 95% chance of each player winning a game in which he served.

And that's in games in which they wouldn't be all that tired.

Wow, that's high.

At Thursday, June 24, 2010 10:09:00 PM,  David Barry said...

Another aspect of this question is: even given the strong-serving, weak-returning nature of the players (especially when so fatigued), how unlikely is 136 consecutive holds of serve? (And there were actually more than that, counting the previous set.)

The page here gives a formula for the probability of winning a service game, given the probability of winning a service point.

Plugging in a probability of a point win as 0.8 gives a probability of a hold of serve at 0.978, and raising that to the power of 136 gives 0.05.

So the marathon final set was moderately unlikely, even allowing for the unusual characteristics of the players in this match.

At Thursday, June 24, 2010 11:23:00 PM,  Jim A said...

I'm curious whether the chances of holding serve has increased over the years, presumably due to better rackets. I know this was a rare event, but at some point if the server is winning 90-95% of games, the sport becomes less interesting to me.

On a related note, it's often said that the team/player with the best "true talent" would always win a game of infinite length. This game was perhaps as close to that as we'll ever see, yet it was still extremely even. Amazing.

At Friday, June 25, 2010 6:23:00 PM,  Craig said...

Didn't see your post before I wrote up a little Matlab simulation to test it out. Results seem pretty similar (thankfully.)

http://bit.ly/dnQmqa+

Nice job!

At Sunday, June 27, 2010 9:55:00 PM,  plutoman said...

You can find my attempt at calculating this, as well as some background to the match here

I got a probability in "only" the 300-to-1 range, although my assumptions are probably rubbish. Garbage in, garbage out as they say. One thing for sure, calculating the odds of very low probability events that have never happened before isn't easy.

At Friday, July 16, 2010 4:14:00 PM,  Dave said...

You say "it must have been 6-6 at one point". I'll do you one better. It also must have been 5-5 at one point. Maybe it was 4-4 before that or maybe it was never 4-4, but unless the score was 5-5 at one point the match would have been over.

At Wednesday, July 11, 2012 3:40:00 PM,  Dan Kurt said...

I saw some of the match on TV. It was at an outer court. Both players could not return worth a darn. By the 5th set they were dog tired and as the 5th set continued they were just serving, many times not even trying to return.

The match was a travesty. Bring on the 5th set tiebreak.

Dan Kurt