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?