What are the Odds: Isner-Mahut Redux

Italian translation at settesei.it

We all know what happened when John Isner and Nicholas Mahut played a first-round match at last year’s Wimbledon.  In today’s draw ceremony, it was determined that they will face each other again next week.

What are the odds?  (Thanks to Rick Devereaux for suggesting the question via email.  Judging from some google queries that led people to this site, he’s not the only one that wondered.)

It turns out: Roughly 1 in 142, or 0.7%.  Here’s why:

Before the draw ceremony, we knew that both players were unseeded.  Thus, both Isner and Mahut could have landed in 96 (128 total spots, minus 32 seeds) different places in the draw.  Of those spots, 32 (two-thirds) would’ve pitted an unseeded player against a seed.  If either Isner or Mahut had gone to one of those spots, obviously they could not have faced each other.

Imagine that, rather than randomly choosing players for draw positions, we randomly choose draw positions for players.  In other words, we start by saying, “Where will Isner go?” and then pick a number out of a hat, and determine that he’ll be placed in, say, draw position #101.   After that, #101 is not in the hat, and we move on to Mahut.

Here’s the calculation.  If Isner is assigned a position first, there are 96 places to choose from.  32 of them close the door to a Mahut matchup; 64 of them leave open the possibility of a Mahut matchup.  Thus, there is a 64/96 = 2/3 chance that Isner is assigned a position that leaves the door open.

Next, we assign a position to Mahut.  There are 95 non-seeded positions left (128 minus 32 seeds minus Isner’s spot).  Only one of those 95 spots is a first-round matchup with Isner, so if a matchup is possible, the odds of Mahut being assigned that spot are 1 in 95.

Thus, the probability of a matchup occurring is (2/3)*(1/95) = (2/285), or 1 in 142.5.

(Assuming, of course, that the draw is truly random!)

UPDATE: There are many “calculations” floating around online that end up with different results.  Here are some, along with why they are wrong:

• 127 to 1.  That sounds appealing, since there are 128 men in the draw.  But 127 to 1 is only correct if there is no seeding.  There are only 96 possible places in the draw for unseeded players such as Isner and Mahut, and as described above, not all of them allow two unseeded players to face each other.
• 95 to 1.  Better, since it acknowledges seeding.  But it doesn’t take into account the possibility that Isner or Mahut would draw a seed.
• 16,000+ to 1.  Any number this big is talking about the odds of two specific players facing each other two years in a row.  In retrospect, we know that Isner/Mahut turned out to be very interesting, but in May of 2010, no one was asking the question, “What are the odds of Isner and Mahut drawing each other this year, and then drawing each other next year, too?”  The answer to that question is roughly 20,000 to 1, but it’s not the right question.  It’s a given that Isner and Mahut played each other last year–in terms of probability, there’s a 100% chance that they faced each other a year ago.  Since we’re aware of the history, the relevant question is: What were the odds they would draw each other again?  It’s far-fetched, but not 16,000-to-1 far-fetched.  If they draw each other again in 2012, then we can start talking about 20,000 to 1.