{"id":383,"date":"2011-06-17T15:19:01","date_gmt":"2011-06-17T15:19:01","guid":{"rendered":"http:\/\/heavytopspin.com\/?p=383"},"modified":"2011-06-17T15:19:01","modified_gmt":"2011-06-17T15:19:01","slug":"what-are-the-odds-isner-mahut-redux","status":"publish","type":"post","link":"https:\/\/www.tennisabstract.com\/blog\/2011\/06\/17\/what-are-the-odds-isner-mahut-redux\/","title":{"rendered":"What are the Odds: Isner-Mahut Redux"},"content":{"rendered":"

Italian translation at settesei.it<\/em><\/a><\/p>\n

We all know what happened<\/a> when John Isner and Nicholas Mahut played a first-round match at last year’s Wimbledon. \u00a0In today’s draw ceremony, it was determined that they will face each other again<\/em>\u00a0next week.<\/p>\n

What are the odds?<\/strong> \u00a0(Thanks to Rick Devereaux for suggesting the question via email. \u00a0Judging from some google queries that led people to this site, he’s not the only one that wondered.)<\/p>\n

It turns out: Roughly 1 in 142, or 0.7%<\/strong>. \u00a0Here’s why:<\/p>\n

Before the draw ceremony, we knew that both players were unseeded. \u00a0Thus, both Isner and Mahut could have landed in 96 (128 total spots, minus 32 seeds) different places in the draw. \u00a0Of those spots, 32 (two-thirds) would’ve pitted an unseeded player against a seed. \u00a0If either Isner or Mahut had gone to one of those spots, obviously they could not have faced each other.<\/p>\n

Imagine that, rather than randomly choosing players for draw positions, we randomly choose draw positions for players. \u00a0In 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. \u00a0 After that, #101 is not in the hat, and we move on to Mahut.<\/p>\n

Here’s the calculation. \u00a0If Isner is assigned a position first, there are 96 places to choose from. \u00a032 of them close the door to a Mahut matchup; 64 of them leave open the possibility of a Mahut matchup. \u00a0Thus, there is a 64\/96 = 2\/3 chance that Isner is assigned a position that leaves the door open.<\/p>\n

Next, we assign a position to Mahut. \u00a0There are 95 non-seeded positions left (128 minus 32 seeds minus Isner’s spot). \u00a0Only 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.<\/p>\n

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

(Assuming, of course, that the draw is truly random!)<\/p>\n

UPDATE: <\/strong>There are many “calculations” floating around online that end up with different results. \u00a0Here are some, along with why they are wrong:<\/p>\n