Raonic looked better yesterday than I’d thought he would, a lot more agressive and confident; it would be a very tasty match up with Nadal.

I really enjoyed your blog and will be interested to see the outcome of the polls. I think it will be Sharapova v Bartoli and Azarenka v Kvitova in the Women’s senses. ]]>

What I calculate there is the probability that if we choose only one matchup it will be Mahut vs Isner. But there is more than one matchup in a draw !

So I guess 1/140 is right ðŸ™‚

]]>it’s 1/7632

It’s choose 2 among 128 (total number of 2 player combination) – choose 2 among 32 (total number of 2 seeds combination)

= Choose(128,2) – Choose(32,2)

= 8128 – 496

= 7632

So the proability is 1/7632

]]>I don’t remember my college statistics that well but don’t we take the odds that they would have met last year (even though it was a commonplace occurrence at the time) and then multiply it by the odds that they would have met this year?

My point is, it’s certainly lower than 141/1.

Thanks for the spreadsheet and looking forward to your response.

]]>The chance of one of them being in a non-seed v non-seed match is indeed 64/96 = 2/3, but if one of them draws a non-seed, then by taking one of the non-seed v non-seed slots in the draw, it reduces the chance of the other one getting a non-seed v non-seed slot to 63/95. You are then correct to say the chance of that slot being the right one is 1/63.

Or, cutting out the middle step, the chance of Isner avoiding a seed is 2/3, as it says on your spreadsheet, and the chance of Mahut getting the slot next to him is 1/95. Either way, you get an answer of 2 in 285 = 1 in 142.5.

Since I’m sure the journos will go with “about 1 in 140” if they mention it at all, this is all pretty academic anyway of course! (and at least we both agree that it IS about 1 in 140, which I imagine is a more accurate consensus than the one in the Wimbledon press room ðŸ˜‰ )

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