Salmon wrote:

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An interesting coincidence... in three of the four major tournaments held in Qatar this year (i.e. Doha and the three Futures), two wild cards have faced off in the first round. Something this complex is beyond my abilities, so someone please tell me what's the probability of this happening in 3/4 events?


(Just a reminder: ATP Doha awards just 3 WCs, unlike the Futures).




And well played, Luke Skywalker.

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1.215% by my calculations, including the possibility of 2 WC v WC matches in any of the Futures 1st round.

Prob (WC v WC in ATP Doha 1st Round) = 2/22 = ~9%

Prob (at least 1 WC v WC in Qatar F# 1st round) = 2/22 + ((20/22)*(2/21)) = ~18%

Prob (WC v WC in Qatar F1, F2, F3 and NOT ATP Doha) = ~ 18%*18%*18%*91% = ~0.5%

Prob (WC v WC in ATP Doha, in Qatar F1, F2, and NOT F3) = ~9%*18%*18%*82% = ~0.24%
Prob (WC v WC in ATP Doha, in Qatar F1, F3 and NOT F2) = ~0.24%
Prob (WC v WC in ATP Doha in Qatar F2, F3 and NOT F1) = ~0.24%

So 0.24%*3 + 0.5% is approx 1.22%.

Someone please tell me if that's wrong! 

Josh wrote:

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Salmon wrote:

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An interesting coincidence... in three of the four major tournaments held in Qatar this year (i.e. Doha and the three Futures), two wild cards have faced off in the first round. Something this complex is beyond my abilities, so someone please tell me what's the probability of this happening in 3/4 events?

(Just a reminder: ATP Doha awards just 3 WCs, unlike the Futures).

And well played, Luke Skywalker.

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1.215% by my calculations, including the possibility of 2 WC v WC matches in any of the Futures 1st round.

Prob (WC v WC in ATP Doha 1st Round) = 2/22 = ~9%

Prob (at least 1 WC v WC in Qatar F# 1st round) = 2/22 + ((20/22)*(2/21)) = ~18%

Prob (WC v WC in Qatar F1, F2, F3 and NOT ATP Doha) = ~ 18%*18%*18%*91% = ~0.5%

Prob (WC v WC in ATP Doha, in Qatar F1, F2, and NOT F3) = ~9%*18%*18%*82% = ~0.24%
Prob (WC v WC in ATP Doha, in Qatar F1, F3 and NOT F2) = ~0.24%
Prob (WC v WC in ATP Doha in Qatar F2, F3 and NOT F1) = ~0.24%

So 0.24%*3 + 0.5% is approx 1.22%.

Someone please tell me if that's wrong! 

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Aargh, I get back very late to find something like this waiting for me (yes, I know I didn't have to take the bait, but I'm me, I can't help it! )

I don't get exactly the same answer, but I do get something very similar (1.2241% cf your 1.215%), so we both end up with 1.22% after rounding.

It looks like you over-simplified a bit, but perhaps you get the last laugh if you can simplify that much and still get so close instead of going through all the steps below! 

If you want to compare methods, then looking at the Futures 4-wild card case, I started by using the hypergeometric distribution to work out the probability of W wild cards avoiding seeds and got:

0.0066 - no WCs avoid seeds
0.0843 - exactly 1 WC avoids a seed
0.3162 - exactly 2 WCs avoid seeds
0.4216 - exactly 3 WCs avoid seeds
0.1713 - all 4 WCs avoid seeds

In the 0 and 1 cases, it's impossible for 2 WCs to be paired, obviously.

In the 2 case, you can call the WCs who avoided seeds WC1 and WC2 and say that wherever WC1 ends up in the draw, there's a 1 in 15 (0.0667) chance that WC2 ends up in the same pair. Alternatively, you can do it by the standard approach of enumerating all the combinations (which gives the same answer in the end), but I find my 'quick and dirty' method a bit easier and more intuitive.

In the 3 case, if you fix WC1, there is a 2 in 15 chance that WC2 or WC3 end up in the same pair, but you also have to add in the possibility of neither of them being paired with WC1 but then getting drawn with each other. By similar reasoning to the above, this gives (2/15) + (13/15 x 1/13) = 0.2

Similarly, in the 4 case, the probability of at least 1 pair of WCs is (3/15) + (12/15 x 2/13) + (12/15 x 11/13 x 1/11) = 0.3846

So, overall, the probability of at least 1 WC v WC match in a Futures draw is:

(0.3162 * 0.0667) + (0.4216 * 0.2) + (0.1713 x 0.3846) = 0.1713

I treated the Doha case similarly (obviously it's a bit easier because there are fewer possibilities to consider) and got 0.0870 for that one.

Then I combined those probabilities the same way as you did to get (using unrounded figures throughout instead of the 4 DP figures above) 0.012241, i.e. 1.22% to 3 SF.

I must admit, can't quite fathom your method, but it looks like an impressively quick way to get a very good approximation to the answer!

And now, back to the tennis ... 

paulisi wrote:

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I have a degree involving statistics and that analysis hurt my head.....

Back to the tennis:

Liam and Scott both first on court and both are 3-0 up first set.

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Now that's something I understand ! (the last bit, that is . . .. )

Liam has won 32 points to his opponent's 13.   6-0 1-0 hardly says the half of it . . .

Liam through 6-0 6-1. Taking care of business comfortably. 

DWH wrote:

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Scott's score frozen at 3-1* 15-15 

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Now it's disappeared altogether!