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Given that there are 16 Q's and 8 WC's compared to 104 direct entrants, the probability of a Q/WC drawing one of only 32 of the seeds seems low yet every GS there are always a handful of seeds were get these draws.....has it ever happened that no seed has gotten a Q/WC in Rd 1?
 

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I guess if you consider of the 128 players, these 32 seed can't draw each other, so those 32 1R spots are available to the 96 other players. So there's really a 1 in 5 chance for a Q/WC to draw a seed in the opening round. It's easier then it looks, but obviously not impossible for all to not draw a seed.

With that said I'm sure it must've happened during te 16 seeds days, but I'm not sure about the current 32 seed system.
 

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The chance of this happening is 0.00096%, or 1 in over 100000.
 

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The chance of this happening is 0.00096%, or 1 in over 100000.
How did you come up with that number? I'm sure the maths behind it is correct, I'm just wondering if it takes into account the changing sample size. Two players can't face the same opponent in the first round, after all. Once a WC/Q has been paired off with an unseeded player, the unseeded player is removed from the pot as it were.
 

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I'm getting a 0.003% chance (about 1 in 29,000).

There are 104 direct entrants, and 24 of them need to be chosen to play the qualifiers/WCs. There are 104!/(24!*80!) ways for that to happen.
There are 72 direct entrants that are not seeds. There are 72!/(24!*48!) ways to choose 24 of them to play the qualifiers/WCs.

Divide the second number by the first to get the probability :D

This is assuming none of the seeds and wild cards face each other. It's a lot more complicated otherwise.
 

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I did it like:

Pick 32 out of 96 unseeded opponents (for seeds)
Pick 32 out of 72 unseeded opponents who are not Q/WC

And divide it like that. Is that wrong?
 

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Given that there are 16 Q's and 8 WC's
The numbers vary. The Australian Open has 16 and 8 for men but 12 and 8 for women.
Kworb is right for 16 and 8. Suppose the 32 seeds are drawn first. The first Q or WC can be drawn in 96 places and 64 of them avoid a seed. The second can be drawn in 95 places and 63 of them avoid a seed, assuming the first avoided a seed.
The chance all 24 avoid a seed is 64/96 * 63/95 * ... * 41/73 = 0.0096% = 1/104136.

For 12 and 8 we get 64/96 * 63/95 * ... * 45/77 = 0.0091% = 1/11019.
 

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yeah kworb's way works (and is much simpler :eek: )

Edit. Now I'm looking at it like:

Fill in the 32 seeded players. There are 96 spots to place the 24 Q/WCs, and 64 of them result in not facing a seed. I get the same number as kworb.
 
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