1 in 733
It is soooo easy to make a mistake. I should've divided the total combinations of the fixtures by the permutations within themselves, i.e. AvB, CvD, EvF is the same combination as AvB, EvF, CvD.
So the total possible fixture combinations is ((20 choose 2) * (18 choose 2) *...*(4 choose 2))/10! = 654729075
The total possible fixture combinations featuring only teams playing in the same half is (((10 choose 2) * (8 choose 2) * (6 choose 2) * (4 choose 2))/5!)^2 = 893025
So 893025/654729075 = 0.001363961, i.e. 1 in 733
Posted By: yarmyyarmy on November 16th 2011 at 16:30:41
Message Thread
- fao yarmyyarmy (Other Football) - SCC 28, Nov 16, 15:34:01
- 1 in 733 (Other Football) - yarmyyarmy, Nov 16, 16:30:41
- The prob of it happening in the championship is 1 in 2927 (Other Football) - yarmyyarmy, Nov 16, 16:54:45
- super stuff (Other Football) - SCC 28, Nov 16, 16:58:49
- BOOOOOOOOOOOOOOOOOOOOOOOOOOOOM (Other Football) - SCC 28, Nov 16, 16:34:33
- So there have been around 700 Premierm League fixtures (Other Football) - norway, Nov 16, 16:34:20
- thing is fixture lists aren't random (Other Football) - SCC 28, Nov 16, 16:38:33
- The prob of it happening in the championship is 1 in 2927 (Other Football) - yarmyyarmy, Nov 16, 16:54:45
- so looks like i just needed to remove about half of my formula (n/m) (Other Football) - SCC 28, Nov 16, 15:37:31
- 1 in 733 (Other Football) - yarmyyarmy, Nov 16, 16:30:41
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