St_Tel49 Posted 9 December, 2011 Share Posted 9 December, 2011 (edited) Take the experimental approach. Draw up a grid for a set of 4 teams. (4x4 = 16). Eliminate the fact that they cannot play themselves: 16-4 = 12. We are only interested in the fact that they meet at all - not whether they are home or away therefore: 12/2 = 6. So the chance that any 2 out of 4 teams will meet is 1/6. i.e 1/4x1/3x2. Now increase the teams to 6 and repeat. You will find that the chance of any 2 teams out of 6 will meet is 1/6x1/5x2= 1/15. Extend that to our example it is 1/64x1/63x2=1/2016 that Man U will play Man C.at all, or 1/4032 that Man City will play Man U at home. Edited 9 December, 2011 by St_Tel49 spelling Link to comment Share on other sites More sharing options...
Whitey Grandad Posted 9 December, 2011 Share Posted 9 December, 2011 Take the experimental approach. Draw up a grid for a set of 4 teams. (4x4 = 16). Eliminate the fact that they cannot play themselves: 16-4 = 12. We are only interested in the fact that they meet at all - not whether they are home or away therefore: 12/2 = 6. So the chance that any 2 out of 4 teams will meet is 1/6. i.e 1/4x1/3x2. Now increase the teams to 6 and repeat. You will find that the chance of any 2 teams out of 6 will meet is 1/6x1/5x2= 1/15. Extend that to our example it is 1/64x1/63x2=1/2016 that Man U will play Man C.at all, or 1/4032 that Man City will play Man U at home. But the teams can be in any order so there are 64 positions that any team can be placed so it is 1/63. Link to comment Share on other sites More sharing options...
St_Tel49 Posted 9 December, 2011 Share Posted 9 December, 2011 But the teams can be in any order so there are 64 positions that any team can be placed so it is 1/63. You are of course quite right Link to comment Share on other sites More sharing options...
saint si Posted 9 December, 2011 Share Posted 9 December, 2011 Come on, this isn't that difficult... Imagine looking at Saints' home fixture list and picking a game at random. What's the chances it's the game against the skates? 1 in 23. Ok, now put all the possible opponents that City (or alternatively United) could have had in a list and pick one at random. 1 in 63. Link to comment Share on other sites More sharing options...
Whitey Grandad Posted 9 December, 2011 Share Posted 9 December, 2011 You are of course quite right It's a very interesting question though, and rather pleasant to see mathematics being discussed in such a civilised manner. Link to comment Share on other sites More sharing options...
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