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Was this 'random' draw fixed ?

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mazza47
New Member

France
1 Posts

Posted - 10/11/2012 : 10:20:45

40 teams are taking place in a knock-out competition in which there is no seeding. They all have rankings determined by previous performance. The pairings are decided by a completely random draw, e.g. all the team names are put in a hat and drawn by a neutral party. What are the chances of the 14 top-ranked teams not meeting any team ranked higher than 15th?

Thanks,

Mazza

Ultraglide
Advanced Member

Canada
276 Posts

Posted - 10/13/2012 : 11:02:17

This is not an easy problem. Consider the probability of drawing a team form the top 14 which is 14/40. Now consider the probability of drawing the second team from the bottom 15 which is 15/39. So the probability of drawing the first two teams from the top and bottom group respectively is (14/40)(15/39). So the probability of not getting a match is 1-(14/40)(15/39). Now for the second draw, you have to consider cases: was the first one drawn from the top, middle, or bottom group? From what group was the second one drawn and so on. I don't know where this problem came from but it is a lengthy process if you want to continue.

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