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| T O P I C R E V I E W |
| mazza47 |
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 |
| 1 L A T E S T R E P L I E S (Newest First) |
| Ultraglide |
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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