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Probability : Bracket (Cup) System

Q. The 'cup' system for determining the champion amongst 2^n players consists of drawing lots to arrange the players in 2^n-1 pairs who are to play each other, then repeating this with the 2^n-1 winners of these matches, and so on. The winner and loser of the final match recieve the first and second prizes repectively.
Suppose that each of the players has a defined strength (all different), and the strongest player wins each match. then is it clear that the first prize goes to the strongest of all the players. what is the probability that the second prize goes to the second strongest player?
In the case n=4, what is the probability that the four strongest players all reach the semi-finals ?

Solution Preview

For the first time, this probability is equal to the probability that the final game happens to be between the first and the second strongest persons. Therefore, the second one should not meet the first one any time before the final match.

Let's consider the simplest case, where we have 4 persons. The possibilities that we can group them for the first round are C(3,2)=3 (C means choose). After that, there is only 1 way to have the final match i.e. the runners of the first round. Now, we consider 8 persons. The possibilities that we can group them for the first round are C(7,6)*C(5,4)*C(3,2)=7*5*3. Then for the second round as there are 4 players in the list we have the same situation of the first case i.e. C(3,2)=3 and again for the final it would be 1. ...

Solution Summary

Probablitilies are calculated for events in the bracket system for sporting events.

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