6 football teams each of different level
of play compete in a knockout fashion: the 1st pair chosen at random and in the following rounds the winner is facing another randomly chosen team.
Given that a certain team won three games so-far what is
the probability of winning the 4th game?
In order to win four games, a team must either be in the first pair or the team playing the winner of the first pair. This eliminates four teams right off the bat - teams 4, 5, & 6, and the loser of game 1.
Consider each game an even proposition, the odds for Team 3 to win three games are as follows:
1/6 * 1/2 * 1/2 * 1/2 = 1/48.
Teams 1 and 2 are as follows:
2/6 * 1/2 * 1/2 * 1/2 = 2/48
These added together come to 5/48. So to win the 4th game, it is 5/96, or ~ 5.21%.
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Posted by hoodat
on 2023-03-17 21:44:41 |
Solution
