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?
Back to the ambiguity question. Upon further reflection, the problem is worded as a Bayes question, considering a team with three wins as a starting point. There are only three teams with the possibility of having that opportunity. Or if one considers "the 4th game" to be specific to all teams instead of relative to a team's own situation, then there are only two teams having that possibility.
But for any team finding itself with three wins, the probability of winning their next game will be an even proposition. It is no different from flipping 'heads' three times in a row. Past flips do not affect the outcome of the next flip.
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Posted by hoodat
on 2023-03-18 16:29:59 |
re: Puzzle answer
