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?
Although I still believe that no clarifications are needed, I would like the solvers absolutely accept the following assumptions:
There always be 5 games to be played
In each of them the more qualified team wins , the loser goes home.
Lets name the teams ABCDEF , A being the best and F the weakest
No matter what are the results of drawings, if A is chosen he never stops playing.
Clearly C can win at most 3 games and therefore for him chances of having another game are zero
Only A and B can win 3 in a row and once you perceive it you have D2 puzzle to complete.
I hope someone does it prior to posting the official solution.
Please do.
ALL SOLVERS. READ IT CAREFULLY
