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?

Let’s assume that the better team always wins when two teams play each other. 

There are 15 ways to pick the pair of teams that did NOT participate in the first 3 games, and so 15 ways to pick the 4 teams that DO participate in the first 3 games. 10 of those ways include the #1 team, 4 include the #2 team but do NOT include the #1 team, and only 1 way includes the #3 team without also including 1 or 2.  (It’s easier to see this by counting the number of omitted pairs that contain both 1 and 2 [1], that contain 1 but not 2 [4] and that do not contain 1 [10].)

Clearly only the best of the four teams can win all 3 games, and only if they play in the first game. The best team will play in the first game in exactly half of the configurations for a given set of 4 participants, although this doesn’t really matter — it just must have been true in order to observe a team winning their first 3 games.

If the winning team is the #1 team (which happens 10/15 of the time) then there’s a 100% chance of them also winning their 4th game. Similarly, if the winning team is the #3 team (1/15 of the time) there’s no chance at all that they’ll beat their next opponent since both of the remaining teams are better than them. And if the winning team is the #2 team (4/15) of the time) there’s a 50% chance that their next opponent is the #1 team and they’ll lose and so a 50% chance they’ll win.

Overall then, the winning probability is 10/15 * 1 + 4/15 * 1/2 + 1/15 * 0 = 10/15 + 2/15 = 12 / 15 = 4/5, which agrees with Charlie’s enumeration.

Posted by Paul on 2023-03-17 16:38:28