In the draw for a preliminary round of a soccer competition, each of the 32 seeded teams, placed in Urn A, is paired with one of the 32 unseeded teams, placed in Urn B, to play a single elimination match. The host of each match is determined as follows: seeded teams, as well as unseeded teams, are randomly assigned the numbers 1 through 32. Then, at each of the 32 steps, a ball containing the name of a team from Urn A and a ball containing the name of a team from Urn B are extracted and the host of that match is the team with the lower number. If the same number is drawn, the host will be the unseeded team. What is the probability that the draw produces the maximum possible number of matches that could be hosted by the seeded teams?
The maximum number hosted by the seeded team is 31
32 with 31
31 with 30
2 with 1 so far all hosted by seeded team
1 with 32 hosted by unseeded team
I cannot see any other way of doing this.
If seeded are lined up and each one randomly paired with an unseeded then there wll be 32*31*.....*1 = 32! possible matches
So I think the prob that 31 games will be hosted by the seeded team is 1/32!
To answer your private message question
I put each seeded team number with the unseeded number directly below
32 with 31 etc
the ones left at the end are seeded 1 with unseeded 32
I agree that there are a possible 32! pairings in total, and that the maximum number of possible matches in a draw hosted by the seeded team is 31. This is the same as saying the minimum number of matches hosted by the unseeded team in a draw is 1. Since there are 32 unseeded teams the number of possible draws that result in just 1 unseeded team hosting is 32, so I make the probability p = 32/32! or p = 1/31!.