Answer: Tuesday Teaser #31
For the first problem, let’s consider the properties of the tournament rather than trying to deal with a messy tournament bracket. The tournament is straight knockout. That means that all players except the winner, end up losing one (and only one) match, so we can immediately conclude that if there are 128 players, the total number of matches must be 127. Nice.
For the second problem, let’s consider the position of the first seed in the draw. The draw is random, so each position of the tournament bracket is equally likely. Whatever position the first seed ends up in, let’s now consider the position of second seed. All remaining positions are equally likely, with 63 of them on the same side of the draw as the first seed and 64 on the other side. Given that the results go to seed, the probability of them meeting in the final is therefore 64/127.
If, like me, you originally thought that the answer to this was a half, consider the case when there are only four players. You can write out all 24 possible draws and you’ll see that 16 of them (i.e. 2/3) have the first two seeds on different sides of the draw. This agrees with the logic above; fixing the first seed, there are three possible positions left for the second seed and two of them are on the opposite side of the draw.
In general then, we can conclude that for a tournament with 2n players, the likelihood of the first and second seeds meeting in the final (assuming results go to seed), is n/(2n-1).
For the more involved third problem, let’s use the same logic as above, but extend it to the case with 16 seeds. As above, fix the position of the first seed, and consider the position of the second seed. In order to make the round of 16, the second seed has to be in a position where they won’t play the first seed for at least three matches. There are seven remaining spots where they will be drawn against the first seed in the next three matches, so 120 of the remaining 127 will suffice.
Now consider the position of the third seed. In order to progress to the round of 16, they need to avoid both the first seed and the second seed, so there are now fourteen spots to avoid. This leaves 112 of the remaining 126 positions that will suffice. Continuing in this way until we reach the sixteenth seed, we obtain the following product:
Which computes to a probability of 0.00000302. This is the probability that all sixteen seeds will reach the round of 16, assuming that the results go to seed. This confirms the natural intuitive response that this scenario is extremely unlikely!
