Probability of the 'big guns' staying apart until final?

Question

It is a non-rigorous discussion on probability.

I am reading the book 'How long is a piece of string?' by Rob Eastaway and Jeremy Wyndham. In one of the chapters it talks about sports games and why the 'underdogs' manage to win sometimes. One factor that the authors think is the 'knockout' system.

In the knockout system, contestants are divided into groups. For example, A, B, C and D are joining the competition. We divide them into $2$ groups: (A vs B) or (C vs D). In general, we will have $3$ possible groupings. If A and B are the so-called 'big guns' of the competition, then the probability that they will meet in the next-round, which is also the final, is $\frac23$.

Now then the book claims the following:

In fact, with $N$ teams left , the probability of teams staying apart until the final turns out to be $\frac{N}{2N-1}$.

I don't see that coming. For one thing, the situation itself is unclear, like how many knockouts will there be in every groups?

Now suppose I have $8$ contestants, in which $2$ of them are the 'big guns'. In every round, we have only $2$ groups and the worst-performing contestants in each group will be out. Assume that the 'big guns' never lose, then the probability of the 'big guns' meeting in the final rounds should be $\frac47 \times \frac35 \times \frac23 = \frac{8}{35}$.

Am I misunderstanding the knockout system? Why cannot I get the same answer as the book? Thanks in advance.

Answer 1

I think the book is wrong - slightly.

Imagine the draw for $2N$ (not $N$) teams, and arranging it in the typical sideways pyramid

Pick your favourite team. Then there are $N$ teams they cannot meet until the final, in the opposite half of the draw, and $N-1$ in the same half that they will meet before the final if they meet at all. If slots in the draw are distributed at random then any particular opponent will have probability $N/(2N-1)$ of being in the can-only-meet-in-the-final group. With four teams $N=2$ and the probability is $2/3$ as you calculated.

(It does not matter, of course, whether the entire tournament draw is worked out at the beginning or pairwise draws are carried out at the end of each round)

Answer 2

Julia Hayward's explanation is correct; the book's sole error is the condition of $N$ teams left; it should be $2N$ teams left.

Your computation of $8/35$ is really the probability that two given teams are in separate halves of the draw after each round, if the teams are randomly rearranged in the draw after each round. No real sports tournament that I'm aware of runs things that way, and in any event the value you derived doesn't tie directly back to the probability of meeting before the final in such a tournament.

I would have left this as a comment, but I'm not permitted to do so yet. Since I have to leave it as an answer, I might as well take the opportunity to explore one of the other questions of underdogs winning.

Single-elimination tournaments are on lots of sports-watchers' minds at the moment, since March Madness is about to start. Setting aside the play-in tournament (a small oddity), there are 64 teams in the tournament, arranged in four regions; each region consists of a 16-team bracket, seeded 1 through 16.

Much is often made of the higher-than-expected propensity of some lower seeds (especially, it seems, the 12-seeds) to mount a few upsets. I did some rummaging through NCAA history, and found that it's not so much that they win any individual game more likely than you'd expect, but that the structure of the tournament, and the way that team strengths are distributed, conspire to make it easier for 12-seeds to advance to certain rounds (specifically, the third round, the so-called "Sweet Sixteen") than some higher seeds.

Intuitively, the explanation is that although the 12-seed has a harder time beating its 5-seed opponent than (say) an 8-seed has beating its 9-seed opponent, that difference is more than made up for by the substantial difference between the 12-seed's second-round opponent (likely the 4-seed) and the 8-seed's second-round opponent (even more likely the 1-seed).

I put some details in a blog post here.