Probability that the second-best player finishes second in a single-elimination tournament, given that better players always defeat weaker players? A chess tournament (single-elimination format) has 16 players. Suppose that no two players have the same strength, and that each player always defeats the players weaker than himself/herself (i.e. no draws). The loser of the final round becomes the runner-up. What is the chance that the second-best player turns out to be the runner-up? What if there are $2^n$ players?
I'm not sure how to approach this. Would it be correct to think that the the probability is $\frac{14}{15} \times \frac{6}{7} \times \frac{2}{3}$, since at each round, there is only one person who can cause the player not to advance, and the number of players in each round is halved?
How then, would I approach the follow-up question, where I am supposed to answer this in the general case?
 A: The second best player will always lose when they play the best player. They will come second if they reach the final. They will reach the final if they don't meet the best player prior. Therefore they will come second if they are in the opposite half of the draw to the best player.
If they are seeded then $p=1$.
If they are assigned randomly then $p=\frac{8}{15}$, or for the general case of $2n$, $p=\frac{n}{2n-1}$
A: Let's start by assuming that in total there are 16 players. They are divided into two groups of 8 players, each leading to one of the semifinals.
In this case, what we want is that the only time the best player and the second-best player meet is during the final. That means these two fellas musn't start in the same initial group of 8 players.
Now, the total number of ways these two groups could be formed is given by simply choosing 8 people among the whole collective of 16 players, and then let the remaining 8 form the other group. In  $\mathrm{Total}=\frac{16!}{8! 8!}$ ways.
As shown before, in the first 8 people - group there must be only one of the two strongest. That leaves us with $\mathrm{C}=\frac{14!}{7! 7!}2$ favourable combinations (you randomly choose 7 people out of the 14 least strong, then you pick one of the 2 left).
The probability is given by $\mathrm{p}=\frac {C}{Total}$.
With 16 players, it is equal to $\frac {8}{15}$.
In general, with $2^k$ players $$\mathrm{p}=\frac {2^{k-1}}{2^k-1}$$
