# 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?

• Looks good to me. Observe that the product simplifies to $8/15$. – Jyrki Lahtonen Sep 22 '14 at 7:38

If they are seeded then $p=1$.
If they are assigned randomly then $p=\frac{8}{15}$, or for the general case of $2n$, $p=\frac{n}{2n-1}$
• I understand your answer, but I'm having trouble understanding why my reasoning for the probability is incorrect... Isn't it true that for the round of $n$ players, the probability of the second-best player advancing $\frac{n-2}{n-1}$? And by multiplying this probability for successive rounds, I was under the impression that it would give me an answer... – calvin cheng Sep 22 '14 at 7:16
• Hmm. In a field of sixteen players wouldn't the probability of the best player to be on the other side of the bracket be $8/15$ as opposed to $1/2$? You place the second best player in some random slot. Then there will be 7 open slots on the same half and 8 on the opposite half for the best player to be put on. Thus $8/15$ ways of placing the best work out well for our runner-up. BTW, this is what the OP's calculation simplifies to. Generalization to $2^n$ players is immediate. – Jyrki Lahtonen Sep 22 '14 at 7:33
• Please note that $p = 0.5$ is wrong for the second case. If the strongest player is on one side of the draw, the probability for the second best to be on the other side is $8/15 = 0.5333...$ because one possibility has been taken by the best player. Hence, OP's calculation is correct (but overly complex, I agree). – Traklon Sep 22 '14 at 7:36