Chess Tournament Puzzle: When Do the Top Two Meet?

A knockout tournament poses a surprising question about randomness and ranking:

There are \(2^n\) players labeled \(1, 2, \ldots, 2^n\), ordered by strictly decreasing skill.
The best player is #1, second-best is #2, etc.
The tournament is single elimination, and pairings are random in each round.
Better player always wins.

Question: What is the probability that players 1 and 2 will meet only in the final, not in any earlier round?

Step 1: Understand the Structure

  • The tournament has \(n\) rounds (since \(2^n\) players are halved each time).
  • Player 1 never loses.
  • Player 2 loses only to Player 1.
  • To meet in the final, they must be in opposite halves of the bracket throughout.

Each round’s pairings are randomized, so it’s not a fixed bracket.

Step 2: Key Insight

Think recursively: as long as players 1 and 2 avoid each other in every round until the final, they survive.

So the question becomes:

What’s the chance that 1 and 2 don’t get paired together in any round before the final?

Step 3: Derive the Formula

The total number of distinct matchups in a \(2^n\) tournament tree is:

\[ 2^n - 1 \]

Why? Because each match eliminates one player, and there are \(2^n - 1\) eliminations total.

Now consider: out of all these matches, in exactly one of them will players 1 and 2 meet—assuming they both survive to that point.

If they meet in the final, that match must be the last one (the \(2^n - 1\)-th match).

So: The number of total match trees where 1 and 2 meet only in the final is:

\[ \text{Favorable cases} = 2^{n-1} \]

(Why? Because in a random binary tree with \(2^n\) leaves, the number of ways to position two specific players in opposite halves is \(2^{n-1}\)).

Hence:

\[ P(\text{1 vs 2 only in final}) = \frac{2^{n - 1}}{2^n - 1} \]

Final Answer

The probability that players 1 and 2 meet only in the final is:

\[ \boxed{\frac{2^{n - 1}}{2^n - 1}} \]

Reference