Answer: Second-Best Runner-Up

Key Idea / Intuition

The second-best player wins the runner-up cup if and only if they never meet the best player until the final. This happens precisely when they are placed in the opposite half of the bracket from the best player. Since the bracket has two halves of equal size, and the best player occupies one slot in one half, the question reduces to: what fraction of the remaining slots are in the other half?

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Formal Proof / Solution

Setup. Fix the best player's position (by symmetry, it doesn't matter where they are). There are $8 - 1 = 7$ remaining slots for the second-best player to occupy.

When does the second-best win the runner-up cup?

The second-best player can only lose (and thus get the runner-up cup) by facing the best player — but to receive the runner-up cup, this loss must happen in the final.

Since the best player always wins, the best player will reach the final regardless. The second-best player reaches the final if and only if they never meet the best player before the final, which happens if and only if they are in the opposite half of the bracket.

Counting. With 8 players, each half of the bracket has 4 slots. The best player occupies 1 slot in their half. The remaining 7 slots are distributed as:

• 3 slots in the same half as the best player
• 4 slots in the opposite half

So the probability is:

$P(\text{runner-up cup}) = \frac{4}{7}$

General case with $2^n$ players.

Each half has $2^{n-1}$ slots. One slot is taken by the best player in their half. Remaining slots:

• $2^{n-1} - 1$ in the same half
• $2^{n-1}$ in the opposite half
• Total remaining: $2^n - 1$

$\boxed{P = \frac{2^{n-1}}{2^n - 1}}$

For $n = 3$ (8 players): $P = \frac{4}{7} \approx 0.571$.

Sanity check as $n \to \infty$: $\frac{2^{n-1}}{2^n - 1} \to \frac{1}{2}$ which makes sense: in a huge tournament, the two halves are nearly equally likely, so the second-best ends up in the opposite half roughly half the time.

What's beautiful here: The problem looks like it requires tracking all possible bracket matchups, but the key insight collapses everything to a single observation — the second-best player's fate is entirely determined by which half of the bracket they land in. All the internal randomness within each half is irrelevant, because both the best and second-best players are guaranteed to win every match except against each other.