The Chess Tournament Bracket

63 total games are played across 6 rounds. The elegant proof: each game eliminates exactly 1 player, and 63 players must be eliminated to crown 1 champion from 64. General formula: n players require exactly n-1 games and log₂(n) rounds.

There are two ways to count: the round-by-round method and the elegant elimination argument. The elimination argument is more powerful because it works for any number of players, not just powers of 2. Step 1: Round-by-round: Round 1 — 64 players → 32 games → 32 winners. Round 2 — 32 players → 16 games. Round 3 — 16 → 8 games. Round 4 — 8 → 4 games. Round 5 — 4 → 2 games. Round 6 (Final) — 2 → 1 game. Step 2: Total games = 32+16+8+4+2+1 = 63. This is a geometric series: 32(1 - (1/2)^6)/(1-1/2) = 63. Step 3: Elegant argument: In a single-elimination tournament, every game produces exactly 1 loser. Every player except the champion loses exactly once. So total losers = 63. Total games = total losers = 63. Step 4: This argument works for ANY number of players, not just powers of 2: a tournament with n players always requires exactly n-1 games, regardless of the bracket structure. Step 5: Number of rounds (for powers of 2): each round halves the field. Starting with 64 = 2^6, we need 6 halvings to reach 1. Rounds = log₂(64) = 6. Step 6: General formula: n players → n-1 total games, log₂(n) rounds (when n is a power of 2). Key Insight: The elimination argument is one of the most elegant counting techniques in combinatorics. Instead of counting forward (how many games per round?), count backward (how many players must be eliminated?). Since each game eliminates exactly one player and exactly n-1 must be eliminated, the answer is immediate: n-1 games, always.