Minimum matches in a single-elimination tournament with byes: An elimination tournament has 139 players. Some rounds may include a bye if there is an odd number. What is the minimum number of matches required to determine a single champion?

Introduction / Context:In any single-elimination tournament (knockout), each match eliminates exactly one player. Regardless of how byes are assigned, to go from n players down to 1 champion, we must eliminate n − 1 players, which requires n − 1 matches.

Given Data / Assumptions:

• n = 139 entrants.
• Single-elimination; a bye is not a match and eliminates no one.
• Each match eliminates exactly one player.

Concept / Approach:

• Matches needed = total eliminations = n − 1.

Step-by-Step Solution:

Verification / Alternative check:Construct any bracket: every non-champion loses exactly one match; count of losses equals number of matches. The champion has no loss. Thus total matches equals losers = 138.

Why Other Options Are Wrong:

• 136, 137 undercount; 139 overcounts by one.

Common Pitfalls:

• Thinking byes change the total count; they alter scheduling, not the total eliminations.

Final Answer:138