In a knockout competition (Wimbledon?) where there can be no draws, N teams enter. In the first round there are 'R1' matches and 'b' byes such that 'R1' - n is a power of 2. Thus the remaining rounds halve the participants until there is a winner in final.

eg N=10, then R1=2 and b=6 leaving 8 teams after first round. Then 4 teams are after 2nd, then 2 after 3rd then 1 winner.

Question: What is the total number of matches before there is a champion?

Originally posted by wolfgang59 In a knockout competition (Wimbledon?) where there can be no draws, N teams enter. In the first round there are 'R1' matches and 'b' byes such that 'R1' - n is a power of 2. Thus the remaining rounds halve the participants until there is a winner in final.

eg N=10, then R1=2 and b=6 leaving 8 teams after first round. Then 4 teams are after 2nd, then ...[text shortened]... n 1 winner.

Question: What is the total number of matches before there is a champion?

Hint: Everyone loses once, except the winner of the competition, right?

Originally posted by wolfgang59 In a knockout competition (Wimbledon?) where there can be no draws, N teams enter. In the first round there are 'R1' matches and 'b' byes such that 'R1' - n is a power of 2. Thus the remaining rounds halve the participants until there is a winner in final.

eg N=10, then R1=2 and b=6 leaving 8 teams after first round. Then 4 teams are after 2nd, then ...[text shortened]... n 1 winner.

Question: What is the total number of matches before there is a champion?

I'd suppose it's N - 1 because to find out a winner, everyone else must be eliminated, therefore, 1 match for each loss of every other team. 10 teams, 9 matches, 1 winner, 9 losers.