# The math behind a bracket

Assuming single elimination with no byes, I know that the number of teams in the field is 2^x where x is the # of games the champ needs to win. AND that the field is x-1. My question is: WHY is this true? And I realize the explanation will be glaringly obvious.

2nd statement is not correct and I am not sure what your question is

Because the field is halved in each round.

Nope.

Work backwards. If it’s single elimination for X teams, then at the end of the tournament, every team but one (the champion) will have a loss. Thus, there are X-1 games (since each game doles out one loss). As for the number of games the champ needs to win, that completely depends on whether or not you have an even factorial of 2 teams (2, 4, 8, 16, 32, 64, etc.) or not. If you don’t, then your champ could certainly have come from a seeding where they got a bye.

If you want a proof, I would suggest Induction as the method you want to use.

ETA: and your initial statements are not correct.

2^x should be the number of teams. x is the number of games needed by the winning team to win the tournament. It works for x=0 too.
(2^x) - 1 is the total number of games in the tournament.

So in a 63 team tournament you need to win 5.9772 games? Or a 33 team tournament requires 5.0443 wins? No, 2^x comes up with the maximum number of slots for a tournament of x tiers.

You misspelled “power”.

Spot on otherwise, though.

No. 5.984375 games on average. Google “Bye.”

I know what a bye is, I mentioned them in post #4. I’m merely stating that his formula doesn’t give the number of teams, rather the number of maximum slots. As for the numbers, I got mine from Wolfram Alpha. Maybe solving for 2^x=63 has a different result on Google, but I doubt it.

:smack: Thanks.

I hate to argue about nothin’ but …
If you’re using 2^x=63 to calculate average games won by winner, you do need to study up on “Bye.”

I was assuing 2 raised to a power, and the power itself, are whole #s.

The OP specified single elimination with no byes, which implies that the number of teams must be a power of 2.

If byes were assigned completely at random, this would be accurate.

If I assume each of the 63 teams has the same likelyhood of winning the tournament, the expected number of wins for the tournament winner is
(626 + 15)/63 = 5.984127

If I assume the team with a Bye has twice the chance of winning as all the others, the expected number of wins for the tournament winner is
(626 + 25)/64 = 5.96875

Using 2^x = 63 gives me 5.97728.

What am I missing?

I goofed. The base 2 logarithm of the number of teams is a lower bound for the number of games the winner has to play, but it’s not exactly right.