What's the fewest number of games it would take to name one champion out of six teams?

[b0mbrman]

1 Way: 7

Round 1
A) 1A wins over 1Ax
B) 1B wins over 1Bx
C) 1C wins over 1Cx


Round 2
D) 1A vs. 1B, 1A wins
E) 1C vs. 1Ax
F) 1Bx vs. 1Cx, 1Cx loses

At this point:
Game (E) is what matters

Suppose the transitive property holds.

If 1C wins, then
1A > 1Ax, 1B, 1Bx
1C > 1Ax, 1Cx
1B > 1Bx, 1Cx
1A plays 1C and the winner is better than five other teams

If 1Ax wins, then
1A > 1Ax, 1B, 1Bx, 1C, 1Cx
1Ax > 1C, 1Cx
1B > 1Bx
1A is already better than five teams, but we need a championship game...So 1A v 1Ax again

 

[b0mbrman]

Another way: 5

Round 1
A) 1A wins over 1Ax
B) 1B wins over 1Bx
C) 1C wins over 1Cx

Winner with the biggest margin of victory gets a bye in round two
D) 1B wins over 1C

Championship
E) 1A wins over 1B

 

[Minjin]

Without looking at any other responses, I'm going to say 10.

I set up two trees: a winners/normal and a 2nd chance bracket. To make my example easier the seeds will all finish in the order that was expected.

First bracket of normal tree has all the teams in seeded order. 1v2, 3v4, 5v6. 2nd bracket will have winner of each (always lowest number for ease sake). This means 3v5 and 1 gets a bye. Next bracket has 1v3 and 1 wins.

Meanwhile, in the second chance bracket, you have 2,4, and 6. Highest seed always gets the bye, so 2 waits while 4v6. 2nd bracket brings in the 2nd round of losers. So we have 2v3 and 4v5. Next bracket will be 2v4 with the winner 2 going on to face 1 again for the real championship.

9 normal games and one championship game: 10

When I wrestled, I think this is how we did it but its been a while.

 

[thesurge]

The "fairest" way (without pre-ranks) would be every team playing each other (other
algorithms require seeds). Therefore, the answer is 15 or 6C2. Winner is person with best record (of course tied records would imply winner matches. Also, the only possible tied records would be that everyone has the same record. Then, you just repeat the algorithm until someone gives.).

Note: This is only for unseeded teams.

 

[imported_Baloo]

5

 

[b0mbrman]

Originally posted by: Garet Jax

Originally posted by: kstu
7 is fairest

divide into 2 sets of 3. round robin in each group to determine two finalists (with tiebreaker in case each team goes 1-1), then finalists face for the championship

This is the fairest.

The least fair is 0. Stick all names in a hat and randomly draw one.

What if the top two teams end up in the same initial group of three?

 

[kstu]

Originally posted by: b0mbrman

Originally posted by: Garet Jax

Originally posted by: kstu
7 is fairest

divide into 2 sets of 3. round robin in each group to determine two finalists (with tiebreaker in case each team goes 1-1), then finalists face for the championship

This is the fairest.

The least fair is 0. Stick all names in a hat and randomly draw one.

What if the top two teams end up in the same initial group of three?

It's fair in the sense that the champion will be the best team out of the six. There's no guarantee that the finalist who loses in the championship game is the second best team. Like you said, the second best team may very well face the best team in the round robin and get eliminated.

 

[sciencewhiz]

Originally posted by: TuxDave
If we turn the question into a logic question where the better team will ALWAYS beat a worse team, LAST game played HAS to be #1 vs #2, and no two teams are equal.

If the best team always wins, the scenario of 5 games like the NFL playoffs is the best. That's not real life, however.

What if you say that the better team will win X% of the time, and and then you optimize for the minimum number of games while maximizing the chance that the #1 team wins the championship game. Run a monte carlo simulation with X as your variable to determine the best result.

I'd be really curious if the 2x3 team round robin or the double elim tournament works best, or if there is another option.

 

[LeiZaK]

5

 

[Ilmater]

8

1 team plays all 5 others - 5 games

Top 4 seeds advance, with tie-breaker being strength of schedule.

4 teams play in Semis - 2 games

Championship game - 1 game

 

[Ilmater]

Originally posted by: SpanishFry
5, just like NFL playoffs

Can't do that. NFL has bye teams, and you can't have bye teams unless you have a regular season before-hand.

 

[homercles337]

15 to determine the top two (permutation, everyone plays everyone once), then one for the championship for #1 and #2.

 

[teckmaster]

you could do it in 4 games. just set it up where the 2 winning teams that scored the most runs play for the championship.

 

[apepooooop]

Originally posted by: homercles337
15 to determine the top two (combination, everyone plays everyone once), then one for the championship for #1 and #2.

But why not have each team play each other, multiple times?

In interest of time, i would say round 1 with three games. then round 2 with three games (each possible combination of the previous winners), and a final championship match. 7 games.

 

[DrPizza]

If you look at it from a betting point of view, where you'd deal in probabilities, there is no transitive property.

That is, if the probability of A beating B is > .5
and the probability of B beating C is > .5
and the probability of C beating D is > .5
it doesn't necessarily follow that the probability of A beating D is > .5
In fact, it could be more probable that D would beat A.

It all depends on how the teams match up against each other.

Even TuxDave's response - assume the better team always wins - wouldn't necessarily determine the outright best team. I can come up with a set of 4 dice such that for each pair, one is definitely better than another, but none of them are better than the rest.

 

[Pugnax]

5. 5 teams must lose. Therefore 5 games must be played.