Tournament Permutation Problem

[DaveC426913]

TL;DR

Looking for a solution to play off ten darts teams on five boards in only 3 sessions.

I've got ten teams, five boards, and only three nights to figure out who's the best team.

We've had our regular season, so we have the teams ranked. (i.e. 1st place in regular season gets to play against last place, etc.)

Can we get a fair score out of 10 teams in three nights?

(Is that enough information? It's been so long. Curse you, Covid!)

 

[mfb]

What do you call a "fair score"?

Assuming each session only allows one competition per board: The largest possible connected set of teams (A played B, B played C, ...) is 2³=8. Your 10 teams will be split into at least two groups who had no games against any team of the other group.
You can use the existing ranking as assistance, but then you'll have to figure out how much you want to rely on that and how much impact these three sessions should have. Any choice will favor some teams and disfavor others. "We use the existing ranking" is the least ambiguous option, but obviously that's not what you want.

 

[jedishrfu]

You could have the teams shoot darts at each other and the last one standing wins. :-)

 

[DaveE]

https://officepoolstop.com/Brackets.aspx

 

[pasmith]

Single elimination appears to require 4 rounds, so unless someone plays twice in one night that won't fit.

You could use a Swiss pairing, which has the advantage of being non-elimination so everyone gets 3 matches.

 

[Office_Shredder]

Assuming each session only allows one competition per board: The largest possible connected set of teams (A played B, B played C, ...) is 2³=8. Your 10 teams will be split into at least two groups who had no games against any team of the other group.

This doesn't sound right to me. If the first round is 1 plays 2, 3 plays 4 etc, then the next round is 2 plays 3, 3 plays 4,.., 10 plays 1, then 1 played against 2 played against 3 played against 4 played against 5 etc. You only need two rounds to get a connected set.

 

[pbuk]

You could use a Swiss pairing, which has the advantage of being non-elimination so everyone gets 3 matches.

Yes a Swiss pairing system is designed for exactly this situation. The only disadvantage is that pairings cannot be decided in advance which can make it impractical where teams have to travel to each other.

 

[pbuk]

Can we get a fair score out of 10 teams in three nights?

Yes, but you might have a tie ($\lceil \log_2 n \rceil$ rounds are required to avoid this).

 

[DaveC426913]

Yes a Swiss pairing system is designed for exactly this situation. The only disadvantage is that pairings cannot be decided in advance which can make it impractical where teams have to travel to each other.

That should be all right. Most of us can climb the flight of stairs (to the lounge, with the fifth dart board) without too much forewarning. (We're all in the same club.)

 

[mfb]

This doesn't sound right to me. If the first round is 1 plays 2, 3 plays 4 etc, then the next round is 2 plays 3, 3 plays 4,.., 10 plays 1, then 1 played against 2 played against 3 played against 4 played against 5 etc. You only need two rounds to get a connected set.

Oh right, I forgot the "with a unique winner" condition. In your example two teams winning both matches don't have a comparison. It's impossible to beat single-elimination or equivalent systems there.