Can't figure this one out... ... ...

[vangogh228]

How close can you come to solving this?

Eight card-playing people are going to play a game of bridge (2-person teams), with four people each at two tables. They will play seven rounds and rotate partnerships after each round so that each player plays with every other player as a partner exactly once. But, they also want to have each player play each other player exactly twice as opponents. How do you set up the partnerships/opponents each round ot accomplish this?

I can get the partnerships part, but the opponents part... I don't know... is it impossible? How close can you come? Can it be done?

 

[Kraj]

I don't think this is possible. Since each player must partner with each other player exactly once, you get 29 team combinations. Divide that by seven rounds and you have one team combination that won't be made.

 

[TessB]

They will play seven rounds... they also want to have each player play each other player exactly twice as opponents.

If you play with only two tables, then to play someone exactly twice would only limit the game to 4 rounds, wouldn't it? Or am I missing something?
Better bring our bridge expert, Pat in.

 

[selenau837]

This sounds like a question that I had last night on my math final.

My brain is fried and I don't have my notes, but it appears to be a permuation equation (sp?) instead of a combination equation, since the order is important. (nPr) per table times two, times 7 rounds. *shurgs* Don't know. Perhaps I will copy this and try it out and send it to my math prof to see if he can figure it out. Maybe get some extra credit. *laughs*

We didn't get very far into statistics/ probability, but thought I would give it a try.

Anyways, I am interested in finding out the answer.

 

[FoFa]

I hate math, but minus yourself there are 7 other people, but 2-3 boards = 1 round. After each round the partners change. 7 people, 7 rounds, if done properly you could have each other player for a partner, and play each person twice (each round you have 1 partner and two opponents). So the math seems to work at least.

 

[Kraj]

Hmmm.... this might actually be possible... but its beyond my desire to attempt. My guess is it's all about mixing the 4-player match-ups so no combination is repeated more than twice, but how to do that mathematically I do not know.

 

[BarryMK]

Shouldn't Pat be responding to this one?

 

[allan57]

If a have read the question right, then is this not some form of round-robin competion ie

Round 1: Player 1 & 2 vs 3 & 4
Round 2: Player 1 & 3 vs 2 & 4
Round 3: Player 1 & 4 vs 2 & 3
Round 4: Player 1 & 5 vs 8 & 7
Round 5: Player 1 & 6 vs 5 & 7
Round 6: Player 1 & 7 vs 6 & 8
Round 7: Player 1 & 8 vs 5 & 6


Allan

 

[StevenS]

Solution?

I think this is the solution. First i tried to solve the partnership part, then i switched the teams around untill i came up with this one: (can someone check if it's correct?)

Round 1: 1&2 vs. 5&6; 3&4 vs. 7&8
Round 2: 1&3 vs. 2&4; 5&7 vs. 6&8
Round 3: 1&4 vs. 5&8; 2&3 vs. 6&7
Round 4: 1&5 vs. 3&7; 2&6 vs. 4&8
Round 5: 1&6 vs. 4&7; 3&8 vs. 2&5
Round 6: 1&7 vs. 2&8; 3&5 vs. 4&6
Round 7: 1&8 vs. 3&6; 2&7 vs. 4&5


Steven

 

[vangogh228]

FANTASTIC !! It works !!

Thanks !! I have been trying to figure this out on and off for a long time. I tried (I thought) every possible combination. Thanks again.