# Math Help - One against Two combinations

1. ## One against Two combinations

In a tournament often a competitor has to play against all other opponents one at a time (BTW I forgot the formulas to calculate how these pairings are being made if anybody has them I would be thankful if you provide them)

But now I have a tournament I'm planning where each player has to play against TWO opponents at a time.
There will be 12 competitors in each group.
How can I achieve a schedule where each player will play against all other opponents just ONCE without any player playing together with the same 'opponent/partner' against any opponnet?

2. Originally Posted by SoftwareTester
How can I achieve a schedule where each player will play against all other opponents just ONCE without any player playing together with the same 'opponent/partner' against any opponnet?
It’s not possible. If there are 12 players, each player will have 11 opponents to play against or partner with, and it is impossible for the player to play/partner every opponent exactly once. If you don’t want the player to play/partner any opponent more than once, there must be some opponent that player does not play/partner. If you want the player to play/partner every opponent, there must be some opponent that the player has to play/partner more than once.

Mathematically:

Let S be the set of all the players in the group. Then a match corresponds to a subset of S containing exactly 3 elements (players). Note that it doesn’t matter who are the partners and who the solo player; all that matters are which three players are involved.

Suppose $a\in S$ is a player. There are $^{11}C_2=55$ 3-element subsets of S containing a; let $A_1,\ldots,A_n\ (n\leq55)$ be a list of these subsets. In other words, $A_1,\ldots,A_n$ are n matches in which a plays.

(i) If you want a to play/partner each of the other 11 players no more than once, each of the other 11 elements of S must belong to only one of the subsets $A_1,\ldots,A_n$. Since 11 is an odd number, it won’t be possible to fit all 11 of them in; there will always be at least 1 element left over. Your best option (i.e. having a play/partner as many opponents as possible) is to have $n=5$ and 1 element in S not in any $A_i$.

(ii) If you want a to play/partner each of the other 11 players at least once, then all the other 11 elements of S must go into $A_1,\ldots,A_n$ somehow. Again it’s because 11 is an odd number that this won’t be possible; this time at least 1 element must belong more than one of the subsets $A_1,\ldots,A_n$. Your best option this time (i.e. having a double-play/partner as few opponents as possible) is to have $n=6$ and 1 element in S (other than a) belonging to $A_i$ and $A_j$ where $i\ne j$.

3. Is my interpretation right there should be an ODD number of players in total.
So if another player (nr 13) would be added it would be possible to arrange them in such a way each of them would play against eachother and with eachother just once?

Also : how do I generated those pairings?

The post by Jane is very useful to me as it makes me not look further for something being impossible, but it doesn't solve my practical problem .
In my case it will be best if I will play as well (and not only fill in scorecards and organize things) making it 13 players in total so the 'odd number 11' will be changed into the 'even number 12' and (the way I understood) by doing this making it posisble to let each player play against/with each other player.
My problem now is to generate those unique sets. It would be great help if someone could tell me HOW to generate those unique sets