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I am trying to put together a Munchkin game tournament where I am assuming I will have 16 people coming to my tournament. As part of that, I want to have as many games as possible where people are not playing against the same player twice. (So, every combination is unique.)

I can easily come up with these groups for three rounds of the tournament. Is it possible to do more? If so, can you point me in the right direction? Unfortunately, I think I am confusing my combinations and permutations and it's leaving me stumped.

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    I'm sure my explicit answer can fixed, but Gerry's answer is better.2011-06-08

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Identify the players with elements of ${\bf Z}_2^4$ (that is, with ordered 4-tuples of zeros and ones, where addition of 4-tuples is done component-wise and using mod $2$ arithmetic). There are $5$ subgroups of $4$ elements each such that no two subgroups have any element in common other than $(0,0,0,0)$. Each subgroup, together with its cosets, gives you one round of your tournament, so this is a way to go $5$ rounds.

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    @Gerry, I understand and agree with your decision. You wanted to kee$p$ it as sim$p$le as possible and not introduce any more machinery than is absolutely necessary.2011-06-08
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This is nothing more than Gerry's answer made explicit. (I apologize if I've made any mistakes or typos... Any mistakes are mine, not Gerry's!)

Label your 16 players with the letters A,B, ..., P. Then your rounds should have the following games of 4 players:

Round 1 {A,B,C,D} {E,F,G,H} {I,J,K,L} {M,N,O,P}

Round 2 {A,E,I,M} {B,F,J,N} {C,G,K,O} {D,H,L,P}

Round 3 {A,F,K,P} {B,E,L,O} {C,H,I,N} {D,G,J,M}

Round 4 {A,J,O,H} {B,I,P,G} {C,L,M,F} {D,K,N,E}

Round 5 {A,G,L,N} {B,H,K,M} {C,E,J,P} {D,F,I,O}

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These kinds of problems can be difficult; see the discussion Math Games: Social Golfer Problem about scheduling threesomes as well as foursomes playing concurrently, a topic called resolvable Steiner systems. The case of 16 people in foursomes over 5 days is shown (both as a listing of 4-sets and as a strange diagram) in nearly the middle of that page/article.

Day 1: ABCD EFGH IJKL MNOP Day 2: AEIM BFJN CGKO DHLP Day 3: AFKP BELO CHIN DGJM Day 4: AGLN BHKM CEJP DFIO Day 5: AHJO BGIP CFLM DEKN 
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    Interesting that this solution matches Gerry's/Jeff's with just days 4 and 5 swapped. No relabeling required.2011-06-08
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See Creating combinations that have no more than one intersecting element.