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Math Help - Beer Olympics

  1. #1
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    Beer Olympics

    My friend and I are hosting a game of "Beer Olympics". We have 8 events (Beer Pong, Darts, Frisbee, etc.). Lets call these Event1, Event2, Event3, etc.

    We also have 16 teams of two people. Lets call them TeamA TeamB TeamC, etc.

    Our dream is that every team will play against every other team, and each team will compete in each event. Cause that sounds the most fun right?

    So we tried to come up with a solution, or prove that no solution existed.

    We quickly found that it's impossible for an even grid of 2x2.. Try it! (Yet it is always possible, AND trivial, to solve for an odd numbered grid) And we were content with it being unsolvable.

    Then a friend of ours (With a masters in mathmatics) came along and suggested that 2x2 may be a special case. Our hunch is that it is generally impossible, but we can't seem to prove it. Any ideas?
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  2. #2
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    Re: Beer Olympics

    What do you mean by "even grid of 2x2"? Off-hand, I would say that since you want every team to play every other team in every event, the fact that there are multiple events does not complicate the problem, since each event is just going to be a copy of the first event. So examine one event. Every team playing every other team exactly once sounds an awful lot like a complete graph to me. The vertices are teams, and the edges are games. If there are n teams (16 in your case), there will be n(n-1)/2 games played. So in your case, each event will have 120 games played. Multiply by 8 events and you get a whopping 960 games!

    So I would say there's a solution no matter how many teams you have. Or am I missing something here?
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  3. #3
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    Re: Beer Olympics

    An even grid of 2x2 just means that there are 2 events and 4 teams, so in round one it's teamA and teamB playin event 1, also team c and team d are playing event 2. The problem comes when there are matrix is even. When it's odd say 3x3 it's no prollem
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    Re: Beer Olympics

    Oh; so you're saying that only one event can be played at a time?
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  5. #5
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    Re: Beer Olympics

    Quote Originally Posted by Ackbeet View Post
    Oh; so you're saying that only one event can be played at a time?

    There are eight events. There are eight rounds. There are 16 teams.
    All eight events will be played every round.
    Two teams will compete in each event (teamA vs. teamB)
    Each team can only compete in one event per round (can't be in 2 places at once).
    Each team will never play the same game twice.
    Each team will never compete against the same opponent.

    Like I said, we have a hunch that it is impossible, but we would like to be able to say so definitively. Additionally, if it is impossible, we would like to find a "best" solution that involves a sort of "buy" round.
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    Re: Beer Olympics

    Quote Originally Posted by birdmw View Post
    There are eight events. There are eight rounds. There are 16 teams.
    All eight events will be played every round.
    Two teams will compete in each event (teamA vs. teamB)
    Each team can only compete in one event per round (can't be in 2 places at once).
    Each team will never play the same game twice.
    Each team will never compete against the same opponent.

    Like I said, we have a hunch that it is impossible, but we would like to be able to say so definitively. Additionally, if it is impossible, we would like to find a "best" solution that involves a sort of "buy" round.
    With those conditions, it is definitely impossible. The condition that there only be eight rounds precludes that every team will compete against every other team. Think about it: with eight rounds, and a particular team only allowed to play against one team per round, they'll only get to play against a total of eight other teams. But there are a total of sixteen teams. Hence, 7 teams will not have been played against when the eight rounds are completed.

    Make sense?
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  7. #7
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    Re: Beer Olympics

    Quote Originally Posted by Ackbeet View Post
    With those conditions, it is definitely impossible. The condition that there only be eight rounds precludes that every team will compete against every other team. Think about it: with eight rounds, and a particular team only allowed to play against one team per round, they'll only get to play against a total of eight other teams. But there are a total of sixteen teams. Hence, 7 teams will not have been played against when the eight rounds are completed.

    Make sense?
    Slight misunderstanding here but we are getting closer.
    Each team need not encounter each other team.
    We only wish to avoid duplicate matchups ie team A will play against team B at most once, but never encountering them is ok.
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    Re: Beer Olympics

    Quote Originally Posted by birdmw View Post
    Slight misunderstanding here but we are getting closer.
    Each team need not encounter each other team.
    We only wish to avoid duplicate matchups ie team A will play against team B at most once, but never encountering them is ok.
    Oh. You did mention in the OP that

    Quote Originally Posted by birdmw
    Our dream is that every team will play against every other team, and each team will compete in each event.
    Is that no longer a driving force in your problem?
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  9. #9
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    Re: Beer Olympics

    It's cool, putting math into words, and then interpreting it is a lot like tieing shoelaces in the dark with one hand.
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  10. #10
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    Re: Beer Olympics

    So... nobody?
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