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Howdy all - I think I have the answer to this question, but I wanted to make sure. Given a sports season of, say 12 teams and 12 games, how would you calculate how many possible ways the season can be played out (i.e. the various win/loss combinations).

I'm viewing this as a combination problem, so for each round, you have 6 games with the teams playing against each other. So you'll have 2^6 combinations for each round of games. Take that outcome and raise it to ^12 for the total number of possible win/loss combinations for the league.

Does that logic make sense?

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    Is this a game that doesn't allow a tie?2011-04-13

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As Hans Lundmark says, with 12 teams there should be 11 rounds of 6 matches if each team plays every other team once.

Your logic is correct: There are $2^6$ possible sets of results each round, and multiplying the possibilities for each round gives a total of $\left(2^6\right)^{11}$ possible sets of results.

An alternative approach giving the same answer is that there are 66 matches in the season, so $2^{66}$ possible sets of results.

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    As @Henry pointed out, what matters is the number of games. So for example if there are $12$ teams and each plays $12$ games (meaning it plays some team more than once), then the total number of games is $(12)(12)/2$. Thus the number of possibilities (if it is basketball, no ties) is $2^{72}$.2011-04-13