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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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This is maybe irrelevant, but if you have 12 teams, the natural number of rounds would be 11 rather than 12 (since each team should meet each of the other 11 teams once). – Hans Lundmark Apr 13 '11 at 14:53
Is this a game that doesn't allow a tie? – joriki Apr 13 '11 at 15:05
up vote 3 down vote accepted

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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Ok, but what if all teams do not play each just once, and some play each other twice (say if it is a 14 week season?) My thought is is that doesn't affect the outcome, because for each week of games there's still the 2^6 combinations per round. – chrismanderson Apr 13 '11 at 20:08
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 $2^{72}$. – André Nicolas Apr 13 '11 at 20:23

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