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So there are six teams (let's say: 1,2,3,4,5,6), and they pair up to face each other, (so three games in total). In each game, one team either wins or their is a tie.

Let's set up the teams and their game possibilities like this:

1 and 2: W L T 3 and 4: W L T 5 and 6: W L T

I would like to determine the total number of possibilities from these games. (Technically, the loss option doesn't really count because obviously if one team wins, the other loses)

One possibility is that all games end up to be wins FOR THE FIRST TEAMS (ex: teams 1,3,5 would be the 'first' teams and teams 2,4,6 would be the 'second' teams.

Thanks in advance!

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    $\begingroup$ Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6? $\endgroup$
    – rubberchicken
    Feb 25, 2014 at 1:04
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    $\begingroup$ Because if so, then there are 3 outcomes for each game (odd team wins, tie, even team wins) and 3 games for $3^3=27$ possibilities. $\endgroup$
    – rubberchicken
    Feb 25, 2014 at 1:06
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    $\begingroup$ But if not, there are $\frac{6!}{{2^3}3!}$ ways to pair all the teams up at once (do you see why?) $\endgroup$
    – rubberchicken
    Feb 25, 2014 at 1:09
  • $\begingroup$ @rubberchicken, that is pretty slick that I did not think of while trying to understand the problem $\endgroup$
    – Satish Ramanathan
    Feb 25, 2014 at 1:55
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    $\begingroup$ How does this question have 13k views (and no votes)? $\endgroup$
    – Jakub Konieczny
    May 2, 2017 at 23:24

2 Answers 2

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Here is the "slick" way to solve it: there are 3 outcomes for each game (either the odd team wins, they tie, or the even team wins), and there are 3 separate games, so since each game is independent of the other, there are $3^3 = 27$ possible outcomes.

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Answer:

If I understand your question properly, the below will be the answer

The total number of winning combinations ( no ties) = (135,136,145,146,235,236,245,246) = 8

The total number of 1 game tie and rest of the winning combinations = (1-2)T - (3,5),(3,6),(4,5),(4,6) and similarly for the rest of the two games tieing. = 4*3 = 12

The total number of two games tie and rest of the winning combinations = (1-2)T,(3,4)T - (5)(6)W and two such combinations for (1-2)T,(5,6)T and (3,4)T,(5,6)T = 3*2 = 6

All three games could be a tie and the total number is = 1

Summing all = 8+12+6+1 = 27.

I really do not know if there is a slick way to solve this without full enumeration.

Thanks

Satish

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