Outcome possibilities with three teams and three outcomes for each game

probability combinatorics permutations combinations

48,203

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.

48,203

Ol' Reliable

Updated on March 31, 2020

Comments

Ol' Reliable over 3 years

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!
- rubberchicken over 9 years
  
  Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
- rubberchicken over 9 years
  
  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.
- rubberchicken over 9 years
  
  But if not, there are $\frac{6!}{{2^3}3!}$ ways to pair all the teams up at once (do you see why?)
- Satish Ramanathan over 9 years
  
  @rubberchicken, that is pretty slick that I did not think of while trying to understand the problem
- Jakub Konieczny over 6 years
  
  How does this question have 13k views (and no votes)?