Basic permutation calculation - brain is broken so is Google
July 30, 2022 10:47 PM Subscribe
I can't believe I can't do this simple calculation. I also can't believe I can't even word the question properly to get a relevant answer from Google.
Four teams play each other once. No ties are possible. What is the number of possible outcomes.
TIA.
TIA.
If I'm understanding your question correctly, then you need six games for four teams to play one another once each (each of the four teams plays three games corresponding to its three possible opponents, so 4x3=12, but then you divide by two to avoid double counting, getting six games total).
Since each of these games has two possible win-lose outcomes, there are 26=64 possible outcomes overall.
posted by kickingtheground at 11:03 PM on July 30, 2022 [4 favorites]
Since each of these games has two possible win-lose outcomes, there are 26=64 possible outcomes overall.
posted by kickingtheground at 11:03 PM on July 30, 2022 [4 favorites]
I'm also not a math person, but I came up with 12 as well. (I'd initially missed where you said no ties are possible.) Here's how I charted it out as I can't figure out how to make a table on AskMe, though I'm sure the math folks (and the people who've taken a math class in the last 35 years) would have an equation.
That said, it's possible I'm completely missing something. ;-)
1) A plays B, A wins
2) A plays B, B wins
3) A plays C, A wins
4) A plays C, C wins
5) A plays D, A wins
6) A plays D, D wins
(A can't play A)
7) B plays C, B wins
8) B plays C, C wins
9) B plays D, B wins
10) B plays D, D wins
(B can't play B, and B has already played A in the A the above examples)
11) C plays D, C wins
12) C plays D, D wins
(C can't play C, and C has already played both A and B in the above examples)
(D can't play D, and has already played A, B, and C in the above examples)
posted by The Wrong Kind of Cheese at 11:04 PM on July 30, 2022
That said, it's possible I'm completely missing something. ;-)
1) A plays B, A wins
2) A plays B, B wins
3) A plays C, A wins
4) A plays C, C wins
5) A plays D, A wins
6) A plays D, D wins
(A can't play A)
7) B plays C, B wins
8) B plays C, C wins
9) B plays D, B wins
10) B plays D, D wins
(B can't play B, and B has already played A in the A the above examples)
11) C plays D, C wins
12) C plays D, D wins
(C can't play C, and C has already played both A and B in the above examples)
(D can't play D, and has already played A, B, and C in the above examples)
posted by The Wrong Kind of Cheese at 11:04 PM on July 30, 2022
Response by poster: OMG I didn't word my question properly. I'm very sorry. There are eight teams. There are four matches.
posted by little eiffel at 11:18 PM on July 30, 2022
posted by little eiffel at 11:18 PM on July 30, 2022
There are 6 games to play:
A vs B
A vs C
A vs D
B vs C
B vs D
C vs D
Each of these 6 games has 2 possible outcomes. So there are 2^6 = 64 possible outcomes of win-loss combinations over all the games.
I agree with kickingtheground on this one. I think the other folks are saying 12 because there are 6 games, each of which can be won or lost... but if I understand correctly, you are seeking the total number of possible outcome combinations across all six games. (An example of one possible outcome in this set: A beat B, lost to C, and beat D; B lost to C and beat D; D beat C).
posted by cnidaria at 11:22 PM on July 30, 2022 [1 favorite]
A vs B
A vs C
A vs D
B vs C
B vs D
C vs D
Each of these 6 games has 2 possible outcomes. So there are 2^6 = 64 possible outcomes of win-loss combinations over all the games.
I agree with kickingtheground on this one. I think the other folks are saying 12 because there are 6 games, each of which can be won or lost... but if I understand correctly, you are seeking the total number of possible outcome combinations across all six games. (An example of one possible outcome in this set: A beat B, lost to C, and beat D; B lost to C and beat D; D beat C).
posted by cnidaria at 11:22 PM on July 30, 2022 [1 favorite]
Response by poster: No there are not six games I can't even write or form a thought it seems.
There are 8 teams.
They play each other once. Not each team plays each other team. There are only four matches.
posted by little eiffel at 11:24 PM on July 30, 2022
There are 8 teams.
They play each other once. Not each team plays each other team. There are only four matches.
posted by little eiffel at 11:24 PM on July 30, 2022
Best answer: Just saw your update after I posted... 8 teams and four matches?
Do you mean the matches would be:
A vs B
C vs D
E vs F
G vs H
?
In this case, you'd have 2^4 = 16 possible outcomes. All that matters is you have four matches each with a win/loss (binary) outcome.
Let me know if I'm interpreting your question correctly.
posted by cnidaria at 11:24 PM on July 30, 2022 [7 favorites]
Do you mean the matches would be:
A vs B
C vs D
E vs F
G vs H
?
In this case, you'd have 2^4 = 16 possible outcomes. All that matters is you have four matches each with a win/loss (binary) outcome.
Let me know if I'm interpreting your question correctly.
posted by cnidaria at 11:24 PM on July 30, 2022 [7 favorites]
Response by poster: The Canadian football league. Each week there are four matches between eight teams. For only that one week -- what is the possible number of outcomes?
posted by little eiffel at 11:27 PM on July 30, 2022
posted by little eiffel at 11:27 PM on July 30, 2022
Response by poster: Cnidaria has the answer! Thank-you! And thanks to everyone who responded. I was losing it there. Computer almost went out the window.
posted by little eiffel at 11:28 PM on July 30, 2022 [2 favorites]
posted by little eiffel at 11:28 PM on July 30, 2022 [2 favorites]
It's pretty easy to get all snarled up in thinking about permutations when what you're really asking about is combinations and/or independent binary possibilities and vice versa.
There are 8 teams.
They play each other once. Not each team plays each other team. There are only four matches.
No ties are possible. What is the number of possible outcomes.
Given a pre-established pairing into four matches, such as you'd find on any given week in a football league fixture, there are indeed 24 = 16 possible outcomes from that pairing.
But the number of possible outcomes across all possible pairings is going to be bigger, and to work out exactly what it is requires specifying carefully what counts as a distinct outcome.
There are many different ways to pair our eight teams for their four matches, and each such pairing offers its own set of possible outcomes that might be considered distinct from any other. For example, team A winning over team B might be held to be a different outcome from team A winning over team H, or might be considered the same because team A wins regardless.
If a distinct outcome consists of some set of four teams winning, and we don't care who they beat in order to do that, then the number of possible outcomes is just the number of ways to select four teams out of eight - there are four matches, and ties are not possible, so there will always be exactly four winners.
There's a standard formula for choosing K items from N, which in this case will come out to 8! / (4! * 4!) = 40320 / (24 * 24) = 70 possible combinations of four winning teams when eight teams play four matches.
But if a distinct outcome is some specific set of teams beating some other specific set of teams - if, for example, team C beating team A makes an outcome distinct from one in which it beat team H - then there are many more cases that will be considered distinct. Each of our winners could have beaten one of four other teams in order to do that; having done that, only three teams are available to have lost to other winners; for each of those, only two teams could have lost to other winners and the last loser is then forced. So there are 4! = 24 different ways in which any given set of winners could possibly have won, making 70 * 24 = 1680 possible four-match outcomes in total.
It should also be possible to arrive at the same result by multiplying the number of possible outcomes for any given pairing by the number of possible pairings, so let's have a crack at that. How many ways exist to pair eight teams for four matches?
Let's start with a smaller league, with only two teams, and pair those up into matches. Clearly there's only one way to do that.
If we had four teams we could start making pairs by picking any one of them; doesn't matter which. That team has to play somebody, and there are three left to pick from. Having picked one of those there are now only two teams left and they have to play each other. So four teams gives us three possible pairings all told.
With six teams: first team we pick has to play somebody, and there are five choices for that. For each of those five we need to generate all pairings for the four remaining teams, and as worked out above there are three ways to do that, so six teams yields 5 * 3 = 15 possible pairings.
Eight teams, following the same pattern, will have 7 * 15 = 105 possible pairings.
Any one of those pairings will yield 24 = 16 possible win outcomes, so the total number of possible win outcomes will be 16 * 105 = 1680 which is the same as the result from the other method and that's a great relief, to be honest. Combinatorics hurts me in my thinkamus.
posted by flabdablet at 5:51 AM on July 31, 2022
There are 8 teams.
They play each other once. Not each team plays each other team. There are only four matches.
No ties are possible. What is the number of possible outcomes.
Given a pre-established pairing into four matches, such as you'd find on any given week in a football league fixture, there are indeed 24 = 16 possible outcomes from that pairing.
But the number of possible outcomes across all possible pairings is going to be bigger, and to work out exactly what it is requires specifying carefully what counts as a distinct outcome.
There are many different ways to pair our eight teams for their four matches, and each such pairing offers its own set of possible outcomes that might be considered distinct from any other. For example, team A winning over team B might be held to be a different outcome from team A winning over team H, or might be considered the same because team A wins regardless.
If a distinct outcome consists of some set of four teams winning, and we don't care who they beat in order to do that, then the number of possible outcomes is just the number of ways to select four teams out of eight - there are four matches, and ties are not possible, so there will always be exactly four winners.
There's a standard formula for choosing K items from N, which in this case will come out to 8! / (4! * 4!) = 40320 / (24 * 24) = 70 possible combinations of four winning teams when eight teams play four matches.
But if a distinct outcome is some specific set of teams beating some other specific set of teams - if, for example, team C beating team A makes an outcome distinct from one in which it beat team H - then there are many more cases that will be considered distinct. Each of our winners could have beaten one of four other teams in order to do that; having done that, only three teams are available to have lost to other winners; for each of those, only two teams could have lost to other winners and the last loser is then forced. So there are 4! = 24 different ways in which any given set of winners could possibly have won, making 70 * 24 = 1680 possible four-match outcomes in total.
It should also be possible to arrive at the same result by multiplying the number of possible outcomes for any given pairing by the number of possible pairings, so let's have a crack at that. How many ways exist to pair eight teams for four matches?
Let's start with a smaller league, with only two teams, and pair those up into matches. Clearly there's only one way to do that.
If we had four teams we could start making pairs by picking any one of them; doesn't matter which. That team has to play somebody, and there are three left to pick from. Having picked one of those there are now only two teams left and they have to play each other. So four teams gives us three possible pairings all told.
With six teams: first team we pick has to play somebody, and there are five choices for that. For each of those five we need to generate all pairings for the four remaining teams, and as worked out above there are three ways to do that, so six teams yields 5 * 3 = 15 possible pairings.
Eight teams, following the same pattern, will have 7 * 15 = 105 possible pairings.
Any one of those pairings will yield 24 = 16 possible win outcomes, so the total number of possible win outcomes will be 16 * 105 = 1680 which is the same as the result from the other method and that's a great relief, to be honest. Combinatorics hurts me in my thinkamus.
posted by flabdablet at 5:51 AM on July 31, 2022
(I don't know if it's relevant to your question, and it doesn't happen very often, but it is possible for a regular-season CFL game to end in a tie.)
posted by box at 9:47 AM on July 31, 2022
posted by box at 9:47 AM on July 31, 2022
NEVER MIND THIS POINTLESS COMMENT, I HAD AN OLD WINDOW OPEN BY ACCIDENT AND DIDN’T SEE THAT THIS QUESTION HAD BEEN THOROUGHLY ANSWERED, OOPS. But I'm not supposed to delete comments, I understand, so...
Let me see if I understand. There are 4 matches. In each match 2 teams play each other - so a total of 8 teams are taking part. So
A versus B
C versus D
E versus F
G versus H
Are the matches, and they are set - you aren't shuffling which team plays which, there's just one "round" of matches. You know that there cannot be a tie, so each match has 2 possible outcomes. You want to know the total possible number of results from this "round."
Is that a correct assessment of the question?
posted by branca at 3:26 PM on July 31, 2022
Let me see if I understand. There are 4 matches. In each match 2 teams play each other - so a total of 8 teams are taking part. So
A versus B
C versus D
E versus F
G versus H
Are the matches, and they are set - you aren't shuffling which team plays which, there's just one "round" of matches. You know that there cannot be a tie, so each match has 2 possible outcomes. You want to know the total possible number of results from this "round."
Is that a correct assessment of the question?
posted by branca at 3:26 PM on July 31, 2022
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The Xs in the chart are because they can’t play themselves
With a chart like this each matchup happens twice so I assigned one a Win and one a Loss, the two possible outcomes if there are no Ties
Then just count the Wins and add to the Losses
. . 1 . 2 . 3 . 4
1 . X . L . L . L
2 . W . X . L . L
3. W . W . X . L
4. W . W . W . X
posted by nouvelle-personne at 10:54 PM on July 30, 2022