Outcome possibilities with three teams and three outcomes for each game The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How many ways are there for $2$ teams to win a best of $7$ series?What is the probability of winning.Probability for incomplete informationTwo chess players, A and B, are going to play 7 games. There are three possible outcomes for each game, A wins, A loses, or Tie8 teams and X amount of games, Need to play each game and each teamBiggest number of teams with 16 wins in a tournamentProbability that no team in a tournament wins all games or loses all games.How many Olympic games do we need for 8 teams to never play the same team or same game twice?A 13-lot lottery gameThe probability that all 8 teams loses at least one game and wins at least one game in a tournament?
One-dimensional Japanese puzzle
Does Parliament hold absolute power in the UK?
Why can't devices on different VLANs, but on the same subnet, communicate?
Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?
Are there continuous functions who are the same in an interval but differ in at least one other point?
Circular reasoning in L'Hopital's rule
Single author papers against my advisor's will?
Did the new image of black hole confirm the general theory of relativity?
For what reasons would an animal species NOT cross a *horizontal* land bridge?
Why not take a picture of a closer black hole?
Do I have Disadvantage attacking with an off-hand weapon?
Word to describe a time interval
Is an up-to-date browser secure on an out-of-date OS?
Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?
Using dividends to reduce short term capital gains?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
What force causes entropy to increase?
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
Presidential Pardon
What was the last x86 CPU that did not have the x87 floating-point unit built in?
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
Make it rain characters
Button changing its text & action. Good or terrible?
Why doesn't a hydraulic lever violate conservation of energy?
Outcome possibilities with three teams and three outcomes for each game
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How many ways are there for $2$ teams to win a best of $7$ series?What is the probability of winning.Probability for incomplete informationTwo chess players, A and B, are going to play 7 games. There are three possible outcomes for each game, A wins, A loses, or Tie8 teams and X amount of games, Need to play each game and each teamBiggest number of teams with 16 wins in a tournamentProbability that no team in a tournament wins all games or loses all games.How many Olympic games do we need for 8 teams to never play the same team or same game twice?A 13-lot lottery gameThe probability that all 8 teams loses at least one game and wins at least one game in a tournament?
$begingroup$
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!
probability combinatorics permutations combinations
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
$endgroup$
add a comment |
$begingroup$
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!
probability combinatorics permutations combinations
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
$endgroup$
1
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
1
$begingroup$
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 '14 at 1:06
1
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 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 '14 at 1:55
1
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24
add a comment |
$begingroup$
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!
probability combinatorics permutations combinations
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
$endgroup$
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!
probability combinatorics permutations combinations
probability combinatorics permutations combinations
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
asked Feb 25 '14 at 0:58
Ol' ReliableOl' Reliable
1643414
1643414
1
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
1
$begingroup$
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 '14 at 1:06
1
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 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 '14 at 1:55
1
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24
add a comment |
1
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
1
$begingroup$
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 '14 at 1:06
1
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 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 '14 at 1:55
1
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24
1
1
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
1
1
$begingroup$
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 '14 at 1:06
$begingroup$
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 '14 at 1:06
1
1
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 at 1:09
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 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 '14 at 1:55
$begingroup$
@rubberchicken, that is pretty slick that I did not think of while trying to understand the problem
$endgroup$
– Satish Ramanathan
Feb 25 '14 at 1:55
1
1
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
$endgroup$
add a comment |
$begingroup$
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
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
protected by Community♦ Mar 31 at 6:54
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
$endgroup$
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.
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
edited Feb 26 '14 at 22:22
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
answered Feb 26 '14 at 1:07
rubberchickenrubberchicken
32616
32616
add a comment |
add a comment |
$begingroup$
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
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
$begingroup$
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
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
$begingroup$
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
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
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
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
answered Feb 25 '14 at 1:38
Satish RamanathanSatish Ramanathan
10k31323
10k31323
add a comment |
add a comment |
protected by Community♦ Mar 31 at 6:54
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
1
$begingroup$
Are we saying that 1 will definitely face 2, 3 will face 4, and 5 will face 6?
$endgroup$
– rubberchicken
Feb 25 '14 at 1:04
1
$begingroup$
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 '14 at 1:06
1
$begingroup$
But if not, there are $frac6!2^33!$ ways to pair all the teams up at once (do you see why?)
$endgroup$
– rubberchicken
Feb 25 '14 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 '14 at 1:55
1
$begingroup$
How does this question have 13k views (and no votes)?
$endgroup$
– Jakub Konieczny
May 2 '17 at 23:24