Permutations and Combinations

R

RahzelPoker721

New member
Joined
Sep 21, 2015
Messages
2
All, I have a question that may not make sense but here goes:

28 teams
14 Scheduled Matches
14 Winners Only
14 Losers Only
0 Ties

How do I calculate the total number of possible outcomes for all 14 games. Basically I'd like to count all combinations where A loses to B as well as B losing to A and so on through the whole list of teams.

* Note teams only play their scheduled match for that day. Example A will play B only. A will not play C or E or D etc....


A vs B
B vs C
D vs E
F vs G
H vs I
J vs K
L vs M
N vs O
P vs Q
R vs S
T vs V
W vs X
Y vs Z
AA vs BB
 
A lot of this problem depends on how, exactly, the winners of the matches are determined. Is it by a fair coin flip (that is, either team has a 50-50 shot of winning)? Is one team more likely to win than the other? If some teams are more likely to win than others, is it always the same for every pairing of teams? Like if team A wins 2/3 of the time, then will, say, team C also win 2/3 of the time?
 
RahzelPoker721 said:
All, I have a question that may not make sense but here goes:

28 teams
14 Scheduled Matches
14 Winners Only
14 Losers Only
0 Ties

How do I calculate the total number of possible outcomes for all 14 games.
Click to expand...

This is 28 people on 14 teams of two each.
\(\displaystyle 2\left(\dfrac{28!}{2^{14}(14!)}\right)\) WHY?
 
Last edited:
RahzelPoker721 said:
All, I have a question that may not make sense but here goes:

28 teams
14 Scheduled Matches
14 Winners Only
14 Losers Only
0 Ties

How do I calculate the total number of possible outcomes for all 14 games. Basically I'd like to count all combinations where A loses to B as well as B losing to A and so on through the whole list of teams.

* Note teams only play their scheduled match for that day. Example A will play B only. A will not play C or E or D etc....


A vs B
B vs C
D vs E
F vs G
H vs I
J vs K
L vs M
N vs O
P vs Q
R vs S
T vs V
W vs X
Y vs Z
AA vs BB
Click to expand...


Game AB = 2 outcomes (A wins or B wins)
Game CD = 2 outcomes
.
.
.
Game AABB = 2 outcomes


2*2*2*2*2*2*2*2*2*2*2*2*2*2
or 2^14
 
Followup

Hopefully this explanation makes more sense. There are 15 total NFL games coming up. I just wanted to know how calculate all of the possible outcomes. I want to intentionally exclude ties from the calculation. A team can only win or lose.

Example in one combination of the 15 games Dallas beats Atlanta and Denver loses to Detroit, while in another combination of the 15 games the results are reversed.

Sunday Sept 27, 2014 NFL Week 3


Atlanta vs Dallas
Buffalo vs Miami
Chicago vs Seattle
Cincinnati vs Baltimore
Denver vs Detroit
Indianapolis vs Tennessee
Jacksonville vs New England
New Orleans vs Carolina
Oakland vs Cleveland
Philadelphia vs N.Y.J.
Pittsburgh vs St. Louis
San Diego vs Minnesota
San Francisco vs Arizona
Tampa Bay vs Houston
Kansas City vs Green Bay
 
You must log in or register to reply here.
Top