Alright, I'm having a really hard time figuring this out even though I think it should be pretty basic. My question is:
how many unique ways can you pair up a group 16 people so that each person is in a pair?
Eg. if the question was how could you pair up 4 people (ABCD), the answer is 3. You can have AB/CD, AC/BD or AD/BC. Obviously CD/AB is the same as AB/CD so that "different" combination doesn't count. Brute force becomes pretty well impossible for 6 people or more though and I can't seem to google any formula which helps me out.
any ideas?
Thanks.
Lemmie take a crack at it.
Note that you will have 8 pairs. So imagine that you have 8 slots, and in each slot, you will put a pair. Now, in the first slot, how many ways can you get a pair to put in that slot? (from the 16 people, choose 2 to make a pair). For the second slot? (choose a pair from the remaining 14 people). Continue this way until the seventh slot, where you will have 4 people left. In the seventh slot, you can put a pair in there ways. At this point, you will only have a pair left. So you can put pair in the last slot.
But now, these pairs can be permuted, so divide by 8! Hence, the number of pairings are:
Which will give you the same number as the previous answer.
ah, now I see my problem and why I though it'd be much lower. I'm actually asking the wrong question.
To be more precise, the question I'm asking is (and sorry for the frivolity here!) what are the chances of a tied score happening in my fantasy australian rules football league? there's 16 people in a league who are randomly paired up each week. So far there's been 10 weeks of competition and in that time teams have scored the same score in the same week four times which is the number I'll use as the chance that any two teams will score the same score in a particular week (this isn't exactly right, but it's the best way of doing it because calculating the objective chance is impossible given the number of variables). I was looking for the right denominator in this equation though, and 22m (the previous answer x10 weeks) is not that number because I was asking the wrong question.
So the question I'm actually asking is something like - given a 40% chance that two teams will score the same points in any particular week, what is the chance that those two teams will be matched up against each other?
Does that make sense? Sorry to lead people down the garden path with my first question!
To be honest I don't really understand your question. But if you're looking for the probability that two randomly selected teams are paired together, should be 1/15. (Because each team can be paired with one of 15 other teams, and each possibility is equally likely.)
Sorry but that didn't help. Maybe it's because I haven't played fantasy football. Question: When you wrote in post #7:
Did you mean 16 teams instead of 16 people?
Okay so let me see if I'm getting this right though. You have 16 people, label them {A,B,...,P}. You've preselected two of them, say, A and B. There is a 40% chance that A and B will both wear blue jumpers. You are given that all 16 people will be paired randomly. What is the chance that A and B will be paired together? Since the color of their jumpers is not a factor in how they are paired, the answer is just 1/15.
So as you can see, I still don't know what you're asking.