Probability question: names in hats

[robvalue]

Hehe

It's a fucker. I wish that was the answer, I'd have had some sleep in the last 10 years

[Whateverist]

Can you point out any errors in my reasoning? I think it's right.

Meanwhile I'll post another probability problem I found very hard to solve for your amusement. (After failing to find the answer after a couple hours each working on it by ourselves a few of us put our heads together and did answer it, but I wouldn't take a lot of credit personally.)

[Excited Penguin]

I would have to know how to work with probability, and I have no fucking clue.

[brewer]

If you had chosen pulling names out of a vagina you might have got my attention.

[robvalue]

But that's what I'm saying too.  So why shouldn't the probability that the last guy draws his own name be the product of everyone else successfully drawing any name but his?

If I want to know what the probability of drawing (in the dark) a pair of black socks in a drawer that contains 3 black and 2 red socks.  I can find this two ways.  Lets start by naming each individual sock b1, b2 and b3 for the black socks and r1 and r2 for the red ones.  Now list all the possible combinations (reversing the order wouldn't affect the outcome):

1-b1,b2
2-b1,b3
3-b1,r1
4-b1,r2
5-b2, b3
6-b2,r1
7-b2,r2
8-b3,r1
9-b3,r2
10-r1,r2

Only combinations 1, 2 and 5 include a pair of black socks so the P(pair of black socks) is 3/10

The other way to calculate the probability is to take the product of the probability the first draw is black and the probability the second draw is also black (after the first draw):

3/5 • 2/4 = 6/20 = 3/10

So I think multiplying the probability of that each person draws any name except person ten's should get it done.  Am I missing something?

Yeah, but the probability isn't fixed for each person. It depends on previous picks. I explained how even person 2's probability isn't fixed, but depends on what person 1 takes. It alters the pool of numbers he has, affecting the probability. It matters whether or not his number is still there, because it effectively disappears from the selection, giving less numbers, if it is there. So it requires more of a breakdown into possibilities.

[Chas]

Pro tip:  It is not about probabilities, it is about combinatorics.

For 2 players (A, B), there is 1 legal order, and 0 ways for B to end up with 'B'.  P= 0.

For 3 players (A, B, C), there are 3 legal orders  and 1 way for C to end up with 'C'.  P= 1/3.

<Recomputing>

Farting around with the combinations, I come up with n! combinations of which n!/2 are legal, and of those (n-2)! end with the last guy stuck.

So the probability P is (n-2)! / (n!/2)  -> 2/n(n-1).

For 10 players, that comes out to P = 1/45.

[robvalue]

You're right Chas, but the problem is that the combinations don't all produce the same probability.

I think the solution involves both counting combinations and being able to count up how many there are for each probability. I got stuck into it at one point, but I didn't quite get there.

It's the "not your own name" that throws the wrench in, and the probability for each person changes depending on whether their number has already come up or not.

I did the example for 4 people to show this. There's 2 legal combos, but they have different probs:

For 4 players:

2, 3, 1, 4 = 1/3 * 1/3 * 1/2 = 1/18
or
3, 1, 2, 4 = 1/3 * 1/2 * 1/2 = 1/12

Total = 5/36

[robvalue]

If I have cocked up and the solution is much more simple than I'm making out, I'll be happy to be corrected!

But I really can't see any straightforward way of counting all the probabilities like you would normally do with such problems.

[TheRealJoeFish]

I've been playing around with this for a little bit now. It's clear that it's not as straightforward as a pure plug-and-chug probability problem or a simple combinatorics problem. I'm looking at it from a couple different angles, but I don't really have time to post about them, sadly. I've played with permutations and countings of subsets and such. I'll keep thinking about it, because this does seem totally wack yo.

[robvalue]

Nice!

Yeah, it got right in my head for a while. I spent a long time wandering around, trying to grasp little ideas of how to count these things.

I tried many ways of creating iterative formulae, to carry across the whole tree. But I failed, and life will never be the same again.