Tough probability riddle

http://www.able2know.org/forums/viewtopic.php?p=2987106#2987106

Ugly. here are the solutions per the rules for group sizes 1 -> 6 people

S(1) = 100% (problem collapses)
S(2) = 0%
S(3) = 1/3
S(4) = 2/6 = 1/3
S(5) = 7/14 = 1/2
S(6) = 14/28 = 1/2

Fun, fun, fun!

Einstein's stumped. I've been reading the other thread.

Even if N was the square of the speed of light in Roman numerology there surely must be a chance of you getting your own coat. I'll admit the chance may be infinitessimal but it's there.

If nobody chose your coat only your arithmetic could make it disappear.

A party of 4 yields a 3-1 chance first up, then a 2-1 chance next and finally an evens chance. A $ on a treble yields 23-1. Just under 4%.

It's probably been set in a party cloakroom because the teachers don't like mentioning gambling and card games to their charges but, of course, it is on those that it is based.

g_day, you're assuming that the first man grabs the wrong coat. The problem doesn't specify that everyone grabs the wrong coat, only that they grab a random coat.

Suppose the "wrong coat" they grabbed was the one that looked more expensive and better fitting than the one they came in with.

One can hardly imagine Oliver Hardy grabbing Stan's coat and going out into a north-eastern ice-storm in it.

In modern weather conditions one imagines that fur coats were grabbed first and so the question might resolve it self into the gender mix at the party. It could even have a class factor.

Was it a gay party, on the sword side I mean, or a gaggle-of-geese one.

That's it!

It's an asexual party. Of course. Teachers tend to be a bit hung up about sexual parties.

Let required probability be p(n) for n>1
There are two possibilities (i) first person A gets his own coat, prob.=1/n
(ii) first person A randomly takes someone else coat but not the coat of last person,for this n-2 possible choices out of n. Now the person B whose coat was taken by A comes at kth (k=2,3,4 . . .,n-1) position (its probability 1/(n-2)). For him there will be n-k+1 coats available including the coat of A and of the last person. In this case we can rename the coat of A as the coat of B and B is randomly selecting one of the coat out of n-k+1 coats
=> p(n) = (1/n) +[(n-2)/n]*sum from 2 t0 n-1 of [{1/(n-2)}*p(n-k+1)]
= (1/n) +[1/n]*sum from 2 t0 n-1 of p(k)
=> np(n) = 1+sum from 2 t0 n-1 of p(k) --------(1)
=>(n+1)p(n+1) = 1+sum from 2 t0 n of p(k) --------(2)
from (2)-(1), (n+1)p(n+1) - np(n) = p(n)
=>(n+1)p(n+1) = (n+1)p(n)
=>p(n+1) = p(n)
As p(2)=1/2, p(n) = 1/2 for n>1

I believe it is ((N-2)!(N-2) + 1)/((N-1)!(N-1) + 1)
This, however this is not needed if the first person chooses his/her jacket