j*****4
发帖数: 292 | 1 The members of a company decide to give each other presents in the following
way. Everybody brings a present, which is put with the others,mixed and dis
tribued at random to the people. what's the probability that nobody gets his
own presents? |
f**x
发帖数: 109 | 2 complement+inclusion&exclusion |
j*****4
发帖数: 292 | 3 more details?
【在 f**x 的大作中提到】
: complement+inclusion&exclusion
|
j*****4
发帖数: 292 | 4 Got a link.
http://en.wikipedia.org/wiki/Derangement
【在 j*****4 的大作中提到】
: more details?
|
s*********k
发帖数: 1989 | 5 Say 3 ppl(N=3), the prob. you get your own present back is 1/3(N=3).
Then the prob no one get own present = 1- prob (1 ppl get own back)
-prob(2 get own back) -prob(3 get own back); p(N-1)=p(N)
That is it, right? |
b*****t
发帖数: 10 | 6 E_i=ith member gets his/her own present.
p = P(E_1 U E_2 U...U E_n)at least one get his/her own present.
then P(no one get own present)=1-p
Then solve p by inclusion/exclusion principle.
say P(E_1 U E_2 U E_3)=(n-3)!/n! and there are Choose(n,3) combinations.
so finally,
p=1-1/2!+1/3!+...+...(-1)^(1+n)1/n! |
