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Possible Duplicate:
Number of permutations where n ≠ position n

I have the following exercise:

Suppose that four guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave.

Determine the probability that no guest will receive the proper hat.

I'm not sure how to find out that no guest will receive the proper hat... Can someone help me with an intuitive explanation from which I could start out?

Community
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upabove
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    With only four people you can just list and count the derangements and divide by $4!$, or twenty-four. But more generally, this would be a duplicate of http://math.stackexchange.com/questions/14666/number-of-permutations-where-n-position-n or http://math.stackexchange.com/questions/17320/derivation-of-the-partial-derangement-rencontres-numbers-formula – anon Oct 16 '11 at 10:16
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    Since there are in total only 24 possible orderings of the hats, the number of combinations that satisfy the requirement will be smaller. Therefore, it's not much work to try to write them down explicitly. To do it systematically, maybe writing down a decision tree helps? – Raskolnikov Oct 16 '11 at 10:22

2 Answers2

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Hint: If you want the practical answer just list the different patterns ($4!=24$) and see which match or not, perhaps starting 

ABCD - match 4
ABDC - match 2

etc. and ending 

DCBA - no matches

and then count the results.

Henry
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This is called the derangement problem.

Let $A_g$ denote the event that guest $g$ is given back the proper hat. One asks for the probability of the intersection $B$ of every $(A_g)^c$, hence $1-\mathrm P(B)$ is the probability of the union of the $A_g$. By the exclusion-inclusion principle, $$ 1-\mathrm P(B)=\sum_{k\geqslant1}(-1)^{k-1}\sum_{|G|=k}\mathrm P(A_G), $$ where for every subset $G$ of the set of guests, $A_G$ is the intersection of every $A_g$ such that $g$ is in $G$.

For every $G$ with $|G|=k$, there are $(n-k)!$ ways to give back their hats to the $(n-k)$ guests not in $G$ and one way to give back their hats to the $k$ guests in $G$ hence $\mathrm P(A_G)=(n-k)!/n!$. There are ${n\choose k}$ such subsets $G$ hence $$ 1-\mathrm P(B)=\sum_{k=1}^n(-1)^{k-1}{n\choose k}\frac{(n-k)!}{n!}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k!}. $$ Finally, the probability that no guest receives the proper hat is $$ \mathrm P(B)=\sum_{k=0}^n\frac{(-1)^{k}}{k!}. $$ Note that $P(B)\to\mathrm e^{-1}$ when $n\to\infty$.

Did
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