$5$ guests leave their coats in the hall. Each takes a fur coat at random when they leave. What's the probability that every coat was someone else's? Five guests leave their coats in the hall. Each takes a fur coat at random when they leave. What is the probability that all the coats were someone else's?
I answered it in the following manner: $P=\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}\times 1=0.2$. However, I know that the correct answer is $0.37$, not $0.2$. 
Please help me understand what I'm getting wrong.
 A: Here is where your approach goes wrong. The first person $A$ grabs the wrong coat with a probability of $\frac{4}{5}$, that's correct. But what if person $A$ ended up grabbing the coat from the second person, person $B$? Then person $B$ has a probability $1$ to not grab his/her own coat!  Of course, had person $A$ grabbed someone else's coat, say $C$, then you are right it would be $\frac{3}{4}$ ... but we already see this is not true in all cases.
Same for considering a third person: If either of $A$ or $B$ had grabbed $C$'s coat, then $C$ is certain not to grab their own goat rather than with a probability of $\frac{2}{3}$. Also note that if $A$ and $B$ grab each other's coat, then the calculations for the three remaining people not grabbing their own coat are going to be quite different then when $A$ and $B$ grabbed a different coat, but of different people than themselves.
Finally note that you say that $E$ grabs a different coat with a probability of $1$, after $A$ through $D$ have each grabbed a different coat. Not true: Maybe $A$ grabbed $B$'s coat, $B$ grabbed $C$'s, $C$ grabbed $D$'s, and $D$ grabbed $A$'s. Now it is in fact impossible for $E$ to not grab their own coat!
So, all kinds of different scenarios can happen, all effecting the subsequent propabilities in complicated ways.  Indeed, there is no really simple formula for this. Well, there is a formula, but it's not straightforward: As the Comment says: look at derangements.
A: As noted in the comments, this question falls under the category of 'derangements'. To cut to the answer of your question, it should go like this: 
$$ \frac{!5}{5!} = ~44/120$$
Note the notation !5 can be calculated as below. See reference here. 
$$
!n = \left\lfloor{\frac{n!}{e} + \frac12}\right\rfloor
$$
And to the finale, since the scope is relatively small I used a short python program to list of derangements. So assume the 5 people are ABCDE and none can take his/her fur we have those 44 ways.
1     (B, A, D, E, C) A takes B, B takes A, C takes D etc..
2     (B, A, E, C, D)
3     (B, C, A, E, D)
4     (B, C, D, E, A)
5     (B, C, E, A, D)
6     (B, D, A, E, C)
7     (B, D, E, A, C)
8     (B, D, E, C, A)
9     (B, E, A, C, D)
10    (B, E, D, A, C)
11    (B, E, D, C, A)
12    (C, A, B, E, D)
13    (C, A, D, E, B)
14    (C, A, E, B, D)
15    (C, D, A, E, B)
16    (C, D, B, E, A)
17    (C, D, E, A, B)
18    (C, D, E, B, A)
19    (C, E, A, B, D)
20    (C, E, B, A, D)
21    (C, E, D, A, B)
22    (C, E, D, B, A)
23    (D, A, B, E, C)
24    (D, A, E, B, C)
25    (D, A, E, C, B)
26    (D, C, A, E, B)
27    (D, C, B, E, A)
28    (D, C, E, A, B)
29    (D, C, E, B, A)
30    (D, E, A, B, C)
31    (D, E, A, C, B)
32    (D, E, B, A, C)
33    (D, E, B, C, A)
34    (E, A, B, C, D)
35    (E, A, D, B, C)
36    (E, A, D, C, B)
37    (E, C, A, B, D)
38    (E, C, B, A, D)
39    (E, C, D, A, B)
40    (E, C, D, B, A)
41    (E, D, A, B, C)
42    (E, D, A, C, B)
43    (E, D, B, A, C)
44    (E, D, B, C, A)

