Combinatorics question having trouble combining two sample spaces $3$ rabbits are playing outside their individual holes. An eagle comes and they all randomly go into a hole with
one rabbit per hole. What is the probability that no rabbit went to its own hole? Assume that all configurations
are equally likely
I'm assuming my sample space would be 
$$|S|=3^3 = 27 \text{ possible combinations.}$$
Then, the possibilities that a rabbit is not in their corresponding hole is
$$\{(X,Y)| \text{where X = rabbit, Y = hole and } X \neq Y \} = \{\{(1,2),(2,3), (3,1)\}, \{(1,3),(2,1),(3,2)\}\}$$
How do I combine this?
Would the answer be $\frac{2}{27}$?
 A: The probability that the first rabbit does not go down its own hole is $1-(\frac13) = \frac23$. The probability that the second rabbit does not go down its own hole is $1 - (\frac12) = \frac12$. The probability that the third rabbit does not go down its own hole is $1$ (since its hole is already taken by one of the other two rabbits). $\frac23 ⋅ \frac12 ⋅ 1 = \frac13$.
A: Except the size of sample space, your approach is correct. Since it is given that rabbits go holes with one rabbit per hole, the size of sample space should be $3\cdot2\cdot1 = 3! = 6$. So the answer should be $\dfrac{2}{6} = \dfrac{1}{3}$ instead of $\dfrac{2}{27}$.
A: Let's enumerate the hole $1,2,3$ and the rabbit $1,2,3$.
Your problem comes down to the rearranging $1,2,3$.
You pick $1$ hole out of the $3$ for the first rabbit, then $1$ hole out of the $2$ left for the second rabbit then the remaining is assigned to the last rabbit. . There are $3\times2\times1=3!=6$ re arrangement. 
What re arrangements are such that no rabbit ends up in its hole:  $2,3,1$ and $3,1,2$.
The probability is $\frac 26 = \frac 13$
