Secret Santa Perfect Loop problem 
*

*(n) people put their name in a hat.  

*Each person picks a name out of the hat to buy a gift for.  

*If a person picks out themselves they put the name back into the hat.

*If the last person can only pick themselves then the loop is invalid and either
. start again
. or step back until a valid loop can be reached.


What is the probability that if n is 33 that the chain creates a perfect loop?
An example of a perfect loop where n is 4:  


*

*A gives to B  

*B gives to C  

*C gives to D.  

*D gives to A.    


An example of a valid but not perfect loop where n is 4:


*

*A gives to B  

*B gives to A  

*C gives to D.  

*D gives to C.    


You are asking for the chance of a single cycle given that you have a derangement.  For $n$ people, the number of derangements is the closest integer to $\frac {n!}e$  To have a cycle, person $1$ has $n-1$ choices, then that person has $n-2$ choices, then that person has $n-3$, etc.  So there are $(n-1)!$ cycles.  The odds are then (just about) $\frac e{n}$
