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) $\frac e{n}$

The Diophantine equation $x_1^6+x_2^6+x_3^6=z^2$ where exactly one $(x_i)\equiv 0{\pmod 7}$.

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Polynomial has roots in any ring $\mathbb{Z}/n\mathbb{Z}$ but not in $\mathbb{Z}$

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Why is the total variation of a complex measure defined in this way?

Image of unit ball dense under continuous map between banach spaces

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$X \cong \operatorname{Proj} \oplus_n H^0(X, O(n)) $