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Disclaimer: There are some similar questions on the site already, but they all involve drawing again if you draw yourself. This question will not.

In a Secret Santa draw, there are 19 participants. Every one is given one of the 19 names. There is a chance of getting oneself as Secret Santa.

The question is: After the names have all been handed out, what are the chances that none of the participants got themselves as Secret Santa?

My thoughts

$\displaystyle P(\text{first person gets someone other than themselves}) = \frac{18}{19}$ trivially. But now one of the other participants are drawn. That person has a 100% certainty of getting someone other than themselves. The other 17 have a 17/18 chance of getting someone else. Thus

$\displaystyle P(\text{second person gets someone other than themselves})\\ = \overbrace{\frac{1}{18} \cdot 1}^{\text{one person has already been drawn, can only draw someone else}} + \overbrace{\frac{17}{18} \cdot \frac{17}{18}}^\text{these 17 could potentially draw themselves}$

Now, two people have drawn someone other than themselves, and there are two others with 100% of getting someone other than themselves. Thus

$\displaystyle P(\text{third person gets someone other than themselves}) = \frac{2}{17} \cdot 1 + \frac{15}{17} \cdot \frac{16}{17}$

In general: $\displaystyle P(\text{nth person gets someone other than themselves}) = \frac{n-1}{20-n} + \frac{\left[19-2(n-1)\right]\left[19-n\right]}{(20-n)^2}$

In what I believe is an over-(or under-)complication, something has gone amiss here, because summing over this expression quickly yields a result which is greater than 1.

Have I overlooked a detail? Or have I simply overcomplicated the thing from the start?

Alec
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    The key concept here is Derangement – lulu Dec 04 '19 at 19:03
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    It is hard to follow your calculation...the probability that any given person gets a name other than their own is $\frac {18}{19}$. And why would you sum these probabilities anyway? – lulu Dec 04 '19 at 19:05
  • @lulu - I added overbrace comments to one of the probabilities. Does it clear up my thought process? – Alec Dec 04 '19 at 19:08
  • I think you are trying to compute conditional probabilities at each stage, which you ought then to multiply at the end...but this is a very hard method to carry out. After $2$ people have been assigned names you need to consider the possibility that one or both of them might have been given the name of the other. The cases branch out badly. Recursive methods work much more easily. – lulu Dec 04 '19 at 19:09
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    What you want to count are derangements, giving an answer of $\frac{!n}{n!}\approx \frac{1}{e}$. In your calculation, you are missing lots of cases (e.g. if person 2 draws person 1, there are no longer 2 others with probability 1 of drawing someone else) – Brian Moehring Dec 04 '19 at 19:10
  • Oof, I see your point @lulu and Brian. If a person draws someone who has already drawn... Yeah, that gets real ugly. Unfortunately, derangement is a completely new concept to me. I would accept any answer which demonstrates it. – Alec Dec 04 '19 at 19:13
  • The link I cited in my first comment gives a very clear introduction to the concept. – lulu Dec 04 '19 at 19:14
  • The usual way which students are first introduced to the concept (often before they ever even hear the word "derangement") is via inclusion-exclusion. Try running inclusion-exclusion over the events "The $i$'th person was assigned themselves" – JMoravitz Dec 04 '19 at 19:14
  • See this answer. – JMoravitz Dec 04 '19 at 19:16
  • I've written an answer based on the helpful suggestions of looking into the concept of derangements. I believe I've arrived at the right conclusion. Any chance someone could give it a look? – Alec Dec 04 '19 at 19:44
  • Good answer (+1). Note that the numerical value is indeed close to $\frac 1e$, as predicted by @BrianMoehring – lulu Dec 04 '19 at 19:50
  • Awesome. Thanks so much for the assist! – Alec Dec 04 '19 at 19:51

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From some helpful comments, I've been directed to look at derangements. https://en.wikipedia.org/wiki/Derangement

This problem can be reduced to finding the number of permutations of $1..19$ such that no number is mapped to itself. Such an outcome is called a derangement.

The number of derangements for $n$ objects is $!n = n!\sum\limits_{i = 0}^{n}\frac{(-1)^i}{i!}$. This is then the number of desired outcomes, whereby the number of total possible outcomes is $n!$.

Thus, we get the probability for no one drawing themselves as $\displaystyle P = \frac{!19}{19!} = \frac{19!\sum\limits_{i = 0}^{19}\frac{(-1)^i}{i!}}{19!} = \sum\limits_{i = 0}^{19}\frac{(-1)^i}{i!} \approx 36.8\%$

Alec
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