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There are n candidates in a conference, the organizers have created name tags for each of them but accidentally gave them out at random.

I am taking distribution as X = how many candidates got the correct name tags (correct means their own name tag)

X: 0 1 2 3 4 5 . . . . . n P: ? ? ? ? ? ? ? ? ? ? ? ?

n = no. of candidates who got their own name tag which will of course be equal to no. of candidates

Is it possible to find it for n candidates or will have to go individually? (By individually I mean by taking n = 2 (it will mean there are only 2 candidates), it's pretty easy to find with n = 2,3 but after that things get complicated.)

AnthonyML
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Sahil Bagdi
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    how about recurrence relationship, however if the first person gets the right tag, then it seems easy that the n-1 remaining reduces to m-1 getting the correct tag from n - 1 passes, however if the first person gets the wrong tag then his tag becomes useless and we are looking for m from n-1 tags with 1 useless tag, the next person choosing either gets their tag, the useless tag, or one of the other tags (by 'useless' I mean it can never be allocated to the correct person now). so you need P(n, m, l) the prob of m correct tags for n people, where l tags have become useless. – Cato Oct 28 '22 at 16:02
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    See this answer. Give a reply if the answer has helped. – callculus42 Oct 28 '22 at 16:20
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    @callculus42 Thanks, It was very complicated but really helpful. – Sahil Bagdi Oct 29 '22 at 09:48
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    @SahilBagdi It pleases me, that it has helped. If some questions are still open about this issue, feel free to ask. – callculus42 Oct 29 '22 at 09:59

2 Answers2

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We want to know the probability that a random permutation of $n$ objects has exactly $k$ fixed points, where $0 \le k \le n$. In terms of the original problem statement, this is the probability that exactly $k$ of the candidates get their correct name tag.

A permutation with no fixed points is called a derangement. It is well-known that the number of derangements of $n$ objects is $$D_n = n! \sum_{i=0}^n (-1)^n \frac{1}{i!}$$ See the Wikipedia article on derangements for a proof of this formula using the inclusion-exclusion principle.

There are $n!$ permutations of $n$ objects, all of which we assume are equally likely. We would like to count the number with exactly $k$ fixed points. There are $\binom{n}{k}$ ways to pick the $k$ objects, and then the permutation of the remaining $n-k$ objects must be a derangement. So the number of permutations with exactly $k$ fixed points is $$\binom{n}{k} D_{n-k}$$ and the probablity of such a permutation is $$\frac{1}{n!} \binom{n}{k} D_{n-k} = \frac{1}{k!} \sum_{i=0}^{n-k} (-1)^i \frac{1}{i!}$$ after a little algebra.

awkward
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Since the probability of any individual member getting their correct name tag is $\frac{1}{n}$, the total expected value considering all $n$ members is $n\frac{1}{n}$ = $1$.

mathlander
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    Yes, that is the correct answer for expected candidates to get their own nametag. But I actually wanted the individual probability for the Distribution X. – Sahil Bagdi Oct 29 '22 at 09:47