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Let's say an astrologer is tasked with guessing the star signs of 12, for him, unknown students. The astrologer knows that all 12 star signs are represented among the 12 students, i.e. a single star sign will belong to one and only one student.

The astrologer can only guess one star sign per person. Additionally, he will have to assign each star sign to someone, i.e. each single star sign cannot be assigned to more than one student.

Given this premise, what is the probability that the astrologer guesses the correct star sign for at least 4 students?

Edit: I am not a probability student, I am just curious about how well the astrologer fared in this video. I started by thinking that the probability of randomly guessing each one correctly would be (1/12)(1/11)(1/10)...(1/2), but I'm not even sure if that is correct and I don't know where to start to answer my question.

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    $\begingroup$ This looks like an assignment for a class that has just discussed inclusion-exclusion formulas (en.wikipedia.org/wiki/…). It is easier to calculate the probability of the astrologer guessing correctly for at least 0, at least 1, at least 10, or at least 11 students, and maybe you can work from there in one direction or the other. $\endgroup$
    – Matt F.
    Jul 20, 2023 at 13:53
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    $\begingroup$ This looks like a HW question. Those are OK here, but you should show what you have tried so far. $\endgroup$
    – Peter Flom
    Jul 20, 2023 at 14:01
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    $\begingroup$ Research partial derangements, also known as the problème des rencontres. $\endgroup$
    – whuber
    Jul 20, 2023 at 14:04
  • $\begingroup$ Your $(1/12)(1/11)(1/10)...(1/2)$ is correct for the probability of randomly guessing each one of the 12 correctly $\endgroup$
    – Henry
    Jul 20, 2023 at 14:16

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Assuming there is no information about the students that is known to correlate with their star-sign (allowing this information to be used in the guess), this is an instance of a famous antique probability problem, which is variously called the "problem of coincidences" (or "problème des rencontres" in French), the "hat-check problem", the "secret-Santa problem", or the "classical matching problem". In the general version of this problem there are $n$ distinct things arranged in a random order and you have to guess the true order of the things. We are then interested in the distribution of the number of correct "matches". (This problem can also be framed in many equivalent ways that sound slightly different to what I have described but are probabilistically equivalent.)

The distribution of the number of matches in this problem (your correct guesses) has been derived and discussed by a number of authors, going back to Mortmont (1708, 1713), and it is closely related to derangments and partial derangments. You can find a detailed analysis of this distribution and a useful generalisation of it in O'Neill (2023). The distribution has probability mass function given by:

$$\text{Match}(k|n) = \frac{!(n-k)}{k!(n-k)!} = \frac{1}{k!} \sum_{i=0}^{n-k} \frac{(-1)^i}{i!} \quad \quad \quad \quad \quad k = 0,...,n.$$

You can compute the probabilities from the matching distribution using the various probability functions in the stat.extend package. The cumulative distribution function can be computed with the pmatching function. In your problem you have $n=12$ things and you want to compute the probability of at least four matches, so you can use the following code:

library(stat.extend)
1 - pmatching(3, size = 12)
[1] 0.01898819

We see from this that the probability of at least four correct guesses is $0.01898819$, so it will happen around one in 52 or 53 times over the long-run. If you prefer, you can compute the probabilities for each possible number of correct guesses using the dmatching function to get the same answer.

You might also find it useful to note that the linked paper also examines a generalisation of this distribution, where we allow for some non-zero probability that the person can correctly guess an outcome from contextual information. This would allow you to examine probabilities that would arise if your guesser has some other contextual information about the students that is related to star-signs, that allows them to guess "better than random".

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Not sure I got it, but I lay out my process, which might help. Let the set integers from 1 to 12 be the zodiac signs. The space of guesses is $12!$, i.e. the permutations of this set. A guess is correct when the number falls in the same place as it was originally.

Now, you want to know in how many cases every guess is wrong, one is correct, two are correct, and three are correct. Those four probabilities should be summed and subtracted from 1 to get your answer.

The probability that no sign is assigned to the right person is a derangement, and the number of derangements can be computed as a sub factorial $ !n = n!/e$ rounded to the closest integer. What about the others?

The probability of having a single guess right is to have a derangement of $11$ and a correct guess, and you can have 12 different guesses, so you need to multiply by that. I think this can be easily extended to 2 and 3 correct guesses to finally get your answer.

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