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I recently watched a video by Cut on YouTube called "Guess my Zodiac Sign." It's at the following link: https://www.youtube.com/watch?v=bfkwAAi4t2M

So, at the end, two of the people got $4/12$ correct and it got me wondering what the expected value is. They were saying they did well, but I'm not convinced that they did better than average.

I will rephrase this question. Say that I have $12$ distinct cards and $12$ pots; each card corresponds to exactly one pot. Assuming we place the cards randomly in the pots, what is the expected value of correct guesses?

I have attempted this question, but, I must admit, my probability skills are very rusty. There are $12!$ ways one can place the cards in the pots. I started by trying to count the number of ways one could get 0 correct. Going to the first pot, there are $11$ incorrect choices, but then with the second pot it branches. It's possible that we put pot 2's card in pot 1, so there are again $11$ choices. It's also possible that we did not do that, so there are 10 choices. I can see how we would calculate the total number of ways to get 0 correct from there, but it seems it will be very tedious. Is there are a faster/more clever way? I feel like I'm overthinking this.

It would be interesting to see this done in general as well with $n$ pots and $n$ cards, however I would be completely satisfied with the case $n=12$.

Thank you!

user501794
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  • Also note the comment under that (linked) question that, from linearity of expectation, it follows very easily that the expectation is exactly $1$. (The expectation $\mathbb E(X_i)$ for each $X_i=\begin{cases}1&i\text{'th card went into its pot}\0&i\text{'th card did not go to its pot}\end{cases}$ is merely $1/12$ and our "number of cards in their pot" is nothing else but $X_1+\ldots+x_{12}$, so the expectation is $12\times(1/12)=1$. –  Dec 29 '20 at 12:56

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What you are looking for is the number of permutations of n with k fixed points, they are also known as partial derangements. You have to find them for k=0 to k=12. For example with k=0 you are going to count the number of derangements. Check this out: Derivation of the Partial Derangement (Rencontres numbers) formula

saman saffari
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