If n friends go to a diner, and when leaving, each of them take a hat randomly, what is the probability of "nobody gets his original hat"? My book says it goes to 1/e, but I don't have more explanation... Can you help me?
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1Start with this: https://en.wikipedia.org/wiki/Derangement – Feb 09 '20 at 07:14
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Some related posts: What is the probability that no letter is in its proper envelope?, Why is the Derangement Probability so Close to $\frac{1}{e}$?, Number of permutations of $n$ elements where no number $i$ is in position $i$, How many fixed points in a permutation. – Martin Sleziak Feb 09 '20 at 09:19
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On one hand from inclusion-exclusion principle it follows, that the number of ways to distribute the hats in a way, that nobody gets the original one is $!n := n! \sum_{i = 0}^n \frac{(-1)^i}{i!}$.
On the other hand $e^x = \sum_{i = 0}^n \frac{x^i}{i!}$.
Thus $\lim_{n \to \infty} \frac{!n}{n!} = \frac{1}{e}$
Chain Markov
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It will be $P(E)=\frac{D_n}{n!}$ where $ D_n $ is dearangement of n objects at n places . For more about dearangement you can listen to podcast https://podcasts.google.com?feed=aHR0cHM6Ly9hbmNob3IuZm0vcy9lZjA0NTQ4L3BvZGNhc3QvcnNz&episode=NTU3YmQyZTQtYjVlOC00MTlkLTliN2UtNjhlYjU5MzFiNGIw Or. Listen to inclusion and exclusion principle , Dearrangement And Number of onto functions. on TuneIn http://tun.in/tjwNzd
And as $n \rightarrow \infty \,\,,\frac{D_n}{n!}=\frac{1}{e}$
Rajan
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