Calculamos a probabilidade de $n$ pessoas com chapeus diferentes irem para festa tusca e após o voo de chapeus (folklore de tusca é com chinelos) ninguém recuperar seu próprio chapeu. Observe que as pessoas jogam seus chapéus para o ar e recuperam algum chapéu aleatório.
Você sabia que essa probabilidade tende a $\frac{1}{e}$ quando número de pessoas fica grande! Essa probabilidade é quase 37%. Veja aqui.
De fato, seja $A_i$ representar as configurações nas quais $i$'ésima pessoa acerta seu chapeu. Então $|A_i| = (n-1)!$ e $A_{ij}, i \neq j$ configurações as duas pessoas $i, j$ acertarem seus chapeus: $|A_{ij}| = (n-2)!$ … Observe que estamos buscando a cardinalidade de complemento da $\bigcup_{i=1}^{n} A_i$ e para tal vamos calcular $|\bigcup_{i=1}^{n} A_i|$: $$ |\bigcup_{i=1}^{n} A_i| = \sum_{i=1}^{n} |A_i| - \sum_{i < j} |A_{ij}| + \cdots + (-1)^{n-1} |A_{123\cdots n}| $$ $$ = n (n-1)! - { n \choose 2} (n-2)! + \cdots = n! - \frac{n!}{2!} + \frac{n!}{3!}- \cdots $$ e portanto a probabilidade de ninguem acertar seu chapeu é $$ \frac{1}{n!} (\frac{n!}{2!} - \frac{n!}{3!}+ \cdots) $$ e quando $n$ tende ao infinito a série acima tende a $e^{-1}.$
Agora o problema de Tusca de verdade: $n$ pessoas jogam seus chinelos para o ar e recuperam aleatóriamente dois chinelos (pode ser dois de esquerdo …). qual é a probabilidade de ninguem acertar seu par de chinelos. Qual é limite desta probabilidade quando $n$ tende ao infinito?
Neste caso temos $2n$ chinelos e a número de configurações de $i$'ésima pessoa acertar o par de chinelos é: $\frac{(2n-2)!}{2^{n-1}}$. A probabilidade que $i$- ésima pessoa pegar seu par correto de chinelos é: $$\frac{1}{n(2n-1)}.$$ Interprete o número como probabilidade ou use $\frac{\frac{(2n-2)!}{2^{n-1}}}{\frac{(2n)!}{2^{n}}} = \frac{1}{n(2n-1)}.$
Agora calculamos a probabilidade de que $i, j$ acertem seu par de chinelos: é igual $2^2 \frac{(2n-4)!}{(2n)!}$…
A probabilidade de que alguem acerte seu par de chinelos é:
$$ \sum_{k=1}^{n} (-1)^{k+1} 2^k \binom{n}{k} \frac{(2n-2k)!}{(2n)!} $$
conjectura: Quando $n$ tende ao infinito essa probabilidade tende a zero.
Observação: o evento de ninguem acertar o chinelo esquerdo não é independente de ninguem acertar o pé direito.
Discussion
I asked ChatGPT:
Question: suppose 100 persons with hats go to a party and after the end of the party they change the hats randomly. what is the probability of nobody take his own hat?
Answer: The problem you've presented is a classic example of the Derangement problem. The Derangement problem is a combinatorial problem that asks how many ways it is possible to arrange n distinct objects such that none of the objects are in their original position.
In this case, we have n = 100 people, and we want to find the probability that none of the people get their own hat back. Let's call this probability P.
We can use the principle of inclusion-exclusion to compute P. The principle of inclusion-exclusion says that the number of elements that belong to at least one of several sets can be computed by adding the sizes of the sets, then subtracting the sizes of the intersections of the sets, then adding the sizes of the intersections of the intersections, and so on.
Let A_i be the event that person i gets their own hat back. We want to compute P, which is the probability that none of the events A_i occur. We have:
P = 1 - P(at least one A_i occurs)
To compute P(at least one A_i occurs), we can use the principle of inclusion-exclusion. We have:
P(at least one A_i occurs) = sum over i of P(A_i) - sum over i<j of P(A_i and A_j) + sum over i<j<k of P(A_i and A_j and A_k) - … + (-1)^100 P(A_1 and A_2 and … and A_100)
The probability that person i gets their own hat back is 1/100, since there are 100 hats and each person is equally likely to get any of the hats. Therefore, P(A_i) = 1/100 for all i.
The probability that two specific people, say i and j, both get their own hats back is (1/100) * (1/99), since person i is equally likely to get any of the hats, and person j is equally likely to get any of the remaining 99 hats. Therefore, P(A_i and A_j) = (1/100) * (1/99) for all i<j.
Similarly, the probability that three specific people, say i, j, and k, all get their own hats back is (1/100) * (1/99) * (1/98), since person i is equally likely to get any of the hats, person j is equally likely to get any of the remaining 99 hats, and person k is equally likely to get any of the remaining 98 hats. Therefore, P(A_i and A_j and A_k) = (1/100) * (1/99) * (1/98) for all i<j<k.
Continuing in this way, we can compute all the probabilities P(A_i and A_j and … and A_k) for all subsets of {1, 2, …, 100} of size at most 100. Note that P(A_1 and A_2 and … and A_100) is just the probability that all 100 people get their own hats back, which is 1/(100!) by the same argument as before.
Substituting all these probabilities into the equation for P(at least one A_i occurs) and simplifying using the principle of inclusion-exclusion, we get:
P(at least one A_i occurs) = sum over k from 1 to 100 of ((-1)^k / k!) * (100 choose k)
Therefore,
P = 1 - P(at least one A_i occurs) = 1 - sum over k from 1 to 100 of ((-1)^k / k!) * (100 choose k)
Using a calculator or computer program to compute the sum,