For the sample space of inserting n balls into n urns let...

For the sample space of inserting $n$ balls into $n$ urns let each sample point be an $n$-tuple $\left(x_1, x_2, \ldots, x_n\right)$, where $x_j$ represents the number of the ball put into the $j$ th urn (sometimes, unromantically, called a pot). Thus, each component is a number from 1 to $n$ and no two components are equal. Then the event $A_k=\left\{\left(x_1, x_2, \ldots, x_n\right)\right.$ $\left.\in \Omega: x_k=k\right\}$. Prove that $P\left[A_k\right]=1 / n$. Thus, the probability of a man getting his own hat does not depend on whether he gets to make the first, second, or even last choice.

Key Concepts

Permutations

A permutation is an arrangement of all the elements of a set in a specific order. In this problem, each n-tuple representing the placement of n balls into n urns is a permutation of the numbers 1 through n. This concept is fundamental in combinatorics and probability because it provides the structure for counting the total number of outcomes when all elements are arranged with no repetition.

Uniform Distribution on Permutations

The assumption that each permutation is equally likely implies that the sample space is uniformly distributed. This is crucial because it allows us to compute probabilities simply by counting the number of favorable outcomes and dividing by the total number of outcomes, without needing to consider any weighting or biases among the outcomes.

Fixed Point in a Permutation

A fixed point in a permutation is an element that remains in its original position. In the context of the problem, the event A_k (where the kth element is equal to k) is an example of a fixed point. Studying fixed points is important in probability theory and combinatorics because it leads to interesting results like the hat-check problem, where the probability of a fixed point is easily determined due to the symmetry in the permutation.

Factorial Counting

Factorial counting refers to the calculation of the number of ways to arrange a set of objects, which is given by n!. In the problem, the total number of outcomes is n! and the number of outcomes that satisfy the event A_k (where the kth ball gets fixed into the kth urn) is (n-1)!. This ratio, (n-1)!/n!, directly leads to the probability 1/n.

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