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Chapter 1: Problem 32

Suppose all \(n\) men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Show that the probability that none of the \(n\) men selects his own hat is $$ \frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-+\cdots \frac{(-1)^{n}}{n !} $$ Note that as \(n \rightarrow \infty\) this converges to \(e^{-1}\). Is this surprising?

Short Answer

Expert verified
The probability that none of the n men selects his own hat is given by the alternating sum of the reciprocals of factorials: $$ P(\text{no man selects his own hat}) = \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots \frac{(-1)^n}{n!} $$ As n approaches infinity, this probability converges to \(e^{-1}\).

Step by step solution

01

Derangement Concept

A derangement, in the context of permutations, is a permutation with no fixed points, i.e., there is no element that remains in its original position after permuting. In this exercise, we will be finding the probability that none of the n men selects his own hat, which translates to finding the derangement of n elements.
02

Counting Total Permutations and Derangements

There are n! (factorial n) possible ways to arrange the hats for n men. Let D_n denote the number of derangements for a given n. There are various ways to compute D_n, but we will use the principle of inclusion and exclusion.
03

Inclusion-Exclusion Principle

According to the principle of inclusion and exclusion, the total number of derangements can be calculated by alternating sum of the factorials: $$ D_n = n! - \binom{n}{1}(n-1)! + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + \cdots + (-1)^n\binom{n}{n}(0)! $$
04

Simplifying the Derangement Formula

To find the probability that none of the n men selects his own hat, we will divide the number of derangements by the total number of permutations (n!). $$ P(\text{no man selects his own hat}) = \frac{D_n}{n!} $$ Plugging in the derangement formula from step 3, we have: $$ P(\text{no man selects his own hat}) = \frac{n! - \binom{n}{1}(n-1)! + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + \cdots + (-1)^n\binom{n}{n}(0)!}{n!} $$
05

Cancelling n! from the Derangement Formula

We can simplify this further by cancelling n! from both the numerator and denominator: $$ P(\text{no man selects his own hat}) = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!} $$ So the probability that none of the n men selects his own hat is: $$ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots \frac{(-1)^n}{n!} $$ As the exercise notes, this probability converges to \(e^{-1}\) as n approaches infinity.

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Most popular questions from this chapter

For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that $$ P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \cdots E_{n-1}\right) $$

If \(P(E)=0.9\) and \(P(F)=0.8\), show that \(P(E F) \geqslant 0.7\). In general, show that $$ P(E F) \geqslant P(E)+P(F)-1 $$ This is known as Bonferroni's inequality.

Suppose we have ten coins which are such that if the \(i\) th one is flipped then heads will appear with probability \(i / 10, i=1,2, \ldots, 10\). When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m .\) Assume that in the count of the votes all possible orderings of the \(n+m\) votes are equally likely. Let \(P_{n, m}\) denote the probability that from the first vote on \(A\) is always in the lead. Find (a) \(P_{2,1}\) (b) \(P_{3,1}\) (c) \(P_{n, 1}\) (d) \(P_{3,2}\) (e) \(P_{4,2}\) (f) \(P_{n, 2}\) (g) \(P_{4,3}\) (h) \(P_{5,3}\) (i) \(P_{5,4}\) (j) Make a conjecture as to the value of \(P_{n, m}\).

Let \(E\) and \(F\) be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event \(E\) or event \(F\) occurs. What does the sample space of this new super experiment look like? Show that the probability that event \(E\) occurs before event \(F\) is \(P(E) /[P(E)+P(F)]\). Hint: Argue that the probability that the original experiment is performed \(n\) times and \(E\) appears on the \(n\) th time is \(P(E) \times(1-p)^{n-1}, n=1,2, \ldots\), where \(p=P(E)+\) \(P(F)\). Add these probabilities to get the desired answer.
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