Probability of sending wrong email to all 5 people

Question

Say I have to write 5 emails to 5 people: Al, Ben, Charlie, Dan, Eli. Each email starts with the person's name (i.e., "Dear Al").

Mistakenly, as it was late at night, I wrote the To section to the wrong person (example below).

Question: what is the probability that all emails will reach to the wrong person?

Is this type of questions relates to the Inclusion–exclusion principle? If so, how would you solve it?

Answer 1

Permutations in which every person gets the wrong email are called derangements, which you can read about on that wiki page. The answer is the number of derangements of $5$, $!5 = 44$, divided by the number of permutations $5! = 120$. So $$ \frac{!5}{5!} = \frac{44}{120} = \frac{11}{30}. $$ For more detail you should refer to counting derangements, but you can use inclusion-exclusion to derive $$ !n = n! \sum_{i=0}^n \frac{(-1)^i}{i!}; $$ in particular \begin{align*} !5 &= 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \\ &= \frac{120}{2} - \frac{120}{6} + \frac{120}{24} - \frac{120}{120} \\ &= 60 - 20 + 5 - 1 \\ &= 44. \end{align*} Also there is an elegant formula for $n \ge 1$, $$ !n = \left\lfloor{\frac{n!}{e} + \frac12}\right\rfloor $$ So you could also have gotten the answer by taking $120/e \approx 44.1455329406$, rounding to $44$, and dividing by $120$.

SEE ALSO: 1, 2, 3, 4, on mathSE.

Answer 2

Yes, one would need to use the inclusion-exclusion principle. The are 5! ways of sending the emails (universal set). There are |A| = 4! ways on which the 1st person receives the right mail, |B| = 4! ways in which the second person does, |C| = 4! ways in which the third person does ...... By the inclusion exclusion principle the number of ways in which at least one of them gets the right mail is [5(4!)] - [10(3!)] + [10(2!)] - [5] + [1] = 76. Therefore, the number of ways in which all of them get the wrong mail is 5! - 76 = 44. So the probability would be 44/120 = 11/30

Answer 3

Here is a solution without using the Exclusion-Inclusion principe:

There are only two kinds ways everyone ends up with the wrong email: Either you get a cycle of 5 (e.g. The emails to ABCDE get sent to BCDEA respectively), or you have a cycle of 3 plus a cycle of 2 (e.g. ABC get sent to BCA and DE to ED)

How many 5 cycles are there? Starting with A, the next is one of 4, the third of 3, etc. ... So that is 4! =24

How many 3+2 cycles? There are ${{5}\choose{3}}=10$ ways to divide up the 5 into 3 and 2, and the 3 cycle can be done in 2 ways, so that's 20.

Total: 24+20 = 44 ways for all of them to get a wrong email.

So, probability is $\frac{44}{5!}= \frac{44}{120}=\frac{11}{30}$