Dog marking mailboxes, conditional probability problem

Question

There are $n$ houses in the neighborhood. A dog took to marking 1 mailbox a day, choosing a house randomly (it doesn't matter if he marked it before or not).

What is the expected number of clean mailboxes after $m$ days?

This seems like a simple problem, but I'm stumped - my probability skills are very rusty. I'm trying to brush up on it, thus this problem. I don't know the answer though.

I tried to list every possibility for each day and see a pattern, but I haven't solved the problem yet.

• Day 1. Dog marks new mailbox with probability $1$. Number of marked mailboxes $=1$.

P(new)$=1$.

• Day 2. A dog can mark a new mailbox, or the old (i.e. marked) mailbox.

P(new,new)$=\frac{n-1}{n}$. Number of marked $=2$.

P(new,old)$=\frac{1}{n}$. Number of marked $=1$.

• Day 3.

P(n,n,n)$=\frac{n-1}{n}\frac{n-2}{n}$. Number of marked $=3$.

P(n,n,o)$=\frac{n-1}{n}\frac{2}{n}$. Number of marked $=2$.

P(n,o,n)$=\frac{1}{n}\frac{n-1}{n}$. Number of marked $=2$.

P(n,o,o)$=\frac{1}{n}\frac{1}{n}$. Number of marked $=1$.

• Day 4.

P(n,n,n,n)$=\frac{n-1}{n}\frac{n-2}{n}\frac{n-3}{n}$. Number of marked $=4$.

P(n,n,n,o)$=\frac{n-1}{n}\frac{n-2}{n}\frac{3}{n}$. Number of marked $=3$.

P(n,n,o,n)$=\frac{n-1}{n}\frac{2}{n}\frac{n-2}{n}$. Number of marked $=3$.

P(n,n,o,o)$=\frac{n-1}{n}\frac{2}{n}\frac{2}{n}$. Number of marked $=2$.

P(n,o,n,n)$=\frac{1}{n}\frac{n-1}{n}\frac{n-2}{n}$. Number of marked $=3$.

P(n,o,n,o)$=\frac{1}{n}\frac{n-1}{n}\frac{2}{n}$. Number of marked $=2$.

P(n,o,o,n)$=\frac{1}{n}\frac{1}{n}\frac{n-1}{n}$. Number of marked $=2$.

P(n,o,o,o)$=\frac{1}{n}\frac{1}{n}\frac{1}{n}$. Number of marked $=1$.

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I've also done Day 5. So far the probability sequences for $l$ marked mailboxes $P_l$ are:

$$P_1=1,\frac{1}{n},\frac{1}{n^2},\frac{1}{n^3},\frac{1}{n^4},\dots$$

$$P_2=0,\frac{n-1}{n},\frac{3(n-1)}{n^2},\frac{7(n-1)}{n^3},\frac{15(n-1)}{n^4},\dots$$

$$P_3=0,0,\frac{(n-1)(n-2)}{n^2},\frac{6(n-1)(n-2)}{n^3},\frac{25(n-1)(n-2)}{n^4},\dots$$

$$P_4=0,0,0,\frac{(n-1)(n-2)(n-3)}{n^3},\frac{10(n-1)(n-2)(n-3)}{n^4},\dots$$

So far I see the pattern for $P_1$ (trivial), for $P_2$ (coefficient is the sum of powers of $2$), but that's it.

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I'm sure there is an easier way to solve this problem, I would appreciate some hint.

Answer 1

We can write the number of clean mailboxes as $$N_C=\sum_{i=1}^nX_i$$ where $X_i=1$ if mailbox $i$ is clean after $m$ days and $X_i=0$ if it isn't. The expected value of this is $$ E(N_C) = \sum_{i=1}^nE(X_i) = \sum_{i=1}^nP(\mbox{mailbox $i$ is clean after $m$ days}).$$ Since the mailboxes are completely interchangable, we can simplify this to $$ E(N_C) = n P(\mbox{mailbox $1$ is clean after $m$ days}).$$

Now we just need to calculate the probability that mailbox $1$ (or any givne mailbox) is clean after $m$ days. On a given day it has a $(1-1/n)$ chance of not being marked, so the probability of keeping this up for $m$ days is $$P(\mbox{mailbox $1$ is clean after $m$ days}) = \left(1-\frac{1}{n}\right)^m$$ so the expected number of clean mailboxes after $m$ days is $$E(N_C) = n\left(1-\frac{1}{n}\right)^m.$$

Answer 2

For mailbox $i\in\{1,2,\dots,n\}$ define a Bernouilli random variable $X_i$ that takes value $1$ if it is clean after $m$ days, and takes value $0$ otherwise.

Then $X:=X_1+\cdots+X_n$ is the number clean mailboxes after $m$ days.

With linearity of expectation and symmetry we find:

$$\mathbb EX=\mathbb EX_1+\cdots+\mathbb EX_n=n\mathbb EX_1=n\Pr(X_1=1)=n\left(1-\frac1{n}\right)^m$$