The Collector’s Problem

Question

I have the following Question in my text book , where the Answer is given. I have got my own alternative Answer. I want to know whether that alternative Answer is Correct & Equivalent to the text book Answer.

Suppose that each package of bubble gum contains the picture of a baseball player, that the pictures of $r$ different players are used, that the picture of each player is equally likely to be placed in any given package of gum, and that pictures are placed in different packages independently of each other.
The problem now is to determine the probability $p$ that a person who buys $n$ packages of gum ($n$ ≥ $r$) will obtain a complete set of $r$ different pictures.

The textbook answers as follows:

For i = 1, ..., r, let $A_{i}$ denote the event that the picture of player i is missing from all n packages. Then $ \bigcup \limits_{i=1}^{r}A_{i} $ is the event that the picture of at least one player is missing. We shall find $Pr \left( \bigcup \limits_{i=1}^{r}A_{i}\right) $

Since the picture of each of the r players is equally likely to be placed in any particular package, the probability that the picture of player i will not be obtained in any particular package is $\frac{r-1}{r}$. Since the packages are filled independently, the probability that the picture of player i will not be obtained in any of the n packages is $\left [ \frac{r-1}{r} \right ] ^n$, hence $$ Pr \left( A_i \right) = \left ( \frac{r-1}{r} \right ) ^n $$ Now consider any two players i and j. The probability that neither the picture of player i nor the picture of player j will be obtained in any particular package is $\frac{r-2}{r}$. Therefore, the probability that neither picture will be obtained in any of the n packages is $\left [ \frac{r-2}{r} \right ] ^n$. Thus, $$ Pr \left( A_i\bigcap A_j \right) = \left ( \frac{r-2}{r} \right ) ^n $$ If we next consider any three players i, j, and k, we find that $$ Pr \left( A_i\bigcap A_j \bigcap A_k \right) = \left ( \frac{r-3}{r} \right ) ^n $$ By continuing in this way, we finally arrive at the probability $Pr \left( A_i\bigcap A_j \bigcap... \bigcap A_r \right)$ that the pictures of all r players are missing from the n packages. Of course, this probability is 0. $$ Pr \left( \bigcap \limits_{i=1}^{r} A_i \right) = r\left( \frac{r-1}{r} \right)^n - \binom{r}{2} \left( \frac{r-2}{r} \right)^n + ... + \left(-1 \right)^r \binom{r}{r-1}\left(\frac{1}{r} \right)^n = \sum_{j=1}^{r-1}\left( -1\right) ^{j+1} \binom{r}{j}\left(1- \frac{j}{r} \right)^n $$ Since the probability p of obtaining a complete set of r different pictures is equal to $$ 1 − Pr \left( \bigcup \limits_{i=1}^{r}A_{i}\right) $$ it follows from the foregoing derivation that p can be written in the form: $$ p=\sum_{j=0}^{r-1}\left( -1\right) ^{j}\binom{r}{j}\left(1- \frac{j}{r} \right)^n $$ I just used another method to solve this. Each package has r choices, so the total outcomes would be $r^n$ and then we first select r packages out of n to have a complete set of r pictures, there are $\binom{n}{r}$ combinations, and then arrange those r packages with r! choices. The remaining n-r packages each have r different arrivals. so my calculation is : $$ \frac{ \binom{n}{r}r!r^{n-r}}{r^n} $$

I don't know whether it's correct or not. I can understand the procedure provided by the textbook, so are these two answers the same?

Answer 1

Your way is not right. It will under-count & over-count certain Cases.

Let $n=8$ & $r=5$.
You want to put the $5$ in some order , then rest $3$ you want to put in-between those $5$.

Issue 1 : you are stating that you will put the $5$ in some order , Eg $12345$ , then the rest $3$ have $5$ Positions , but those actually have $6$ Positions : $.1.2.3.4,5.$ where $.$ indicates a Position.

Issue 2 : when you put $2$ or more (eg $x,y,z$) in the Same Dot Position in-between , Eg $123xyz45$ , you have to count many ways !
$123xyz45$ , $123xzy45$, $123yxz45$ , $123yzx45$ , $123zxy45$ , $123zyx45$ : $6$ ways when we put $3$ in-between in Same Dot Position.

Issue 3 : when you put a number at a Dot Position in-between , the neighbors might be the Same , hence the count will go wrong.
Eg put the $5$ like this : $12345$
then put Element $3$ in-between like this : $123345$
Is this Element $3$ inserted at 3rd Position or at 4th Position ?
Both are Same , hence you are over-counting those.
With $123345$ , insert Element $3$ again to get $1233345$ : Which $3$ was inserted now ?
With $1233345$ , insert $3$ again to get $12333345$ : Which $3$ was inserted now ?

Issue 4 : when you put the $5$ like this , $12345$ , & then put Element $3$ like this , $123453$ , it might have been counted like this too : $12453$ , with $3$ inserted later between $2$ & $4$.

Issue 5 : when you have $12345$ & then insert $4$ & $5$ like this : $123454$ , $1234545$ , it might also occur like this : $12354$ , $123545$ , $1234545$. You might have counted the Same Eventual Way too many times.

These over-counting & under-counting Issues are not easy to rectify with your way. There are too many Cases to check !

The text book Answer is Direct , without these Issues !