How long does it take to collect all coupons?

This post discusses the coupon collector problem, a classical problem in probability.

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The Coupon Collector Problem

The problem is usually stated as a coupon collector trying to collect the entire set of coupons. For example, each time the coupon collector buys a product (e.g. a box of breakfast cereal), he receives a coupon, which is a prize that can be a toy or a baseball card or other interesting item. Suppose that there are Image may be NSFW.
Clik here to view. different types of coupons (prizes) and that the coupon collector wishes to collect the entire set. How many units of the product does the coupon collector have to buy in order to collect the entire set?

A simplified discussion of the coupon collector problem is found in another blog post. This post is more detailed discussion.

This blog post in another blog discusses the coupon collector problem from a simulation perspective.

As shown below, if there are 5 different coupons, it takes 12 purchases on average to get all coupons. If there are 10 different types of coupons, on average it would take 30 purchases. If there are 50 different types of coupons, it would take on average 225 purchases collect the entire set. The first few coupons are obtained fairly quickly. As more and more coupons are accumulated, it is harder to get the remaining coupons. For example, for the 50-coupon case, after the 49 coupons are obtained, it takes on average 50 purchases to get the last coupon.

Suppose that the coupon collector does not want to collect the entire set and only wishes to collect Image may be NSFW.
Clik here to view. distinct coupons where Image may be NSFW.
Clik here to view. $r \le n$ . It turns out that this special case only requires a minor tweak to the case of collecting the entire set. Our strategy then is to focus on the main case. The special case will be discussed at the end of the post.

We first consider the main case that the coupon collector wishes to collect the entire set. The problem can be cast as a random sampling from the population Image may be NSFW.
Clik here to view. $S=\left\{1,2,3,\cdots,n \right\}$ . Selecting a number at random from Image may be NSFW.
Clik here to view. with replacement is equivalent to the coupon collector receiving a coupon. Let Image may be NSFW.
Clik here to view. X_n be the minimum number of selections such that each number in Image may be NSFW.
Clik here to view. is picked at least once. In this post we discuss the probability function of Image may be NSFW.
Clik here to view. X_n as well as its mean and variance.

Another interpretation of the problem is that it can be viewed as an occupancy problem, which involves throwing balls into Image may be NSFW.
Clik here to view. cells at random. The random quantity of interest is the number of balls that are required to be thrown such that each cell has at least one ball. Clearly this formulation is identical to the coupon interpretation and the random sampling interpretation. The angle of occupancy problem is useful since we can leverage the formulas developed in this previous post. A description of the occupancy problem is given here.

Regardless of the interpretation, the goal is obtain information on the random variable Image may be NSFW.
Clik here to view. X_n , the minimum number of random selections from Image may be NSFW.
Clik here to view. $S=\left\{1,2,3,\cdots,n \right\}$ in order to have the complete set of Image may be NSFW.
Clik here to view. distinct values represented in the sample.

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Mean and Variance

The mean and variance of Image may be NSFW.
Clik here to view. X_n are easier to obtain. So that is where we begin. The key is to break up Image may be NSFW.
Clik here to view. X_n into a sum as follows:

Image may be NSFW.
Clik here to view. $X_n=C_{1}+C_{2}+C_{3}+\cdots+C_{i-1}+C_{i}+\cdots+C_{n} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0)$

where Image may be NSFW.
Clik here to view. $C_{i}$ is the additional selections from Image may be NSFW.
Clik here to view. $S=\left\{1,2,3,\cdots,n \right\}$ to get a number that is distinct from the Image may be NSFW.
Clik here to view. i-1 distinct numbers that have been chosen. For example, Image may be NSFW.
Clik here to view. C_3 is the number of random selections to get a number that is distinct from the two distinct numbers obtained up to that point.

Note that each Image may be NSFW.
Clik here to view. C_i involves repeated sampling until some criterion is reached, thus resembling a geometric random variable. Indeed they are. As the sampling continues and as more distinct values are obtained, it is not as easy to obtain a new number. After Image may be NSFW.
Clik here to view. i-1 distinct numbers have been obtained, the probability of drawing a new distinct number is Image may be NSFW.
Clik here to view. $p=\frac{n-(i-1)}{n}$ . As geometric random variables, each Image may be NSFW.
Clik here to view. C_i has the following mean and variance.

Image may be NSFW.
Clik here to view. $\displaystyle E[C_i]=\frac{1}{p}=\frac{n}{n-(i-1)}$

Image may be NSFW.
Clik here to view. $\displaystyle Var[C_i]=\frac{1-p}{p^2}=\frac{n(i-1)}{[n-(i-1)]^2}$

where Image may be NSFW.
Clik here to view. $i=1,2,3,\cdots,n$ . Note that the random variables Image may be NSFW.
Clik here to view. C_i are independent. The value of Image may be NSFW.
Clik here to view. C_i does not depend on how many trials it takes to draw the previous Image may be NSFW.
Clik here to view. i-1 distinct numbers. The following gives the mean and variance of Image may be NSFW.
Clik here to view. X_n .

Image may be NSFW.
Clik here to view. $\displaystyle E[X_n]=\sum \limits_{i=1}^n E[C_i]=\sum \limits_{i=1}^n \frac{n}{n-(i-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Image may be NSFW.
Clik here to view. $\displaystyle Var[X_n]=\sum \limits_{i=1}^n Var[C_i]=\sum \limits_{i=1}^n \frac{n(i-1)}{[n-(i-1)]^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The expectation Image may be NSFW.
Clik here to view. E[X_n] can be rearranged as follows to give more information.

Image may be NSFW.
Clik here to view. $\displaystyle E[X_n]=n \ \biggl[\frac{1}{n}+ \cdots + \frac{1}{3}+ \frac{1}{2} + 1\biggr]=n \ H_n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

The quantity Image may be NSFW.
Clik here to view. H_n is the partial sum of the harmonic series. Note that Image may be NSFW.
Clik here to view. $H_n \rightarrow \infty$ as Image may be NSFW.
Clik here to view. $n \rightarrow \infty$ . Thus Image may be NSFW.
Clik here to view. $E[X_n] \rightarrow \infty$ as Image may be NSFW.
Clik here to view. $n \rightarrow \infty$ . The quantity Image may be NSFW.
Clik here to view. H_n can be interpreted as the average number of units of product that are required to purchase per coupon. The following table lists out the expected values for selected values of Image may be NSFW.
Clik here to view..

Table 1

Image may be NSFW.
Clik here to view. $\begin{array}{cccccc} \text{Number of} & \text{ } & \text{Expected Number of Trials} & \text{ } & \text{Expected Total Number} & \\ \text{Coupons} & \text{ } & \text{per Coupon} & \text{ } & \text{of Trials (rounded up)} & \\ n & \text{ } & E[H_n] & \text{ } & E[X_n] & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 1 & \text{ } & 1.0000 & & 1 & \\ 2 & \text{ } & 1.5000 & & 3 & \\ 3 & \text{ } & 1.8333 & & 6 & \\ 4 & \text{ } & 2.0833 & & 9 & \\ 5 & \text{ } & 2.2833 & & 12 & \\ 6 & \text{ } & 2.4500 & & 15 & \\ 7 & \text{ } & 2.5929 & & 19 & \\ 8 & \text{ } & 2.7179 & & 22 & \\ 9 & \text{ } & 2.8290 & & 26 & \\ 10 & \text{ } & 2.9290 & & 30 & \\ 20 & \text{ } & 3.5977 & & 72 & \\ 30 & \text{ } & 3.9950 & & 120 & \\ 40 & \text{ } & 4.2785 & & 172 & \\ 50 & \text{ } & 4.4992 & & 225 & \\ 60 & \text{ } & 4.6799 & & 281 & \\ 70 & \text{ } & 4.8328 & & 339 & \\ 80 & \text{ } & 4.9655 & & 398 & \\ 90 & \text{ } & 5.0826 & & 458 & \\ 100 & \text{ } & 5.1874 & & 519 & \\ \end{array}$

Table 1 gives an estimate on how long to expect to collect the entire set of coupons for selected coupon sizes. The third column gives the expected total number of purchases to obtain the entire coupon set. The second column gives an estimate of how many purchases on average to obtain one coupon. For the 50-coupon case, it takes on average about 4.5 purchases to obtain one coupon. However, it does not tell the whole story. To get the 50th coupon, it takes on average 50 trials. Note that Image may be NSFW.
Clik here to view. $E[C_{50}]=50$ in the 50-coupon case in formula (1). In a simulation of the 50-coupon problem, it took 54 trials to obtain the 50th coupon. To get the 49th coupon, it takes on average 50/2 = 25 trials.

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The Occupancy Problem

We now view the coupon collector problem as an occupancy problem in order to leverage a formula from a previous post. Suppose that we randomly throw Image may be NSFW.
Clik here to view. balls into Image may be NSFW.
Clik here to view. cells. Let Image may be NSFW.
Clik here to view. $Y_{k,n}$ be the number of empty cells as a result of randomly assigning Image may be NSFW.
Clik here to view. balls into Image may be NSFW.
Clik here to view. cells. The following gives the probabilities Image may be NSFW.
Clik here to view. $P(Y_{k,n}=j)$ where Image may be NSFW.
Clik here to view. $j=0,1,2,\cdots,n-1$ .

Image may be NSFW.
Clik here to view. $\displaystyle P(Y_{k,n}=0)=\sum \limits_{i=0}^{n} (-1)^{i} \binom{n}{i} \biggl(1-\frac{i}{n}\biggr)^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

Image may be NSFW.
Clik here to view. $\displaystyle P(Y_{k,n}=j)=\binom{n}{j} \sum \limits_{i=0}^{n-j} (-1)^{i} \binom{n-j}{i} \biggl(1-\frac{j+i}{n}\biggr)^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

where Image may be NSFW.
Clik here to view. $j=1,2, \cdots, n-1$ .

The notation Image may be NSFW.
Clik here to view. $\binom{n}{j}$ is the binomial coefficient, which is the number of ways to choose Image may be NSFW.
Clik here to view. objects out of Image may be NSFW.
Clik here to view. objects where order does not matter. The calculation is defined by Image may be NSFW.
Clik here to view. $\displaystyle \binom{n}{j}=\frac{n!}{j! (n-j)!}$ .

The formula (5) gives the probability of having Image may be NSFW.
Clik here to view. empty cells. In throwing Image may be NSFW.
Clik here to view. balls into Image may be NSFW.
Clik here to view. cells, the probability of having Image may be NSFW.
Clik here to view. occupied cells is then Image may be NSFW.
Clik here to view. $P[Y_{k,n}=n-w]$ .

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The Probability Function

We now discuss the probability function of the random variable Image may be NSFW.
Clik here to view. X_n , namely Image may be NSFW.
Clik here to view. P[X_n=k] for Image may be NSFW.
Clik here to view. $k=n,n+1,n+2,\cdots$ . The event Image may be NSFW.
Clik here to view. X_n=k means that all Image may be NSFW.
Clik here to view. cells are occupied after throwing Image may be NSFW.
Clik here to view. balls with the first Image may be NSFW.
Clik here to view. k-1 balls landing in Image may be NSFW.
Clik here to view. n-1 cells. Putting it in another way, there are zero empty cells after throwing Image may be NSFW.
Clik here to view. balls and there is 1 empty cell after throwing the first Image may be NSFW.
Clik here to view. k-1 balls. This can be stated using the notation in the preceding section on the occupancy problem as follows:

Image may be NSFW.
Clik here to view. $P(X_n=k]=P[Y_{k,n}=0 \text{ and } Y_{k-1,n}=1] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$

Consider the following derivation.

Image may be NSFW.
Clik here to view. $\displaystyle \begin{aligned} P(X_n=k]&=P[Y_{k,n}=0 \text{ and } Y_{k-1,n}=1] \\&=P[Y_{k,n}=0 \ \lvert \ Y_{k-1,n}=1] \times P[Y_{k-1,n}=1] \\&=\frac{1}{n} \times \binom{n}{1} \sum \limits_{i=0}^{n-1} (-1)^i \binom{n-1}{i} \biggl[ 1-\frac{1+i}{n} \biggr]^{k-1} \\&=\sum \limits_{i=0}^{n-1} (-1)^i \binom{n-1}{i} \biggl[ 1-\frac{1+i}{n} \biggr]^{k-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) \end{aligned}$

where Image may be NSFW.
Clik here to view. $k=n,n+1,n+2,\cdots$ .

Rather than memorizing the probability function in (7), a better way is to focus on the thought process that is inherent in (6).

One comment about the calculation for (7). The summation for Image may be NSFW.
Clik here to view. P[X_n=k] has Image may be NSFW.
Clik here to view. terms. A given probability may involve multiple values of Image may be NSFW.
Clik here to view., e.g.

Image may be NSFW.
Clik here to view. $P[X_6>15]=1-P[6 \le X_6 \le 14]$

Unless the number of values for Image may be NSFW.
Clik here to view. is very small, the calculation should be done using software. Microsoft Excel is an excellent way to perform the calculation. The calculations for the examples below are programmed in Excel.

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Examples

Example 1
Suppose that a fair die is rolled until all 6 faces have appeared. Find the mean number of rolls and the variance of the number of rolls. What is the probability that it will take at least 12 rolls? What is the probability that it will take more than 15 rolls?

Using the notation developed above, the random variable of interest is Image may be NSFW.
Clik here to view. X_6 . Its mean and variance are:

Image may be NSFW.
Clik here to view. $\displaystyle E[X_6]=6 \ \biggl[ 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \biggr]=6 \times 2.45 = 14.7 \approx 15$

Image may be NSFW.
Clik here to view. $\displaystyle Var[X_6]=\sum \limits_{i=1}^6 \frac{6(i-1)}{[6-(i-1)]^2}=38.99$

The following is the probability function for Image may be NSFW.
Clik here to view. X_6 .

Image may be NSFW.
Clik here to view. $\displaystyle P[X_6=k]=\sum \limits_{i=0}^{5} (-1)^i \binom{5}{i} \biggl[ 1-\frac{1+i}{6} \biggr]^{k-1}$

where Image may be NSFW.
Clik here to view. $k=6,7,8,\cdots$

For each Image may be NSFW.
Clik here to view., the quantity Image may be NSFW.
Clik here to view. P[X_6=k] requires 6 calculations. Performing the calculations in Excel, the desired probabilities are:

Image may be NSFW.
Clik here to view. $P[X_6 \ge 12]=1-P[6 \le X_6 \le 11]=1-0.356206419=0.643793581$

Image may be NSFW.
Clik here to view. $P[X_6 > 15]=1-P[6 \le X_6 \le 15]=1-0.644212739=0.355787261$

Even though the average number of trials is 15, there is still a significant probability that it will take more than 15 trials. This is because the variance is quite large. Image may be NSFW.
Clik here to view. $\square$

Example 2
An Internet startup is rapidly hiring new employees. What is the expected number of new employees until all birth months are represented? Assume that the 12 birth months are equally likely. What is the probability that the company will have to hire more than 25 employees? If the company currently has more than 25 employees with less than 12 birth months, what is the probability that it will have to hire more than 35 employees to have all 12 birth months represented in the company?

The random variable of interest is Image may be NSFW.
Clik here to view. $X_{12}$ . The following shows the mean and probability function.

Image may be NSFW.
Clik here to view. $\displaystyle E[X_{12}]=12 \ \biggl[ 1 + \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{12} \biggr]=37.23852814 \approx 38$

Image may be NSFW.
Clik here to view. $\displaystyle P[X_{12}=k]=\sum \limits_{i=0}^{11} (-1)^i \binom{11}{i} \biggl[ 1-\frac{1+i}{12} \biggr]^{k-1}$

where Image may be NSFW.
Clik here to view. $k=12,13,14,\cdots$

Performing the calculation in Excel, we obtain the following probabilities.

Image may be NSFW.
Clik here to view. $P[X_{12} > 25]=1-P[12 \le X_{12} \le 25]=1-0.181898592=0.818101408$

Image may be NSFW.
Clik here to view. $P[X_{12} > 35]=1-P[12 \le X_{12} \le 35]=1-0.531821149=0.468178851$

Image may be NSFW.
Clik here to view. $\displaystyle P[X_{12} > 35 \ \lvert \ X_{12} > 25]=\frac{P[X_{12} > 35]}{P[X_{12} > 25]}=\frac{0.468178851}{0.818101408}=0.572274838$

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A Special Case

We now consider the special case that the coupon collector only wishes to collect Image may be NSFW.
Clik here to view. distinct coupons where Image may be NSFW.
Clik here to view. $r \le n$ . Of course, Image may be NSFW.
Clik here to view. is the total number of distinct coupon types. Let Image may be NSFW.
Clik here to view. $X_{n,r}$ be the minimum number of purchases such that Image may be NSFW.
Clik here to view. distinct coupons have been obtained. In the random sampling interpretation, Image may be NSFW.
Clik here to view. $X_{n,r}$ would be the minimum sample size such that Image may be NSFW.
Clik here to view. distinct elements have been chosen from the sample space Image may be NSFW.
Clik here to view. $S=\left\{1,2,3,\cdots,n \right\}$ . The mean and variance of Image may be NSFW.
Clik here to view. $X_{n,r}$ follow from the same idea. Each Image may be NSFW.
Clik here to view. $X_{n,r}$ is the independent sum of geometric random variables as in (0).

Image may be NSFW.
Clik here to view. $X_{n,r}=C_{1}+C_{2}+C_{3}+\cdots+C_{i-1}+C_{i}+\cdots+C_{r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)$

where Image may be NSFW.
Clik here to view. $r \le n$ .

Thus Image may be NSFW.
Clik here to view. $E[X_{n,r}]$ and Image may be NSFW.
Clik here to view. $Var[X_{n,r}]$ would be like (1) and (2) except that the summation is through Image may be NSFW.
Clik here to view. instead of Image may be NSFW.
Clik here to view..

For the probability function of Image may be NSFW.
Clik here to view. $X_{n,r}$ , we only need to tweak the thought process expressed in (6) slightly. For the event Image may be NSFW.
Clik here to view. $X_{n,r}=k$ to happen, exactly Image may be NSFW.
Clik here to view. cells are occupied after throwing Image may be NSFW.
Clik here to view. balls with the first Image may be NSFW.
Clik here to view. k-1 balls landing in Image may be NSFW.
Clik here to view. r-1 cells. In other words, there are exactly Image may be NSFW.
Clik here to view. n-r empty cells after throwing Image may be NSFW.
Clik here to view. balls and there are exactly Image may be NSFW.
Clik here to view. n-r+1 empty cells after throwing Image may be NSFW.
Clik here to view. k-1 balls. The following expresses this condition in terms of the occupancy problem, i.e. similar to (6).

Image may be NSFW.
Clik here to view. $P[X_{n,r}=k]=P[Y_{k,n}=n-r \text{ and } Y_{k-1,n}=n-r+1] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$

Here’s the important components that need to go into Image may be NSFW.
Clik here to view. $P[X_{n,r}=k]$ with the first one coming from the occupancy formula (5).

Image may be NSFW.
Clik here to view. $\displaystyle P[Y_{k-1,n}=n-r+1]=\binom{n}{n-r+1} \sum \limits_{i=0}^{r-1} (-1)^i \binom{r-1}{i} \biggl[ 1-\frac{n-r+1+i}{n} \biggr]^{k-1}$

Image may be NSFW.
Clik here to view. $\displaystyle P[Y_{k,n}=n-r \ \lvert \ Y_{k-1,n}=n-r+1]=\frac{n-r+1}{n}$

Multiply the above two probabilities together produces the desired probability Image may be NSFW.
Clik here to view. $P[X_{n,r}=k]$ for Image may be NSFW.
Clik here to view. $k=r,r+1,r+2,\cdots$ .

Image may be NSFW.
Clik here to view. $\displaystyle P[X_{n,r}=k]=\binom{n-1}{r-1} \sum \limits_{i=0}^{r-1} (-1)^i \binom{r-1}{i} \biggl[ 1-\frac{n-r+1+i}{n} \biggr]^{k-1} \ \ \ \ \ \ \ \ (10)$

Note that when Image may be NSFW.
Clik here to view. n=r (collecting the entire set of coupons), formula (10) would be identical to (7). The following example demonstrates the calculation.

Example 3
Consider the 6-coupon case discussed in Example 1. Suppose that the coupon collector is interested in collecting Image may be NSFW.
Clik here to view. r=4 coupons. What is the expected number of purchases to get 4 coupons? What is the probability that it will take more than 6 purchases to get 4 coupons? What is the probability that it will take more than 8 purchases to get 4 coupons? Compare these results with Example 1.

The random variable of interest is Image may be NSFW.
Clik here to view. $X_{6,4}$ . The mean is:

Image may be NSFW.
Clik here to view. $\displaystyle E[X_{6,4}]=6 \biggl[ \frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3} \biggr]=5.7 \approx 6$

Note that it is much faster to obtain 4 coupons than the entire set of six. The following gives the probability function for Image may be NSFW.
Clik here to view. $X_{6,4}$ .

Image may be NSFW.
Clik here to view. $\displaystyle P[X_{6,4}=k]=\binom{5}{3} \sum \limits_{i=0}^{3} (-1)^i \binom{3}{i} \biggl[ 1-\frac{3+i}{6} \biggr]^{k-1}$

where Image may be NSFW.
Clik here to view. $k=4,5,6,\cdots$

Performing the calculations in Excel gives the following probabilities.

Image may be NSFW.
Clik here to view. $P[X_{6,4} > 6]=1-P[12 \le X_{12} \le 25]=1-0.74845679=0.25154321$

Image may be NSFW.
Clik here to view. $P[X_{6,4} > 8]=1-P[12 \le X_{12} \le 35]=1-0.928712277=0.071287723$

The first probability shows there is still a good chance that it will take more then the mean number of trials to get 4 coupons. The wait time is much less than in Example 1 since the probability of wait time more than 8 is fairly small. Image may be NSFW.
Clik here to view. $\square$

___________________________________________________________________________

Moment Generating Function

One distributional quantity that is easy to obtain is the moment generating function (MGF) for the random variable Image may be NSFW.
Clik here to view. X_n (the case of collecting the entire set of coupons) as well as for the random variable Image may be NSFW.
Clik here to view. $X_{n,r}$ (the partial case). Since both of these random variables are the independent sum of geometric random variables, their MGFs would simply be the product of the individual geometric MGFs. The following gives the results.

Image may be NSFW.
Clik here to view. $\displaystyle M_{X_n}(t)=\prod \limits_{i=1}^n \frac{[n-(i-1)] e^t}{n-(i-1) e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)$

Image may be NSFW.
Clik here to view. $\displaystyle M_{X_{n,r}}(t)=\prod \limits_{i=1}^r \frac{[n-(i-1)] e^t}{n-(i-1) e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)$

___________________________________________________________________________
Image may be NSFW.
Clik here to view. $\copyright \ 2017 \text{ by Dan Ma}$

How long does it take to collect all coupons?

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