# Coupon collector problem with partial collection of a specific set of coupons

I am very new in probability and combinatorics and have a naive question around a variation on the coupon collector problem with partial collection.

Lets assume we have a box with 45 coupons labeled 1-45. Now in this case I would like to adjust the CCP such that I can calculate the expected value (amount of draws necessary) to collect 10 specific items. For example item 1-10. How do I adjust my model such that I can calculate the amount of draws necessary to collect each item n times.

I assume that I have to adjust CCP2 in following post (Coupon Collector's Problem with Partial Collections and Coupon Packages) to include the probability that I catch one item is 10/45.

All tips and tricks are welcome!

• – Henry Jan 12 at 23:46

Let $X_n$ be the number of draws it takes to draw the $n^{th}$ desired item from the box after having drawn $X_{n-1}$ ($X_1$ is just the number of draws to draw the first item). For example $X_5$ is the number of draws it takes to draw the fifth item after the fourth item.
Then the number of draws, $X$, necessary to draw all the desired items satisfies, $$X=\sum_{n=1}^{10} X_n.$$ And consequently, $$E[X]=\sum_{n=1}^{10} E[X_n].$$
The probabiltiy of drawing the first desired item on each draw is $p_1=\frac{10}{45}$ (I'm assuming the drawn coupon is replenished which is the coupon collector problem that I'm familiar with). So $X_1 \sim \text{Geometric} (\frac{10}{45})$. Can you find the form of the rest of the $X_n$ to compute the above sum? If you want to collect each of the 10 desired items $k$ times, do you see how you can extend the work provided here?
• Since I'm closing another question as a duplicate of this one, I'll flesh out the hint: The expected number of trials with success probability $\frac{10}{45}$ until the first success is the reciprocal $\frac{45}{10}$. After you've drawn the first desired item, there are $9$ unseen desired items left, so the probability of drawing a new desired item is now $\frac9{45}$; and so on. – joriki May 19 at 20:53