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The problem:

There are n different kinds of coupons, and we need to get every kind of coupon at least once to win. Each time we get new coupons we get them in a batch of k coupons where k ≤ n (the same coupon can appear multiple times in the batch of k coupons).

Let Xk,n be a random variable that denotes the number of batches we need to buy to collect all the n different coupons.

I already know and understand how to calculate the expected value when k=1 (that is the classic coupon collector problem), but have difficulty calculating E[Xk,n] for k > 1.

If it helps, I've also made some simulations and estimations, and came up with this table: https://i.stack.imgur.com/QsKnJ.jpg

Henry
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David Simka
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  • Note that the formula derived in my answer to the duplicated question reproduces the numbers in your table. – joriki Jun 18 '16 at 00:11

1 Answers1

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Let $k=2$. Let $E(m)$ be the expected number of pairs needed to complete $n$ coupons when you already have $m$ coupons.
The chance of ending with $m+1$ coupons comes from three cases: First new, second old;First old, second new;Both the same new coupon. $$E(m)=1+\frac{(n-m)(n-m-1)}{n^2}E(m+2)+\frac{2m(n-m)+(n-m)}{n^2}E(m+1)+\frac{m^2}{n^2}E(m)$$ Then you build the matrix to get numerical answers; or solve the algebra to get exact answers.
For example, $E(n-1)=1+\frac{(n-1)^2}{n^2}E(n-1)=\frac{n^2}{2n-1}$

Empy2
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  • If $n=4$ then $E(0)=4^2\left(\frac47−\frac6{12}+\frac4{15}−\frac1{16}\right)$, where the numerators come from Pascal's triangle and the denominators are $16−m^2$. I think that gives the general form for $X_{2,n}$ – Empy2 Jan 06 '16 at 03:11
  • Is there a general formula for calculating T(m, i) where m is the number of coupons we already have (out of the n total coupons) and T is the probability of ending up with i coupons after drawing a batch of k coupons. – David Simka Jan 08 '16 at 19:34