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In the coupon collector process, the goal is to assemble a collection of $n$ distinct coupons, while we get a random coupon at each time.

I am looking at a generalization of this problem, where at each time we get a batch of $k$ random coupons (with repetitions) at once, for some $k\in\mathbb N^+$.

Assume that we have already collected $N\le n$ distinct coupons and let $y\in\mathbb N$, what is the probability that we get $y$ new and distinct coupons in the next batch (i.e., we will have $N+y$ distinct coupons after that batch)?

Does this has a close-form formula?

For example, if $n=10, k=3$ and we have so far collected $N=6$ coupons, the probability of collecting another (exactly) $y=1$ coupon is

$$ (4/10)\cdot(7/10)^2 + (6/10)\cdot(4/10)\cdot (7/10) + (6/10)^2\cdot (4/10) = 0.508.$$

Here, I looked at it as if the samples in the batch were ordered 1,2,3, and the summands corresponds to getting a new coupon at the first/second/third sample.

This approach doesn't seem to allow computation over a reasonable number of arguments.

Is there a better way to evaluate it?

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The probability that all coupons you draw are among the $N$ coupons already seen and $y$ particular new coupons is $\left(\frac{N+y}n\right)^k$. Conditional on this, the probability that all of the $y$ particular new coupons are drawn is, by inclusion–exclusion,

$$ \sum_{j=0}^y(-1)^j\binom yj\left(1-\frac j{N+y}\right)^k\;. $$

Thus, the probability to draw exactly $y$ particular new coupons is

$$ \sum_{j=0}^y(-1)^j\binom yj\left(\frac{N+y-j}n\right)^k\;. $$

There are $\binom{N-n}y$ ways to select these $y$ new coupons, so the probability to draw exactly $y$ new coupons is

$$ \binom{N-n}y\sum_{j=0}^y(-1)^j\binom yj\left(\frac{N+y-j}n\right)^k\;. $$

In your example, this is

$$ \binom{10-6}1\sum_{j=0}^1(-1)^j\binom 1j\left(\frac{6+1-j}{10}\right)^3=4\left(\left(\frac7{10}\right)^3-\left(\frac6{10}\right)^3\right)=0.508\;, $$

in agreement with your result.

You may also be interested in Coupon Collector Problem with Batched Selections and Coupon Collector's problem, version with multiple coupons in a box, which are about the same modification of the coupon collector’s problem.

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