Probability when drawing and only replacing non-successes

Question

This is a slight variant on a normal probability of getting k successes in n draws type question that I can't seem to find guidance for.

Suppose we have a standard 52-card deck and we want to calculate the probability of drawing k number of Hearts in x amount of rounds. Each round, we draw 5 cards at a time, and keep the successes (in this case any Heart), putting the non-successes back in and shuffling the deck before drawing another 5. How do we calculate say, the probability of drawing 3 Hearts in 5 rounds of this?

I know to use a hypergeometric distribution for thinking about drawing without replacement regardless of success, but I was wondering if there's a distribution (and corresponding pmf for calculating probabilities) for this type of case. I can answer this kind of question fine with simulation, but I'm very curious to know how to approach this mathematically. I can somewhat envision a tree with "compounding" hypergeometric distributions, but I'm having trouble generalizing/formalizing it.

Answer 1

Say we draw $h_1$ hearts the first time, $h_2$ cards the second time, and $h_3$ cards the third time. The probability of this is $$ {{13\choose h_1}{39\choose 5-h_1}\over{52\choose5}} {{13-h_1\choose h_2}{39\choose 5-h_2}\over{52-h_1\choose5}} {{13-h_1-h_2\choose h_3}{39\choose 5-h_3}\over{52-h_1-h_2\choose5}} $$ For a particular number $h$ of hearts, you have to sum the foregoing over all $h_1+h_2+h_3=h.$

It doesn't look like there's much possibility of simplification to me.

EDIT

Computation gives these probabilities:

 0 0.010872236580
 1 0.067246432045
 2 0.177867796098
 3 0.265039719299
 4 0.246774086435
 5 0.150563606796
 6 0.061421735291
 7 0.016794089546
 8 0.003039824908
 9 0.000354147345
10 0.000025286774
11 0.000001018818
12 0.000000019931
13 0.000000000134

Answer 2

Above answer is the first thing that comes to mind, however, since there are 5 rounds, we should use $h_1 + h_2 + h_3 + h_4 + h_5 = 3$, this gives total 35 combinations of getting 3 hearts in 5 rounds, and each case's probability can be calculated similarly to what above answer suggests.

Another approach would be to make a tree chart of different possibilities and keep summing and multiplying the probabilities. First-round can result in 0, 1, 2, 3, 4 5 hearts and each with a probability of:

$$\frac{13C_i * 39C_{(5-i)}}{52C_5}$$ for i in [0,5] where last three cases are favourable.

For the first three cases, you go to the next round and calculate the winning probabilities. Repeat this process till 5th round.

Sorry for the bad formatting, I'm not sure how to use the latex here.

Edit: corrected formatting. 'Above Answer' refers to the answer by 'saulspatz' below.