Calculate probability of sampling cards while discarding some We assume there is a fixed amount $n$ of playing cards available. If the whole deck is available, then $n = 52$.
We are given three cards at a time and must choose two and discard one. All three cards can be seen. The discarded card is removed from the deck. This process is repeated $m$ times. So, after $m = 4$ draws, for example, we have a total of $4 \cdot 2 = 8$ cards in our hand.
How can we calculate the probability that we can at least get three hearts (or any other suit) in $m$ draws? We know the distribution of suits in $n$, of course.
My first intuition was to calculate the probability that at least three hearts appear when sampling $m \cdot 3$ out of $n$ cards. The problem I am facing with this is (besides my amateurish knowledge of probability theory): Consider one draw contains three hearts, and all other draws do not. Then, the previously mentioned probability would not be correct, since we cannot obtain three hearts.
 A: If there are $n$ cards, $k$ of them are hearts, and you sample $3m$ cards, three at a time, then you will fail if either


*

*No hearts appear.

*Exactly one heart appear.

*Exactly two hearts appear.

*Exactly three hearts appear, and they all appear in the same sample.


Case 1-3 are calculated using hypergeometric probability, and the probability for a specific one of them to happen is
$$
\frac{\binom{k}{i}\binom{n-k}{3m-i}}{\binom{n}{3m}}
$$
where $i$ is the relevant number of hearts (i.e., $0$ for case $1$, and so on).
As for case $4$, the probability that exactly three hearts will appear is again hypergeometric like above, but this time you need to refine it a bit, since you will only fail if all the hearts that appear do indeed appear in the same sample.
If exactly three hearts appear in the total sample, then the probability that those three hearts appear in the same sample is equal to $\frac{m}{\binom{3m}{3}}$. The reason is that if you were to pick out three places in the sample where the hearts appear, then there are $\binom{3m}{3}$ ways to do that. But since there are $m$ samples, there are only $m$ ways to do it so that the hearts all appear in the same sample.

For a concrete example, let's say you have the whole deck, and you are allowed three samples (so you're sampling $9$ cards). Then the probabilities for cases 1-3 are


*

*$\dfrac{\binom{13}{0}\binom{39}{9}}{\binom{52}{9}}=0.0576$

*$\dfrac{\binom{13}{1}\binom{39}{8}}{\binom{52}{9}}=0.2174$

*$\dfrac{\binom{13}{2}\binom{39}{7}}{\binom{52}{9}}=0.3261$


Again, case four is a bit different. Now, we still have the probability of there being exactly three cards in the sample, that's equal to $$\frac{\binom{13}{3}\binom{39}{6}}{\binom{52}{9}}=0.2536$$
but of these, we need to find the fraction of heart distributions that have all three hearts in the same sample, which is $\frac{3}{\binom{9}{3}} = \frac{3}{84} = 0.0357$. So we need to multiply these two together to find the total probability of case 4, and we get $0.2536\cdot 0.0357 = 0.0091$.
So, the probability of failing (assuming you're not stupid in discarding cards) is
$$
0.0576+0.2174+0.3261 + 0.0091 = 0.6102
$$
which means the probability of success is
$$
1-0.6102 = 0.3899
$$
A: Reformulation of the problem:
Here $4$ hands are filled with cards from a deck of $n$ cards containing
exactly $k$ hearts. $3$ cards are placed in every hand. So $12$ of the $n$ cards are placed in hands.
What is
the probability that these $12$ cards contain at least $3$ hearts, and
this in such a way that more than one hand contains hearts? 
Looking for the probability of the complement of this event we have a look at the following events:
Exactly $0$ hearts are among the $12$ cards. Probability: $$\binom{12}{0}\binom{n-12}{k}\binom{n}{k}^{-1}=\binom{n-12}{k}\binom{n}{k}^{-1}$$
Exactly $1$ heart is among the $12$ cards. Probability: $$\binom{12}{1}\binom{n-12}{k-1}\binom{n}{k}^{-1}=12\times\binom{n-12}{k-1}\binom{n}{k}^{-1}$$
Exactly $2$ hearts are among the $12$ cards. Probability: $$\binom{12}{2}\binom{n-12}{k-2}\binom{n}{k}^{-1}=66\times\binom{n-12}{k-2}\binom{n}{k}^{-1}$$
Exactly $3$ hearts are among the $12$ cards and they are all in
the same hand. Probability: $$4\times\binom{3}{3}\binom{3}{0}\binom{3}{0}\binom{3}{0}\binom{n-12}{k-3}\binom{n}{k}^{-1}=4\times\binom{n-12}{k-3}\binom{n}{k}^{-1}$$
Factor $4$ is there because for the $3$ hearts there are $4$ hands available.
The final anwer is $1$ minus the summation of these probabilities.
