Does mulligan matter? I'm a TCG player, and in the game in question you work with a 50 card deck and with the limit of 4 copies of a single card, the game proceeds as follows:

*

*Draw 5 cards from your deck

*Decide if you like your hand or put the 5 cards into your deck again and shuffle, then draw 5 again.

*After steps 1 or steps 1&2 you add 5 cards from your deck to your life deck.

*With 40 cards left in your deck you draw 1 card each turn.
So my questions are?

How can I calculate the probability of 1 copy of an specific card being in the first 5 cards of my 40 card deck if I decide to mulligan my hand first?
 A: Without the Mulligan rule
The probability without the Mulligan rule is hypergeometric distributed.
The distribution is not easy to comprehend, but the mean distribution of the number of cards is intuitively $5 \frac{4}{50} = 4/10$ if you draw 5 cards from a deck of 50 containing 4 of the cards that you are looking for.
Roughly speaking you will also get the same probability of having 1 card as that mean number of cards. But there is slight some discrepancy:
For a 50 card deck drawing 5 cards with k of the cards inside of it is the following:
                cards inside
                k = 1    k = 2    k = 3    k = 
 0 cards        0.9      0.808    0.724    0.647  
 1 card drawn   0.1      0.184    0.253    0.308
 2 cards drawn  0.0      0.008    0.023    0.043
 3 cards drawn  0.0      0.000    0.001    0.002
 4 cards drawn  0.0      0.000    0.000    0.000

So the more cards that you put into the deck, the more likely they end up into your first hand, but it is not a linear relationship. With more cards you do not get a proportional increase of the probability of the card ending up in your deck.
With the Mulligan rule
If you are allowed to redraw then you will get the following probability:
Let $p_{single}$ be the probability to draw the card without Mulligan and let the decision to Mulligan fully depend on whether or not you get the card (for a game this is not realistic and you should consider multiple measures instead of only the presence of a single card) then the probability with Mulligan is
$$p_{Mulligan} = p_{single} + (1-p_{single}) p_{single} = 2 p_{single} - p_{single}^2$$
If $p_{single}$ is small (close to zero) then the Mulligan option will almost double the probability of drawing the card.
If you really want a specific card in your first hand then with $k=4$ the probability to draw at least a copy is $1-0.647 = 0.353$ and the probability to draw at least a copy with Mulligan is $0.581$. It is not exactly double but probably enough close for a rule of fist. Just remember that the closer to $1$ you want to go, the more difficult it gets (and you need to adapt for that).
