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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:

  1. Draw 5 cards from your deck
  2. Decide if you like your hand or put the 5 cards into your deck again and shuffle, then draw 5 again.
  3. After steps 1 or steps 1&2 you add 5 cards from your deck to your life deck.
  4. 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?

Alyosha OL
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    Whether or not returning the cards to the deck & shuffling changes the probability of an event depends entirely on What is the event for which you are computing the probability? – Sycorax Oct 24 '22 at 02:21
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 24 '22 at 03:11
  • Yes, step 2 does obvioudly change probabilities of a 'good' hand. If you have a good at the beginng in half of the cases, you will keep that, only if it is Bad, you get another chance for a good or a bad hand. This will leave you with an above average hand in 50% + 50%*50% = 75%. – Bernhard Oct 24 '22 at 10:36
  • @Bernhard This is basically correct, if the probabilities of Good and Bad hands are exactly 0.5. But there's no reason for the probabilities to be precisely 0.5, and the five cards taken out of the deck could conceivably influence whether a hand is good (for instance, if it removes a key card for a strategy, then a Good hand could be a Bad hand without it). Moreover, if the probability of a Good hand is 0 (or 1), mulligans do not make a difference. – Sycorax Oct 24 '22 at 13:09
  • I see that OP has edited their question. This is an improvement! I'll re-open the question, but future readers and answer-writers should note that the answer depends on the number of copies if the card that are in the deck, and having 1 or more copies is a different event than exactly one copy. – Sycorax Oct 24 '22 at 14:31
  • Are you interested in the probability of exactly 1 or at least 1? – Demetri Pananos Oct 24 '22 at 14:49

1 Answers1

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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).

Sextus Empiricus
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  • Sidenote: I am amazed that you are interested in these statistics. I have myself recently re-started playing MTG but, as a statistics junky, have not yet felt an urge to analyse it statistically (because it is so dreadful complicated). – Sextus Empiricus Oct 24 '22 at 15:31