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I'm creating a combo deck, and I'm interested in the probability of having a certain combo on the first turn.

For this combo, I need 4 different cards. There are 4 copies of each of these cards in the deck. The deck has 60 cards.

What is the probability of seeing these 4 cards in the starting hand of 7 cards?

I read How do you calculate the likelihood of drawing certain cards in your opening hand?, but my question is a little different and I don't know how to calculate this probability.

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    My guess is that besides these 4 cards you also need a land or other mana source right? if that is the case you also need to put in the equation the amount of lands you have.
    – Ivo
    Feb 26, 2015 at 10:15
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    @IvoBeckers yeah you're right, i need a land .. so become 5 cards in start hand. i'm playing 18 lands
    – Ilario
    Feb 26, 2015 at 10:26
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    I think that this question is on topic. It's based on mathematics but also has practical usage in Magic. Someone with a solid understanding of magic may be more thorough.
    – Rainbolt
    Feb 26, 2015 at 14:16
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    wow!!! two down vote in a question about a game! and without comment too..! please smile
    – Ilario
    Feb 26, 2015 at 18:21
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    This question is on topic. Just because you don't know how to do the math, doesn't mean I don't.... :)
    – John
    Feb 26, 2015 at 20:26

1 Answer 1

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You can do this calculation using the multivariate hypergeometric distribution. The setup is as follows: The deck of 60 cards consists of: 4 cards of type A, 4 cards of type B, 4 cards of type C, 4 cards of type D, and 44 cards of type E (other).

Your criteria are that a hand of 7 cards contains at least 1 card of type A, at least 1 card of type B, at least 1 card of type C, and at least 1 card of type D.

For a given hand arrangement, you can calculate the probability using the formulas in the link. As an example, the probability of the hand (1 card of type A, 1 card of type B, 1 card of type C, 2 cards of type D, and 2 cards of type E) is:

(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 2) * (44 choose 2) / (60 choose 7) = ~0.000941.

Note that this probability is for this specific hand, and there are many that meet your requirements. You will want to make a table of all realizable hands and sum up the probabilities. (or alternatively a table of all hands that don't meet the criterion, and subtract the sum from 1).

=== table of realizable hands ===

  • 1,1,1,1 of (A, B, C, D), 3 of other

(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 1) * (44 choose 3) / (60 choose 7) = ~0.00879

  • 1,1,1,2 of (A, B, C, D), 2 of other [4 variants]

4 * [(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 2) * (44 choose 2) / (60 choose 7)] = ~0.00376

  • 1,1,2,2 of (A, B, C, D), 1 of other [6 variants]

6 * [(4 choose 1) * (4 choose 1) * (4 choose 2) * (4 choose 2) * (44 choose 1) / (60 choose 7)] = ~0.000394

  • 1,2,2,2 of (A, B, C, D), 0 of other [4 variants]

4 * [(4 choose 1) * (4 choose 2) * (4 choose 2) * (4 choose 2) / (60 choose 7)] = ~0.00000895

  • 1,1,1,3 of (A, B, C, D), 1 of other [4 variants]

4 * [(4 choose 1) * (4 choose 1) * (4 choose 1) * (4 choose 3) * (44 choose 1) / (60 choose 7)] = ~0.000117

  • 1,1,2,3 of (A, B, C, D), 0 of other [12 variants] 12 * [(4 choose 1) * (4 choose 1) * (4 choose 2) * (4 choose 3) / (60 choose 7)] = ~0.0000119

Total Sum of above = ~0.01307

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    60 - 16 = 44, not 46. I changed the numbers, but you may want to give a new result.
    – Toon Krijthe
    Feb 26, 2015 at 10:18
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    @Ilario There is a reason combo decks tend to run a lot of card filtering and tutoring
    – diego
    Feb 26, 2015 at 13:36
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    @Rainbolt The question is about the chances to see 4 specific cards in the first 7 you draw, it doesn't ask about anything else including mulligans, card draw, or tutoring. Therefore there is no reason for the answer to include information about that, if they wanted to include it they could, but I don't feel it necessary (especially for card draw/tutoring since the question doesn't say the deck is running any)
    – diego
    Feb 26, 2015 at 15:45
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    @Rainbolt, I fully disagree. Not only goes it give the answer, it explains how to get it. You act as if the question seeked deck-building help, but that's not the case.
    – ikegami
    Feb 26, 2015 at 21:43
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    @Rainbolt Wow, that's probably the most liberal use of the "not useful" flag I have seen lately on SE.
    – xLeitix
    Feb 27, 2015 at 6:52

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