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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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Ilario
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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 '15 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 '15 at 10:26
  • I guess you should better ask it at mathSE, because this question is about probability and chances – Novarg Feb 26 '15 at 13:04
  • @Novarg also question in link that i've posted is on probability, but no one said that the question is in the wrong section ... – Ilario Feb 26 '15 at 13:07
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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 '15 at 14:16
  • You probably also want to take into account mulligans. But I still think the odds are going to be ridiculously against you. Even getting a specific two card combo isn't all that likely. – bwarner Feb 26 '15 at 14:29
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    wow!!! two down vote in a question about a game! and without comment too..! please smile – Ilario Feb 26 '15 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 '15 at 20:26
  • If I would have to guess, the reason why people downvote (I didn't) may be that the probability of this happening seems so low that it sounds like a brainteaser more than a concrete, directly applicable in-game question of strategic relevance. – xLeitix Feb 27 '15 at 06:50
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    @xLeitix thanks for your comments man. But before doing this question, i didn't know the probability :) so I thought someone here could help me :) – Ilario Feb 27 '15 at 08:48
  • Probability drops to about 0.97% when you need one each of 4, 4, 4, 4, 18. – CodesInChaos Feb 28 '15 at 12:19

1 Answers1

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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

Hao Ye
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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 '15 at 10:18
  • @HaoYe so i've < 1% to draw this start hand?? – Ilario Feb 26 '15 at 13:09
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    @Ilario There is a reason combo decks tend to run a lot of card filtering and tutoring – diego Feb 26 '15 at 13:36
  • @diego you're right i've some cards for tutor my combo, but this question its only when i can do the combo on the firs turn... I thought they were a little bit higher odds .. – Ilario Feb 26 '15 at 13:39
  • To put diego's statement into a question: Does this answer does take into account card filtering, drawing, and tutoring? If not, then I'd say it's not very comprehensive. – Rainbolt Feb 26 '15 at 14:15
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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 '15 at 15:45
  • @diego You're right - the answer technically addresses the question as it was written by the OP. But the answer has almost no practical usage, so I clicked the "This answer is not useful" button. – Rainbolt Feb 26 '15 at 19:49
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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 '15 at 21:43
  • @Ilario, Slightly less than one in a thousand (0.1%), actually. – ikegami Feb 26 '15 at 21:46
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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 '15 at 06:52
  • I think everyone here misunderstood me. The downvote button says "This answer is not useful" when you hover over it. I did not flag anything. @xLeitix Look at the available answer flag reasons and tell me your last comment was not a total lie. Seen lately? How could you have "seen lately" something that doesn't even exist? – Rainbolt Feb 27 '15 at 14:04
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    I wrote a simple simulation which confirms your result. – CodesInChaos Feb 28 '15 at 12:01