<p>You can do this calculation using the <a href="http://en.wikipedia.org/wiki/Hypergeometric_distribution#Multivariate_hypergeometric_distribution">multivariate hypergeometric distribution</a>. 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).</p>

<p>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.</p>

<p>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:</p>

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

<p>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).</p>

<p>=== table of realizable hands ===</p>

<ul>
<li>1,1,1,1 of (A, B, C, D), 3 of other</li>
</ul>

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

<ul>
<li>1,1,1,2 of (A, B, C, D), 2 of other [4 variants]</li>
</ul>

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

<ul>
<li>1,1,2,2 of (A, B, C, D), 1 of other [6 variants]</li>
</ul>

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

<ul>
<li>1,2,2,2 of (A, B, C, D), 0 of other [4 variants]</li>
</ul>

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

<ul>
<li>1,1,1,3 of (A, B, C, D), 1 of other [4 variants]</li>
</ul>

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

<ul>
<li>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</li>
</ul>

<p>Total Sum of above = ~0.01307</p>

CC BY-SA 3.0
2015-02-26T10:18:07.397Z
60 - 16 = 44, not 46. I changed the numbers, but you may want to give a new result.
Toon Krijthe
CC BY-SA 3.0
2015-02-26T13:09:18.673Z
@HaoYe so i've < 1%  to draw this start hand??
Ilario
CC BY-SA 3.0
2015-02-26T13:36:06.640Z
@Ilario There is a reason combo decks tend to run a lot of card filtering and tutoring
diego
CC BY-SA 3.0
2015-02-26T13:39:19.450Z
@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
CC BY-SA 3.0
2015-02-26T14:15:39.827Z
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
CC BY-SA 3.0
2015-02-26T15:45:49.210Z
@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
CC BY-SA 3.0
2015-02-26T19:49:55.057Z
@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
CC BY-SA 3.0
2015-02-26T21:43:36.943Z
@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
CC BY-SA 3.0
2015-02-26T21:46:33.977Z
@Ilario, Slightly less than one in a thousand (0.1%), actually.
ikegami
CC BY-SA 3.0
2015-02-27T06:52:29.717Z
@Rainbolt Wow, that's probably the most liberal use of the "not useful" flag I have seen lately on SE.
xLeitix
CC BY-SA 3.0
2015-02-27T14:04:47.887Z
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
CC BY-SA 3.0
2015-02-28T12:01:57.690Z
I wrote a simple [simulation](https://gist.github.com/CodesInChaos/9226bad330dd1b80db24) which confirms your result.
CodesInChaos
CC BY-SA 3.0
2015-02-26T10:05:13.470Z
2015-02-27T06:51:30.860Z
2015-02-27T06:51:30.860Z
Hao Ye
<p>I'm creating a combo deck, and I'm interested in the probability of having a certain combo on the first turn.</p>

<p>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.</p>

<p>What is the probability of seeing these 4 cards in the starting hand of 7 cards?</p>

<p>I read <a href="https://boardgames.stackexchange.com/questions/7850/how-do-you-calculate-the-likelihood-of-drawing-certain-cards-in-your-opening-hand">How do you calculate the likelihood of drawing certain cards in your opening hand?</a>, but my question is a little different and I don't know how to calculate this probability.</p>

CC BY-SA 3.0
2015-02-26T10:15:35.570Z
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
CC BY-SA 3.0
2015-02-26T10:26:50.997Z
@IvoBeckers yeah you're right, i need a land .. so become 5 cards in start hand. i'm playing 18 lands
Ilario
CC BY-SA 3.0
2015-02-26T13:04:38.380Z
I guess you should better ask it at mathSE, because this question is about probability and chances
Novarg
CC BY-SA 3.0
2015-02-26T13:07:17.530Z
@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
CC BY-SA 3.0
2015-02-26T14:16:11.810Z
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
CC BY-SA 3.0
2015-02-26T14:29:43.960Z
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
CC BY-SA 3.0
2015-02-26T18:21:23.407Z
wow!!! two down vote in a question about a game! and without comment too..! please smile
Ilario
CC BY-SA 3.0
2015-02-26T20:26:39.997Z
This question is on topic.  Just because you don't know how to do the math, doesn't mean I don't....  :)
John
CC BY-SA 3.0
2015-02-27T06:50:27.787Z
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
CC BY-SA 3.0
2015-02-27T08:48:36.933Z
@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
CC BY-SA 3.0
2015-02-28T12:19:50.393Z
Probability drops to about 0.97% when you need one each of 4, 4, 4, 4, 18.
CodesInChaos
<magic-the-gathering><probability>
What's the probability of having a combo on the first turn in MTG?
Ilario
# What's the probability of having a combo on the first turn in MTG?

(Asked by: username_0 on 2015-02-26)

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?](https://boardgames.stackexchange.com/questions/7850/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.

**Tags:** magic-the-gathering, probability

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1Beckers yeah you're right, i need a land .. so become 5 cards in start hand. i'm playing 18 lands  
> **username_2**: I guess you should better ask it at mathSE, because this question is about probability and chances  
> **username_0**: @username_2 also question in link that i've posted is on probability, but no one said that the question is in the wrong section ...  
> **username_3**: 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.  
> **username_4**: 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.  
> **username_0**: wow!!! two down vote in a question about a game! and without comment too..! please smile  
> **username_5**: This question is on topic.  Just because you don't know how to do the math, doesn't mean I don't....  :)  
> **username_6**: 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.  
> **username_0**: @username_6 thanks for your comments man. But before doing this question, i didn't know the probability :) so I thought someone here could help me :)  
> **username_7**: Probability drops to about 0.97% when you need one each of 4, 4, 4, 4, 18.  

---

## 1 answers

### Answer by username_8 on 2015-02-26. Score: 13 | Accepted answer

You can do this calculation using the [multivariate hypergeometric distribution](http://en.wikipedia.org/wiki/Hypergeometric_distribution#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

> **Comments:**
> **username_9**: 60 - 16 = 44, not 46. I changed the numbers, but you may want to give a new result.  
> **username_0**: @HaoYe so i've < 1%  to draw this start hand??  
> **username_10**: @username_0 There is a reason combo decks tend to run a lot of card filtering and tutoring  
> **username_0**: @username_10 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 ..  
> **username_3**: To put username_10'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.  
> **username_10**: @username_3 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)  
> **username_3**: @username_10 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.  
> **username_11**: @username_3, 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.  
> **username_11**: @username_0, Slightly less than one in a thousand (0.1%), actually.  
> **username_6**: @username_3 Wow, that's probably the most liberal use of the "not useful" flag I have seen lately on SE.  
> **username_3**: 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.** @username_6 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?  
> **username_7**: I wrote a simple [simulation](https://gist.github.com/username_7/9226bad330dd1b80db24) which confirms your result.  

---

