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I'm playing with a deck in Magic: The Gathering which can be seen here: http://tappedout.net/mtg-decks/21-05-15-robots/

The deck has sixty cards. I'm trying to find the probability of pulling 5 specific cards in my opening hand of 7 cards in any order. There are 6 copies of one card (either memnite or ornithopter), 4 copies each of the next three (springleaf drum, and ensoul artifact, mox opal), and 17 copies of the last card (any land card).

What is the Percentage Chance of me pulling the cards I need?

In summary: 7 card hand need five specific cards in any order. Number of copies of each card:

-6 copies (Ornithopter/Memnite)

-4 copies (Springleaf Drum)

-4 copies (Ensould Artifact)

-4 copies (Mox Opal)

-17 copies (any land card) - alternatively the land card could be replaced by a second mox opal

Thanks for your help, also if you could explain the thought process and work that'd be great. As I understand it'll probably use factorials and maybe hypergeometric probability but I'm not quite sure how.

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  • $\begingroup$ What exactly constitutes a "good" starting hand? You say 5 specific cards, but you only have four named cards and lands. You want to get at least one ornithopter, at least one springleaf, at least one ensould artifact, at least one mox opal, and at least one land for your starting hand? $\endgroup$ – JMoravitz May 22 '15 at 2:51
  • $\begingroup$ Well there are many other good starting hands but the one I named is among the best. I named 5 cards, the fifth which you didn't notice is any land card (of which there are 17 copies - 4 Darksteel Citadel, 4 Glimmervoid, 4 Inkmoth Nexus, 4 Blinkmoth Nexus, and 1 Island). I also mentioned that the land could be replaced by a second mox opal but I don't care so much if that part is not factored into the probabilitiy, since 2 mox opals are not as good as a mox opal and a land. $\endgroup$ – Benjamin May 22 '15 at 2:53
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Imagine if you will for a moment that for each specifically named card, you have a unique id number on it. I.e. you have an Ornithopter1, an Ornithopter2, an Ornithopter3, etc...

As there are 60 cards total in the deck and your handsize is 7 cards, the number of different starting hands is $\binom{60}{7}$.

Our ideal hand contains at least one of each of our five card categories, abbreviated hereout as $A,B,C,D,E$, which have 6,4,4,4, and 17 copies available respectively. There are then 25 other cards (not of the aforementioned types) in the deck.

We can approach the counting required directly or via inclusion-exclusion. Both methods at first glance will require similar amounts of work, so I will go directly.

$\begin{array}{c|c|c|c|c|l} ~A&B&C&D&E&\text{total hands}\\ \hline 1&1&1&1&1&6\cdot 4\cdot 4\cdot 4\cdot 17\cdot \binom{25}{2}\\ 2&1&1&1&1&\binom{6}{2}\cdot 4\cdot 4\cdot 4\cdot 17\cdot 25\\ 1&2&1&1&1&6\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17\cdot 25\\ 1&1&2&1&1&6\cdot 4\cdot\binom{4}{2}\cdot 4\cdot 17\cdot 25\\ 1&1&1&2&1&6\cdot 4\cdot 4\cdot \binom{4}{2}\cdot 17\cdot 25\\ 1&1&1&1&2&6\cdot 4\cdot 4\cdot 4\cdot \binom{17}{2}\cdot 25\\ \color{red}{2}&\color{red}{2}&\color{red}{1}&\color{red}{1}&\color{red}{1}&\color{red}{\binom{6}{2}\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17}\\ 2&1&2&1&1&\binom{6}{2}\cdot4\cdot\binom{4}{2}\cdot4\cdot17\\ \vdots\\ 3&1&1&1&1&\binom{6}{3}\cdot4\cdot4\cdot4\cdot17\\ \vdots\\ \end{array}$

6*4^3*17*25*12 + 15*4^3*17*25 + 6*6*4^2*17*25*3 + 6*4^3*17*8*25 + 15*6*4^2*17*3 + 6^3*4*17*3 + 6^3*4*17*3 + 6^2*4^2*17*8*3 + 20*4^3*17+6*4^3*17*3 + 6*4^3*17*8*5 = 4479840 + 625600 = 5105440

The probability then is: $$\frac{5105440}{\binom{60}{7}}\approx 0.0132$$

or about one in seventy-five starting hands.

How to calculate one of the rows:

Taking the red row for example, we are in the situation where there are 2 of card A, two of card B, one of card C, one of card D, one of card E, and 0 cards of an unlisted type. By multiplication principle, pick which specific copies of card A you use: $\binom{6}{2}$ number of choices, pick which specific copies of card B you use: $\binom{4}{2}$ number of choices, pick which specific copy of card C you use: $\binom{4}{1}=4$ number of choices, etc... for a total number of $\binom{6}{2}\binom{4}{2}\cdot 4\cdot 4\cdot 17$ different hands in this specific case.


Note: I did not account for the cases where you have no lands and two Mox Opal yet, but those can be rather easily added in if you feel like stretching your muscles a bit. Just add the cases where $D\geq 2$ and $E=0$ and $A,B,C\geq 1$. This also does not take into account mulligan rules where you might be allowed to ditch a starting hand if it was particularly bad.

In particular with $A=B=C=1, D=2$ and $E=0$ it will add another $6\cdot4\cdot4\cdot\binom{4}{2}\cdot\binom{25}{2} = 172800$ additional "acceptable" hands, bumping the probability up a little higher. The other cases will add a rather insignificant amount in comparison.

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  • $\begingroup$ Thanks for the answer! And for laying it out so understandably. The only possible problem I see is that you said "There are then 35 other cards (not of the aforementioned types) in the deck", but there would only be 25, there would be 35 of the aforementioned types and 25 other types. One part that I don't exactly understand are the numbers in the parenthesis such as the 60 over 7 and 6 over 2 $\endgroup$ – Benjamin May 22 '15 at 3:37
  • $\begingroup$ @Benjamin oh, I did add wrong... Silly arithmetic. Replace all occurances of 35 with 25 then., it will unfortunately affect the probabilities by a noticable margin. In the meantime, the notation $\binom{n}{k}$ is the combination "n choose k" and is the same as writing $\frac{n!}{k!(n-k)!}$. The number of ways of selecting $k$ objects out of $n$ total. $\endgroup$ – JMoravitz May 22 '15 at 3:38
  • $\begingroup$ Even so, intuitively, it makes sense for the probabilities to fall somewhere in this range. Essentially, it is like asking for a straight of either A-10 or K-9 in 7-card stud. While possible, wizardofodds.com/games/poker cites them as being under a percent as well. $\endgroup$ – JMoravitz May 22 '15 at 3:42
  • $\begingroup$ @Benjamin corrected the calculations caused by that arithmetic error. It dropped to about one in 75 starting hands now. $\endgroup$ – JMoravitz May 22 '15 at 3:51

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