To date, there are 11,986 unique cards released for Magic The Gathering. There are certain rules specifying constraints on deck building:

  • A deck must have a minimum of 60 cards.
  • A deck may not have more than four of any particular card.

(Please note that we are ignoring land cards)

I have come up with the following expression. Is it right? If so, what is its order of magnitude (wolframalpha fails to evaluate it).

$$\large\large\sum_{n=60}^{47944} {47944 \choose n}$$

  • 1
    $\begingroup$ Probably you should pay no attention to someone who confesses not to know what a land card is. But the expression cannot be right. It assumes that somehow each of the $11986$ cards occurs in each of four suits. $\endgroup$ – André Nicolas Oct 2 '11 at 22:55
  • $\begingroup$ Your expression is not quite right because it treats the four copies of each card as distinguishable. For instance, you're counting four different decks, each containing a different one of the four Goblin Sharpshooters, but these four should all count as the same deck. $\endgroup$ – joriki Oct 2 '11 at 23:01
  • $\begingroup$ Do you include rare lands? $\endgroup$ – JeremyKun Oct 2 '11 at 23:01

The number of decks with less than $60$ cards is negligible compared to the total number of decks with any number of cards, so the number you're looking for is well approximated by that total number. This is straightforward to calculate, since there are $5$ possibilities for each of the $11,986$ unique cards (you can have $0,1,2,3$ or $4$ of each), so there are $5^{11986}\approx10^{8378}$ different decks.


The expression you have is the number of ways to pick any number from 60 to 47944 cards from a collection of 47944 cards. In particular, we know that $\sum_{i = 1}^n\binom{n}{i} = 2^n$, so your number is "close" to $2^{47944}$, if we agree that in comparison the first 60 terms of that sum are small. This number has 14433 digits. If you want an exact answer, look at the following in Mathematica:

Apply[Plus, Table[Choose[47944,i],{i,0,59}]]

And subtract this from $2^{47944}$.

On the other hand, since the four repeated cards you're picking from are identical, it does not matter how you pick them. So this will not be the right answer. What you want instead is to assign a number 0-4 to each of the 11986 cards (representing inclusion and multiplicity in a deck), and exclude any assignments which give fewer than 60 total cards. This gives a closer upper bound of $5^{11986}$, which has 8378 digits.

  • $\begingroup$ Since $2^{47944}\approx 3.820\times 10^{14432}$ and $\sum_{i=0}^{59}\binom{47944}i\approx 1.015\times 10^{196}$, the subtraction makes very little difference. $\endgroup$ – Brian M. Scott Oct 2 '11 at 23:13
  • $\begingroup$ Right, but he can come up with an exact answer if he wants one. The expression is an upper bound. $\endgroup$ – JeremyKun Oct 2 '11 at 23:36

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