Tl;DR: You have winning cards. To win, you must be able to play those cards, and have them in your hand. Your hand is randomly drawn. When might you win? How could find the answer to this (very complex) problem?

(I realize that my question isn't very formal, but I'm not exactly sure how to correctly pose the question. Feel free to edit. Also, draw3cards or Boards & Games don't like math questions)

The decklist

(If you know magic, you should skip the next section, and probably the one after that, too)

Basic Rules

  • You have a 60-card deck, which has just been shuffled.
  • At the beginning of the game, you draw 7 cards. From then on, each turn you draw one card.
  • All cards except lands require mana to play. Mana is drawn from your mana pool. A land card on the field can be tapped (used) to add a mana of a color to your mana pool. (Mana either has a color, or is uncolored. Islands for blue mana, Swamps for black).
  • Colored mana can always be turned into uncolored mana.
  • Each turn, you may put no more than one land from your hand onto the field. It can immediately be tapped for mana.

Combo explanation

  • You have your Myr. As the decklist shows, you have Silver and Leaden myr, which can be tapped for Blue or Black mana, respectively. They cost two mana each, of any color.
  • In the same vein, Alloy Myr grants one mana of any color, and Palladium Myr grants two uncolored mana
  • You also have the Myr Galvanizer. For one mana and being tapped, it untaps all other Myr.
  • If you have one Galvanizer, and your mana-myr can be tapped for more than one, you can get an extra mana
  • If you have two Galvanizers, and your mana-myr can be tapped for more than one, you can have infinite mana
  • If you have infinite mana, you can win with Exsanguinate (which requires two black mana - so if you have two Galvanizers and a Palladium Myr, but only one swamp, you can't win with Exsanguinate)
  • ...or you could win with Blue Sun's Zenith, which requires 3 blue mana. It can also be useful without infinite mana, since it allows you to draw cards (one card for every mana of any color you pay beyond 3 blue)
  • You also have three shapeshifters - Cackling Image (one mana of any color, two blue mana), Cryptoplasm (same), and Evil Twin (two uncolored, one black, one blue). All of these can make a copy of a creature on the field
  • You also have Diabolic Tutor, which gives you any card you want for two black mana, two uncolored mana
  • The other cards are irrelevant.


Given that you try to get mana-myr on the field first, what chances are there of winning on what turn? I realize that this is a very tricky question that probably involves quite a bit of work, so an explication of how to get an answer is almost as good an answer itself.

Also really useful would be a generalization of this.

For extra points, ideas about finding the optimal number of each card (keeping the irrelevant defensive cards, and no more than 70 cards) would be very appreciated.

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    $\begingroup$ Is there a way that you can generalize your question (say, "I have 60 cards with 7 X's, 10 Y's, 30 Z's, and the rest irrelevant. What are the odds that...?")? This would help you get a better understanding of any solutions given, and would probably help those who would answer. $\endgroup$ May 19, 2012 at 1:52
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    $\begingroup$ The best way to answer such a question is probably through simulation. Write a computer program that implements such a strategy, run it with a large number of randomly shuffled decks, and count what fraction result in a win. It will probably be quite difficult to produce an exact value for the probability, and even more difficult to get one in closed form. $\endgroup$ May 19, 2012 at 2:04
  • $\begingroup$ @TheChaz: Part of my question (as I state in the parentheses) is how to generalize this. I guess I'll edit it to make this more obvious, however. Several people tell me that a big part of answering questions is posing them well, and that's part of what I'd like help with. $\endgroup$ May 19, 2012 at 2:26
  • $\begingroup$ @NateEldredge: I was really hoping for a strict definition, because that would be faster to calculate, and far more interesting, then a plain simulation. However, the idea of at least getting an approximate method would be very nice. If I don't get that, I'll write a simulator. $\endgroup$ May 19, 2012 at 2:28
  • $\begingroup$ I think this question is off-topic here, as it deals too much with the particulars of M:TG. $\endgroup$ May 19, 2012 at 2:46

2 Answers 2


This question is almost certainly impossible to answer precisely, because Magic the Gathering can simulate an arbitrary Turing machine. In particular, there is a small but non-zero probability that randomly chosen hands will force the players into an infinite loop that cannot be predicted even with complete information.

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    $\begingroup$ A turing machine couldn't happen, since I don't have the mentioned cards (and my opponent is assumed to do nothing but discard cards). How could the game reach an infinite loop? $\endgroup$ May 19, 2012 at 23:24

I'm not sure this question is possible to answer at this time. Similarly to the fact that the number of winnable solitaire games is not known, there are some problems with this question. Maybe I can explain a little some difficulties with an answer.

1) Since you're going to win the game using one of several specific combinations, it is perhaps not so difficult to determine the probability of drawing the cards needed for any one given combination in a set number of turns. By the time the last card is drawn, you have completed the combination. However, I am not sure you can write down a way to express that as you are working on combination (a), you pull enough cards that it now becomes favorable to work on combination (b) instead.

2) M:TG is not as straight forward as "draw one card per turn until you have sufficient cards to win." The order you play these cards will matter, and since you only get 20 life, you have to keep yourself afloat with whatever defenses you have available until then. Although I'm not familiar with these cards in particular, I highly doubt any of them reads "give up x number of y mana, you win the game." So, in short, you have to go through n turns to gather the cards, and, assuming you can play them as soon as you get them without losing them, defend them until the combination wears down your opponent. This brings me to the third issue.

3) Your opponent exists. I know that sounds snarky, but you have only two options as for how to go about playing your Exodia combination.

a) Play nothing but lands and mana giving cards until you can lay down your whole combo at once. This is obviously bad - you'll just die before you get them out, or have low chances of surviving.


b) Play the cards as you go (or in some collection of short bursts, i.e. maybe two at a time). In this case though, you also have to consider that there are now probabilties of drawing/having any defensive cards needed to stop your opponent from killing your combo before you get it out. In this case, we would have to know the content of the other person's deck to calculate the probability in this case.

I guess what I'm getting at is, even though it is obvious from elementary probability that such a value exists, it has too many factors to be computed by hand. You would be best off running a simulation of all possible ways the cards can be drawn in both yours and your foe's deck, and estimate the probabilities this way instead. This is how we approximate the number of games of Solitaire that can be won, and that game is much simpler, with only 52 cards instead of two 60 card decks.

Hope that helps.

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    $\begingroup$ This assuming that your opponnent does nothing. $\endgroup$ Feb 23, 2014 at 1:31
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    $\begingroup$ I am OP. The question was assuming that the other player does nothing. $\endgroup$ Feb 23, 2014 at 13:57
  • $\begingroup$ You are correct, and I am silly. $\endgroup$ Feb 23, 2014 at 17:01
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    $\begingroup$ If you have to assume your opponent does nothing, then it's not a very good deck :P. $\endgroup$
    – mjqxxxx
    Feb 25, 2014 at 14:55

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