A Magic: the Gathering deck is seldom larger than 60 cards, around 40% of which are lands.

I have cards and effects that "mill" my opponent, meaning it takes cards from the top of their deck and places them in the discard pile. Some of these effects mill a certain number of cards, and others mill until they mill a land card (or multiple land cards), then they stop. At times, both of those effects happen "at once", and I get to decide the order in which they happen. I have other cards that get better the more cards are in their discard pile.

My question is this: given that I can see how many lands are on the field (and thus deduce approximately how many lands are left), is there a way for me to determine which has the larger benefit: either milling a certain number or milling until they reach a (certain number of) land(s)? Does this depend on the number of cards being milled or on the number of lands to be reached before milling stops, or both, or is there even any way to determine?

  • $\begingroup$ Do you know the number of remaining cards in your opponent's deck? $\endgroup$ – BallBoy Jan 25 '18 at 16:43
  • $\begingroup$ Yes, either by estimating or counting. For estimating, at any given point it's ~60 minus the number of lands in play (usually one per turn) minus the number of cards in play minus the number of cards they have in their hand. $\endgroup$ – John Doe Jan 25 '18 at 16:46

One way to approach this is to ask what the expected number of cards milled will be of each option. If you have a card that mills $m$ cards, regardless of type, then the expected number is $m$. The mill-until-land card is a little trickier, but there is some good discussion of an equivalent problem here. The upshot is as follows: if your opponent has $\ell$ land cards left among $n$ total cards left, the expected number of cards through the $k$th land card (including that card) is $k\left(\frac{n-\ell}{\ell+1}+1\right) = k\left(\frac{n+1}{\ell+1}\right)$. The quick argument for this is that the "average" deck has the $\ell$ land cards equally spaced, and the remaining $n-\ell$ cards in $\ell+1$ evenly-sized piles before, after, and between the land cards.

  • $\begingroup$ Thanks! I realize I'm shooting for the moon, but can there be any accounting for randomly spaced land cards left in the deck? $\endgroup$ – John Doe Jan 25 '18 at 16:49
  • $\begingroup$ @JohnDoe I'm not sure I follow exactly what you mean. $\endgroup$ – BallBoy Jan 25 '18 at 16:50
  • $\begingroup$ You said "l land cards evenly spaced". The decks are shuffled before the game; I was just wondering if there was some way to account for the random spacing, as opposed to the equal spacing, in the equation. $\endgroup$ – John Doe Jan 25 '18 at 16:51
  • $\begingroup$ @JohnDoe If we don't know anything about the random spacing, the even spacing is the average possibility -- all the land cards could be on top, or all the land cards could be on bottom, or anything in between, but on average, they'll be evenly spaced. (That's, informally, the mathematical concept of expected value -- if you played a game where you flip a coin and win \$4 for heads but lose \$2 for tails, we can say that on average you win \$1, even though this never actually happens.) $\endgroup$ – BallBoy Jan 25 '18 at 16:54
  • $\begingroup$ @JohnDoe This is valid assuming the shuffling orders the cards completely randomly, so any ordering is equally as likely as any other ordering. If we have more information about how the rest of the deck might be ordered, it might be possible to account for that as well. $\endgroup$ – BallBoy Jan 25 '18 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.