The original idea for this installment was to cover a whole game of Return to Ravnica Limited and cover the odds of drawing cards from your own deck along with predicting your opponent’s behavior, but the number of things that came up just with the opening hand was high enough that I decided to commit a whole piece on just hitting land drops based on your starting seven. This article has a lot of technical details, but most of them can be glanced over, and the results (which are the tables included later on) can still be appreciated. The tools we learn this week can be applied to later weeks and other game situations as well, so this week’s piece is a rather important piece in the puzzle. You can download the spreadsheet here.

A Dash of Basic Probability Theory

Week 1 started with covering the hypergeometric pmf, but I left out some basic probability theory (as it wasn’t relevant at the time) that I’ll need to cover before starting today. Let’s say we have Event A: “The plane I take back to the U.S. crashes;” the probability that the plane crashes is P(A), whereas the probability it doesn’t crash is P(A). Since the plane must either crash or not crash (let’s assume any positive outcome, such as the plane becoming lost in the Bermuda Triangle, counts as it crashing), P(A) = 1 − P(A). Making this more general, the probability of some event is always equal to 1 minus the probability of all other events. Another example, which will become more relevant later, is that P(drawing at least two lands in five cards) = 1 − [P(drawing zero lands in five cards) + P(drawing one land in five cards)].

Opening Hands

From Week 1, we can use the hypergeometric pmf to find the probability of having a certain number of lands in our opening hand. For example, in a deck of forty cards with seventeen lands, the probability of having n lands in our opening seven is:

Which gives us the following table:

Given how many lands are in our opening hand, how can we find out how likely it is to hit our j^-th lands drop on turn j (for example, playing our fourth land on turn four)? The answer is relatively simple once we pose the question in a more precise manner, which is: What is the probability of having drawn at least j lands by turn j given how many lands were in our initial seven-card hand? The derivation of this equation is relatively simple, and I will go through it step-by-step here (using our sample deck above with forty cards and seventeen lands).

First, we’ll define the following variables:

• d = cards left in deck (d would equal 33 after our initial seven-card hand)
• C = number of lands left
• L = number of lands desired
• k = number of cards drawn
• s = number of lands in our seven-card hand

With these variables defined, we can start by stating the very obvious: If s > j, the probability of having at least j lands is 1. With this, we now know that the cases we’re trying to derive an equation for are cases where s < j, which is to say we’re looking for P(draw at least j−s lands in either k = j [on the draw] or k = j − 1 [on the play] draw steps|s). From the basic probability covered at the start of the article, we know this probability is equivalent to 1 − [P(draw zero lands|s) + P(draw one land|s)  . . . + . . . P(draw j − s − 1 lands|s)].

We will start with the simplest case of j − s = 1:

When j − s = 1, we are simply looking for P(draw at least one land in k draw steps|s) = 1 − P(draw zero lands|s). Since drawing at least one lands requires calculating the probability of drawing one, two,  . . . , k lands, it’s simpler if we just find the probability of drawing zero instead (in addition, the formula we are looking to derive here is the solution to P(drawing L cards from a subset of C-over-k draw steps)). Once again, we’ll start with the simplest case of j – s = 1, which is also when k = 1. Let’s say that we’re on the draw and keep an initial seven with zero lands; what is the chance we play a land on our first turn? It’s quite simple. This is simply 1 – P(not drawing a land), where P(not drawing a land) is the number of nonland cards left in the deck (16) divided by the number of cards left in the deck (33), or:

What about if we set k = 3 (keeping j – s = 1)? This is once again fairly simple; it’s 1 minus the probability of the first card not being a land (16 ÷ 33) times the probability of the second card not being a land (15 ÷ 32) times the probability of the third card not being a land (14 ÷ 31):

Factorial notation was introduced to make the statement more general, and it’s left to the reader to work out that the statements are equivalent. We can make this expression even more general (since the equation will no longer depend on a specific s) by noting that 33 is just d, and 16 is d – C, therefore giving us:

With the simplest case worked out, it’s time to consider a more complicated case: j − s = 2, s = 2, and k = 3; the j − s value means the probability we desire is 1 – [P(drawing zero lands|s = 2) + P(drawing one land|s = 2)]. P(drawing zero lands), as you can see by the equation above, does not depend on what j − s is, so we can just use the expression we derived above, whereas the probability of drawing one land is more complicated. We can draw the land as the first (P = 15 ÷ 33), second (P = 15 ÷ 32), or third (P = 15 ÷ 31) card we draw, knowing the other two must be nonland. We therefore have the following scenarios:

Draw land as the first card:

Draw land as the second card:

Draw land as the third card:

Due to multiplication being commutative, we know all three of these probabilities are the same, therefore the order of the cards we draw doesn’t matter. Since we know the order doesn’t matter, using what we learned from Week 1, we can see that this is finding the number of combinations of drawing three cards and having exactly one of them be a land, which is expressed as . We can therefore solve P(drawing one land) as:

Just as we did above, we can now recognize these numbers as the result of operations on L, C, d, and k and can express it in the most general form:

And that is the general probability to draw a desired number, L, from a set of C cards over k draws. (Please take your time understanding where this equation came from, as it’s a rather important result). Using this formula, we create the following table (the probabilities in each cell are P(playing land j on turn j) given you kept a hand with that many lands):

Table 1: On the Play

Table 2: On the Draw

The numbers under each column are P(play land j on turn j) for the deck overall and are not dependent on the starting hand. You can edit the tab labeled “Opening Hand – 1 Color Deck” to see how the numbers look for decks in other configurations.

A Simple Mulligan Rule

Let’s say we want to set up a simple rule of when to mulligan: If P(I play my third land on turn three) < P(On average, I play my third land on turn three with one fewer card), I mulligan. The results look like this (you would mull any hand in for which the result has red text) (any hand with P(Lands in Opener) = 0 can be ignored for obvious reasons):

Results When on the Play

Results When on the Draw

The quick takeaway from this experiment is that you can be much more aggressive when keeping land-light hands on the draw (obviously).

Exercise: Say we have a Constructed deck with twenty-four lands and eight 1-drop accelerators. We want to mulligan any hand where P(having four lands in play by turn four) < P(having four lands in play by turn four with one fewer card in hand) when on the draw.

– Chris Mascioli

@dieplstks on Twitter