Probability

by MTGViolet

Earlier, we talked about randomness. For the time being, we’re going to assume that the deck is going to be truly randomized and see how we can use probability theory to help us better our game. I touched on this earlier, but I intend to get a little bit more in depth here now, and talk about something concrete, namely mulliganning.

When should you mulligan? Well, obviously if you have no lands in your opening hand, you should probably mulligan. Likewise, if you have only spells, that is also not great. If you only have spells that are too expensive to play before your opponent runs you over, that would be bad too, but harder to quantify. For now, let’s just focus on the ratio of spells to lands, and worry about mana costs and colors later.
Optimally your first few turns you should make a land drop. For the sake of argument, we’ll say that your first three turns you absolutely want to make your land drop, so by turn three you want to make sure that the probability of drawing three lands is as high as possible. To do this, we build a deck of sixty lands – kidding! We want to have spells to cast as well. So, by turn three, if you’re on the play, you’ll have drawn (disregarding any shenanigans) nine cards.Of those nine cards, you’ll want at least three spells and three lands. The other three can be either spells or lands, both are about equally good at this point. So what is the probability of this happening?

Let’s say the deck consists of L lands and S non-land spells. L+S=60, because we’re not playing Commander (or with Battle of Wits). The chance that, out of 9 cards, you will draw 3 lands, is something that we call an Urn Problem, because it is analogous to drawing a certain number of colored balls out of an urn. In this particular case, we are drawing balls without replacing them, which gives us a distribution known as the hypergeometric distribution. The probability of drawing k(=3) lands out of n(=9) cards from a deck consisting of N(=60) cards, m(=L) lands, is equal to

P(L)=(C(m,k)*C(N-m,n-k))/C(N,n)
=(C(L,3)*C(60-L,9-3))/C(60,9)

where C(n,k) is the polynomial coefficient of n over k, also spoken as “n choose k”. The probability for drawing 3 spells in 9 cards is the same, but substituting S for L.
So, now we have to find out the probability that we draw 3 spells, right? Well yes… And no. You see, when we figure out how likely it is that we draw 3 lands out of 9 cards, we are simultaneously drawing 6 spells. What we actually have to do is figure out how likely it is to draw 6 lands and therefore three spells. And also 4 lands and 5 spells, and 5 lands and 4 spells. We already have the tools to do so, we just have to plug in the numbers and add them together.
What we are left with is…. A very long equation that we can solve for L (though it wouldn’t be easy), but we have e advantage of experience on our sides. Since the average Magic deck will have between 19 and 26 lands, we can just plug in the numbers and see which one gives us the highest probability… But I leave that as an exercise for the reader. 😉

Exercises:

1. Plug in the numbers and find out what the optimum number of lands is in this situation.
2. How does the situation change if you are on the daw? What is the optimal number of lands then?

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