@NoTeef: You are very right. For some reason I thought to complicated and I am so used to solve problems like that by simulation.

@premiersoupir: Indeed this episode inspired me to write this article!

By the by, here’s a plug for MTGO Academy contributor Marshall Sutcliffe’s podcast. In an episode from a few months ago, Marshall’s co-host, Brian Wong, does a fine job treating probabilities and the question of splashing cards. How to optimize colored source count for a one card splash, two card splash, etc.? http://lrcast.com/limited-resources-189-mana-bases/

I have one comment about the first graph. I think that graph can also be generated from the hypergeometric distribution. Please correct me if I’m wrong (I’m much better at being wrong than at probability).

Here’s what I was thinking. A player wants 3 lands by turn 3. If on the play, the player gets two draws meaning we can just look at the probability of drawing three lands in a “starting hand” of 9 cards. That means the probability of drawing 3 lands (and thereby playing 3 lands) is ncoosek(L,3)*nchoosek(40-L, S+D-3)/nchoosek(40,S+T), where L is the number of lands in the deck, S is the number of cards in a starting hand (7), and D is the number of draws (2). For L=16, this comes out to be 25.6%. To obtain the final result, we have to multiply this result by the probability that the player draws a nonland card next, which, for these numbers, is 18/31.

To completely generalize with a bunch of confusing letters, the probability of playing exactly X lands in a row (i.e. X lands by turn X) is :

nchoosek(L,K)*nchoosek(C-L,S-1)/nchoosek(C,X+S-1)*(C-L-S+1)/(C-X-S+1)

where L is the number of lands in the deck, C is the number of cards in the deck, and S is the size of the starting hand. This might be able to be further simplified… (with logic or with identities).