#boringmath

When I first started playing Magic, I thought I could use my knowledge of probability to get an advantage over other players, which didn't work as well as I hoped. But probability can be helpful for some players, or interesting for knowledge's sake, so I'll share my mathematical findings here.

Land Curves

(Forty card decks. The red is less than 50% chance of drawing that many lands or more, green is better than 90% chance.)

Here's a nifty picture with many bright colors! But what in the heck do all those numbers mean?

Here we see the first problem with using calculation: The numbers are very difficult to translate into real strategic decisions. I know I can have a 79.5% chance of drawing between two and four lands in my opening hand, but without something to compare it to, that number is meaningless. How important is drawing the right number of lands anyway? What will other decks be doing?

Underneath the main graph I put some numbers that might help with mulliganning decisions, and that brings up the second problem with calculations: any decision the player can make during the game will throw off the numbers wildly if we can't predict the player's choice. The choice to mulligan is extremely difficult to make optimally in some cases, and if we can't predict whether the player will mulligan, how can we predict what cards will be in his or her hand later?

Magic is designed to be extremely resistant to number-crunching.

I think calculations like this might be useful for pro players trying to precisely tune their decks, or mid level players trying to learn new archetypes, but playtesting several actual games will almost always help you understand more. And for very new players, using general rules of thumb like "Use 17 lands, 2 colors" is much better.

Calculations for drawing a certain kind of land or combination of lands might be more helpful but are much more difficult. I might post some other graphs later. I can help you find numbers for your deck if you are curious enough to ask.

Here are some graphs for other numbers of lands,

More than 17 lands

Less Than 17 lands

We came up with the formula total power = n + (n-1)(2^n - 1). It seems our formula is correct, but I want to prove it's true for every number using induction.

Before we get to the induction part, I want to start the proof with a lemma. A lemma is just like a real theorem, but it's so simple to prove it throws itself off the cliff with no effort. Difficult theorems that would never be proven by themselves can be tricked by the lemmas into jumping off with them.

My lemma is "When n or more Beastmasters attack, the nth trigger will correspond to a body of size 2^(n-1), and will add 2^(n-1) to every other body." This is super obvious and easy to prove. Every other Beastmaster will have a trigger, and every trigger will exactly double the size of every body that hasn't triggered, so that nth body will be doubled n-1 times.

Now on to the main theorem. We need to start with an assumption about the number n, then using that, prove the assumption is true for n+1. Our assumption is that n attacking Beastmasters will have a total power of n + (n-1)(2^n - 1) after all the triggers go off.

Now for the case for n+1, With the help of our lemma, we know that new Beastmaster will be 2(n+1-1) = 2^n, and will give 2^n power to all the other Beastmasters. The total power will be:

n + (n-1)(2^n - 1) (power of the n old Beastmasters)

+ (2^n) (power of the last Beastmaster)

+ (n)(2^n) (power added to the others by the last trigger)

total: n + (n-1)(2^n - 1) + (2^n) + (n)(2^n)

Simplifying it:

n + (n-1)(2^n - 1) + (n+1)(2^n)

n + (n-1)2^n -(n-1) + (n+1)(2^n)

n-n+1 + (n-1+n+1)2^n

1 + 2n*2^n = 1+ n*2^(n+1)

Which looks a bit off... Ah. I can rewrite the original formula as n - (n-1) +(n-1)(2^n) = 1 + (n-1)(2^n). Replace n in our original formula with n+1 and you'll get the expression we just calculated.

Now we know that if we assume our formula is correct for any n, it's true for n+1. So, we just need to knock down the first domino: one Beastmaster will obviously have a total power of just one, and our formula gives 1 + (1-1)(2^1) = 1+ 0*2 = 1, so it works!

This wasn't extremely rigorous, but I think it shows beyond a shadow of a doubt that the formula is correct.

An asker on Matt Tabak's tumblr gave a formula for total power when attacking with multiple Wild Beastmasters, which they just made up and tested for several values. I was a little annoyed-- don't they teach mathematical induction in schools any more? You can create a formula and prove it's correct. This is how I would do it.

First of all, let's nail down the basics. Every time a Wild Beastmaster is declared as an attacker, it triggers an ability that grants to every other attacking creature +X/+X, where X is its power. They start as 1/1s, so you will usually want to buff the Beastmaster; but even without anything extra, attacking with more than one can generate a lot of power.

Now look at a few examples to see the general patterns. Attack with three Beastmasters, A, B, and C and stack the triggers to go off in that order. Their powers will be 1,1,1, then "1",2,2, then 3,"2",4, then 7,6,"4", total power is 17. Each trigger will add 1, then 2, then 4 to two others.

The total power before any triggers will be n, the number of Beastmasters. The first trigger adds (n-1) total power, the next 2(n-1), then the last 4(n-1). Sum((n-1)*2^i) with i from 0 to n-1. This simplifies to (n-1) * Sum(2^i) with i from 0 to n-1 and (n-1)*(2^n - 1). Let's check: for n=3, we get 2*(8-1) = 14. Oops, this is just the added power, not including the base 1 power for each. The real formula should be n+(n-1)(2^n - 1).

Let's check for n=5. First I'll add up each Beastmaster's power step by step.

1,1,1,1,1, then "1",2,2,2,2, then 3,"2",4,4,4, then 7,6,"4",8,8, then 15,14,12,"8",16, then 31,30,28,24,"16", and the final total is 61+28+40=129. Not too shabby for a bunch of 1/1s.

Our formula gives 5+(4)(2^5 - 1)=5+4(31) = 5+124 = 129. Perfect.