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# Math Magic - Thomas Edgar (6/3/2010)

Greetings! My name is Tosus, and I am a student of mathematics at the University of Tübingen in Germany. I've been playing Magic off and on since Ice Age, and one of the things hat keeps drawing me back to the game, despite all the attempts the outside world has made to intrude on my favorite hobby, was how it stimulates the math cells in my brain. Each game is a math puzzle, every hand an equation to be solved, every play a formula. Understanding the underlying mathematical principles can not only give you a deeper understanding of the game, it can also make you a better player.Don't believe me? Let's look at deck building, for starters. Every deck needs a solid mana base, and to help figure out how many lands of what type to include, each player has their own personal formula, even if they don't express it that way. Whether it's, ´´this is an aggressive deck, so I'll play fewer lands,´´ or ´´I want to hit my land drops each turn to help my Landfall abilities go off, so I'll add two more fetch lands to the mix,´´ are examples of players applying different formulae to their process of deck building.

Usually, players figure out their mana base intuitively, without writing down massively long equations and taking two years of probability theory, which works fine for the most part. For better players, this intuition yields the optimal result more often than not, but as a mathematician, that's not enough for me. I want to know what the actual mathematically optimal build is, and have it backed up by equations and functions and proofs. What can I say, math geeks are weird like that.

Math enters into the game in a lot of other places as well, such as deciding when to attack, which cards to fetch with Fact or Fiction

Fact or Fiction | |

Set: Invasion4Cost: BlueColor: InstantType: URarity: 57Number: Terese NielsenArtist: Reveal the top five cards of your library. An opponent separates those cards into two face-up piles. Put one pile into your hand and the other into your graveyard.Text: |

Let's start with a light exercise. Imagine a theoretical variant of magic where each player builds their decks according to the following rules:

You must have one card each with a converted mana cost of 1-6

No card may have an activated ability that costs or produces mana

You can't draw cards other than your one card per turn, nor can you play more than one land per turn (this included tutor and Rampant Growth

Rampant Growth | |

Set: Classic Sixth Edition2Cost: GreenColor: SorceryType: CRarity: 246Number: Tom KyffinArtist: Search your library for a basic land card and put that card into play tapped. Then shuffle your library.Text: |

Basically, we want to draw one card per turn, play a land, and then play a spell with that land, with no funny business that would mess up this mathematical exercise (that includes you, Living Wish

Living Wish | |

Set: Judgment2Cost: GreenColor: SorceryType: RRarity: 124Number: Eric PetersonArtist: Choose a creature or land card you own from outside the game, reveal that card, and put it into your hand. Remove Living Wish from the game.Text: |

Exercise to the reader: what deck would you build in his format, assuming players didn't lose from being decked?

The obvious question is how many lands to include? For the purposes of this article, let's assume the deck is mono-colored, and we are only using basic lands. We would want to be able to cast our 6-cost spell, obviously, while not having too many lands keeping us from drawing our spells (no, we’re not allowing for landfall abilities. Let's keep it simple!). It would seem our optimal land count would be six. Let's see how well we fare with six lands, and see if that really is the optimal number.

One way to define an ´´optimal amount of lands´´ is by saying that we ´´have the highest possible probability of both drawing a spell, and having enough mana to cast it.´´ one of the fundamental sources of tension in Magic is exactly this problem. Before Zendikar, most of the time your card draw would either give you one, but not the other. Say you have three plains on the battlefield, and are facing down a horde of zombies that will kill you next turn. You need the Day of Judgment

Day of Judgment | |

Set: Zendikar4Cost: WhiteColor: SorceryType: RRarity: 9Number: Vincent ProceArtist: Destroy all creatures.Text: |

In line with our definition, let's go ahead and define a ´´game state´´ that expresses this. (For those who are familiar with probability theory, what I'm doing here is defining a subset in the game state space, a massive vector space that I will define in a later article)

Let C_t be defined as, ´´I have drawn exactly one card with converted mana cost of exactly t´´ and L^t as ´´I have drawn exactly t lands´´. I define A_t as meaning, ´´I have drawn both a card of converted mana cost of exactly t and exactly t lands,´´ or ´´both C_t and L^t´´.

You may have noticed I use the word ´´exactly´´ a lot in the definitions. That is because in math-ese, saying ´´I have one apple´´ is the same as saying ´´I have at least one apple´´, so we use ´´exactly´´ to avoid confusion.

To look at the probability of something occurring, we use the letter ´´P´´ for probability. In our case, to define the probability of something happening after drawing n cards, well say ´´P_n´´. So, for example, the probability of drawing a land with seven cards is

P_7(L^1)=?

Quick, what is the answer to the equation above. Go ahead, I'll wait.

Figure it out? He answer is actually quite simple when you realize the trick. If we draw seven cards, even if we drew each and every one of our spells, were still left with one card, which must be a land [Editor’s note: Remember, we have a 12 card deck with 6 spells and 6 lands]. Therefore,

P_7(L^1)=1,

or 100%, if you prefer. No matter what, you will always draw a land in your opening hand! Even better, this follows for the rest of your draws as well:

P_8(L^2)=1,

P_9(L^3)=1,

And so on. In this variant, you will never miss a land drop. How fun!

The deciding factor then on whether you can do anything this turn therefore becomes, what is the probability I will draw gas on this turn, or in symbols, what is

P_7(C_1)=?

For your first turn, for example? (note that I am assuming you are playing first. If you are drawing first, you're even better off, but I will get to that later)

One way of determining the probability is by asking what the probability is that you do NOT draw your one-drop in your opening hand, then taking the difference to one from that number. For example, if you're rolling a six-sided die, and you lose if you roll a six, then the chance of you winning is

1-1/6=5/6.

In our case, you ´´lose´´ if you don't draw your one drop. To figure out the probability, we multiply the probabilities of drawing a card other than your Elite Vanguard

Elite Vanguard | |

Set: Magic 2010WhiteColor: CreatureType: Human SoldierSub Type: URarity: 9Number: Mark TedinArtist: 2Power: 1Toughness: |

11/12*10/11*9/10*8/9*7/8*6/7*5/6 = 5/12

Subtracting that from one gives us

P_7(C_1)=1-5/12=7/12=58.3%

or just over half. So the chance of drawing both your Plains

Plains | |

Set: Oversize CardsLandColor: LandType: XRarity: Tony RobertsArtist: T: Add W to your mana pool.Text: |

Elite Vanguard | |

Set: Magic 2010WhiteColor: CreatureType: Human SoldierSub Type: URarity: 9Number: Mark TedinArtist: 2Power: 1Toughness: |

Moving along, using the same method, you can see that the chances of drawing your spell for that turn is:

P_8(C_2)=2/3 = 66.7%

P_9(C_3)=3/4 = 75%

P_10(C_4)=5/6 = 83.3%

P_11(C_5)=11/12 = 91.7%

P_12(C_6)=1 = 100%

Since by your sixth turn you'll have drawn your whole deck.

If you're on the draw, things look even better for you. You have the same chance of drawing your gas spell, one turn earlier.

Exercise to the reader: determine the probabilities P_t(A_s) for the second players turns.

That's about all for today. I hope I've introduced some basic concepts to you, and gotten you to look at some new concepts. If you have any questions, feel free to contact me at dergeek@gmail.com, or follow me on twitter at Twitter.com/tosus, or comment on this article at the forums. Before I go, let me get you started thinking about the next article:

Exercise to the reader: examine the variant where instead if six spells each player has eight, one of each CMC. How are the probabilities different? What other factors do you have to consider?

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