This post will be heavy in math (interesting to me, probably not as interesting to you) and not all that heavy in Magic theory (interesting to you). If you're a new player wondering about why certain things are taken as canon in deck construction, this is a pretty good start, though. Also, without math, Magic would be chess with pretty pictures!

When we deal with probabilities in Magic, most often we're talking about things like drawing a land vs. a spell, the probability of drawing 0, 1, or more copies of a card in your opening hand depending on how many copies you're playing, and match win probabilities. The simplest probabilities to deal with are questions like "What is the chance this next card is a land?" These questions are easy to deal with because the result can only go one of two ways. In this case, "Is it a land?," the only answers are "Yes, it is a land" and "No, it isn't a land." These are binary outcomes. Calculating the probability of both outcomes in a binary outcome is pretty easy, since we only have to do the math on one side of the outcome.

Mathematically, probability is the likelihood that a given outcome will occur. It's expressed as a number between 0 and 1, with 0 being the probability of an outcome that cannot occur, and 1 being the probability of an event that is sure to occur. The sum of the probability of all outcomes of an event is 1. So if an event has binary outcomes, if we're able to determine the probability $p$ of one of the outcomes, then the outcome of the other outcome is $(1-p)$.

Independent and Dependent Trials

One of the most important things to consider when dealing with probability of repeated events is whether the results of one trial affects the probability of the following trial. For example, if you roll a fair die, the result of that roll will not affect the next roll. You're just as likely to roll any of the sides of that die no matter what the previous roll was. This is an independent trial. If you pull raffle tickets from a hopper and discard each ticket after it's drawn, then the probability of any ticket being drawn changes. A ticket that hasn't been drawn has its probability of being selected in the next draw slightly increased, and a ticket that has been drawn has its probability reduced to 0. This is a dependent trial.

If we were to talk about casino games, then games like roulette, craps, slots, keno, and single deck card games like Four Card Poker and Let It Ride are independent trials. The result of any given game doesn't affect the probability of outcomes of future games. If a roulette table has 10 straight trials that land on black, the 11th trial is not more or less likely to land on any space than the 10th trial (although that doesn't necessarily mean the roulette wheel is fair). Games like blackjack and baccarat that use multiple decks from a shoe and don't shuffle between each hand are dependent trials. Because the composition of the deck changes between hands, certain outcomes are more or less likely to occur.

The interesting thing is that some trials appear to be dependent when they're not. For example:

I have 5 playing cards face down. One is an ace and the rest are jacks. You are given two options on how to reveal 2 of the 5 cards.

  • You can reveal one card, then if it isn't the ace, you can reveal a second card from the 4 that remain face down. or;
  • You can select 2 cards, and the other 3 cards are removed. You then turn both cards face up at the same time.

With the first method, the probability of revealing the ace with the first card is $\frac{1}{5}$. If you don't hit with the first card, there's 1 ace in 4 cards. Better odds, right? But the actual probability of revealing the ace with the second card depends on the result of the first card. A fifth of the time, there's no chance the second card is an ace (and you wouldn't even look) because you revealed the only ace with the first card. The  probability then is the probability of not revealing the ace with the first card, $\frac{4}{5}$, times the probability of revealing the ace with the second card, $\frac{1}{4}$:

$$\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{5}$$

The second card has the same probability as the first card of being an ace, and this makes sense from a common sense perspective.

Combinations, Permutations, and Probability

What about the second method? Common sense tells us the probability is the same as the first method, but how do we prove that? Since probability is the ratio of outcomes that end in the desired result to the total number of outcomes, we can just figure out how many ways there are to choose 2 cards from 5 and how many of those ways include the ace.

There are two ways to look at the number of ways you can choose $x$ things from a pool of $y$ items. We can either care about the order the items are chosen or not care. If we care about the order (for example, if the first item drawn is the winner of a raffle, and the second item drawn gets a consolation prize, and so forth), then we want to figure out how many permutations there are:

$$_yP_x=\frac{y!}{(y-x)!}$$

$y!$ means $y$ factorial, the result when multiplying all the integers from 1 to $y$. When we don't care about the order the items are chosen, we want to figure out how many combinations there are:

$$_yC_x=\frac{y!}{x!(y-x)!}$$

In our example above, there are 5 objects, and we're selecting 2 of them, so there are:

$$_5C_2=\frac{5!}{2!3!}$$
$$_5C_2=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot3\cdot2\cdot1}$$
$$_5C_2=\frac{5\cdot4}{2}$$
$$_5C_2=10$$

10 ways to select 2 cards from 5. Now, we need to figure out how many ways you can choose an ace in those two cards. There are two subsets of cards amongst the main set, a subset of 1 (the ace) and a subset of 4 (the jacks). Our two cards have to have 1 card from the ace subset and 1 card from the jack subset. $_1C_1$ is easy, there's only 1 way to select 1 card from a set of 1. $_4C_1$ is also easy. There are 4 jacks, and you're selecting one of them, so there are 4 possible outcomes. $\frac{4}{10}=\frac{2}{5}$, which is the same probability as selecting the cards one at a time.

Applying Probability to Magic

Players apply probabilities the most in deck construction. How many lands do I run? How many of each type? How many copies of this card or that card? How much does a 41st/61st card hurt me? In other instances, probability affects decisions in-game. When do I crack that fetchland? How much less likely am I to draw a land if I get a land out of my deck now? Some things to consider:

What is my optimal opening hand composition? If you're playing an aggressive deck, you likely want 2-4 lands in your opener. The more controlling your deck, the less acceptable 2 lands becomes and the more acceptable 5 lands becomes. A mid-range deck might only be able to accept only the most stellar of 2 or 5 land hands, and might only be comfortable with 3-4 land hands. Let's take a look at three common ways you configure your limited deck and see how those configurations mesh up with our desired starting hand compositions. As above, the probability of getting $a$ lands in $b$ cards when your deck starts with $x$ lands and $y$ cards is:

$$\frac{_xC_a\cdot_{(y-x)}C_{(b-a)}}{_yC_b}$$

Doing the math for 16, 17, and 18 lands, we get:

Common sense tells us that having fewer lands skews the curve of expected lands down towards 0 lands. What math shows us is that adding that land or two pushes things towards the center faster than it spreads it out on the top end. As a result, you have more truly unacceptable hands (hands with 0,1,6, or 7 lands) with 16 lands in your limited deck than you do with 17, and less still with 18. So why is common convention to play 17 lands in a limited deck and not 18? How much less acceptable is a 5 land hand than a 2 land hand? Let's take a look at where we're at on turn 5. We're comparing the following deck compositions (all with 33 cards in deck):

  • 15 lands in deck (17 lands, 2 land hand)
  • 12 lands in deck (17 lands, 5 land hand)
  • 16 lands in deck (18 lands, 2 land hand)
  • 13 lands in deck (18 lands, 5 land hand)

On the play, we'll get to see 4 cards; on the draw, we'll see 5. Let's take a look at where we expect to be on the play.

The green sections highlighted are hands that give you 4-6 lands and 5-7 spells on turn 5 on the play. You can see that 17 lands in your deck gives you about a 5% worse chance of having a salvageable hand than 18 lands when you keep a 2 lander on the play. But you get about 5.5% extra chance to salvage a 5 land hand on the play. You can also see that there's a fairly large gulf between the salvageability of a 2 land hand vs. a 5 land hand. How about on the draw?

Once again, the difference in salvageable hands (in this case, 4-7 lands and 5-8 spells) is there between 2 land hands and 5 land hands, but is much more pronounced in the 18 land deck. And the likelihood of salvageable hands is more with 17 lands than 18 (by about 1%). So what we see here is:

  • 2 land hands are more salvageable than 5 lands hands;
  • 17 land decks give you more 2 land hands and less 5 land hands than 18 land decks, but 18 land decks give you more hands in the 3/4 land sweet spot;
  • You salvage more hands overall in the 17 land deck than the 18 land deck.

If you have other math questions, let me know in the comments or tweet me @StillHadThese!