Because math isn't always boring

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### Mulligans, part 2

Now, finally, for some practical application!

We spent the last several weeks learning how to calculate things like how likely a given mix of cards in your hand is and developing a metric to decide how long we have to wait until we draw a given number of lands and so on. Now to put that all into play.

The nice thing about this sort of math is, you can figure it out all in advance. In fact, I’m thinking of, after my move (I’m moving, by the way), writing a small app that will let you type in the stats of your deck and then automatically make sort of a spreadsheet that helps you make mulliganing decision.

The crux of the matter is, as you might have already suspected, comparing the expected time to draw a given number of lands (we’ll say four, for the sale of argument) based on the cards you already have in hand to the more vague number you would get if you throw your cards back and draw six more. If the number for the six-card-hand is lower, you mulligan. If it’s higher, you throw it back.

Which works wonderfully… Until you sit on a seven-land opening hand and are pleased as punch that the number can’t get any lower.

…okay, not really. We can’t just look at the number of lands, we have to look at the number of non-lands as well. So, we’ll perform the same calculations, but with non-lands instead of lands. Just switch out the numbers, and compare them again. This can all be done beforehand and easily memorized before you sit at the table. You might even want to modify these a bit. Say, you want to include other mana sources in with the lands, that’s fine. Say there are some non-land cards you’d per not to have in your opening hand, just subtract them from the second set. At the end of the day, your gut is going to play into this a lot, however once you have the charts, and the flowchart of mulliganing decisions, you can do yourself one big favor.

Stick to them. Don’t go by your gut, stay with the raw numbers. Why do this, do you ask? Is is analysis so bulletproof that you can’t do any better? Well, no, not by a long shot, but what I’ve seen time after time, and I’m even guilty of this myself, is players will second guess themselves. I’ll go, “oh, well, I only have one land, but two of my spells cost two, so I just need one more to go off…” and the use myself for being mana screwed for the whole game. No, once you get used to playing by the quote unquote rules, then you can start modifying it for those corner cases where the rules may say one thing, but the cards themselves give you slightly better odds.

Of course, this only covers mono-colored decks. When dealing wi multiple colors, things get a bit mo complex, but not really more complicated. The same principles apply, just evaluating them takes a bit more work. Again, a simple program could figure these out and even help you with your deck building. But that, as they say, is a topic for another day.

Exercises

• Take your favorite deck and draw up a mulligan choice flowchart for it
• Write the author of your favorite math-related magic blog and bug him about writing the app he mentioned.

### Mullligans, part 1

Hi Mathmagicians!

First off, some admin stuff. I’ve decided that, due to things like school starting up for me soon, I’m going to take this blog from weekly to biweekly. Which is honestly about as much magic as you might want anyway. On the off weeks I might post interesting links I find, but the articles will be posted once every other week from now on.

So, on to this week’s topic. When you should – and shouldn’t – mulligan.

Last week we learned about expectancies, and that’s the tool we’ll use for deciding when to throw back your hand or not. Now, if you just look at raw numbers, it seems clear that, if you have fewer cards in your starting hand, it’s going to take longer to draw your four or five lands that you want to make your chances of winning as high as possible. But doing so neglects an important aspect of making your mulligan decision, namely that you actually do get to see what’s in your opening hand. (If you didn’t, your mulligan decision would be truly easy – you just plain never would)

Let’s say you draw your opening hand and you see you have one land. Using the math from last week, the expected turn for you to draw your fourth land is:

E=1*Probability of drawing your fourth land on turn 1 + 2*Probability of drawing your fourth land on your second turn +…

What we are looking at here is a conditional expectation, (I keep saying expectancy, but the usual term is expectation. I learned all these terms in German, so sometimes things get lost in translation, sorry) which means that, just as it sounds, were taking the condition that we have a one-land opening hand into account when calculating our expectation. The formula for that is:

`E[X|Y=y]=∑x*P(X=x|Y=y)`

The | in the formula can be read as “under the condition that”, as in “the probability that you draw your fourth land on turn four under the condition that your opening hand has one land”. To calculate this, you take the probability that x and  y are both true, and divide that by how likely y is. In other words, you know that y is pretty unlikely to happen, so you eliminate that from the equation to see how unlikely x is to happen assuming y already has.

Got it? Don’t worry, it makes more sense when you actually do the math. Or at least I think so.

So, in our example, the probability of drawing one land in your opening hand is about 10.5%. The probability of drawing your fourth land on turn 4 after starting out with only one land is the same as saying each card draw is a land (by turn four, on the play, you draw exactly three cards) is 24/53*23/52*22/51, or about 8.6%, which is exactly the conditional probability P(X=4|Y=1). If we wanted to know the probability of drawing one land in our opening hand, followed by three more lands in our first three draws, we would multiply the two probabilities above to get 0.9%. That’s where the formula comes from.

To continue our reasoning above, it’s clear that the first three terms in our sum are zero (it’s impossible to draw three lands with two or fewer draws), so we start at turn four, which we calculated above. For turn five, if we want exactly four lands, we can either go Land Land Land Spell, or Land Land Spell Land, or Land Spell Land Land, or Spell Land Land Land, where the probabilities of all those events are the same. If this seems familiar at all, it should: It’s exactly the same math we used to determine the probability of  drawing a given number of lands in your opening hand, just adjusted for different variables. In this case, we have:

Number of cards in deck: 53 (we already drew our opening hand)
Number of Lands in deck: 24 (one is in our opening hand)
Number of cards drawn: Turn number – 1
Number of successes wanted: 3

Plugging those numbers into the formula gives us a probability function of

`C(24,3)*C(29,t-4)/C(53,t-1)`

and an expectation of

`E[X|Y=1]=∑_(t=4..33)t*C(24,3)*C(29,t-4)/C(53,t-1)≈18.28`

Yes… it’s that really high number again, but it’s okay, for the simple reason I stated above: The actual scale of that number isn’t as important as the fact that we HAVE a number.

Figuring out the conditional expectation of the various opening hands available is now a simple matter of plugging in a few different variables into a simple equation and pasting that into WolframAlpha. You can do that beforehand and write the numbers down.

The other nice thing about these numbers is we can use them in comparison to what might happen when you mulligan… but more on that next week… er, next fortnight.

Exercises:

• Take your favorite deck. Calculate the conditional expectations for the different opening hands.
• How can you calculate the analogous situation for spells versus lands.

### Delay

Hello mathmagicians,

This week’s post will be a bit late. I wasn’t feeling well today, but hopefully it won’t be too delayed.

See you soon!

### Probability, part 3

So, now we’ve defined all these wonderfully confusing terms. How do they help?

Potentially quite a bit, but let’s start by getting some cold, hard numbers. We’ll start with our original 35/25 deck, and figure out a few expectancies. We’ll start with the expected turn by which we’ll have drawn 4 lands.

The probability that we’ll have drawn exactly four lands on turn one, i.e. after having drawn seven cards, is, as we saw last week,

`L[4](1) = (C(m,k)*C(N-m,n-k))/C(N,n)=(C(25,4)*C(35,3)/C(60,7)≈0.214`

So the first term of the sum is 1*0.214. For the second term, we draw one more card, and therefore one more spell, for a probability of

`L[4](2) = (C(m,k)*C(N-m,n-k))/C(N,n)=(C(25,4)*C(35,4)/C(60,8)≈0.258`

and a term of 2*0.258 = 0.516. Adding these two terms together gets us 0.214+0.516=0.730. We go on creating these terms, with the probabilities of

`L[4](n) = (C(25,4)*C(35,n+2)/C(60,n+6)`

Which we can multiply by n and then add all up together. We do this for all values of n… okay, no, not really. Once we’ve drawn our 39th card, even if we started out by drawing all 35 non-lands, the next four cards will have to be lands, so any further draws will put us over the exact number of lands we are looking for, namely four. That means we can stop adding up terms after n=33, which happens to coincide with the last number for which C(35,n+2) is defined. Using our handy-dandy calculator (WolframAlpha is your friend!), we arrive at an expectancy of

`E(L[4]) = Σ n * (C(25,4)*C(35,n+2)/C(60,n+6) ≈ 10.647.`

In other words, we expect to have drawn exactly 4 lands on average between turns 10 and 11. If that number seems high, well, it is, and there’s a few reasons for that, the biggest being that we only have used positive numbers for n. If we accept negative values of n (in other words, we take into account the cards drawn before the start of the game), we arrive at the value 3.894 – a much more reasonable (and realistic) value. So why use the first value at all?

Well, firstly it’s just a mathematical construct that will help us in making mulligan decisions, not an actual representation of  what turn we’ll draw four lands by, and secondly shut up! I goofed, all mathematicians are entitled to a false start or two!

Ahem, anyway, there actually is a good reason for using the first expectancy, and that is because the expectancy I calculated actually doesn’t assume we know anything about our opening hand. What we need to do is look at how the expectancy changes based on the contents of our opening hand. To do this, we have to use something called the conditional expectation.

Let’s say the opening hand has three lands in it already. So what is the probability then that we draw our fourth land in the second turn? Well, the library still contains 22 lands and 31 spells, so the probability is a straight 22/53 ≈ 0.423. The probability that we have four lands in the next turn is the probability that you draw first a land and then a spell plus the probability you draw a spell and then a land, or

`22/53*31/52+31/53*22/52=2*(22*31)/(53*52) ≈ 0.495`

And if this is starting to look familiar to you, then good, you might just be a mathematician after all. You see, once we take out the three lands and four spells out of the library, all we’re basically doing now is drawing balls out of an urn again, just this time it’s an urn that has 22 white and 31 black balls in it. Once again, we can calculate an expectancy of

`E(L'[4])= Σ n * (C(22,1)*C(31,n-1)/C(53,n) ≈ 8.413.`

Again, remember, this is just a mathematical construct, it doesn’t mean you have to wait 8 turns to draw another land. But what IS interesting is when you compare that to the expectancy if you start with 2 lands and 5 spells.

`E(L''[4])= Σ n * (C(23,2)*C(30,n-1)/C(53,n) ≈ 84.15`

Which is, in case you didn’t notice, a considerably larger number. In other words, if you have an opening hand with two lands in it, on average you’ll have to wait quite a bit longer to draw up to four lands than you will if you start with three lands.

Shocking, I know! But we’ll get into how this can actually be of decision-making help to us next week.

Exercises

1. What is the expectancy, without looking at your opening hand, of drawing 3 lands? Of five?
2. What are the expectancies of drawing four lands for the other numbers of lands in your opening hand?

### Probability, part 2

Okay, so all this theory is fun, but how can it make you a better Magic player? Sure, the last post helps with deckbuilding – a little – but most of those numbers have been really well explored by now, and vary more based on other factors than math.

So let’s try and tackle something a little more concrete: The decision on when, and when not, to mulligan.

I suspect this topic might span more than one post, so bear with me.

Whenever you’re tackling a problem like this, it’s good to define your terms. We all know what a mulligan is, it’s discarding your starting hand based on the information you gain from looking at it in the hopes of a better one. What does “better than ” mean? Well, it can mean any number of things, depending on your deck. It could mean you’re likely to draw your combo pieces, or maybe you want to maximize copies of a certain card. But in general, I think that a successful opening hand means that, by the time the game gets underway, you’ll both draw enough lands to cast your spells, and enough actual spells to cast. Again, for each deck this is going to be different; fast decks are going to want more spells, more control-ish or combo decks might need more consistent land drops.

I’m going to introduce a bit of a new concept here, namely the random variable. Which doesn’t mean that you pick a letter at random instead of x, it’s actually a very specific kind of function that has to do with assigning a certain event to a number. For example, the number of lands in your opening hand is a random variable, call it X. You can represent the probability of certain numbers of lands in your opening hand by P(X=3), for example.

A good metric to measure the quality of an opening hand is how long it takes to be reasonably sure of drawing a certain number of lands, and the same for spells. For a baseline, let’s look at a deck with 35 spells and 25 lands. We’ll define our random variable L[n] as the number of turns it takes to draw n lands and, along the same lines, S[n] as the number of turns it takes to draw n non-land spells. Remember that random variables are a number that are associated with a certain probability, in the sense that we are interested in things like what P(L[5]=3)  is.

Actually, let’s go ahead and calculate that. Using our formula from last week, we have

P(5)=(C(m,k)*C(N-m,n-k))/C(N,n)=(C(25,5)*C(35,5)/C(60,10)≈0,229,

assuming we’re playing first which, for the sake of argument, I’ll be assuming here throughout this series of articles. As a reminder, C(a,b) is the binomial coefficient, sometimes also called “a choose b”.

What is interesting for us to help us decide when to mulligan is the average number of turns we’ll need to draw a given number of lands – we’ll say, for the sake of argument, 4, since most decks want to make their first four land drops pretty consistently. When talking about random variables, this “average” is called the expectancy.

When you calculate the average of a set of numbers, for example, you add up the numbers, and then divide by n, the number of numbers in the set. The expectancy is very similar, if you look at the numbers in the set earlier all to have the same probability of occurring, namely 1/n.

When we calculate the expectancy, we again add up all the numbers that can occur in the random variable, but we multiply each one first by the probability it has of occurring. So, in our example above, we take the number of turns “3”, and multiply it by the probability that by turn 3 we have drawn exactly 3 lands, which is 22,9% (-ish). we do this for all conceivable numbers, which would be infinitely many, but luckily we can pretty much stop counting after 60, since by that point you’d have drawn all of your deck anyway. The formula for this is written:

`E(L[n]) = Σ (P(L[n]=X) * X)`

and the result is simply a number, in our case one that lies somewhere between 0 and 60. As a reminder, Σ is used to denote a summation, in other words we add up all the terms that appear after it. I left off the range because it should be clear that we’re adding over all integers, although in our case we can stop after 53. Of course, actually calculating this can be a bit of a pain, but there are a few shortcuts.

I’ll delve into those, and how we can use the expectancy to help our Mulliganing decisions, next week.

Exercises:

1. Given a deck with a 35/25 land ratio, what is the probability that, on turn four, you will have drawn exactly four lands? What about at least four lands?
2. How are the random variables L[n] and S[m] connected? Can you make a formula to calculate one based on the other?
3. What is the expetency for the random variable of “number of lands drawn in your opening hand” for a deck with a 35/25 land ratio?

### Probability

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?

### Strictly Better (or not)

Often when comparing two cards people will use the term “strictly better,” as in, “Lightning Bolt is strictly better than Shock.” To understand what players mean by this, we need to make a brief excursion into the field of mathematics known as “Game theory”

Game theory is a field of mathematics that deals with strategic decision making, something that Magic players do all the time. In game theory, a lot of the discussion revolves around so-called “dominating strategies.” A strategy is said to “dominate” another strategy, if said strategy will always lead to a better outcome.

The concept of “strictly better” stems from this idea. A card is said to be “strictly better” than another card if in every possible circumstance having that card will lead to a better outcome than having the second card. In the example above, Lightning Bolt could be considered “strictly better” than Shock because they have the same mana cost and speed, but while Lightning Bolt deals 3 damage to a creature or player, Shock only deals 2.

In almost every situation, you will prefer to deal three damage, even if two would be enough to kill the creature in question, if you’re paying the exact same amount of mana either way.

But here is where Magic is tricky. With over 12,000 unique cards, and a near (but not actually) infinite variety of play situations, saying that having a given card will always be better than another is almost always doomed to failure.

Say, for example, you have a Chandra, the Firebrand with six loyalty counters on it, so that you can activate her ultimate ability to deal six damage to your opponent, and you know your opponent, who is at 8 life, has a Faith’s Shield in hand. If you have a Lightning Bolt in your hand, casting it will bring your opponent down to 5 life, which will then allow him to casts Faith’s Shield and use the Fateful Hour ability, preventing you from being able to kill him or her with Chandra this turn. If you have Shock instead, casting it will bring your opponent to six life, meaning you can activate Chandra’s ultimate to finish him or her off.

Or your opponent uses Mindslaver and casts it on yourself. Pretty much any card’s “strictly better”-ness can be disproved with Mindslaver, honestly. Which is why, in Magic, the term has grown to near-uselessness. Instead, a different definition has to be used, one that focuses on the attributes of the card itself, rather than the theoretical background of the phrase.

A card can instead be said to be “strictly better” than another, if each attribute of the card – speed, (mana) cost, power, toughness, abilities, etc. – is better than the other.

A list of examples can be found here.

As I said, this isn’t strictly better in a mathematical sense, but it is in a way that is more useful for Magic players.

Exercises:

1. Find more examples of cards that are “strictly better” in the Magic sense.
2. Find situations where the examples you discovered, and the ones mentioned on the wiki, actually turn out to be worse than the alternative. For purposes of this exercise, ignore effects like Mindslaver and Sorin Markov’s ultimate ability.

### Infinity

There is no such thing as infinity in Magic.

I know, this seems a bit unfair, since infinity is such a cool concept, but sadly the rules specifically state, anytime an infinite number of things happen, you just choose a number for how many times they happen, and then it ends. Unless you can’t, then you break the game as badly as if you had just divided by zero. This also is in the rules, and is usually referenced by the old resolving-a-O-ring-with-only-two-other-O-rings-on-the-board example.

What this means is, assuming you and your opponent both have a way of producing infinite mana, he or she will never be able to cast a Fireball big enough that you can’t save yourself with an Alabaster Potion.

But what if we got rid of the rules on infinities? What if you really could produce an infinite amount of mana, not merely an arbitrarily large amount?

Well, first of all we have to make sure we actually can make an infinite amount of mana. Most “infinite mana combos” involve taking a finite number of steps that end up where you started from but with some number of extra mana in your mana pool. If you wanted infinite mana, you’d have to perform these steps an infinite amount of times, which isn’t actually possible. Luckily, the magic rules state that if you can demonstrate a loop like this, it’s enough to point out the sequence and, assuming your opponent can’t or doesn’t want to do anything to interrupt it, name a number of times you want to perform it. So let’s make things easy on us and just say that we modify that rule to allow for that number to be “infinite”.

So, back to where we were. Say we have an infinite amount of mana, and our opponent casts an infinitely large Fireball. Can we save ourselves with an Alabaster Potion?

Well, this is where things get tricky. Infinity is strange in that regular math doesn’t actually work on it. You might think, sure, if you can prevent an infinite amount of damage, then you’re safe, but it doesn’t actually work like that.

Take the set of integers (0,1,2,…), for example. There are an infinite amount of numbers. Now subtract all the even numbers from that set, and you’re left with 1,3,5,…., which is still an infinite amount of numbers. Infinity minus infinity is not defined. So that’s our first problem, and it exemplifies one of the main reasons Magic doesn’t allow for infinities.

But let’s say that we redefine “infinity” in a way that we actually can resolve such a situation. Say that we state that, whenever we create an infinite amount of something, we have to define a set of infinite integers and use that to mark that particular infinity. so, in the example above, the Fireball player defines his or her infinity as the set of integers. Then all the Alabaster Potion player needs to do is define the same set for his or her infinity, and they’l live.

Let’s take another example. Say we still have infinite mana, and a way to recur Gelatinous Genesis. We cast it once for X=1, creating 1 1/1 Ooze. Then we cast it for 2, then for three, and so on, an infinite number of times (again, modifying the rules so that we can do this without actually taking an infinite amount of time). Your opponent has a Deepfire Elemental. Can your opponent kill all your oozes?

Well, the answer to this is… maybe. The problem is, for each value of X you have to choose between one of X creatures to target, which doesn’t sound that bad, until you realize that you, as a person, have to choose which one of those creatures to target. An infinite number of times. And there’s no way to shortcut that, at least not in the Magic rules. Luckily, math comes to the rescue here… sort of.

There is a principle called the Axiom of Choice, which basically says that if you have to make an infinite number of decisions for something, like choosing an ooze, you can. The funny thing is, math doesn’t need the Axiom of Choice. (Most) math works just fine without it, though some proofs get much trickier, which is why a lot of times in some fields of mathematics, you’ll see the phrase “Assuming AC,…” simply because without that phrase the proof just plain doesn’t work, though that doesn’t actually mean that the result isn’t any less true without the AC.

So, let’s add the Axiom of Choice to the comprehensive rules. Now you can kill all the oozes, but will that save you? Maybe, maybe not, because the Ooze player can just cast it again, and you can kill them, and they can cast them, and you can kill them, and so on, ad infinitum. There’s no easy way to resolve this situation, it doesn’t fit the unbreakable-loop rule, so the game goes on forever or until one player gets tired and falls asleep. Then the other player gets the last word in, and wins the little face-off.

…and that’s why we don’t allow “infinities” in Magic.

I’ll have more to say on the subject for sure, but I hope you enjoyed this little excursion.

Exercises:

1. Say instead of Deepfire Elemental above, you have a way to recur Pernicious Deed. Your opponent has cast his or her infinite number of Oozes and passed the turn to you. Can you kill them all? Do you need AC to do so?
2. What other situations can you think of where infinity breaks the game?

### Randomness, part 3

First off, I’d like to apologize for not posting this week. This Wednesday, which is when I usually post these, happened to be my thirtieth birthday, and I was out of town. So this week’s post is a bit delayed. I’ll try to get back on schedule for next week.

If you read last week’s article, you may despair that you can ever truly randomize your deck. Can your deck ever truly be random, if every shuffle is deterministic?

Well, yes, and no. The answer veers a bit out of mathematics and into philosophy (and a bit of biology).

When we talk about randomization, we might be content with a definition of “the deck is sufficiently random if neither player can be reasonably expected to know the (absolute or relative) location of any card or cards.” In other words, if the human brain isn’t capable of tracking the cards’ positions as you shuffle, that’s good enough for us. But what are the limits of the human brain?

Since I’m not a neurobiologist (shocking, I know), I can’t say for sure, but I’d say that, for a large percentage of the population, a pile shuffle would already meet this definition. For professional players, even a single riffle shuffle might be enough to confound their tracking skills, or at the most two. Then there are the Vegas card counter types – and, before you say anything, they do exist, at every level of play – who can track cards through an arbitrary number of ideal riffle shuffles without much effort. So clearly we can’t rely on the fallibility of the human brain to consider a deck shuffled.

luckily, there is another answer. I mentioned the riffle shuffle above. In an ideal riffle shuffle, we split the deck in half and into two piles (using a 12 card deck as an example)
``` 1 7 2 8 3 9 4 10 5 11 6 12 ```

Then we alternate the cards from one pile and then the other, ending up with the order

`1 7 2 8 3 9 4 10 5 11 6 12`

And the corresponding permutation

`(2 7 4 8 10 11 6 9 5 3)`

1 and 12 are mapped to themselves, so we don’t need to mention the, in the permutation. (already we should be suspicious of this form of shuffling)

Of course, you might be asking yourself, how is this any more random than a pile shuffle? Well… It’s not, honestly. But here we save ourselves through human fallibility.

The average magic player won’t be able to cut a sixty card deck exactly in the middle. When letting the cards fall, allowing one card to alternate with the other exactly over sixty cards is incredibly hard. When we riffle shuffle, we are actually choose a random riffle-ish shuffle from the hundreds of thousands of possible permutations over sixty cards. Granted, most riffle shuffles will still have the problem of keeping the top and the bottom cards in the same place, but that’s why we cut after shuffling, to remove even that regularity. And, last but not least, and this is important. when your opponent presents his or her deck to you, you should always not just cut it, but shuffle it. that way, even if your opponent is a Vegas card shark who has mastered the art of the ideal riffle shuffle (more common than you might think), you can still be assured he or she will have a sufficiently random deck, because you’ll be sure to follow what you’ve learned here to shuffle the deck in a way that even your opponent can’t know the location of any of the cards.

Of course, these cards will still not actually random. They are deterministically distributed in your deck in a non-random way. We just don’t know what the order is until we draw the cards themselves. We could link the cards to a random number generator, bringing true quantum randomness into play, but, as I defined it above, it’s enough for us as magic players to just not know where the cards are, they needn’t be truly random. Once the deck is properly shuffled, we can treat it as though it were truly randomized for the rest of these articles.

Exercises

• What is the permutation for an ideal riffle shuffle over a sixty card deck?
• After how many ideal riffle shuffles will the deck be in the same order as it started?
• What other forms of shuffling do you know? How would you describe their potential for randomness? You don’t have to do so mathematically, just describe it.
• ### Shuffling, part 3

This week we’re tackling a mathematical concept that is going to be important as we continue to look at how Magic can be broken down mathematically: the group.

A group can be anything really – a set of objects, a vector space, a geometrical construct, a system of numbers – as long as it has a set of elements and an operation that takes two of those elements and uses them to create a third (also known as a binary relation). We usually use the multiplication symbol * to signify the relation.

Furthermore, the relation has to follow the following three rules.
1) the relation has to be associative, or a*(b*c) = (a*b)*c
2) there exists a neutral element, which we call 1, so that 1*a=a=a*1, and
3) every element has an inverse element, or there is always a b so that a*b=1=b*a

Let’s look at shuffling. Shuffling is associative, in that if we combine two shuffles into a single operation, it doesn’t matter whether we combine the first two or the second two. There’s a neutral element, where we just don’t shuffle (the equivalent of tapping on the top of the deck), and every shuffle has an inverse, as you can well imagine (you just do the shuffle backwards).

So shuffling is a group,which is very interesting, for a few reasons. For one, let’s take the example of pile shuffling from above. If you do the exact same shuffle enough times, you will eventually get back to your starting point. The interesting thing is, this is true no matter what kind of shuffle you use. If you don’t change up your shuffling, you will eventually end up at your starting point.

Every element of a group has what’s called an order, which is the number n, so that if you take the element to the power of n, you get 1. But, you might think, what if there is no such n? What if the inverse to x isn’t a power of x at all?

Well, let’s say it’s not. Let’s call the inverse of x “y”, and say that, no matter how often you multiply x with itself, you never get y (because otherwise x^(n-1)=y).

So what do you get for the powers of x? Well, you can’t get y, we already said that. And you can’t get 1, that’s kind of the point. But there are only a finite number of different shuffles possible. A very large number, granted, but finite. If ever two powers of x are the same, say x^a=x^b, then if we decide by the smaller of the two (say a) we have x^(a-a)=x^(b-a). Since x^0=1, we’d have 1=x^(b-a), which we said isn’t the case. So no two powers of x can be the same.

Ah, but there are only finite many different elements that powers of x can be, eventually they have to be two of the same. So therefore, there must be an n, so that x^n=1. (more specifically, the order of x is the smallest such number)

So what does this mean? Well, if you perform the exact same action enough times in a row, you’ll end up where you started from. So, performing the exact same action – no matter how random it may seem – is actually not random at all.

So does this mean that you can never randomize your deck? Is it actually impossible to follow the rules? Well, no, but stay tuned to find out why.

Excises: 1)how many elements does the group S of possible shuffles of a deck of sixty cards have? Start with smaller deck sizes
2) what’s the order of the eight-pile pile shuffle on a deck of sixty cards?