# Magic Math: Liliana, the Last Hope

Last week, I studied Deploy the Gatewatch and Lupine Prototype from a quantitative perspective. Today, I’ll analyze another Eldritch Moon card that made an impact on the SCG Open in Columbus last weekend.

# Liliana, the Last Hope

Any 3-cost planeswalker is valuable because of their low mana cost, and Liliana is no exception. Her abilities are quite powerful, too. The +1 takes out a fresh Thalia’s Lieutenant or Mausoleum Wanderer, the -2 fits well in delirium deck, and the ultimate takes over the game quickly.

## How often will you hit a creature off of her -2?

If you already have a creature in your graveyard, then you are guaranteed to bring one back, obviously. If you don’t have a creature in the bin yet, then the probability of a blind hit depends on the number of creatures in your deck. Let’s take a look at a few well-performing decks from last weekend.

# GB Delirium

### Todd Stevens, 20th Place at SCG Standard Open

If your graveyard is empty when you activate Liliana‘s -2, then the probability of hitting at least 1 creature in this 20-creature deck is equal to 1 – [ (59-20)/59 * (58-20)/58 ] = 0.567 or 56.7%. So it’s a bit of a gamble, but it still draws you over half a card in expectation.

And even if you miss, you may get closer to delirium, which is super important for this deck due to Gnarlwood Dryad, Grim Flayer, Mindwrack Demon, and Ishkanah, Grafwidow. For that reason, I like the clean split of 4 planeswalkers, 4 enchantments, 4 instants, and 4 sorceries in this deck.

Note: As always, the math is trivial, but the question of what probability you’re actually interested in is not. The interpretation of the above probability is that you remove Liliana from the deck and put her ability on the stack for the remaining 59-card deck. If the deck is random, you could remove any number of cards from the top of the deck, and you would still have the same probability. Hence, it applies to any turn in the game, and therefore I like to use these numbers to gain insight for deck building purposes.

There are alternative approaches, but they all have their downsides. If you consider the scenario where you remove 3 lands and some other cards from the deck before activating Liliana, then you run into the issue of only being applicable to a very specific situation. So I definitely don’t like that interpretation. I do like approaches where you determine a probability conditional on having drawn at least 1 Liliana and having at least 3 lands on the battlefield after mulligans. These are relevant, but they cannot be easily determined analytically. Ultimately, I think my approach strikes a good balance between simplicity and applicability.

# Sultai Control

### Ali Aintrazi, Top 4 at SCG Standard Open

Thanks to 1 Ishkanah, Grafwidow and 1 Emrakul, the Promised End, this deck cares about having different card types in the graveyard, but the main role of Liliana in this deck is to peck off 1-toughness creatures (which, if Liliana is popular, are quickly becoming poor inclusions in Standard decks) and to return big threats in the late game. And thanks to 4 copies of Languish, this deck can easily reach the late game.

Still, a blind turn-3 Liliana activation could easily happen in a control mirror. For this 14-creature deck, it would provide a creature with probability 1 – [ (59-14)/59 * (58-14)/58 ] = 0.421 or 42.1%. This feels more like a gamble, so you’re probably better off waiting until you have a creature in your graveyard.

# BW Angel Control

### Ronnie Ritner, Top 8 at SCG Standard Open

This deck doesn’t care about delirium and has only 10 creatures, so Liliana seems like she might be out of place, but then again, if creatures like Selfless Spirit are popular, then Liliana can mow them down. And in the late game, you might just be able to return Bruna, which in turn returns Gisela, and meld your way into victory.

A blind activation, however, would provide a creature with probability 1 – [ (59-10)/59 * (58-10)/58 ] = 0.313 or 31.3%. This is notable because 313 is the license plate number of Donald Duck’s car.

Let’s take a closer look at Liliana‘s ultimate ability, which is like a math exercise by itself.

Suppose that, on turn 1, you have Z Zombies on the battlefield and activate Liliana’s ultimate. Assuming that what is dead may never die, how many Zombies do you have at the end of turn T?

Let’s take a look at what happens if you start with 0 Zombies.

Turn 1 – You get 0+2=2 Zombies
Turn 2 – You gain 2+2=4 Zombies for a total of 2+4=6 on the battlefield
Turn 3 – You gain 6+2=8 Zombies for a total of 6+8=14 on the battlefield
Turn 4 – You gain 14+2=16 Zombies for a total of 14+16=30 on the battlefield

If you denote the number of Zombies you have at the end of turn T by N(T), then you can describe N(T) recurrently via N(T)=2N(T-1)+2 for any T>1 with N(1)=2.

From this, it is easy to find that you can write N(T)= 2^T + 2^(T-1) + … 2 for any positive integer T, and if you have ever worked with binary numbers, then you will readily recognize that expression as N(T)=2^(T+1)-2. In words, to get the number of Zombies on the battlefield at the end of turn T, you have to raise 2 to the power T+1 and subtract 2.

This is a nice example of exponential growth, and it gets out of hand quickly. On turn 24, you will have more Zombies than the population of the state Texas. On turn 28, your Zombie horde will exceed the population of the entire United States of America. And on turn 32, the amount of Zombies on the battlefield is larger than the population of the world. Make sure you bring enough tokens.

But I haven’t answered the original question in full yet. What happens if you already have Z Zombies on the battlefield when you activate the ultimate? The derivation of the general formula is similar, except that you double your original Z Zombies every turn. It comes out to N(T)=2^(T+1)-2+(2^T)Z. So if you started with 2 Zombies on the battlefield when activating Liliana’s ultimate, you would have more Zombies than the population of the state Texas on turn 23.

Suppose that player A has the Melira “infinite life” combo with 47 cards left in their deck. Player B untaps, draws down to 45 cards, sweeps the board with Anger of the Gods, and uses Liliana to get an emblem. How much life does player A need to deck player B?

This question was posed to me on Twitter by Grischa Baumann. The answer can be easily determined from the formula with Z=0 starting Zombies: it’s N(1)+N(2)+…+N(45) = 1.40737488355* 10^14. In words, that’s 140 trillion 737 billion 488 million 355 thousand and some change.

So next time you gain “infinite” life, don’t settle for a million. A billion or a trillion are also not going to be big enough. To be safe, pick a googol (10 to the power 100) or a googolplex (10 to the power googol). No one will beat you then.

But Frank, what happens if you have Endless Ranks of the Dead and a Doubling Season on the battlefield as well?

All right, next topic.

# The SCG Open Metagame

The top tables of the Open were dominated by Bant Company, GW Tokens, and WR Humans. Yes, some new cards saw play, such as Liliana, Gisela, Thalia, Ishkanah, Grim Flayer, Mausoleum Wanderer, and Spell Queller, but overall the metagame looked eerily similar to the old Standard.

I was really looking forward to taking a well-performing deck with Grapple with the Past (which is like a cross between Impulse and Regrowth) or any other interesting card to study from a numerical perspective, but sadly I did not find any among the top-performing lists. I’ll just leave you with the statement that a turn-2 Grapple with the Past is somewhat similar to Oath of Nissa, for which you can find the analysis here.

I still have hope that the Pro Tour will feature more innovation. Eldritch Moon is filled to the brim with powerful cards that could have a huge impact on Standard, and even though the first tournament has been a little disappointing, there are a lot of options to be explored.

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