A billion or a trillion are also not going to be big enough. So next time you gain “infinite” life, don’t settle for a million. In words, that’s 140 trillion 737 billion 488 million 355 thousand and some change. This question was posed to me on Twitter by Grischa Baumann. How much life does player A need to deck player B? Player B untaps, draws down to 45 cards, sweeps the board with Anger of the Gods, and uses Liliana to get an emblem. Suppose that player A has the Melira “infinite life” combo with 47 cards left in their deck. 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.
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.
LILIANA THE LAST HOPE FULL
Make sure you bring enough tokens.īut I haven’t answered the original question in full yet.
And on turn 32, the amount of Zombies on the battlefield is larger than the population of the world. On turn 28, your Zombie horde will exceed the population of the entire United States of America. On turn 24, you will have more Zombies than the population of the state Texas. This is a nice example of exponential growth, and it gets out of hand quickly. 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. 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.įrom 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. Turn 4 – You gain 14+2=16 Zombies for a total of 14+16=30 on the battlefield Turn 3 – You gain 6+2=8 Zombies for a total of 6+8=14 on the battlefield Turn 2 – You gain 2+2=4 Zombies for a total of 2+4=6 on the battlefield Let’s take a look at what happens if you start with 0 Zombies. Assuming that what is dead may never die, how many Zombies do you have at the end of turn T? Suppose that, on turn 1, you have Z Zombies on the battlefield and activate Liliana’s ultimate. Let’s take a closer look at Liliana‘s ultimate ability, which is like a math exercise by itself.
LILIANA THE LAST HOPE LICENSE
This is notable because 313 is the license plate number of Donald Duck’s car. And in the late game, you might just be able to return Bruna, which in turn returns Gisela, and meld your way into victory.Ī blind activation, however, would provide a creature with probability 1 – = 0.313 or 31.3%. 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. Sultai Control Ali Aintrazi, Top 4 at SCG Standard Open Ultimately, I think my approach strikes a good balance between simplicity and applicability. These are relevant, but they cannot be easily determined analytically. 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. So I definitely don’t like that interpretation. 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. There are alternative approaches, but they all have their downsides. Hence, it applies to any turn in the game, and therefore I like to use these numbers to gain insight for deck building purposes. 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. 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. Note: As always, the math is trivial, but the question of what probability you’re actually interested in is not. For that reason, I like the clean split of 4 planeswalkers, 4 enchantments, 4 instants, and 4 sorceries in this deck. So it’s a bit of a gamble, but it still draws you over half a card in expectation.Īnd 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. 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 – = 0.567 or 56.7%.
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