As I regularly write about the mathematics of Magic, I occasionally receive math-related questions on Twitter. I don’t always have time to respond to every single one of them, but I compiled some of the more interesting ones from the last few months.
It all started with a clip that literally called out for me.
#SomeoneCallFrankKarsten needs to be a thing. Can we make this a thing?
— Ross Merriam (@RossHunneds) March 1, 2018
I ran the numbers, and I found that the probability of Matt Nass going from one to four Hollow One (accounting for Street Wraith possibilities) is 0.0061%, or once every 16,294 such Cathartic Reunions.
To do this calculation, I started from the point where Cathartic Reunion goes on the stack, i.e., after the first Street Wraith is cycled. I defined success as casting four Hollow Ones that turn. The most likely way to do this is by drawing three Hollow Ones straight away, which occurs with probability 3/50*2/49*1/48, i.e., 0.0051%. The next most-likely way to do this is what actually happened (two Hollow Ones and one Street Wraith on top, followed by another Hollow One), which occurs with probability 3/50*2/49*3/48*3*1/47, i.e., 0.0010%. I multiply by 3 because there are 3 ways to order two Hollow Ones and one Street Wraith. Finally, there are a few extremely rare scenarios with multiple Street Wraiths.
If you wonder about Burning Inquiry and Goblin Lore, then see my article on the topic. As I showed in the comment section of that article, if you don’t mulligan one-landers, then you’ll hit the jackpot of three or four Hollow Ones on turn 1 about once per 200 games on average.
@karsten_frank Have you run the numbers on how often a T1 burning inquiry is likely to leave the opponent with zero lands in hand? Assuming an average keep with 2-3 lands.
— Kurt Berger (@kmberger44) April 20, 2018
If an opponent keeps 2 lands and has a 53-card deck with 22 lands remaining, then a turn-1 Burning Inquiry leaves them with zero lands with probability 1.64%. It’s 0.16% if they keep a 3-lander.
If you’re wondering how I obtained 1.64%, I simply added up two possibilities:
- Their top 3 cards contain 0 lands, which occurs with probability 19.2%, and then they discard 2 lands at random, which occurs with probability 6.7%. Multiplying yields 1.28%.
- Their top 3 cards contain 1 land, which occurs with probability 43.7%, and then they discard 3 lands at random, which occurs with probability 0.8%. Multiplying yields 0.36%.
"Somebody call Frank Karsten"
— Brian Braun-Duin (@BraunDuinIt) April 20, 2018
Richard Hagon asked this question around the release of Dominaria. But given that Core Set 2019 is the latest set for the publication of this article, I redid the analysis for M19. I interpreted “one of the whole set” to mean one of every card, not a full x4 playset. I include the 10 dual lands but disregard foils. A quickly coded simulation yields the following:
In expectation, you need to open 445.8 boosters (or 74.3 Sealed pools) to get one of the whole set. If you want a full x4 playset of everything, then you need 1,240 boosters on average.
While this made for a fun mathematical exercise, cracking packs is not the most efficient way to complete your collection, especially when you have everything except for one missing mythic. Fortunately, the ChannelFireball store has almost all of them in stock.
You can calculate it to some extent. Your mulligan decision should be based on a comparison of the following two things:
- The probability of winning the game when you keep your 6-card hand. This is hard to calculate, but you can often make a reasonable guess based on experience, intuition, and your odds of drawing a land in time. Check out this article to see how I conceptually approach this.
- The probability of winning the game when you mulligan your 6-card hand. This number heavily depends on such factors as the matchup, the presence of Hazoret the Fervent or cheap card advantage spells in your deck to smooth out mulligans, whether you’re on the play or on the draw, and so on. But you can theoretically calculate it if you have enough data. If you trust the estimates from the previously mentioned article, then your win probability after another mulligan will usually be around 27%, plus or minus a few percentage points.
@karsten_frank Hey Frank, I have an (i think) interesting question. In vacuum, what's better, draw a random card or the +1 of Karn? Asuming the opponent never gives you a nonland card, and you never want a land. Hope you can help me!
— Jorge Martinez (@jorgemartinezt7) May 13, 2018
If you only have one +1 activation, then drawing a random card is always better than giving your opponent a choice between two random cards.
This changes when you factor in Karn’s -1. If he sticks around for multiple turns and all you need to win is one copy of a small subset of cards (say, one of your four Cranial Platings) then Karn will be better than a planeswalker that merely draws cards. After all, if you keep activating Karn’s +1 until you have exiled one of the cards you’re looking for (or are lucky enough to hit two at once) then you’re seeing two extra cards per turn for most of the turns.
In terms of expected number of activations needed before you find the card you want, Karn is faster than Ob Nixilis Reignited if and only if less than half of your deck is a successful hit. This is assuming that your deck is infinitely large and that life totals don’t matter. To obtain this cutoff point, I derived the number of required Ob Nixilis activations and the expected number of required Karn activations.
"Someone call @karsten_frank" moment here. Wondering what are the odds for Karn's +1 to hit at least a land in function of lands and spells remaining in your deck (costructed mainly). Can we get any help with an awesome as always article maybe? 🙂
— Matteo Capozucca (@Zukkattack) June 4, 2018
The exact odds depend on the contents and size of your deck at the moment you activate Karn, Scion of Urza. To calculate them, start by determining the probability of hitting two spells. For example, if you have 20 lands and 25 spells in your 45-card deck, then your top card is a spell with probability 25/45 and the second card is also a spell with probability 24/44. Multiplying the two together yields 0.303, which means that the complementary probability of hitting at least one land is 0.697.
But generally speaking, for a typical 25-land deck you’ll hit at least one land in your top two cards about 2/3 of the time.
@karsten_frank Can you check my work here? I’m a little rusty on my math but I think it’s sound.
The criteria for keeping a 7 is “if no 1 drop, mulligan”. Not accounting for hands with a 1 drop but no land. pic.twitter.com/22FluiCDo6
— 🔥 Anna Jane, Roaring Flame 🔥 (@annajanemtg) June 5, 2018
This question pertained to Modern Humans, and my answer was “yeah, that should be accurate.” But it’s worth sharing this question because it illustrates two useful insights.
First of all, under reasonable mulligan strategies, an opponent who keeps their hand will indeed hold a key 4-of card about half of the time.
Second, it’s worth analyzing mulligan strategies when building a deck. The simplification of keeping if and only if you have a 1-drop resonates with me, even if it’s a bit too aggressive. When I played Gifts Ungiven decks back in the day, I almost always mulliganed if I didn’t have at least two lands and one of the 16 “essentials”: Sakura-Tribe Elder, Kodama’s Reach, Sensei’s Divining Top, and Gifts Ungiven. When I play Affinity nowadays, I also almost always mulligan if I don’t have at least two mana sources and one of my 16 “big cards”: Arcbound Ravager, Cranial Plating, Steel Overseer, and Master of Etherium. And when I win the die roll with a Goblin Chainwhirler deck, then I often mulligan if I don’t have one of my 15 1-drops or 2-drops. There are exceptions of course, but this is a useful perspective.
In my experience, if your deck has fewer than 15-16 “essentials” then you should either (a) add more essentials to your deck, (b) be willing to keep good hands without one of the essentials, or (c) ensure your deck mulligans well and is not reliant on having a critical mass of cards. For Humans, by the way, I believe it’s (b) in the sense that you can keep good hands without a 1-drop.
It's easy to use the Hypergeometric Calculator to figure out how likely he is to have 3 forests or 3 islands or a champion or a djinn, but I'm not sure how to combine those probabilities.
— Carl A. Sagan (@saganite) June 26, 2018
To do this properly, you need the multivariate hypergeometric distribution. I intend to write a more detailed introduction to hypergeometric calculations later this year, but for now I can give you the end result.
For the subset of games where you drew Steel Leaf Champion by turn 6 on the play, you’re 50.1% to have drawn fewer than 3 green sources by then. The probability for Tempest Djinn is analogous. Since you’ll often be stuck with them in hand, I would recommend against running these triple-colored creatures in a deck with only nine sources of that color.
Thanks to everyone who submitted a question! I can’t promise to run the numbers for everyone, but if you’d like me to do another mailbag column, then keep the questions rolling.