Unfinity Math: _____ Goblin and Comet, Stellar Pup



Unfinity, the latest Magic: the Gathering set, features flavorful and humorous cards in a retro-futuristic space carnival setting. It’s intended to celebrate the sillier side of the game, with effects that can’t work under Magic’s normal rules. In line with the quirky nature of the set, I always like to analyze cards from a wacky mathematical perspective, and Unfinity contains several cards that grabbed my attention. Omniclown Colossus is one of them, but the main part of this article will focus on two cards that are relevant for Eternal formats (Vintage, Legacy, Commander and Pauper).

Unfinity is the first so-called Un-set to introduce cards without a silver border or an acorn stamp at the bottom, and these cards will be legal in Eternal formats. Two of them not only fascinated me from a mathematical perspective but also impact the competitive Legacy format.

With this in mind, let’s take a closer look at Omniclown Colossus, Comet, Stellar Pup and _____ Goblin.

Omniclown Colossus

Omniclown Colossus

Omniclown Colossus, whose flavor text reads “How to calculate π isn’t important” is an easy way to get mathematicians riled up. What do you mean not important?! If humans didn’t know how to calculate π, then buildings might collapse, planes might fall out of the sky and medical procedures might fail. And for Magic purposes, what if you control Soulfire Grand Master or Toralf, God of Fury and need to adjust life totals? How would you determine that seven times π is approximately 21.99 while eight times π is approximately 25.13 without calculating it? Madness.

So clearly, calculating π can be quite important. Besides, what if a creature is enchanted with Druid’s Call? This results in a partial Squirrel, I guess, but how does that work? Do you need partial tokens? Can you chump block with it? Can Dracoplasm eat a bunch of them to become whole? Of course, these questions are why Pie-roclasm wouldn’t work under Magic’s normal rules, but I can answer the last one.

Since π is an irrational number, it cannot be expressed as a fraction of two integers. This means that you’d never be able to assemble a complete Squirrel from any number of (π - 3) Squirrel tokens. The power/toughness of Dracoplasm can never become an integer, no matter how such tokens you have. In other words, you’d always have Squirrel parts left over. Lugubrious, yet mathematically fascinating.

Comet, Stellar Pup

Comet, Stellar Pup

Comet evokes the memory of Laika, a Soviet space dog who in 1957 became the first animal to orbit the earth. The designers of the Magic card were further inspired by the notion that dogs don’t always listen to you, which resulted in the die roll mechanic. As a result, Comet is the first planeswalker with only one loyalty ability.

Comet has high starting loyalty, and outcomes all look promising for competitive Legacy:

  • Digging up a bone – The 3 ability, which happens 1/6 of the time, is reminiscent of Wrenn and Six when you return a fetchland, albeit at the cost of a loyalty counter. Still, the flexibility to return Expressive Iteration, Brainstorm or Pyroblast to your hand is worth a lot.
  • Attacking – The 4 or 5 ability, which happens 1/3 of the time, reminds me of the -2 abilities on Grist, the Hunger Tide, The Wandering Emperor or Liliana of the Veil. Destroying a creature at the cost of two loyalty is a fine deal, and the ability to finish off your opponent can win games as well. The only downside is that Comet doesn’t listen to you and might deal damage at the wrong times.
  • Doing tricks – The 6 ability, which happens 1/6 of the time, will feel like cheating. Getting two more activations will bury your opponent in value, and the opportunity to chain together multiple sixes on one turn is even better. Good dog, Comet.

The ability to roll again is what fascinated me from a mathematical perspective, so I ran the numbers. In the questions below, I largely focus on expected outcomes in a single turn, which you can interpret as long-run averages.

Expected loyalty gain per activation? It’s zero. Perfectly neutral. The probability to gain a certain amount of loyalty is equal to the probability to lose that much loyalty.

Expected number of activations? Ignoring loyalty changes, the expected number of Comet activations in one turn can be determined analytically. Let’s denote this number by A. Then A = 1 + 1/6 * 2A. Solving yields A = 1.5. So on average, you’ll be rolling 1.5 dice per turn. This calculation ignores the theoretical possibility to hit a string of fives and sixes that reduce Comet’s loyalty to zero while it still has activations remaining, but if you’re just rolling one die per activation, then such a rare sequence of rolls is negligible.

Expected number of Squirrels? Ignoring loyalty changes, let’s denote the expected number of Squirrels created in one turn, when starting with a single Comet activation, by S. Then S = 2/6 * 2 + 1/6 * 2S. Solving yields S = 1. So on average, you’ll be adding one Squirrel to the battlefield per turn.

Expected amount of damage? For the amount of damage, it gets a bit trickier because that depends on Comet’s current loyalty count. While not impossible to determine analytically, I found it easier to code a simulation in Python and run it 10 million times. Based on this, the expected amount of damage when starting with a five-loyalty Comet and one activation is 2.75. If you add damage from hasty Squirrels, assuming that the opponent has no blockers, then the expected total amount of damage in one turn is 3.75. That’s a pretty fast clock. Not bad!

Total damage distribution and probability to win the game? The same Monte Carlo simulation helps to determine this. In a turn where you start with a five-loyalty Comet and one activation, the most common outcomes when counting damage from both the 1-2 ability and the 4-5 ability are: zero damage (which occurs with 17.2 percent probability), two damage (35.3 percent), five damage (34.3 percent) and 10 damage (4.2 percent). But there’s also a 0.75 percent probability to hit 20 or more damage in total. So 0.75 percent of the time you cast Comet, you’ll roll a sufficient sequence of sixes and win the game on the spot!

Evaluating these numbers, I believe that Comet could see competitive play in Legacy. Due to the random nature of the rolls, it may not be superior to Minsc & Boo, Timeless Heroes in its base form. However, Comet could be a valid option if you want additional four-mana planeswalkers or if you’re playing a base red-white deck. Best of all, Comet will truly shine if you add cards that grant a die roll advantage.

Barbarian Class Wyll, Blade of Frontiers Pixie Guide

The numbers I’ve run so far assume that you have zero die roll advantage effects on the battlefield, but such cards naturally change everything. How likely does Comet win the game with one, two or more Pixie Guides (or equivalents) on the battlefield?

Their effects stack, but they don’t leave you with a choice - you have to pick the highest roll, even if you don’t want to. With N Pixie Guides on the battlefield, the probability P(6) that your highest roll is a six is equal to 1 - 5/6 ^ (N + 1). This equates to 30.6 percent for one Pixie Guide and 42.1 percent for two Pixie Guides.

Using this, the probability that your highest roll is a five is equal to [1 - P(6)] * [1 - 4/5 ^ (N + 1)], the probability that your highest roll is a four is equal to [1 - P(6) - P(5)] * [1 - 3/4 ^ (N + 1)], and so on. That’s just doggo math. Summing the resulting probabilities, multiplied by the loyalty change corresponding to each option, yields the expected loyalty loss or gain per activation.

As it turns out, with one to four die roll advantage effects, the expected loyalty gain becomes negative. Intuitively, you can expect to alternate between rolling five and six most of the time, which means alternating between -2 and +1 loyalty. Meanwhile, the loyalty-positive Squirrel effect becomes increasingly unlikely. This means that, given enough time, Comet will sink to zero loyalty almost surely.

When you have five or more die roll advantage effects, you’re rolling a six often enough to offset the loyalty loss from the other abilities. At that point, sequences that diverge to infinity occur with positive probability. You might still fail if you start out with a few unlucky rolls, but the more activations you get, the more likely you are to go infinite. At least Comet is a “may” ability.

Other stats, such as the expected number of activations, Squirrels or damage in one turn, can all be obtained via the aforementioned computer simulation. The same holds for the probability to deal more than 20 total damage in one turn, i.e., to win the game on the spot. The following table has a summary of the results.

Pixie Guides
E[Loyalty gain]
P[roll 6]
E[Activations]
E[Squirrels]
E[Damage]
P[Win the game]
0
0.00
16.7%
1.50
1.00
2.75
0.75%
1
-0.50
30.6%
2.53
0.56
6.36
5.86%
2
-0.50
42.1%
4.69
0.35
12.4
16.4%
3
-0.35
51.8%
8.95
0.22
23.8
29.6%
4
-0.16
59.8%
22.3
0.18
77.3
42.7%
5
0.02
66.5%
Infinite
Infinite
Infinite
54.0%
6
0.17
72.1%
Infinite
Infinite
Infinite
63.2%

With five or more die roll advantage effects, there is always a chance of missing or stopping, but there is also a non-zero probability of never returning to the starting state. This means that the expected values lose their practical meaning and become infinite. Basically at that point, there’s a non-zero chance that Comet allows you to keep rolling until the heat death of the universe. You might be in for a long game, but at least you’ll have the goodest boy in the Magic Multiverse on your side.

_____ Goblin

_____ Goblin

Stickers, which are attached to nonland permanents you own, represent brand new territory in Magic. The way they work in Constructed is that you must bring 10 unique sticker sheets out of the 48 different ones available. Out of those 10, all of which must be different, you randomly select three at the start of every game. Those are then your three available sticker sheets for that game.

If you’re interested in adding _____ Goblin to your deck, then what is the optimal collection of sticker sheets? To answer this, I ran the numbers. In the table below, you’ll find all sheets where at least one of the name stickers has three or more vowels. Stickers are sorted in descending order of their maximum vowel count.

Sticker
First vowels
Second vowels
Third vowels
Max
Playable Delusionary Hydra
3
6
2
6
Unassuming Gelatinous Serpent
3
5
1
5
Unsanctioned Ancient Juggler
5
3
2
5
Ancestral Hot-Dog Minotaur
2
1
4
4
Eldrazi Guacamole Tightrope
3
4
3
4
Misunderstood Trapeze Elf
4
2
1
4
Narrow-Minded Baloney Fireworks
4
4
3
4
Phyrexian Midway Bamboozle
4
3
3
4
Unglued Pea-Brained Dinosaur
2
3
4
4
Cursed Firebreathing Yogurt
2
3
3
3
Deep-Fried Plague Myr
2
3
1
3
Demonic Tourist Laser
3
3
2
3
Elemental Time Flamingo
2
2
3
3
Geek Lotus Warrior
1
2
3
3
Happy Dead Squirrel
2
2
3
3
Night Brushwagg Ringmaster
1
2
3
3
Notorious Sliver War
3
2
1
3
Slimy Burrito Illusion
2
3
3
3
Space Fungus Snickerdoodle
2
1
3
3
Squishy Sphinx Ninja
3
1
2
3
Sticky Kavu Daredevil
2
2
3
3
Trained Blessed Mind
3
1
1
3
Trendy Circus Pirate
2
2
3
3
Unhinged Beast Hunt
3
2
1
3
Unique Charmed Pants
3
2
1
3
Unstable Robot Dragon
3
1
2
3
Vampire Champion Fury
3
3
2
3
Wrinkly Monkey Shenanigans
2
3
3
3
Yawgmoth Merfolk Soul
3
2
2
3
Zombie Cheese Magician
3
1
2
3

So, there’s one sticker sheet with a six-vowel word: “Delusionary”. That’s the dream. There are two sticker sheets with a five-vowel word and there are six sticker sheets with a four-vowel word. That’s nine so far. Regardless of which sticker sheet you pick as your final one, you can guarantee access to a four-vowel name sticker every game. This means that the first _____ Goblin you cast will always add four red mana at minimum.

If you bring 10 stickers including the nine best ones and choose three at random, then a hypergeometric calculator shows that you’ll add:

Six mana 30 percent of the time. (You’ll have the six-vowel “Playable Delusionary Hydra” sticker.)

Five mana 40.8 percent of the time. (You’ll have a five-vowel sticker but miss the six-vowel one.)

Four mana the remaining 29.2 percent of the time.

So if you bring an optimal sticker sheet configuration, which could involve the “Trained Blessed Mind” sticker as your tenth if you like silly memes, then _____ Goblin will add 5.008 mana in expectation. That’s better than Seething Song! It might easily find a home in Legacy Goblins where it can be tutored with Goblin Matron and ramp into Muxus, Goblin Grandee.

Conclusion

With the introduction of novel items like sticker sheets, Unfinity continues a trend: the number of additional game pieces to bring to an Eternal event has become larger and larger. You might need sticker sheets, dozens of dice, various tokens, an Attraction board, partial Squirrels, exertion markers, all of the Dungeons, the Monarch, etcetera. How do you even carry all this and keep it organized?

Fortunately, Ultimate Guard has convenient trading card solutions for everything. The Twin Flip’n’Tray Xenoskin has not only room for your main deck and sideboard, but there’s also a middle box for dice and accessories, and there’s an extra compartment that fits your sticker sheets and tokens. It’s all part of one awesome product, and it has my stamp of approval.

Author: Frank Karsten

Frank Karsten has played in nearly 80 Pro Tours, which puts him in the top five most experienced players of all time. After getting inducted into the Pro Tour Hall of Fame in 2009, he finished his PhD in game theory and stochastic processes, then returned to the game of Magic. In the decade since, Frank found competitive success again, mostly with synergy-driven decks like Affinity or aggro cards like Embercleave. He also joined Wizards of the Coast's event coverage team. But he's perhaps best known for applying his mathematical background to Magic-related problems, ranging from mana base construction to metagame analysis."