The Hypergeometric Calculator

Hello Final Fanatics! My name is Gino Grieco and I’m here to hopefully teach y’all a little bit about the theory and practice behind deckbuilding for the FF TCG. A little bit of background on me: I’ve been a regular MTG player for 15 years now, I can speed run multiple Final Fantasy games (V, VIII, and IX), and I’m on a constant quest to secure the position of Top FFIX Nut In The World. I started playing the FF TCG in the spring of 2018 after a friend informed me of the game’s existence during a live stream. At first I was deeply skeptical, as card games based on licensed properties have a somewhat mixed history of success and quality. However, once I started playing the game, I was surprisingly hooked on the mechanics. I knew I would be a sucker for the game’s flavor, but I did not expect the actual gameplay to be so deep or the deck building choices so varied. And so after months of processing, I want to share some of how I think about the game and hopefully impart some of the wisdom I’ve gained.

For this first article I want to share one of the most useful tools I’ve encountered when building MTG decks: the Hypergeometric Calculator (shoutout to Frank Karsten for turning me on to the tool via his MTG articles). Yes, that is its real name. No, it isn’t scary math from high school that’s come back to haunt you. Yes, you can find one via a quick Google search and use it without an advanced calculus degree. The Hypergeometric Calculator is a simple tool that helps answer a very important question for us: if I have a deck of R cards and am looking at a random sample of X cards from that deck and I’m looking for Y number of Z specific cards, how likely am I to successfully draw those cards (if you got through that sentence on the first try, congrats, you’re now an Arithmetician)? For example, let’s say I’m playing a deck centered around Cid (II), and I want to know how likely I am to draw him in my opening hand if I go first. Cid (II) is an incredibly powerful card if you can play him early, and he gets significantly weaker as the game goes on. You’re likely to draw your 1 CP backups instead of fetching them for free with Cid, and the bonus CP he generates is less important when both players have played multiple backups.

So our question looks like this: what is the likelihood that out of 50 cards there is at least one copy of my three Cids in my top six cards (five cards for the initial hand plus one for our first draw)? The answer is 32 percent! We’re 29 percent likely to see exactly one copy of Cid, 3 percent to see more than one copy, and…68 percent to see no copies of our most important card in our opening hand. That’s real bad! But what about if we always mulligan any hand without Cid? That brings our sample size up to 11 cards and our percentages take a corresponding jump: 41 percent, 12 percent, and 47 percent. So we’re now over the 50 percent hump to see our most important card in our opening hand -- but you want better than coin toss odds on finding your best early play. So what if we take things to the next level? Let’s try adding three copies of Thordan VII to help us fetch Cid when he’s not in our opening draw. We now have six hits in our deck, and that ups our probabilities to 40 percent, 39 percent, and 21 percent. We have twice the chance to draw two copies of what we want than our odds of hitting none. Now we’ve got a workable backbone for our deck!

Let’s try another question that comes up a lot in the FF TCG: how many elements should we play, and how many cards of each element? Casting a card of the six main elements in the FF TCG requires at least one CP of its element. You can generate CP by dulling backups or by discarding cards. However, you need to discard in order to start your support group to help your future plays. In gameplay terms, this means you need at least two cards of the same element in order to play any cards of that element: one to generate CP, and the other to put into play. So what are the odds of getting any two cards of the same element in our starting hand on the play, without a mulligan? The math on a mono element deck is pretty simple: 100 percent! There’s one of the big advantages for mono element decks right there, perfect element availability. How about a two-element deck with a 25/25 split? We’re 90 percent to be able to play a card of one specific element of our two, and in that other 10 percent of the time we’ve drawn five/six cards of the other element. So we’ll definitely be able to play something on turn one, but maybe not the card we want. What if we get more ambitious? How about three elements? With a semi-even 17/17/16 split we’re 68 percent likely to be able to play a card from either of our 17 count elements, and 64 percent for our remaining element. Again, we’ll be able to play something, but it’s getting more likely that your play will be driven by chance. How about if we slant our deck? As any good Chemist can tell you, sometimes you need to mess with your ratios to get a smooth brew. How about a 20/20/10 split? Now we’re 78 percent likely to be able to play our main elements and 34 percent to be able to play our splash element. This might be what you want if you have searchers for cards in your splash element, or if your splash element is largely giving you late-game cards that you can afford to wait for.

Once you get up to four elements, even splits become a recipe for inconsistency. A 13/13/12/12 split leaves you with 50 percent/45 percent probabilities on being able to play a card from a given element. With odds like that you’ll be at the mercy of the top of your deck from the first turn on. At five elements and a 10/10/10/10/10 split, you’re 34 percent likely to be able to play a card of any individual element, and you run a real risk of being unable to play anything for long stretches of the game. As you might expect, a six-element deck is pretty darn inconsistent with 9/9/8/8/8/8 splits and 29 and 24 percent odds to play a card of a given element. This configuration has the distinction of being the only one that can leave you with an opening hand with literally no plays!

Let’s finish this off with a bit of a thought experiment…a dream of mine, if you will. Given how bad the math is for six-element decks with even element splits, can we build a reliable six-element deck? I think the answer is yes, though we will need to pull a Kingdom Hearts and embrace the darkness to make it work. If we start with a base of three Chaos and one Kam’lanut to search for him, we’ll have a solid foundation of backups that produce any element we need. We can then add a package of Earth backups and forwards that help us generate whatever elements we need: three Moogle (FFCC), three Leo, one Shantotto, two Kimahri, and two Epitav. Between all of those searchers and multi-color producing characters, we have an 85 percent chance of seeing one of those cards in our initial hand and a 99 percent chance to see one if we mulligan. That means we will have access to at least one of any element from turn two at the latest. So what do we get for all of that effort? How about we dip into the summon synergies that are spread over every element?

This is just an example of the madness we can get into when we keep our probabilities in mind. Since our critical pieces are all Dark or Earth and we slant our deck towards Earth (22 cards in total), we have a very good chance of casting one of our key characters  -- and then we can use that card to fuel our game from there. This might not be the best use of rainbow CP, but I’d love to hear any other suggestions for how to build around this core.

Hopefully y’all found this useful. I know that my deck building improved dramatically once I started taking probabilities into consideration. If there’s one thing speedrunning FF games taught me, it’s that strategy and creativity are important -- but nothing beats the raw power of numbers.

Gino Grieco is a freelance writer. You can read his work on Giantbomb and Waypoint. He co-hosts the "Deep Listens" podcast which can be found here. You can find him on Twitch, Youtube, and Twitter.