1

When Magic the Gathering players create infinite resource combos, they often name numbers such as 1 million, 1 billion, or 1 trillion. Usually this results in Player A creating that many permanents, gaining that much life, etc.

Obviously if Player B can also create an infinite combo to produce a counteracting resource (more attackers than Player A's blockers, more damage than Player A's life, etc.), then the original number can always be out-enumerated. However, it seems plausible to me that Player A could even be in danger if Player B's combo is not infinite. For example, if Player B could double a creature's power every turn for 40 turns, then it could easily wipe away 1 trillion life with a single attack, nevermind that it could have been attacking each turn along the way. Player A might be able to survive attacks and doublings over 120 turn by naming 1 duodecillion, which sounds like more than enough turns to cover any game of Magic. However, if Player B could double their creature's power multiple times per turn, then this still may not be enough. In fact, if each turn Player B could create a new copy of the source which doubles their creature's power, then greater-than-exponential growth could be achieved. Here my intuition is that a Googol life may not be enough to guarantee Player A indefinite safety, while a Googolplex may be, but is impractical to work with.

How quickly is it possible for Player B to deal life loss to Player A, using any non-infinite combination of cards? Of course, I am not only interested in the life case, but also in the cases of creating an arbitrary number of permanents, or other resources, which may have different thresholds for safety. For each type of resource an infinite combo that does not win a game outright can create, what number is so large that it is virtually impossible to regret not naming a larger number?

7
  • 2
    Why would you assume that there is a general number large enough that covers every possible game state? You have to factor in that the stack exists and player B can respond to player A if they're holding priority, and if they're not, it's a non-issue anyway. Also, for non-infinite situations, solving it is a matter of applying simple math (not saying it's necessarily possible to do so in a reasonable timeframe), whereas in general, it seems unanswerable to me. Also, you seem to assume that player A cannot simply start their combo again, which isn't true for all infinite loops. Jul 16 at 6:52
  • 3
    "what number is so large that it is virtually impossible to regret not naming a larger number?" - No such number exists, a finite number can always be exceeded and an infinite number is not allowed by choice (or is a forced draw). This is a non-question in that it will never ever be relevant to the game.
    – Nij
    Jul 16 at 7:05
  • 1
    I suggest reading soniccenter.org/sm/mtg/megacombo.html for an exploration of the limits of non-infinite combos. Note that that article does not provide a practical upper bound, because it applies deckbuilding and gameplay restrictions beyond the ones in the official rules. See also boardgames.stackexchange.com/questions/49938/… for a similar exploration.
    – murgatroid99
    Jul 16 at 8:42
  • 1
    @Nij it is not obvious that every finite number can be exceeded by a bounded combo, which is the sort of thing the poster is asking about. That's why it's phrased in terms of regret: if the only way to beat a number X is by a combo which could get arbitrarily large, then you won't regret choosing X because any other number would have been beaten too. Jul 16 at 9:08
  • 1
    I agree that there must be some upper bound, and since Turing Machines can be constructed in an MTG game it must be possible for a non-infinitely-repeatable sequence of actions to produce as much output as the busy beaver machine with however many states you can construct. So, a busy beaver number is a lower bound on that upper bound, which means that the upper bound is necessarily not computable.
    – murgatroid99
    Jul 16 at 23:51

4 Answers 4

10

There isn't one.

The question you're asking is a bit like asking this:

My friend and I can both eat infinite numbers of hotdogs. What number of hotdogs do I have to eat such that my friend can't then eat a bigger number of hotdogs?

There is no such number. Your friend can always choose a bigger number: they can choose the number you chose plus one, or squared, or times/plus any arbitrary number. I could pick an immense number like Graham's number, and they can just name Graham's number plus one. That's by the nature of how infinity works—there's always a bigger number.

You're proposing a scenario in which both my opponent and I might have access to infinite combos. (Using a finite amount of cards, but that's true of every deck.) If I have an infinite combo to gain an arbitrary amount of life during my turn, but my opponent has an infinite combo to attack with arbitrarily many creatures or deal arbitrary much damage to me, they can always pick a larger number than I picked. That's just how it works.

Once your opponent has an infinite combo they've probably won. The exception is if you can completely negate whatever they can do infinitely: e.g. they can do infinite Grapeshot, but you have Protection of the Hekma. There's no out-infinity-ing them on the defensive side.

7
  • 2
    The question did specify "using any non-infinite combination of cards", so I thought that infinite combos (which I interpret to mean combinations of cards that can produce an arbitrarily large amount of damage/life/permanents/etc.) were ruled out.
    – David Z
    Jul 16 at 22:58
  • 1
    My interpretation of the question is that player A has an arbitrarily repeatable combo (for which any number can be named) and player B has a combo that is not arbitrarily repeatable but may be fast-growing. It then asks whether there is a number that player A can name that exceeds whatever player B can accomplish. It's obvious that you can't just choose a number that cannot be exceeded by any number the opponent could choose, and by my reading the first sentence of the second paragraph of the question directly acknowledges that.
    – murgatroid99
    Jul 16 at 23:43
  • 6
    Essentially if the question isn't asking "what number can I name that won't be out-infinity-ied", I just have no idea what the question's asking and would consider it needs to be reformulated to clarify what's going on. As-is, this is my answer to my reading. Whether or not the combos are truly infinite in practice, they may be arbitrarily large for any theoretical number up to infinity, and at that point they're the same thing. Jul 16 at 23:44
  • 3
    Player A has an infinite combo and player B doesn't have an infinite combo. Is there a number player A could choose that player B can't exceed without an infinite combo? That's the question, as I read it.
    – murgatroid99
    Jul 16 at 23:54
  • 2
    I think this answer is sufficient since it covers the essence of the issue with the question very well, but I'll happily try myself at an alternative phrasing (with more or less the identical conclusion). Jul 17 at 8:49
3

There is no number in general

While there are really big numbers, most of those are likely not allowed under tournament rules because they cannot be explained to the average opponent in a reasonable amount of time.

I assert with minimal proof that the biggest number you could expect the average opponent to understand is a googolplex (if you don't accept this assertion, read the sections below). This is easy to beat with a finite combo.

Here's a 3 card example:

Each upkeep, have Extravagant Replication copy Doubling Season. Each doubling season doubles the number of doubling season tokens you get from Extravagant Replication, and each one in turn doubles the number of zombies you get from Liliana's emblem. All of this compounds every turn. This is superexponential growth. After 3 upkeeps, you have 2059 doubling seasons and 2 * 10625 zombies. After 4 upkeeps, you're already over a googolplex zombies. Also of note, at this point, you have crashed any digital implementation of MTG because even the scientific notation for how many zombies you have has more symbols than there are atoms in the observable universe.


What about 1010googolplex?

Still not high enough. Even if you keep stacking tens, it won't be high enough.

Let's take the combo from the previous section and run it for 30 turns. That's crazy big. Add Paradox Haze to increase the rate that the combo iterates and another Extravagant Replication to copy Paradox Haze each upkeep. Now, your unfathomably large numbers of Doubling Seasons will each double the number of Paradox Hazes you get, and each Paradox Haze will cause the Extravagant Replications and Liliana's emblem to each trigger an additional time on each subsequent turn, causing you to have a super-exponentially increasing unfathomably large numbers of upkeeps every turn, each of which causes super-exponential growth of your mass of Doubling Seasons and zombie tokens. You rapidly get into numbers that need advanced math notation to even represent. By the way, we're only at 5 cards so far. We haven't even started copying Extravagant Replication, or added Devilish Valet, Fungal Sprouting, and Panoptic Mirror.


What if both players are math PHDs?

This way, you can name incomprehensibly large numbers. Then we get into murky territory, but I still think that a reasonable interpretation of game rules says there isn't a valid number that will be high enough.

Some of the biggest mechanisms formally defined numbers result in numbers whose precise values aren't currently known (recommended supplemental reading). So, just being to name a really big number isn't enough; I would claim that your number needs to be computable in order to use it in a game of Magic. And once we have that stipulation, then no number you can name is high enough because you can implement a Turing Machine in Magic. Thus, any computable number you name can be exceeded by my finite combo of a truing machine that calculates your number plus 20 and deals you that much damage.


Beyond computability

In a lot of formats, there is no maximum deck size. If we ignore the "you must be able to shuffle your deck unaided" bit in the rules and have no time limit on the game, then for any number you pick as your life total, I can design a deck that will beat it with a finite combo. As a trivial example, I can make an arbitrarily large deck of Swamps and Relentless Rats which will eventually get there.


Joke Suggestion

Define "Garfield's Number" to be equal to one greater than the largest amount of damage you can deal with a finite combo in any current Magic format, with the sub-definition of the maximum deck size for formats that don't define one is the largest deck than can be shuffled without assistance by any human that has or will ever live. Set your life total to be that. If you can get a tournament judge to accept this as a valid number, congrats.

11
  • A few MtG formats have limited deck sizes, where at least your highlighted phrase would not apply. In addition, the conclusion that there is no general number just because there is no maximum deck size (meaning that if there were a maximum deck size, the question could more or less easily be answered) doesn't seem to be feasible, going by the other answers. Jul 19 at 18:43
  • @TheThirdMan fair point; I've de-emphasized the maximum deck size
    – Zags
    Jul 19 at 18:51
  • Please don't advance the idea of using "advanced math notation" to pick a number in a game of Magic. It's most likely obfuscation of free information and thus against the tournament rules.
    – Hackworth
    Jul 19 at 22:04
  • @Hackworth would you force someone to write out 10^100 longhand? 10^10000? 10^1000000? At some point writing it longhand is slow play, and were only at high school notation. You pretty quickly get into numbers whose decimal representation doesn't fit on the largest storage device imaginable
    – Caleth
    Jul 20 at 8:36
  • @Caleth If you can't write out a number without a defintion/notation that the average player wouldn't know, then the MTR forbids it on communication grounds.
    – Hackworth
    Jul 20 at 11:01
3

There isn't a specific number that generally applies, or a way to phrase it, other than "high enough".

Assuming player B has a non-infinite combo which they will fire after/in response to player A's infinite, but non-repeatable-after-finished combo, and the two combos are able to interact with each other (in all other scenarios, this is a non-issue, and @doppelgreener's answer applies), then it will always depend on player B's specific combo, and the answer always will be "you have to choose a number high enough so that player B's combo won't be able to top it"... which is rather obvious, and a predicament that's more or less already contained in the question.

There are two situations this can be broken down to:

Player B's combo doesn't involve hidden information

In the event that player A sees all combo pieces, player A can determine the amount of times they need to execute their combo by calculating the exact results of player B's combo. At this point, the limitations to figuring out a number are player A's math skills, time contains (especially in tournament scenarios when you're not only encouraged but required to play at a reasonable pace), or a combination thereof, but there's always a number that can be determined that's high enough to beat player B's combo.

Player B's combo does involve hidden information

In the event that player A can't know how many times player B can repeat their combo, there is no way for player A to know what number to choose - they will have to guess.

0

Hall of Famer Frank Karsten once recommended a googol (10^100) or a googolplex (10^10^100).

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.

The last paragraph is the key one.

Edit: for the record, the largest named number is Rayo's Number, and no non-infinite combo you might meet in Magic is ever likely to exceed it.

6
  • 1
    Adding a step or two beyond the three-card Devilish Valet combo (e.g. a second Fungal Sprouting) exceeds a googolplex, meaning it's not the be-all end-all biggest relevant number one could choose. Jul 17 at 21:15
  • @doppelgreener true, hence I've edited an even larger number in.
    – Allure
    Jul 19 at 15:24
  • 1
    Rayo's Number is not an actual known integer but an equation presented for argument's sake. Jul 19 at 16:26
  • 2
    downvoted because a random phrase by a random person does not necessarily make for a good answer. hall of famer and math expert, sure, but the only way I can interpret the post in question is that it's general practical advice not to just name a million, instead of a solution to the question what the highest number to name would be. As other people said, you could name "the number you said, plus one", and you would still be beat - that applies to large numbers as well. Jul 19 at 18:49
  • 2
    @Allure seconded. We once had a question about using mathematical terms as arbitrarily high numbers, and they are against the tournament rules.
    – Hackworth
    Jul 19 at 22:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .