You are in a Magic: The Gathering commander game. It is now your first main phase and you control these permanents:

  • One Miirym, Sentinel Wyrm: "Whenever another nontoken Dragon enters the battlefield under your control, create a token that’s a copy of it, except the token isn’t legendary if that Dragon is legendary."
  • Exactly $n$ instances of the enchantment Parallel Lives*: "If an effect would create one or more tokens under your control, it creates twice that many of those tokens instead". $n$ is positive

You then cast Astral Dragon, which is indeed a Dragon: "When Astral Dragon enters the battlefield, create two tokens that are copies of target noncreature permanent, except they’re 3/3 Dragon creatures in addition to their other types, and they have flying". Assume that

  • in the ensuing resolution you only target permanents you control and that have names mentioned in this puzzle (Miirym, Sentinel Wyrm; Parallel Lives; Astral Dragon)
  • there is no external interference (instants played by opponents, abilities of other permanents, etc.)
  • you cast no other spells before you proceed to combat.

What is the exact maximum number of 3/3 Dragon creature tokens you can create during this first main phase?

*This is not the only card to have the stated doubling effect. The other two non-legendary ones are Anointed Procession (which I own in real life) and Doubling Season.

The exact answer is a very large number that requires defining auxiliary functions to write down compactly. It is however far less than $2\uparrow\uparrow\uparrow6$.

  • $\begingroup$ "non-legendary permanents like Parallel Lives", so it doesn't have to be Parallel Lives? $\endgroup$
    – justhalf
    Jul 8, 2022 at 10:42
  • $\begingroup$ "you only target permanents you control and that have names mentioned in this puzzle", Parrallel Lives is the only legal target, so why not just say that it copies that $\endgroup$
    – Ivo
    Jul 8, 2022 at 10:57
  • $\begingroup$ Just with n = 1 seems to be very hard to calculate. Let alone a generalisation for higher n $\endgroup$
    – Ivo
    Jul 8, 2022 at 11:27
  • $\begingroup$ @IvoBeckers changing to just PL is a good idea; edited. $\endgroup$ Jul 8, 2022 at 15:07
  • $\begingroup$ Is there any way to make this a self-contained puzzle? In its current state it relies on a bunch of links to a third party website for essential information. $\endgroup$
    – fljx
    Jul 9, 2022 at 8:25

2 Answers 2


Partial answer:

Let's first just see what happens if you have 1 Parallel Lives. When Astral Dragon enters the battlefied you have 2 choices:

  1. You let Miirym's ability resolve first
  2. You let Astral Dragon's ability resolve first

Case 1:

Miirym's ability will put 2 Astral Dragon tokens on the battlefield. There are now 3 Astral Dragon ETB abilities on the stack. They resolve one by one.
The first one will create $2 \cdot 2 = 2^2$ Parallel Lives tokens.
The second one will then create $2 \cdot 2^{1 + 2^2} = 2^6$ Parallel Lives tokens.
The third one will then create $2 \cdot 2^{1 + 2^2 + 2^6} = 2^{70}$ Parallel Lives tokens for a total of $2^2 + 2^6 + 2^{70}$ tokens.

Case 2:
Astral Dragon will create $2 \cdot 2 = 4$ Parallel Lives tokens.
Miirym will create $2^5 = 32$ Astral Dragon tokens, causing also that many triggers to go on the stack.
The first one will then create $2^6$ Parallel Lives tokens.
The second one will then create $2^{70}$ Parallel Lives tokens.
The third one will then create $2 \cdot 2^{1 + 2^2 + 2^6 + 2^{70}}$ Parallel Lives tokens.
As you can see, this will blow up extremely fast. I could easily go on like that, but it will be a huge expression in the end. I'm not sure if there's a compacter notation you can write to get the total.

And this is only for $n = 1$. Maybe there's a better way to write it and to give a generalized notation for it, but I'm not mathematically adept enough for that.

In any case, no matter the $n$, there will only be one choice to make, and that's the choice I demonstrated, and case 2 will be clearly the better choice then.

  • $\begingroup$ Excuse my ignorance, I'm new to MtG. In case 1, wouldn't Miirym's ability generate 4 Astral Dragon tokens due to Parallel Lives? $\endgroup$
    – justhalf
    Jul 8, 2022 at 12:11
  • $\begingroup$ @justhalf without Parralel Lives it would have just be 1 token. Parallel Lives doubles that to 2 $\endgroup$
    – Ivo
    Jul 8, 2022 at 12:14
  • 1
    $\begingroup$ Ah, I somehow swapped the definition between Astral (which creates two) and Miirym (which creates one). Sorry and thanks! $\endgroup$
    – justhalf
    Jul 8, 2022 at 12:15
  • 1
    $\begingroup$ You can use recursive formulation to get the exact formula, btw. $\endgroup$
    – justhalf
    Jul 8, 2022 at 12:16

The recursive function that models this question is: find f(32) given:

f(0) = 5

f(x) = f(x-1) + 2 ^ (f(x-1) + 1)

Explanation of the function:

f(0) = 5 because when the first AD comes into play it makes 2 PLs which is doubled to 4. 4 + 1 = 5. We are looking for f(32) because 2^5 ADs are created by Miirym because we have 5 PLs in play.

f(x) = f(x-1) + 2 ^ (f(x-1) + 1) because the number of PLs after each AD enters = Number of existing PLs + number of new PLs (2 ^ (f(x-1) + 1)). The number of new PLs is (2 ^ (f(x-1) + 1)) because AD creates 2 which is doubled a number of times = existing PLs

I plugged this function into Hypercalc at the suggestion of Gottfried Helms

The answer is 30 PT (3.553934904655 × 10^20)

Which means a power tower of 10s, 30 high with (3.553934904655 × 10^20) on top ie: 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^(3.553934904655 × 10^20)

  • $\begingroup$ So what is the actual answer to the question? $\endgroup$ Sep 17, 2022 at 17:02
  • $\begingroup$ @Rand al'Thor the answer is extremely large. It is at least 2^^33 but that is as far as I have got. $\endgroup$ Sep 17, 2022 at 17:25

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