If I have a Thousand-Year Storm on the battlefield and I cast Epic Experiment, does each spell cast off the experiment get sequentially more storm counts?



For example: If I have a Thousand-Year Storm on the battlefield and I cast Epic Experiment for x=4 and I hit 4 copies of lightning bolt, how many instances of lightning bolt get created total?

My guess is 2+3+4+5=14 lightning bolts?
(1+1)+(1+2)+(1+3)+(1+4)
where the first number in each parenthesis pair is the base card and the second number is the number of storm copies for that base card?


Same question again, but instead of hitting 4 bolts I hit 4 Empty the Warrens?

My guess is 3+5+7+9=26 Empty the Warrens?
(1+1+1)+(1+2+2)+(1+3+3)+(1+4+4)


Same question one more time, but this time I have 4 opponents and I want to know how many times I get punched in the face for attempting this in a multiplayer game?

There will be follow up questions.

Yes, because you are casting them. However, these are all triggers, so you resolve the triggers one at a time, making this a bit easier. Hitting 4 Lightning Bolts would result in 14 Lightning Bolts going through (2+3+4+5). For Empty the Warrens, however, you're copying the spell, so you're really just adding 1 to the effective storm count, since the copy won't storm off, so adds 1 to the number of copies being generated not doubling the copies, resulting in 18 Empty the Warrens going off (3+4+5+6).

Your math on Empty the Warrens is incorrect. You're right to note that the copies don't add to the storm count, but you missed the fact that storm will create N copies and Thousand Year Storm will also create N copies.

Flyheight's equation looks correct to me, except the final sum.
(1+1+1)+(1+2+2)+(1+3+3)+(1+4+4) = 3+5+7+9 = 24

Let's look at the second Warrens as an example. The storm trigger sees two spells cast before it (Epic Experiment and the first Empty the Warrens), so it creates two copies. Then, the Thousand Year Storm trigger sees two instant or sorcery spells cast before it, creating two additional copies.

Eeeeeeeeexcellent.
Figures. Got the math and rules theory right for the equation, can't add for poop.

OK, follow up questions will be a bit more complex.


1)TYS and EE again, this time hitting one Brainstorm and one Lightning Bolt.

Am I correct in saying I get to choose which one goes where on the stack and thus how many of each I get? So I can get either:
a)2 brainstorms and 3 lightning bolts
or
b)2 lightning bolts and 3 brainstorms



2)This one has to do with the order of resolutions.
TYS makes triggers whose counts are decided when the spells are cast, however the TYS triggers don't actually go onto the stack until EE finishes resolving and they all go on the stack at the same time. I believe this means that I should be able to order the resolution of the TYS triggers however I like, independent of the order I cast the spells off EE.

In this example I will hit one windfall, one inspiration, and one cerebral vortex.
According to the stated assumptions above I believe I am able to order the original spells and the TYS triggers as follows:
>Stack Top (resolves first)
>1x Inspiration copy
>2x Windfall copies
>3x Cerebral Vortex copies
>Cerebral Vortex original
>Windfall original
>Inspiration original
>Stack Bottom (resolves last)

Yes, you can cast the spells in whatever order you want.


This is also correct, but just to be specific, each set of TYS copies will resolve before the other TYS triggers. Like so:

1) Epic Experiment resolves, spells are cast, then copy triggers are put on the stack.
2) Inspiration TYS trigger resolves, putting a copy of Inspiration on top of the stack.
3) Inspiration copy resolves.
4) Windfall TYS trigger resolves.
5) Windfall copy #1 resolves.
6) Windfall copy #2 resolves.
7) Cerebral Vortex TYS trigger resolves.

And so on. This is relevant if your opponent has something like Mindbreak Trap. There's no moment where all the copies are on the stack at once.

Thank you for that clarification. I think that's all my questions I have about this interaction for now.