Number of Cards milled

Asked by MindAblaze 4 years ago

I control 2 Halimar Excavator s, a Xenograft naming Ally and an Undead Alchemist . I resolve a Harabaz Druid .

How many cards are Milled in total?

-Fulcrum says... Accepted answer #1

When Harabaz Druid enters the battlefield, you control 4 allies. Both Halimar Excavator s will trigger for four. So eight total.

September 26, 2014 12:31 a.m.

MindAblaze says... #2

If I hit no creatures in the top four cards, you are correct.

September 26, 2014 12:44 a.m.

SimicPower says... #3

Both Halimar Excavator abilities are put on the stack. When the first one resolves, 4 cards get milled. Assume all of those cards are creatures. You get 4 Zombie Ally tokens. Both Halimar Excavator abilities are put onto the stack again, 4 times each. There are now 9 of these triggered abilities on the stack. The next one that resolves will mill you 8 cards, and the cycle continues. If you keep milling creature cards, you will end up milling your entire library.

September 26, 2014 12:59 a.m.

-Fulcrum says... #4

So a minimum of eight cards, and then X * 2 more for each creature card milled where X is the number of allies you control, which increases by 1 for each creature card milled. Seems legit.

September 26, 2014 1:14 a.m.

-Fulcrum says... #5

Can someone convert that to a mathematical formula? I can't, because I suck at math, but I would like to see how that looks in an equation.

September 26, 2014 1:15 a.m.

SimicPower says... #6

Halimar Excavator + Undead Alchemist + Xenograft could be an interesting combo with Labratory Maniac, or any other ally that has abilities that trigger when other allies enter the battlefield. Not the most efficient combo, but could be a fun casual deck.

September 26, 2014 1:26 a.m.

MindAblaze says... #7

It's complicated to calculate because every time a trigger resolves, if you've hit a creature you now have another (minimum) two triggers to resolve before you know X for the original trigger.

September 26, 2014 9:39 a.m.

MindAblaze says... #8

I'm going to mark Fulcrum's first response because it technically answers what I asked, even if it was kind of a trick question.

I'd still like to see a formula though, that would make life easier.

September 26, 2014 9:30 p.m.

BorosPlayer says... #9

Here's the formula where x=number of creature cards hit in the first 8, y=number of creature cards hit in the next wave of milling, z=number of creature cards hit in the third wave of milling, etc.

8+X(X+4)+Y(X+Y+4)+Z(X+Y+Z+4)...

So assume each time 1/3 of the cards are creatures.

8+2(2+4)+4(2+4+4)+13(2+4+4+13)+HugeNumber(2+4+4+13+HugeNumber) and so on

which is more than enough to mill everyone.

September 26, 2014 9:49 p.m.

BorosPlayer says... #10

I'll expand it more.

8+2(2+4)+4(2+4+4)+13(2+4+4+13)+100(2+4+4+13+100)+4100(2+4+4+13+100)

five rounds of milling and an average of approximately 520000 cards milled or 8600 decks.

September 27, 2014 1:18 p.m.

This discussion has been closed