Amy has assembled the Altar of the Brood + Liquimetal Coating + Saheeli Rai combo (this basically equates to “0: each opponent mills one”). Only problem is her opponent has a Progenitus in her library. I know that with something like Emrakul, the Aeons Torn, Amy wouldn't be able to shortcut to just having Emrakul in her opponent's library due to not being able to specify how many times she will need to execute the loop to get there. Is this the same for this case, even though each iteration of the loop will (almost certainly) advance the game state toward this end goal? If so, how would you handle this?

David, I'll take your word for it that this loop variant boils down to your summary (0: mill 1). I'm just not gonna do the “math” to figure that out, right now… ;)

However, this is Yet Another Loop Question (or Yet Another Loophole Quest). Please refer to previous threads; no, this example - albeit with different cards - does not get a different answer.

Here's a couple where I pretty clearly say “you can't do that”:
http://apps.magicjudges.org/forum/post/112379/
http://apps.magicjudges.org/forum/post/32522/

d:^D

Fair enough, Amy can't shortcut through the loop until her opponent just has Progenitus, but that still leaves the question of how to handle this if it comes up. For example, it seems like we can't force Amy to believe that Nicole has a Progenitus in her deck, so she should be able to at least continue the loop until she sees it. If we're using the standard for fragmented loops found in the CR, the player would need to advance the game state at this point. If the player then takes some action outside the loop, can she start back up again after?

Something similar to legacy's four horsemen.

Because the loop isn't 100% guaranteed. We cant say i want to mill u “x”
number of times until the last card in your deck is the Progenitus.

I think there will be a slow play warning if the loop is repeated many
times without a progressive gamestate.

On 5 Oct 2016 08:59, “David Elden” <forum-30432-f0b5@apps.magicjudges.org>
wrote:

Fair enough, Amy can't shortcut through the loop until her opponent just
has Progenitus, but that still leaves the question of how to handle this if
it comes up. For example, it seems like we can't force Amy to believe that
Nicole has a Progenitus in her deck, so she should be able to at least
continue the loop until she sees it. If we're using the standard for
fragmented loops found in the CR, the player would need to advance the game
state at this point. If the player then takes some action outside the loop,
can she start back up again after?

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I understand that they cannot say how many times it will take to get to Progenitus being the only card left in the deck but there is a significant difference between Emrakul and Progenitus in that Progenitus only shuffles itself whereas Emrakul shuffles the entire graveyard. To me this seems more like the Scry example in that we know for certain it will get to the final result of Progenitus being the only card in the library so why couldn't we just shortcut to the point?
Or is it essentially that the shuffle is involved that we cannot?

Edited Matthew Jacques (Oct. 5, 2016 10:41:02 AM)

With scry you can get that result with a mathematically fixed number of iterations. With Progenitus you can't. You might reveal Progenitus as the first card a trillion times in a row. It actually is the same as the Emrakul example there, with the loop having to go an uncertain amount of time (both will eventually get there, although the Progenitus will probably get there faster).

Would it be okay to allow the player to go through it 2-3 (or even 5?)
times with the stated purpose of “milling my opponent enough he has a
chance to deck himself”?

2016-10-05 12:17 GMT-04:00 Jona Bemindt <

Jona

Yeah I began to think about that after I made the post

I believe that by the current rules, if we are asked to intervene, we should stop this loop only when a player mills Progenitus twice in a row, then he is in a loop that doesn't reach a new game state. While he is lucky and Progenitus isn't the top card I would allow the loop to continue, because milling Progenitus after other cards have been milled doesn't get you the same gamestate

Just curious but would this be allowed to work for a while at least? The board state is still changing each time the loop is done as cards are being put in the graveyard. This is not like emrakul where the whole gy gets shuffled back and the player has to repeat it an arbitrary number of times to get what he wants. the player that is milling the opponent with progenitus is still progressing the board state by putting cards from his opponent's deck into the grave. Would slow play not come into effect here until he flips progenitus over twice in a row and fails to advance the board state with the same action?

I.E.
step 1: make combo
step 2: mill lets say 20 cards
step 3: hit progenitus and shuffle back in
step 4: mill 10 cards
step 5: hit progenitus and shuffle back in

Until he hits progenitus immediately after being shuffled back (thereby doing the same action with the same end result twice in a row) I still see this as advancing the board state.

I could also be completely wrong here and just shoving my foot in my mouth. In which case apologies for the long post.

Edit: Exactly what the person above me put. Did not see his post

Edited Mitchell Wetherson (Oct. 6, 2016 10:20:36 PM)

I think it's worth taking a look back at the Four Horsemen combo.

Basically, in that combo, if you want play around graveyard hate, you need to mill until you hit an Emrakul in your own deck. In order to make progress, you need to mill another from a small set of specific cards first. For each time through the loop, you could either make progress or not make progress.

In that combo, the expected number of shuffles required to reach the desired end state was approximately (based on my personal rough math right now, which could easily be wrong) 6. 6 isn't a lot of shuffles to end a game. But, unfortunately, 6 wasn't guaranteed to do the job. The maximum number of iterations is not bounded.

Now let's look at milling an opponent with Progenitus.

For each shuffle, we expect to mill half the deck. That is to reduce the number of cards in the deck to 1 (just Progenits), we expect just north of log2(n) shuffles where n is the number of cards remaining in the opponent's deck at the start of the process. Again, we can call this number approximately 6. And, again unfortunately, 6 is not guaranteed to do the job. There is an unbounded maximum number of iterations.

So, even though the two cases may feel quite different, mathematically they are essentially the same. There is a loop that has a reasonably low expected number of iterations to complete, but is not bounded or deterministic, which means we cannot pick a set number of iterations to arrive at a known result.

And that's why, in both cases, you are not allowed to re-enter the loop once it has been broken by a shuffle.

Thanks, Josh, that's a great conclusion, and that “Four Horsemen” analogy is appropriate.

I've learned, through painful experience, that once one person introduces math into these discussions, it spawns a tangent of “corrections” of that math. Anticipating that, and acknowledging that we already have a good conclusion, I'm wrapping up this thread.

d:^D