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Suppose that my deck contains a Grindstone and 4 copies of Progenitus, and that every other card in my deck shares a color.

I use Grindstone's ability, targeting myself.

Because every card in my deck shares a color, Grindstone's ability will always repeat. Whenever it mills Progenitus, I shuffle my deck. Eventually (after much milling and much shuffling), my deck is reduced to 4 cards. I know that my deck contains only Progenitus, and therefore that all future iterations of Grindstone will have exactly the same result, but my opponent only knows that I have at least 2 copies of Progenitus - the other two cards in my deck could be anything (but are highly likely to be Progenitus).

Hypothetically from my opponent's perspective, I could conceivably mill a card other than Progenitus, which (ignoring the fact that no such cards have ever been printed) could conceivably have a replacement effect which would interrupt the loop. Announcing that this is an infinite loop would require announcing privileged information about my deck.

How do the rules handle this? Am I required to announce that the loop is infinite, even though it contains hidden information, or would the rules allow me to keep milling and shuffling until one of us resigns?

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  • Intentional Draws have always been a part of mtg.
    – Neil Meyer
    Apr 16 at 15:10
  • The only way every other card in your deck shares a color is if someone has a Painter's Servant in play, otherwise your lands have no color and will end the loop.
    – Andrew
    Apr 16 at 20:37
  • @Andrew - Not entirely true; there are cards which are dual-faced with spells on the front and lands on the back. If every land in your deck was such a card, all of the same color, and every other card in that deck was also that color, then that deck would consist of all cards of the same color.
    – Tim C
    Apr 16 at 20:43
  • @TimC Fair, Oops all spells is the exception (though I've never seen an oops all spells that would ever want prog, still is possible)
    – Andrew
    Apr 17 at 16:06
  • @Andrew It would never want Progenitus. Spell-lands are all monocolor, so it would need to generate 8 colored mana from treasures and other mana-fixing to be able to cast Progenitus. This deck has no gameplan except to draw the game.
    – Tim C
    Apr 17 at 16:38

1 Answer 1

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This scenario is not covered by the rules.

Loops of mandatory actions are already an edge case in the rules. This scenario, in which determining whether the loop of mandatory actions is endless depends on hidden information, is an edge case of an edge case. Remember that Magic: the Gathering is Turing Complete, which means that even without invoking hidden information, it is in principle not always possible to determine whether a sequence of mandatory actions will end.


Practically speaking, in a tournament, both players can agree that a game is a draw, even if the game rules do not cause a draw (MTR 2.5). In this scenario, the player performing the combo knows that the game can make no progress, and their opponent can infer that this is the case, so both would reasonably decide to end the game in a draw.

If either player proposes a draw and the other refuses, a judge could look at the library and confirm for both players that the loop is in fact infinite and the game is a draw. In addition, if the player performing the combo refuses a draw, they could get a Slow Play penalty because they know that the game cannot make progress. I can't see a reason that neither player would want a draw, but in that case the game would go to time and end in a draw anyway.

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  • If there are two gennies in your deck it is a mandatory loop. One on the other hand and the grindstone ability resolves with your opponent having one gennie in there library.
    – Neil Meyer
    Apr 16 at 15:09
  • I do remember in 2005 there was an infinite life gain deck in standard where the standard strategy was if you cannot stop the combo play for a draw. The lifegain deck wanted to win so did not want an ID and you could not win so a draw was better than a loss.
    – Neil Meyer
    Apr 16 at 15:15
  • The moment a single grind stone reveals two progenitus, there's no need to look at hidden information to determine the status of the loop as something that cannot be ended, that there are two cards that will continue the loop indefinitely is public information at that point, though proposing a draw as the player who knows they are still in his deck and having a judge confirm is an easy way to shortcut the long process to get to that point.
    – Andrew
    Apr 16 at 20:36
  • Part of the hypothetical in the question was that some other card may be able to break the loop. A card like that does not currently exist, but this answer accounts for it.
    – murgatroid99
    Apr 16 at 20:37

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