I have a Rest in Peace and Helm of Obedience on the battlefield.

I activate the Helm for 1 mana, and my opponent starts milling her entire deck. Assume the only relevant card in her deck is a Blightsteel Colossus. What happens?

My understanding is, she has two options, as there are two replacement effects in play. She gets to chose which one. Based upon her choice, she enters one of two scenarios:

Option 1: She chooses to exile the Blightsteel, and eventually mills her entire deck and loses.

Option 2: She chooses to shuffle the Blightsteel into her deck, and continues milling. Eventually she hits Blightsteel again, she shuffles again, continues. Eventually, Blightsteel is the only card in her deck, and she enters an infinite loop. In this scenario can she call the game a draw? Or is she forced to lose?

  • Because of the wording of the Colossus, i think it's a mandatory redirect the card back into deck instead of sending it to graveyard (exile due to Rest in Peace), mostly due to how it doesn't say you MAY reveal it and shuffle it back, it just says you do reveal it back. Jul 31, 2020 at 20:43
  • and I agree with @GendoIkari that the Helm of Obedience won't go into an infinite loop since it would only logically attempt to mill you once, not mill you, see you're empty, then sees a card was shuffled back and then attempt again. Jul 31, 2020 at 20:45
  • @TylerTrinh I'm not sure if I was clear enough; you say you agree with my comment but it sounds like your'e saying the opposite. I was saying that Helm would cause an infinite loop even if the library does become empty.
    – GendoIkari
    Jul 31, 2020 at 20:47
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    @GendoIkari i think the game will end though because... 104.4b If a game that’s not using the limited range of influence option (including a two-player game) somehow enters a “loop” of mandatory actions, repeating a sequence of events with no way to stop, the game is a draw. Loops that contain an optional action don’t result in a draw Jul 31, 2020 at 20:55

1 Answer 1


Even if your opponent keeps reshuffling the Blightsteel Colossus, eventually their whole deck would be exiled. When their Blightsteel Colossus is the last card in their library, but possibly earlier than that, they may not shuffle it back again and must exile it instead. When your opponent's library is empty, Helm of Obedience cannot continue its process, its resolution finishes, and the game continues to their likely loss.

Rest in Peace prevents any card from entering a graveyard, therefore Helm of Obedience's stop condition would never occur, eventually resulting in a library that only contains the Blightsteel Colossus. If your opponent then continues to shuffle the Colossus back into their library, the game state would no longer change, and the opponent has to choose a different option, i.e. to not shuffle the Colossus, resulting in the colossus being exiled and the library to become empty.

722.3. Sometimes a loop can be fragmented, meaning that each player involved in the loop performs an independent action that results in the same game state being reached multiple times. If that happens, the active player (or, if the active player is not involved in the loop, the first player in turn order who is involved) must then make a different game choice so the loop does not continue.

Once your opponent's library is empty, the process of Helm of Obedience cannot be repeated, its resolution finishes, and the game continues as normal.

Note that the Blightsteel Colossus being the last card in the library is just a special case. The general case is when your opponent shuffles the BC back into their library, and the BC happens to get shuffled to the very top so that Helm reveals it immediately again. Then the same game state has been reached twice (BC on top of X other cards), at which point we have fulfilled the conditions for a fragmented loop, and your opponent may not take the same decision as before, i.e. he has to exile the BC instead of reshuffling it.

Non-deterministic loops (loops that rely on decision trees, probability or mathematical convergence) may not be shortcut. A player attempting to execute a nondeterministic loop must stop if at any point during the process a previous game state (or one identical in all relevant ways) is reached again. This happens most often in loops that involve shuffling a library. ref

You have to play out the loop because you may not shortcut it. In the worst case, if the BC gets shuffled 2nd from top every time, you would have to perform ~X shuffles, where X is the number of cards remaining in the library. Therefore, if this happens in a tournament with time constraints looming, immediately call a judge to discuss the matter and how to proceed.

If time is not a constraint, the result is always the same: Your opponent will be caught in a fragmented loop at some point and has to exile the BC (in fact, also multiple BCs and all other cards that may be shuffled back), along with the rest of their library.

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    A fragmented loop involves choices/actions taken by both players. The player who activated Helm of Obedience isn't making any choices or performing any actions in the loop.
    – GendoIkari
    Jul 31, 2020 at 20:52
  • @murgatroid99 That's what I thought but I cannot find a clear explanation or definition of fragmented loop, and 722.3 isn't completely clear on that point. Also, looking at the basic rules for loops in general, they all appear to be only related to shortcuts; which don't apply at all to this situation. Only a player with priority can propose a shortcut (722.2a), and that doesn't apply here.
    – GendoIkari
    Jul 31, 2020 at 20:58
  • I think the "independent action" part is the closes rules ref here. "Independent" implies "independent of the action that the other player took". But even so, I still am not convinced that anything under 722 is applicable here at all.
    – GendoIkari
    Jul 31, 2020 at 21:02
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    Let us continue this discussion in chat.
    – GendoIkari
    Jul 31, 2020 at 21:10
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    (re my unlikely much more than ~6 shuffles comment: a quick simulation suggests that for a 50-card deck there is a ~6% chance that you would need 10+ shuffles) Aug 1, 2020 at 0:54

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