Can infinite scry 1 be shortcut by searching through your deck?

It appears that with a scry 1 ability and some infinite repeat mechanism, one can cut to the card one desires, keeping the deck in order as a shortcut. But let's say I can't repeat it infinitely for some reason (maybe it has a cost). If my library has 40 cards in it, do I need to be able to scry 1 40 times to shortcut this? How many minimum do I need to do this every time, and do my odds of getting the card I want on top of the deck increase faster as I am able to trigger more scry one repeats?

I don't know much about MTG, so please pardon my vernacular.

  • Did you read the explanation in my answer to that question regarding the details of the scry 1 shortcut?
    – murgatroid99
    Oct 4, 2018 at 4:41
  • 1
    Yes, but does that answer my question? If it does, please explain.
    – user45266
    Oct 4, 2018 at 4:43

1 Answer 1


In order to shortcut some finite number of "scry 1"s as "look at your entire deck, then cut the deck at a specific point to put a chosen card on top", it is enough to be able to scry 1 a number of times equal to twice the number of cards in your library. So, if you have 40 cards in your library, you can use the shortcut if you can scry 1 at least 80 times.

The reason for this is that when you perform that shortcut, you see the order of the cards below the desired card as well as above. In order to get the same information by repeatedly scrying, you would have to put the card you are looking for on the bottom, then cycle through the entire library again to put it back on top. Following this algorithm, in the worst case, if the card you are looking for is at the bottom of your library, you would cycle through your entire library twice, for a total of 2 scrys for each card in your library.

Strictly speaking, you can do it with two fewer scrys. If the card you are looking for is the very last card, then you can just leave it on top of your library at the end of the first step and not cycle through your deck again. And if the card you are looking for is second from the bottom, then in the second step you don't have to look at either of the cards at the bottom to end up with the card that was second from the bottom on top.

  • 1
    Very minor nitpick: If the card you desire is on the bottom you only need to go through your library once. Furthermore, if you know exactly what cards still remain in your library you will know what the bottom card is once you've seen the card on top of the bottom card. This means that if the card you want is directly on top of the bottom card you only need to go through your library once as well. This means that the minimal number of scries you need is twice the number of cards in your library minus two. So if you have 40 cards left you need to have 78 instances of scry 1.
    – Peter
    Oct 4, 2018 at 19:38
  • 4
    I know. That's why I tried to make it clear that the numbers I gave aren't exact. Personally, I think it's easier, both for memory and for explaining the shortcut to have a simple heuristic like "twice the size of the library".
    – murgatroid99
    Oct 4, 2018 at 19:49
  • 1
    You should not use “plus or minus a couple” when it is truly precisely “minus a couple”—even if we aren’t being so precise, an upper bound of twice your library (i.e. “twice your library is sufficient to justify the shortcut”) is useful, while an approximate number that could go either way (as suggested by “plus or minus”) suggests that we should not assume that twice our library is necessarily enough (which it is). In fact, ultimately, the minus two factor is neither complex to calculate nor to explain—why not just include it and be precise? It would make a better answer.
    – KRyan
    Oct 5, 2018 at 2:49
  • 3
    OK, I made the edits
    – murgatroid99
    Oct 5, 2018 at 3:45
  • 1
    The OP is talking about a specific shortcut, described in the linked question, in which you look at the whole deck, then cut to a specific position. This is why I included looking at the whole deck.
    – murgatroid99
    Oct 5, 2018 at 15:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .