15

Do you pay one or two mana to bounce a transformed Delver of Secrets with Repeal? In total I mean, does x have to be equal to 1 or 0?

21

Two mana; for the converted mana cost (CMC) of a transformed card, you have to look at the untransformed side:

711.4b While a double-faced permanent’s back face is up, it has only the characteristics of its back face. However, its converted mana cost is calculated using the mana cost of its front face. If a permanent is copying the back face of a double-faced card (even if the card representing that copy is itself a double-faced card), the converted mana cost of that permanent is 0.

Since Delver of Secrets' converted mana cost is 1, X = 1 and you'll have to pay 2 mana to Repeal it.

Insectile Aberration's side doesn't show a mana cost, and while for some cards that means they have converted mana cost 0, not so for double-faced cards

202.3a The converted mana cost of an object with no mana cost is 0, unless that object is the back face of a double-faced permanent or is a melded permanent.

It used to be different; before Shadows of Innistrad, this exception didn't exist and Insectile Aberration had a CMC of 0. See this link mentioned by @J.Sallé in the comments.


Please note that 'flipped' is another (but similar) mechanic where a card is rotated 180 degrees but its front face stays up, e.g. Akki Lavarunner. The same idea applies there too, though (see rule 709.1c).

  • 11
    I think it's worth noting that the ruling regarding this interaction was changed back in 2016 with the release of Shadows over Innistrad. Prior to that, a transformed DFC's CMC used to be 0. – J. Sallé Aug 26 at 15:38
  • I learned about this from hearing that Isolate can be used on Adanto (flipped side of Legion's Landing). Feels intuitively bizarre to hit a land, and for me a point in favor of the old way. – Vandermonde Aug 26 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.