Jeskas Wille

Jeska's Will
Commander Legends
#187
Jeskas Wille KartenrückseiteJeskas Wille Kartenvorderseite
CT-Mindestpreis
—
CT-Marktpreis
—
US-Marktpreis
€34.44
TYP
Hexerei

FÄHIGKEIT
Bestimme eines. Falls du einen Commander kontrollierst, sowie du diesen Zauberspruch wirkst, kannst du beides bestimmen. • Erzeuge {R} für jede Karte auf der Hand eines Gegners deiner Wahl. • Schicke die obersten drei Karten deiner Bibliothek ins Exil. Du kannst sie in diesem Zug spielen.

FLAVORTEXT
„Sammle deine Kraft. Bereite dich vor." —Jeska zu Radha

ILLUSTRATOR
Izzy

LEGALITÄT
Standard Future Historic Timeless Gladiator Pioneer Modern Legacy Pauper Vintage Penny Commander Oathbreaker Standardbrawl Brawl Competitivebrawl Alchemy Paupercommander Duel Oldschool Premodern Predh Tlr

SPRACHEN
EN FR DE IT JP KR PT RU ES ZH-CN ZH-TW

REGELENTSCHEIDUNGEN
  • 2026-03-20 Once you've chosen both modes for the spell, it doesn't matter whether you continue to control a commander. This is true even if you somehow no longer control a commander as you finish casting the spell.
  • 2026-03-20 Once you've announced that you're casting a spell, players can't take any actions until you've finished doing so. Notably, opponents can't try to remove your commander to change how many modes you may choose.
  • 2026-03-20 You pay all costs and follow all timing rules for cards played with the permission from the last mode of Jeska's Will. For example, if one of the exiled cards is a land card, you may play it only during your main phase while the stack is empty and only if you have an available land play remaining.
  • 2026-03-20 Any of the exiled cards you don't play will remain exiled.
  • 2026-03-20 There's no extra bonus if you control more than one commander.
  • 2026-03-20 Even though the card is named after a specific character, controlling any commander will satisfy its condition.
  • 2026-03-20 The commander you control doesn't have to be your commander.
  • 2026-03-20 Use the number of cards in the target opponent's hand as Jeska's Will resolves to determine how much {R} to add.
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