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Jeska's Will

Jeska's Will

Sorcery

rare

Choose one. If you control a commander as you cast this spell, you may choose both instead. • Add for each card in target opponent's hand. • Exile the top three cards of your library. You may play them this turn.

$37.86 ✦ $41.68View on Scryfall →

Semantic Tags

Impulse Drawmechanical

Exile-based card advantage — exile cards from the top of your library and cast them temporarily. Red's primary draw mechanic.

Rampmechanical

Accelerates mana production beyond the one-land-per-turn baseline. Umbrella tag — see ramp subtypes for land ramp, mana dorks, mana rocks, and rituals.

Format Legality

commanderlegal
oathbreakerlegal
modernnot legal
pioneernot legal
standardnot legal
legacylegal
vintagelegal
paupernot legal
historiclegal
explorernot legal
alchemynot legal

Price history

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Printings (7)

Rulings

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.