If I have 1 Ezreal, and I play Hard Bargain, then my opponent pays 2, do I repeat the cost of Hard Bargain to play it again for 1 or 2? Now what happens if I have 2 Ezreals in this situation? What about 3?
Hard Bargain costs [2] energy and has Repeat [2] (an optional additional cost).
Ezreal, Prodigy has the ability: "Optional additional costs you pay cost [1] or [A] less."
According to the retrieved FAQs and rules:
From FAQ 8415: "Ezreal, Prodigy reduces additional costs that are optional to pay... Repeat and Accelerate are the two main sources of optional additional costs."
From Rule 353.4.d: "Discounts can reduce additional costs, including to 0."
From Rule 353.4.c: "If a discount applies a minimum cost, that minimum applies only to that discount."
Here's what happens in each scenario:
With 1 Ezreal, Prodigy:
- Hard Bargain's Repeat cost is normally [2]
- Ezreal reduces it by [1]
- You pay [1] to Repeat Hard Bargain
With 2 Ezreals, Prodigy:
- Hard Bargain's Repeat cost is normally [2]
- Each Ezreal applies its reduction independently
- You can apply them in any order:
- First Ezreal reduces [2] → [1]
- Second Ezreal reduces [1] → [0]
- You pay [0] to Repeat Hard Bargain (it's free)
With 3 Ezreals, Prodigy:
- Hard Bargain's Repeat cost is normally [2]
- The first two Ezreals reduce it to [0] (as above)
- The third Ezreal's reduction cannot reduce it below [0] (Rule 353.5: "Energy and Power costs can't be reduced below 0")
- You still pay [0] to Repeat Hard Bargain
The key principle: Multiple copies of Ezreal, Prodigy stack their cost reductions (FAQ 6329 confirms this), and you can apply them in any order to minimize the final cost.