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.