Here's a solution to Challenge 1, featuring some Copper strategies.
Kingdom is Banquet, Tower and Masterpiece.
A — a Masterpiece/Tower strategy:
Buy Masterpiece if available $ > 6 and count in supply (Silver) > 1
Buy Colony
Buy Province
Buy Duchy if count in supply (Copper) = 0
Buy Gold if count in supply (Copper) > 0
Buy Masterpiece if available $ >= 5 and count in supply (Silver) = 2
Buy Silver
Buy Estate if count in supply (Copper) = 0
B — simple Banquet/Tower strategy that simply empties the Copper pile and then buys green:
Buy Duchy if count in supply (Copper) = 0
Buy Estate if count in supply (Copper) = 0
Buy Banquet
Buy Copper
C — a Banquet/Tower strategy that has been modified to counter A by using Banquets to gain Masterpieces, and also to get rekt hard by B (it still loses to B without the intentional crippling, but only like 55-45):
Buy Duchy if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Estate if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Banquet if count in supply (Masterpiece) > 0
Buy Masterpiece
Buy Banquet if count in deck (Copper) > count in all opponent's decks (Copper)
Buy Copper
A beats B 98.93-0.83 and beats BMU 88.41-11.09
B beats C 100.0-0.0 and beats BMU 99.46-0.48
C beats A 96.74-2.82 and beats BMU 99.97-0.03.
(using Geronimoo's accurate simulation)
Nice find! To clarify: I assume Banquet's gain preferences are the same as the buy preferences, so B's Banquets always gains 3 Coppers and C's Banquets always gain Masterpiece unless the pile is empty? And does the "intentional crippling" refer to the "if count ..." conditions?
FWIW, I would consider intentional crippling against the spirit of the challenge. If you allowed it, you could solve the challenge 100-0 in any kingdom that contains 3 cards a, b, c such that BM+either of a, b, c beats pure BM:
Player A plays BM+a, except they stop doing anything if card c is gained by the opponent.
Player B plays BM+b, except they stop doing anything if card a is gained by the opponent.
Player C plays BM+c, except they stop doing anything if card b is gained by the opponent.
In my solution, there should theoretically exist a strategy that beats all three (and I don't think a solution exists where that is not the case), but it shouldn't be as easy to find.
Any finite game has a (probabilistic) "optimal strategy" for each player by Nash theory. Dominion has only finitely many (relevantly different) gamestates in any kingdom without tokens or infinite loops (like the kingdom you gave), so in such a game there should indeed always be a strategy that beats all three.