Since we know the puzzle has an answer (or else it couldn't be asked here), the answer must be

blue

Under the assumption there is a unique answer, we can choose our moves rather than picking randomly, since any set of moves must lead to the same result. Simply pull each pair of colors sequentially (YB, RY, RB), reducing the number of all colored discs by 1 after each round of pairs (since each round removes two of each disc and puts one back in). After 1 round we have {5R, 6Y, 7B}, then {4R, 5Y, 6B}, and so on. Eventually, we get to {0R, 1B, 2Y}. From here, our only move is to {1R, 0B, 1Y}, which finally leads us to {0R, 1B, 0Y} and the end of the game.

We can also leverage the existence of an answer by observing that this is

at its core, a parity problem. Since R and Y have equal parity there isn't really any way to distinguish them - if R could be the answer, so too could Y. B has different parity and is unique in this regard, and can be the only unique solution.

Since we know the puzzle has an answer (or else it couldn't be asked here), the answer must be

blue

Under the assumption there is a unique answer, we can choose our moves rather than picking randomly, since any set of moves must lead to the same result. Simply pull each pair of colors sequentially (YB, RY, RB), reducing the number of all colored discs by 1 after each round of pairs (since each round removes two of each disc and puts one back in). After 1 round we have {5R, 6Y, 7B}, then {4R, 5Y, 6B}, and so on. Eventually, we get to {0R, 1B, 2Y}. From here, our only move is to {1R, 0B, 1Y}, which finally leads us to {0R, 1B, 0Y} and the end of the game.

We can also leverage the existence of an answer by observing that this is

at its core, a parity problem. Since R and Y have equal parity there isn't really any way to distinguish them - if R could be the answer, so too could Y. B has different parity and is unique in this regard, and can be the only unique solution. As an extension of this, we can see that when all three colors have equal parity, there is no unique solution - starting from {1R, 1B, 1Y}, we could finish with any of the three colors in the bag, as we can remove any two discs and finish the game in one move.

Since we know the puzzle has an answer (or else it couldn't be asked here), the answer must be

blue

Under the assumption there is a unique answer, we can choose our moves rather than picking randomly, since any set of moves must lead to the same result. Simply pull each pair of colors sequentially (YB, RY, RB), reducing the number of all colored discs by 1 after each round of pairs (since each round removes two of each disc and puts one back in). After 1 round we have {5R, 6Y, 7B}, then {4R, 5Y, 6B}, and so on. Eventually, we get to {0R, 1B, 2Y}. From here, our only move is to {1R, 0B, 1Y}, which finally leads us to {0R, 1B, 0Y} and the end of the game.

Since we know the puzzle has an answer (or else it couldn't be asked here), the answer must be

blue

Under the assumption there is a unique answer, we can choose our moves rather than picking randomly, since any set of moves must lead to the same result. Simply pull each pair of colors sequentially (YB, RY, RB), reducing the number of all colored discs by 1 after each round of pairs (since each round removes two of each disc and puts one back in). After 1 round we have {5R, 6Y, 7B}, then {4R, 5Y, 6B}, and so on. Eventually, we get to {0R, 1B, 2Y}. From here, our only move is to {1R, 0B, 1Y}, which finally leads us to {0R, 1B, 0Y} and the end of the game.

We can also leverage the existence of an answer by observing that this is

at its core, a parity problem. Since R and Y have equal parity there isn't really any way to distinguish them - if R could be the answer, so too could Y. B has different parity and is unique in this regard, and can be the only unique solution.