The following proposal is based on the understanding that the action of “taking one candy from each box” is covered by the condition of “taking two candies from any box”.

1) If W >= R,

1.1) if R is odd, (W, R) is always a winning position.

1.2) if R is even,

1.2.1) if W – R = 0 or 1 (mod 4), (W, R) is a losing position.

1.2.2) Otherwise (W, R) is a winning position.

2) If W < R,

2.1) if W is odd, (W, R) is always a winning position.

2.2) if W is even,

1.2.1) if W – R = 0 or 3 (mod 4), (W, R) is a losing position.

1.2.2) Otherwise (W, R) is a winning position.

If the condition of "taking two candies from any box" means the player "takes two candies from any single box", the game is much simpler and the solution lies already in the previous comments.

]]>So Bob wins if the start position has W-R=0 or 1 (mod 4), and Alice wins if the start position has W-R=2 or 3 (mod 4). No matter what the players are doing. ]]>

(0,0), (1,0), (0,1), (2,0), (1,1), (0,2), (3,0), (2,1), (1,2), (0,3), (4,0), (3,1), (2,2), (1,3), (0,4), (5,0), (4,1), (3,2), (2,3), (1,4), (0,5)…

If the game starts with a pair that is in an odd position in the above list of pairs, the first player (Alice) loses. If the game starts with a pair in an even position in the list above, the first player (Alice) wins. From what I can see, it does not matter whether either or both players make smart moves or not. In fact, there is no smart move they can make to force a win. Of course, though the players can make non-optimal moves, it’s assumed that the moves still need to be as per the rules of the game. Also, they must make a move if they can. The outcome is pre ordained and only depends on which pair they start the game with. In every move, the state of the game toggles from an odd position in the above list to an even position, or vice versa. Player1 wins any game which is in an even position just before his/her turn (any turn), and player2 wins any game in an odd position just before his/her turn (any turn). Thus, neither player can force a win, no matter what he/she or the other player does in any turn.

]]>(0,0), which means 0 red candies and 0 (mod 4) white candies. This is a 2/Bob win, because every 2 turns 4 candies would be eaten, making the boxes (0,0) again, until eventually 2/Bob takes the final 2 pieces

(0,1) would be another 2/Bob win, with similar reasoning to (0,0) except there is 1 candy left over

(0,2) would be an 1/Alice win, with similar reasoning to (0,0) and (0,1) except that after Bob takes his last turn, there are 2 candies for Alice to take leaving Bob with none

(0,3) would be another 1/Alice win, with similar reasoning to (0,2), except there is 1 candy left over

(1,0) either leads to (1,2)/U or (0,1)/2

(1,1) ]]>