but I think you have overstated the probability of the first case. Adding what I think is a missing piece of what you said:
So if a goat is behind A (p=1/3), then Alan will open B and if a car (p=1/2) the game will continue. Becky will open A and find the goat, then will open C and find the keys.
Looking at the six cases, and upper or lower case letters for Alan and Becky's success with your scheme (if I understand it correctly):
1) CKG A B
2) CGK A b
3) KCG A B
4) KGC a B
5) GCK A B
6) GKC a b
Alan and Becky each get their item in four of the six cases, but they BOTH get their item only in cases 1, 3, and 5.
The hint is that they each make their second choice based on what they see in the first. In what you propose, Alan will always chose curtain B for his second pick and Becky will always pick C.