So before any ballot is picked out, there are four possible worlds in exactly what will be drawn:

A(Counterfeit) -> A(REAL)
A(Counterfeit) -> B
B -> A(Counterfeit)
A(REAL) -> A(Counterfeit)

Once a ballot is drawn and it turns out to be A (either real or counterfeit), there is now three possible worlds:

A(Counterfeit) -> A(REAL)                                                       
A(Counterfeit) -> B
A(REAL) -> A(Counterfeit)

So there is a 1/3rd probability that the remaining vote in the envelope is for candidate B.

Sean says:

The answer is correct, although it is my anal conviction that the argument is incomplete.

Suppose there are A votes for candidate A and B votes for candidate B. Let X be the random variable which describes the first ballot drawn, and Y describe the second ballot. Then either X contains the fake ballot, in which case there are (A+B) possibilities for Y, or X contains a real ballot, in which case there is only one possibility for Y (i.e., the fake ballot). Since X must be a vote for candidate A, it follows that there are (A+B)+A total possibilities. Of those possibilities, B of them satisfy the condition "Y is a vote for B." Thus, the probability that Y is a vote for B is B/(2A+B), or in this case, 1/3.

Brandon says:

The answer is correct, although it is my flawless conviction that the argument is incomplete.