Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n.[/b]
I don't think the problem is as easy as everyone seems to think it is. I think the answer is 1/2 only if n=0, which is why the above answers are only conditionally correct. Here's what I think it is:
if they both had n coins, then the probability of Bill getting more heads would be n/(2n+2). This is very easy to show.
So my approach is the following: consider they both throw n coins, and the probability of Bill having more heads is given above. Then consider that Bill has one more coin to throw which of course can be heads with 1/2 probability. Then the overall probability that Bill gets more heads should simply be a weighted average of these probabilities:
(n(n/2n+2) + 1/2)/(n+1).
This only equals 1/2 for n=0, which makes sense. for all other possibilities, it depends on n.
THUD, is this right?