where we have used a well-known identity, .
b) First, let's continue a) by finding and .
By symmetry, . Then
Now consider the case where A has N+1 coin tosses. As before, let X = the number of heads in the first N tosses, and let's say Z is the total number of heads in all N+1 tosses.
We find by conditioning on A's last toss. If A's last toss is a tail, then if and only if . If A's last toss is a head, then if and only if . So
I think that if you substitute the expressions for from (1) and from (2) in the above equation and do a little algebra, you will find that reduces to 1/2.