a) Suppose the players toss N fair coins. Let X be the number of heads tossed by A and Y be the number of heads tossed by B. Then both X and Y have Binomial(N, 1/2) distributions, so

where we have used a well-known identity,

.

b) First, let's continue a) by finding

and

.

By symmetry,

. Then

so

.....(1)

and

....(2)

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.