The old trick in the book

1/3?

No...give some more try ..

1/n.............

1/n.............

Nope...will give u a hint...think of the complement of P.

here it is................(plzz forgive me if it goes horribly wrong)
remove one coin from A's set.
now compare the first no. of heads in first n coins of A and n coins of B.
we get 3 cases(or outcomes)
1)A's n coins have more heads than B's n coins
2)A's n coins have equal no. of heads as B's n coins
3)B's n coins have more heads than A's n coins

we can see by symmetry that P(1)=P(3)
let P(1)=a=P(3) and P(2)=b

=> 2a+b=1

if case1 arises then (n+1)th coin will make no difference and A will surely get more heads.
if case3 arises then also (n+1)th coin will not make any diff. and A will get less heads when compared to B
but real thing happens in case 2
If (n+1)th coin is head(with prob 1/2) then A will have more heads than B.
=> Prob of A having more heads inc. by b/2
=> Total prob. of A having more heads =a+b/2=a+(1-2a)/2
=>a+0.5-a
=>0.5

A and B have same chances for n tosses.... all depends on (n+1)^th toss... so if A gets head there, he gets more heads so prob of getting head in (n+1)^th toss =1/2

@ nishant the statement

the prob of getting more heads=prob of getting less no of heads
is this statement really true ??????
my statement

A having more heads =A having less or equal heads