good question...

Number of ways in which A can win is 126 i guess...bt calculation of probability is quite tough

Is it: 105256

so no one with an answer?

here is the answer....although i am copy pasting it...it is indeed a wonderful question...
P.S:would never come in JEE....

You need to work backwards on this problem.

Let Pmn be the odds of A winning given that
a) it is A's turn to move
b) A has m heads
c) B has n heads

Obviously P50=P51=P52+P53=P54 = 1.0
Obviously P05=P15=P25=P35=P45 = 0.0

Consider P44:
if A now flips a head A wins and the game is over. There is 0.5 chance this will happen. If A flips a tail and B flips a head the game is over and A loses. If they both flip a tail we are at the same state, P44.
SO P44 = 0.5(1.0) + 0.25(0.0) + 0.25(P44) or
P44 = 0.5/0.75 = 2/3

Similarly P43 = 0.50(1.0) + 0.25(P44) + 0.25(P43)
or P43 = 0.50(1.0) + 0.25(2/3) + 0.25(P43)
P43 = 8/9

In general:
Pmn = ( P(m+1)(n+1) + P(m+1)n + Pm(n+1) +Pmn ) / 4
or Pmn = 1/3 * ( P(m+1)(n+1) + P(m+1)n + Pm(n+1) )

You work this all the way back to the calculation of P00 when you will get the answer you want. It's ugly but it works.

This was easy to do in excel with the following results.
The probabilities go from P00 in the top left corner to P55 in the bottom right corner. Columns are increasing n and rows are increasing m.

0.549 0.354 0.173 0.055 0.008 0.000
0.737 0.556 0.332 0.132 0.025 0.000
0.886 0.768 0.568 0.296 0.074 0.000
0.968 0.922 0.815 0.593 0.222 0.000
0.996 0.988 0.963 0.889 0.667 0.000
1.000 1.000 1.000 1.000 1.000 1.000

The final answer is that A to move first has 0.549 chance of getting to 5 heads first.

The exact probability is 10802/(3^9) = 0.5487984555... .