Does Playing first Matters. Part - 1

The first player wins in the following cases.
1) H
2) TTH
3) TTTTH
.....
So the probability is
P(E) = H + TTH + TTTTH ....
with the probability of head and tail being $\frac{1}{2}$ each
P(E) = $\frac{1}{2}$ + $\frac{1}{8}$ + $\frac{1}{32}$ ...
hence P(E) = $\frac{2}{3}$
p + q = 5

Consider each (ordered) pair of of coin tosses. There are 4 equally probable outcomes:

1. heads,heads: the game terminates; the first player wins.

2. heads,tails: the game terminates; the first player wins.

3. tails,heads: the game terminates; the second player wins.

4. tails,tails: the game continues; neither player is any closer to winning or losing than before.

Whenever the game terminates (immediately or after arbitrarily many pairs of tails), the first player wins in 2 cases out of 3.

Therefore, $\frac{p}{q}$ = $\frac{2}{3}$ and p + q = 5.