$P(H_1)=1/2$

The probability of a head second time is the probability of a tail then head:

$P(H_2)=P(T,H)=1/4$

In general:

$P(H_n)=1/2^n$

I think this is called a geometric distribution.

The average (mean) number of times is the sum of the number of times weighted according to their probabilities, ie

av $=\sum _{n=1}^{\infty }n/2^n$

(We want the sum to infinity as it is possible to have to throw the coin any number of times before getting a head.)

To sum this:

$\sum _{n=1}^{\infty }n/2^n=1/2^1+2/2^2+3/2^3+...$

$=(1/2^1+1/2^2+1/2^3+...)+(1/2^2+2/2^3+3/2^4+...)$

$=(1+1/2^1+1/2^2+1/2^3+...)(1/2^1+1/2^2+1/2^3+...)$

$=2\times 1$

$=2$

I'm not completely sure of the validity of what I have just done but the answer seems believable.