Hi rose1,

To make the problem more interesting, let's see if we can actually calculate the average proportion of heads.

If we flip H on the first trial, which happens with probability 1/2, the proportion of heads is 1.

If we flip TH, which happens with probability 1/4, the proportion of heads is 1/2.

In general, if we flip n-1 tails followed by a head, which happens with probability

, the proportion of heads is 1/n.

So the average proportion of heads is the sum of an infinite series:

.

If you know some infinite series, you might know this one:

.

The series for the average proportion of heads is just the case

of this series; so its sum is

, approximately.

The proportion of heads and tails must sum to 1, so the average proportion of tails is

, approximately; so on average there is a larger proportion of heads than tails.