If you flip a coin until you get a head and then repeat this many times, will you tend to have a larger proportion of heads or of tails?

Results 1 to 3 of 3

- Feb 16th 2009, 08:06 AM #1

- Joined
- Feb 2009
- Posts
- 10

- Feb 16th 2009, 10:11 AM #2
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 $\displaystyle (1/2)^n$, the proportion of heads is 1/n.

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

$\displaystyle 1\, (1/2) + (1/2)\, (1/2)^2 + (1/3)\, (1/2)^3 + (1/4)\, (1/2)^4 + \dots$.

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

$\displaystyle -\ln(1-x) = x + (1/2)\, x^2 + (1/3)\, x^3 + (1/4)\, x^4 + \dots$.

The series for the average proportion of heads is just the case $\displaystyle x = 1/2$ of this series; so its sum is $\displaystyle -\ln(1 - 1/2) = \ln(2) = 0.6931$, approximately.

The proportion of heads and tails must sum to 1, so the average proportion of tails is $\displaystyle 1 - \ln(2) = 0.3069$, approximately; so on average there is a larger proportion of heads than tails.

- Feb 16th 2009, 10:40 AM #3
It occurs to me that perhaps the use of infinite series above is overkill, given the simplicity of the original question. So here is an easier way to see that the proportion of heads is greater.

The average proportion of heads and the average proportion of tails must sum to 1. But just by considering the cases H and TH (heads on the first or second flip), we see the average proportion of heads is at least

$\displaystyle 1\, (1/2) + (1/2)\, (1/2)^2 = 0.625$,

so the proportion of heads is greater than the proportion of tails.