December 8, 2009

This is a little problem which somebody (I forgot who it was) told me many years ago. It was wrapped in a story (I guess something about the physics of heads and tails), which I will not try to reconstruct.

Consider a random process x(t) which generates a series of 0s and 1s, but many more 0s because the probability for x(t) = 1 decreases with t as 1/2^t.

Now assume that we encounter this process not knowing 'how far we are already', in
other words we don't know the value of t.
The question is: "What is the probability to get a 1 ?"

Unfortunately there are two ways to answer this question. The first calculates
the 'expectation value', as a physicist would call it, or 'the mean' as a statistician would put it, which is zero.

In other words, we sum over all possible t with equal weight and have to consider
s = sum( 1/2^t ) with t = 1, 2, ... N; It is not difficult to see that
s = 1/2 + 1/4 + ... equals 1.

The answer is therefore prob(1) = s/N = 1/N and because N is infinite (the process never ends) we get prob(1) = 0.

The second answer simply looks at the definition of the process and points out that
prob(1) = 1/2^T, where T is the current value of t. Although we don't know T it must have some finite value and it is obvious that prob(1) > 0.

So which one is it, prob(1) = 0 or prob(1) > 0? And which one do you prefer, the 'equal weight' or the 'unknown T' answer?

added later: A Fatwa has just been issued, which should end this discussion: prob(1) > 0.

Of course, this would raise the heretic question "if prob(1) is greater than zero, what value does it actually have?" but we will not ask that question.

And it also leaves open my question about the lottery ticket (in the comments).