# A Question Related to Chaitin's Omega

Suppose there are two programs in the whole world that halt, and their binary representations are $11001$ and $101$. We know from probability theory that if the experiment is:

1. Flip $N$ fair coins.

Then, the probability of getting any sequence is: $\left(\frac{1}{2}\right)^N$. So the probability of generating $11001$ is $\left(\frac{1}{2}\right)^5$ and the probability of generating $101$ is $\left(\frac{1}{2}\right)^3$.

Now, a line from the article reads:

The probability of randomly choosing one of these programs is $1/2^3 + 1/2^5 = 0.15625$.

How is this calculated? What is the experiment? The formula for union of two events is: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. I am asking this because it is possible for the probability calculation in the article to exceed $1$. For example, suppose if any 3-bit sequence halted, then the probability would be $8*1/2^3 + 1/2^5 = 1.03125$. This would violate one of the axioms of probability.

Thanks.

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