# Does the existence of the universe mean Chaitin's number is more than half?

This may be too philosophical for this stackexchange.

Chaitin's constant $\Omega$ is the probability that a bitstring will halt when run on a universal Turing machine. It is uncomputable due to the halting problem.

My thought is that if $\Omega<\frac{1}{2}$ then most likely nothing would exist. So, since something does exist, it is most likely that $\Omega \geq \frac{1}{2}$.

Is this reasoning valid?

• The World existed long before Turing, and it is not in any way connected to Chaitin constant, so the premise is faulty, besides take two programs, one halts, one loops forever, define dummy move, put it in both programs at the begining. So there are two infinite families of dummy prefixed programs. Now what is the point of betting which one is bigger? Anyway, if you are not sure about validity of question or would like some discussion instead of objective answer, maybe the chat would be a good place? – Evil Aug 29 '17 at 2:47
• Is this two-line solution to a philosophical problem valid? No, of course it isn't. It's far too short to be an explanation of anything. By the way, you might be interested in Jim Holt's book Why does the world exist?. It almost certainly doesn't answer your question but it's about the same general themes. – David Richerby Aug 29 '17 at 8:53
• OK, so I voted to close as not computer science but I've retracted that on seeing that the answer is purely computer science. – David Richerby Aug 29 '17 at 8:57
• The definition of Chaitin's constant is incorrect, and everything following it is predicated on several misunderstandings, one of which is the faulty definition of the constant. – Andrej Bauer Aug 29 '17 at 9:43
• 1. We don't know if universe is Turing-computable. 2. Even if it is, we don't know if our universe will halt or cycle (we don't know Chaitin's constant for our universe). However, both these possibilities sound strange. – rus9384 Aug 29 '17 at 10:02

## 1 Answer

No, your reasoning is invalid. Chaitin's constant depends on the choice of a universal Turing machine, and by appropriately choosing the universal Turing machine, we can fix the first $n$ bits of Chaitin's constant to be whatever we want, for any fixed $n$.

One reason to suspect your reasoning is that the actual argument is missing. Your argument can be paraphrased as follows:

1. $\Omega$ is the probability of something.
2. If $\Omega < 1/2$ then something counterfactual.
3. Hence I suspect that $\Omega \geq 1/2$.

The problematic step here is Step 2. What does $\Omega$ have to do with existence?

• All known physical laws are computable, suggesting the universe is the product of a computation. So, assuming the universe is the product of a computation, it seems the rest of my argument follows. – yters Aug 29 '17 at 16:24
• "All known physical laws are computable" @yters What makes you think that? From what we can tell, quantum mechanics involves infinite precision computation with arbitrary real numbers (including transcendentals), which is well beyond the scope of Turing computation (since there are more real numbers than bit strings). Physics doesn't even seem to be deterministic, let alone computable. – jmite Aug 29 '17 at 18:06
• @jmite, that may be so, but at least Scott Aaronson says a quantum Turing machine is not stronger than a nondeterministic Turing machine. This suggests even quantum reality can be modeled with a Turing machine. – yters Sep 3 '17 at 2:50
• @yters It could suggest that, or it could suggest that a quantum Turing Machine, which is still bounded by finite space and time, is not powerful enough to simulate all of reality. – jmite Sep 3 '17 at 3:36