Also, I expect that Omega appears with certainty, no matter what result the calculator gives.
This could work if you give up control over your own test sheet to the counterfactual you mediated by Omega (and have your own decision control the counterfactual test sheet using counterfactual Omega). That’s an elegant variant of the problem, with an additional symmetry. (In my thought experiment, the you that observed “odd” doesn’t participate in the thought experiment at all, and the test sheet on “even” side is controlled by the you that observed “even”.)
Can’t parse a significant portion of the rest you wrote, but the strategies you consider and consequences of their use are correct for your variant of the thought experiment.
In my thought experiment, the you that observed “odd” doesn’t participate in the thought experiment at all, and the test sheet on “even” side is controlled by the you that observed “even”.
So, what does Omega do in your experiment? What algorithm it follows?
(If my question sounds repetitive, it is because not only I am confused, but I don’t see a way out from the confusion.)
Omega on the “odd” side predicts what the you on “even” side would command to be done with the test sheet on “odd” side, and does that. That’s all Omegas do. You could have a janitor ask you the question on “even” side as easily, we only use “trustworthiness” attribute on “even” side, but need “predictive capability” attribute on “odd” side. An Omega always appears on “even” side to ask the question, and always appears on “odd” side to do the answer-writing.
Thanks, I have automatically assumed that Omega is parity-symmetric.
Edit: So, the strategies lead to:
If Q is even, I get it right in 99% of cases. If Q is odd, Omega changes my answer, and I get it wrong 99% of the time. Success rate = 0.5.
The same reversed. If Q is even, I write down false answer 99% of the time, but if Q is odd, Omega steps in and changes the answer leading to 99% success. Overall 0.5.
If Q is even, I get it right always, and if Q is odd, the result is wrong always. Success rate = 0.5.
If Q is even, I get it wrong always, but if it is odd, I get it right. Also 0.5.
Can it be lifted above 0.5? The ability to write “even” on the even side leads to Omega putting “even” on the odd side. It even seems that the randomness of the calculator is not needed to create the effect.
This could work if you give up control over your own test sheet to the counterfactual you mediated by Omega (and have your own decision control the counterfactual test sheet using counterfactual Omega). That’s an elegant variant of the problem, with an additional symmetry. (In my thought experiment, the you that observed “odd” doesn’t participate in the thought experiment at all, and the test sheet on “even” side is controlled by the you that observed “even”.)
Can’t parse a significant portion of the rest you wrote, but the strategies you consider and consequences of their use are correct for your variant of the thought experiment.
So, what does Omega do in your experiment? What algorithm it follows?
(If my question sounds repetitive, it is because not only I am confused, but I don’t see a way out from the confusion.)
Omega on the “odd” side predicts what the you on “even” side would command to be done with the test sheet on “odd” side, and does that. That’s all Omegas do. You could have a janitor ask you the question on “even” side as easily, we only use “trustworthiness” attribute on “even” side, but need “predictive capability” attribute on “odd” side. An Omega always appears on “even” side to ask the question, and always appears on “odd” side to do the answer-writing.
Thanks, I have automatically assumed that Omega is parity-symmetric.
Edit: So, the strategies lead to:
If Q is even, I get it right in 99% of cases. If Q is odd, Omega changes my answer, and I get it wrong 99% of the time. Success rate = 0.5.
The same reversed. If Q is even, I write down false answer 99% of the time, but if Q is odd, Omega steps in and changes the answer leading to 99% success. Overall 0.5.
If Q is even, I get it right always, and if Q is odd, the result is wrong always. Success rate = 0.5.
If Q is even, I get it wrong always, but if it is odd, I get it right. Also 0.5.
Can it be lifted above 0.5? The ability to write “even” on the even side leads to Omega putting “even” on the odd side. It even seems that the randomness of the calculator is not needed to create the effect.