Thanks, I have automatically assumed that Omega is parity-symmetric.

Edit: So, the strategies lead to:

1. 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.
2. 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.
3. If Q is even, I get it right always, and if Q is odd, the result is wrong always. Success rate = 0.5.
4. 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.