What does it even mean to write an answer on a counterfactual test sheet?

Is it correct to to interpret this as "if-counterfactual the calculator had showed odd, Omega would have shown up and (somehow knowing what choice you would have made in the "even" world) altered the test answer as you specify"?

Viewing this problem from before you use the calculator, your distribution is P(even) = P(odd) = 0.5. There are various rules Omega could be playing by:

• Omega always (for some reason uncorrelated to the parity of Q) asks you what to do iff the calculator shows even, and alters the sheet iff the calculator shows odd. Deciding "write even", your answer is only ever correct if Q is indeed even (regardless of the calculator), hence P(correct) = P(even) = 0.5. Deciding "write odd" is identical to the case without Omega (your answer is just what the calculator says) hence P(correct) = P(calculator is correct) = 0.99. The thing to do is decide to "write odd"
• Same as the above with even and odd reversed. "Leave the counterfactual paper alone" is still the correct answer.
• Some random combination of the previous two uncorrelated to the parity of Q. "Leave the counterfactual paper alone" would still be the correct answer.
• Omega knows the parity of Q, and asks you iff the calculator is correct. Deciding to say "alter the paper", the answer written is always the correct one, P(correct) = 1. "Alter the counterfactual paper" is obviously the correct answer.
• Omega asks you iff the calculator is correct, with probability p (maybe he's somehow unsure of the parity of Q, and asks you iff the calculator's result is the one he thinks is most likely). Deciding "alter the counterfactual paper", the answer written down is the correct one (again, regardless of the calculator) with probability P(correct) = p. As before, "don't alter the paper" gives you P(correct) = 0.99. Hence, answer "alter the paper" iff p > 0.99.

Finding a prior on these possibilities is left to the reader.