You could also make a version where you don’t know what X is. In this case always reject strategy doesn’t work since you would reject k*X in real life after the simulation rejected X. It seems like if you must precommit to one choice, you would have to accept (and get (X+X/k)/2 on average) but if you have a source of randomness, you could try to reject your cake and eat it too. If you accept with probability p and reject with probability 1 - p, your expected utility would be (p*X + (1-p)*p*k*X + p*p*X/k)/2. If you know the value of k, you can calculate the best p and see if random strategy is better than always-accept. I’m still not sure where this is going though.
You could also make a version where you don’t know what X is. In this case always reject strategy doesn’t work since you would reject
k*Xin real life after the simulation rejected X. It seems like if you must precommit to one choice, you would have to accept (and get(X+X/k)/2on average) but if you have a source of randomness, you could try to reject your cake and eat it too. If you accept with probability p and reject with probability1 - p, your expected utility would be(p*X + (1-p)*p*k*X + p*p*X/k)/2. If you know the value of k, you can calculate the best p and see if random strategy is better than always-accept. I’m still not sure where this is going though.