You are agreeing with my solution?
Though in this problem it seems the "average" final number would be 0.625, not 0.75 - since half the time you have an 0.75 expected outcome and the other half a 0.5 expected.
Though in this problem it seems the "average" final number would be 0.625, not 0.75 - since half the time you have an 0.75 expected outcome and the other half a 0.5 expected.
In other words, it isn't sufficient to shoot for the highest score possible.