That's not what he's looking for. He's looking for one (yes or no) answer based on the median method being the best yield of information. He's ignoring that a gambler would not use the best algorithm for gleaning information but the one that yields the best (or even) odds, that's randomness. He's not considering that the method determined the odds and only that he's likely controlling the outcome based on the odds.
I also don't agree with the last bit. If I use randomness I may lose $5 but my chances for that overall would be the same as winning $5.
I'm not sure exactly what you mean there. He's really looking for a game theoretical answer: Median is as good as it gets with random input, and one should actually play once the stats are worked out (I calculate an average of 5.8 tries for a random selection, which is a slight win of 20c), however, using his adversarial strategy, one would consistently lose by selecting the median. If one can guess the adversarial strategy being used, one can turn the tables and win a lot more than the expected 20c. The overall expectation of winning depends on the likelihood of the second strategy being chosen by the adversary.
So (a) you should play the game with the median method if the input is random for the slightly positive expectation (b) you are correct, that randomizing your pivot would defeat the adversarial strategy outlined, (c) if you anticipate the adversarial strategy, you would get a much higher positive expectation, (d) given that you don't know if this particular adversarial strategy is being used, your overall expectation is the weighted probability of the two underlying expectations, where your prior should lean towards the adversarial strategy, given the context of the question.