The article implies that the interviewee assumes that the number is being chosen randomly, when Ballmer could actually be choosing adversarially. However, if the interviewee assumes that Ballmer is being adversarial, then you can pick a different value as your initial guess, which causes the probabilities to shift. Even the OP assumes that the interviewee will start guessing with 50, but, because of the way binary se…
No it doesn't, it's quite clear that Ballmer can be choosing adversarially. The point is that even if Ballmer chooses randomly and the interviewee plays optimally given this, the game still has a negative expectation value, and that is enough to be sure the game is a loser for the interviewee. The post never answers the question "so what is the real expectation value", which is a more difficult question. But I think…
I agree that if you choose your first guess somewhat randomly in the 40-60 range (maybe not a uniform distribution though) Balmer would be forced to choose randomly and you would be back at a positive $0.20 EV. For example, you could flip 6 coins and add the number of heads, then flip another coin to decide whether you add or subtract the number of heads from 50 for your starting guess. But I think you would need to randomize your later guesses a bit also.