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 if the interviewee chooses a number randomly from 40-60 as the first guess and does a binary search from there, Ballmer can't really improve on choosing his initial number randomly.
I think you did the math wrong. The expected value for the guesser is $0.20 if Ballmer chooses randomly. I think Balmer is saying that he can beat you if he chooses adversarially and you choose the expected initial guesses.
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.
The post never answers the question "so what is the real expectation value", which is a more difficult question. But I think if the interviewee chooses a number randomly from 40-60 as the first guess and does a binary search from there, Ballmer can't really improve on choosing his initial number randomly.