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.
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.