Player A does not know what your scheme is, and you are free to change it - in fact, you can change it for each round, and thus defeat any attempt by A to deduce it. In other words, don't have a scheme, forget about f(t), and just pick a number. Any number you pick is, ipso facto, from any number of distributions that are nonzero everywhere (and also, as it happens, from any number that are not), regardless of whether or not you have any of them in mind.
The distribution f(t) does not appear in the explanation of the outcome, and I think it's true to say that the only reason f(t) is mentioned at all is that if you do instead choose from a distribution from which some ranges of numbers are excluded, then the puzzle-poser can no longer say that your odds are strictly greater than 50-50, as player A might always pick from an excluded range (for example, if player B stubbornly insists on only picking positive numbers, and player A is determined on only playing negative numbers.) Without the requirement for f(t) to be nonzero everywhere, I think then either one would have to drop the 'strictly' claim, which makes the result look far less paradoxical (B can sometimes do better than 50-50), or one would have to put some constraints on how A chooses the number-pairs (and, furthermore, those constraints would be dependent on which particular distribution B was using.)
