> 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, likewise, from any number that are not), regardless of whether or not you have any of them in mind.
I'm not sure I agree or understand.
While it's trivially true that all numbers out there are part of some f(t) that is nonzero everywhere, that knowledge isn't helping me.
As an example, if I always pick the same number, then the scheme doesn't work. Perhaps where we disagree is:
> Player A does not know what your scheme is
The beauty of the solution is that you can tell Player A your scheme and still guarantee > 50%. My solution doesn't preclude A from knowing your scheme. So definitely picking the same splitting point over and over will not work.
I take it you probably meant "pick a number, but change the number". Yes, but how? As someone pointed out, using a random number generator on any given machine will not work as player A can always pick two numbers that are close enough to each other that there is no number on the machine in between them.
My concern is that this problem will end up being similar to the well ordering principle, where it's been proven that you can come up with a well ordering on the reals, but it's also been proven that you can never show it to me.
>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
This isn't a minor point. It's trivial to come up with a 50% scheme. This is an interesting solution in that it guarantees > 50% no matter what Player A does.
I think I misunderstood your issue here - I misread 'scheme' as an algorithm for picking splitting numbers.
> Using a random number generator on any given machine will not work as player A can always pick two numbers that are close enough to each other that there is no number on the machine in between them.
On reflection, I am not sure that the method presented in this article can guarantee > 50% in any case where the domain is discretized. Note that the original article does not say what to do if the first number revealed is equal to B's splitting number. If A always chooses adjacent numbers, there are never any 'tiebreaker' splitting numbers - i.e. those that are strictly between A's numbers, and will elicit the correct response regardless of whether B has seen the larger or the smaller of the pair.
I'm not sure I agree or understand.
While it's trivially true that all numbers out there are part of some f(t) that is nonzero everywhere, that knowledge isn't helping me.
As an example, if I always pick the same number, then the scheme doesn't work. Perhaps where we disagree is:
> Player A does not know what your scheme is
The beauty of the solution is that you can tell Player A your scheme and still guarantee > 50%. My solution doesn't preclude A from knowing your scheme. So definitely picking the same splitting point over and over will not work.
I take it you probably meant "pick a number, but change the number". Yes, but how? As someone pointed out, using a random number generator on any given machine will not work as player A can always pick two numbers that are close enough to each other that there is no number on the machine in between them.
My concern is that this problem will end up being similar to the well ordering principle, where it's been proven that you can come up with a well ordering on the reals, but it's also been proven that you can never show it to me.
>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
This isn't a minor point. It's trivial to come up with a 50% scheme. This is an interesting solution in that it guarantees > 50% no matter what Player A does.