4 ms·
One way to resolve this problem is to use a continuous, increasing map f from the reals to say, (-1, 1). Something like arctan(x)/pi. You can then pick a number
by sidedishes 11y ago
One way to resolve this problem is to use a continuous, increasing map f from the reals to say, (-1, 1). Something like arctan(x)/pi. You can then pick a number randomly from (-1, 1), then compare that with f(A), rather than a random real with A.
- idlewords 11y agoBut then it's not really uniform, is it? That innocent instruction to 'pick a random number' is carrying a lot of conceptual baggage.
- sidedishes 11y agoRight, though I don't think uniform is the necessary condition here. We only need a positive chance of picking an R so that f(A) and f(B) are on different sides of R, which isn't as much of a problem if we're picking R from (-1, 1).
- idlewords 11y agoThere's also the question of how the first two numbers were chosen; in other words, what it means that 'the numbers are completely arbitrary'. I can't math real good, but this is the part that smells fishy to me.
- aidenn0 11y agoSo long as you have a non-zero chance of your number "R" being between A and B, you have an advantage.