Yeah, I then tried to back-fill by picking two numbers at random from (0,1). But of course that's just like flipping a coin!
Except for one corner case, which also occurs in the original problem: you can pick exactly the number chosen (with probability 0, of course). Which leads me to wonder if we're not in one of these "integrate a function that's everywhere 0 and get 1" kinds of situations.
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Except for one corner case, which also occurs in the original problem: you can pick exactly the number chosen (with probability 0, of course). Which leads me to wonder if we're not in one of these "integrate a function that's everywhere 0 and get 1" kinds of situations.