Parents method produces a result in finite time with probability infinitesimally close to 1. Yours produces a result in finite with probability infinitesimally close to 0. That is the greatest measurable difference in all of mathematics.
In other words, on method basically always succeeds, the other never succeeds (not "fails", but "never succeeds")
That's funny that would say that since what you are saying is completely wrong. Different techniques would have different probabilities, but none would have what you are describing.
Yeah there's a real difference between a very small but still finite chance, and an infinitesimally small chance. If you're trying to multiply negative infinity by reciprocal infinity, you have to be specific about how big each infinity is. :)
Yep, this is why I wouldn't personally use 0 probability to describe anything in the real world. It technically works, it's just the realm of pure measure theory rather than anything applicable to finitely many trials.
As far as we know, there is no underlying theory of probability for them to be techniques of. So maybe they are equivalent in some sense, but on the face of it, they are separate ideas.
Yeah, that's really different, though. It's essentially a way to simplify counting. But the "converse" of the probabilistic method doesn't work: zero probability doesn't mean inexistence in an infinite set.
Are the probabilities equal? I agree that they're both possible, but I don't have any clear intuition for why they should be the same -- or which would be larger if they're not the same.
The point the parent is making is that 1) represents a class of results rather than a single result, and any single member of that class is equally as likely as 2) or 3). Obviously the class as a whole is more likely than any other single result, but that's a different assertion.
If the probability of producing the desired result in any given attempt is zero, then the probability of producing the desired result given infinite attempts is still zero.
If monkeys are incapable of typing, then the probability is zero.
No. The two methods are qualitatively equally good. In both cases, some outputs have probability floor(2^n/K)/2^n while others have probability ceil(2^n/K)/2^n. The numbers of outputs with each probability are such that they average out to 1/K. The only difference is the distribution of which output numbers have which of these two probabilities.
In other words, on method basically always succeeds, the other never succeeds (not "fails", but "never succeeds")
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