> if something exists with non-zero probability in a finite set there has to be at least one example
I think you mean an infinite set here. If I have a set containing one single coin, and I flip every coin in my set, both heads and tails have a non-zero probability to exist within my flipped set. But there's certainly not an example of both heads and tails in my single-coin set. Indeed for any finite set of coins, there's some chance that either heads or tails never shows up.
However, if I have an infinite set of coins and I flip them all, certainly both heads and tails must exist (although I wouldn't be surprised if even this depends on the axiom of choice or something.)
I think GP did mean finite, just not perfectly clearly how (and also all in missing what the article was focusing on a bit). I.e. if there is some finite set and there is a non-zero probability of something existing in it then there must be at least one _possible_ example of how this can occur - otherwise the probability could not be non-zero. Not in that a single given instance of the finite set must show that result, just that one possibility of what it could contain must have.
What you're saying about the infinite set version of GP's statement depends on what flavor of statistics you're measuring by. E.g. a frequentist approach would give the probability of something that occured 50 times out of \inf as 0 while other approaches may say a small non-0 probability.
Flipping an infinite set of coins is actually an inverse view of what's described above and not necessarily "certainly" even though the probability is 1 ("almost surely" rather than "certainly" is the wording) for the same reasons of a non-empty set not necessarily nudging the probabilities of an infinite sequence. The Wikipedia article explains this better than I could https://en.wikipedia.org/wiki/Almost_surely
Yes, I did indeed mean finite. I was referring to the probabilistic method in combinatorics (https://en.wikipedia.org/wiki/Probabilistic_method). It's a non-constructive technique for proving existence theorems. Of course, this kind of thing works for infinite sets too, but there are some subtleties when it comes to sets of measure zero that kind of obscure the point.
Then you're wrong. I gave you a counterexample to your statement, which you have ignored. If you care about being correct on this, you should ruminate on my comments, because what you are currently saying is wrong.
> Probabilistic method
Neither example on that Wikipedia page shows proving the existence of something from a finite set because it's probability is non-zero.
The first example is actually showing that if the probability is zero from a finite set then there can be no examples. It's an example of #4 below. The critical point is "The number of monochromatic r-subgraphs... is a non-negative integer, hence it must be 0 (if strictly less than 1)." It's the inverse of your statement (and the contrapositive of mine, hence it has the same truth value): "if the probability of something existing in a finite set is 0, there must be no examples". That's very different!
The second example is from an infinite set. It's an example of #1 below The key sentence is: "For sufficiently large n, the probability that a graph has both properties is positive". We're given g and k as parameters to the problem, and we show that there exists some n, above which the probability of a random graph having the property we want is greater than 0. It is NOT proven that the graph we're looking for has exactly n vertices! It's only proven that random graphs of at least n nodes have a positive probability to have the property. The reason the proof works is because there are infinite numbers larger than n! It shows that there are infinite graphs that have non-zero probability to have the property we're looking for, and therefore at least one must actually exist.
Here's a list. Your incorrect claim is counter to #3.
1. Infinite set, nonzero probability: we know at least one example must exist.
2. Infinite set, zero probability: not sure! It could be zero, or any subset of smaller cardinality or measure zero.
3. Finite set, nonzero probability: not sure! It could be zero! My coin flip example shows this.
4. Finite set, zero probability: we know no examples can exist.
Look, I'm not going to argue about this because we're talking about completely different things. Read up on the probabilistic method if you're genuinely interested in what I wrote - Paul Erdos has some famous papers on it.
We're not talking about different things. We're talking about one very specific claim that you made:
> if something exists with non-zero probability in a finite set there has to be at least one example
This is a false claim. See my #1 through #4 above. I strongly suspect if you carefully re-read those Erdos papers, you'll discover that he never relies on my #2 or #3 above to prove anything; only #1 and #4.
If a property P holds over a sample space with positive probability (i.e. the measure of the subset of elements satisfying P is positive), this implies that there is at least one element of the sample space satisfying P. This is easy to prove - it follows formally from sigma-additivity of the measure.
I think you are confusing this with the notion of taking repeated samples from a probability distribution (which is what your coin example is all about).
> i.e. the measure of the subset of elements satisfying P is positive
Question: do any finite sets ever have positive (nonzero) measure? If not, then your sample space is infinite and you're in #4. Haven't you just restricted our conversation only to infinite sets via this assumption? Remember that this entire discussion started because of your claim about finite sets!
That depends on the measure, obviously. It's trivial to construct examples in which finite sets have positive measure - just take the uniform probability distribution over a finite set. Or for a more thematic example, the probability space of random graphs on n vertices. Or really any finite measure you like!
There are also myriad examples of measures over infinite spaces in which there exist finite sets with positive measure - left as an exercise to the reader.
I don't mean this to be rude, but I think you would benefit from reading an undergraduate level textbook on probability theory (or measure theory). Especially before making confident claims about people like Erdos are wielding it.
> if there is some finite set and there is a non-zero probability of something existing in it then there must be at least one _possible_ example of how this can occur
That is not what GP claimed nor does it really make sense. Yes, if there is a non-zero probability of something existing, then it must be possible. That is what non-zero probability means. That has nothing to do with sets. You've just added the word "possible" to make the claim vague enough to be not provably incorrect.
If you have a finite set, and you show that members of that set have a non-zero (but less than 1) probability of having some property, you do not know whether a member of the set actually has that property! Period! Why is this a point of contention?
> a frequentist approach would give the probability of something that occured 50 times out of \inf as 0 while other approaches may say a small non-0 probability.
No. Nobody would say that 50/infinity is greater than 0.
You haven't actually responded to the thing I actually said, nor really to GP's claim. The claim was:
>> if something exists with non-zero probability in a finite set there has to be at least one example
That is false. I gave an example of how it's false.
I'm an ex-PhD in mathematics. We're talking about two entirely separate things - finite numbers of samples from a probability distribution (what you are talking about) vs the measure of a subset of some finite number of objects (the probabilistic method).
I think you mean an infinite set here. If I have a set containing one single coin, and I flip every coin in my set, both heads and tails have a non-zero probability to exist within my flipped set. But there's certainly not an example of both heads and tails in my single-coin set. Indeed for any finite set of coins, there's some chance that either heads or tails never shows up.
However, if I have an infinite set of coins and I flip them all, certainly both heads and tails must exist (although I wouldn't be surprised if even this depends on the axiom of choice or something.)