Vercassivelaunos

And by "all but a certainty" I mean: "In the entire lifetime of the universe, the probability that it has never happened is overwhelming."

The second kind is impossible due to the 2nd law of thermodynamics, which is a statistical law. It boils down to the fact that if we randomly disturb a system with a lot of possible configurations, it is more likely to end up in a state that has more corresponding configurations than in a state that has few corresponding configurations. And calculating the probabilities with anything that is even remotely macroscopic, it is all but a certainty because of the large number of configurations.

But there is proof for the impossibility of both, at least insofar as proof is possible in science. For the first kind, the proof is basically Noether's theorem, which forbids creating energy based on the assumption that physics is time translation invariant. The theorem is mathematical truth and time translation invariance is experimentally verified except on a cosmological scale.

Addendum: Considering my previous comment, my objection is not so much that you are wrong - you are not. But more so that the objection you give to the OP's rigorousness comes at a point in the argument where all the necessary steps to neutralize your objection have usually already been taken and can be assumed.

@lulu I think my main issue is our starting point and the assumptions that come with it. If we start immediately by calculating the probability of a specific event like you do, the existence of a probability space should be assumed because that's where we get the probabilities $p_n$. In that case, convergence of any sum of disjoint probabilities is a given. If, however, the probability space needs to be constructed first, one would start (in the discrete case), with the probabilities of all possible samples and show that their sum is 1. From there, convergence of your sum is again a given.

"I don't think it's obvious that a probability exists for this problem". That's my point: It is obvious in discrete probability spaces as long as the probability space is sound - which you do seem to take as obvious, since you say that $p_n$ obviously exists. If the sum of probabilities of disjoint events does not converge, then their probabilities weren't valid probabilities in the first place.

And that's a fine definition. But the issue you are mentioning, namely wether that definition is sound, is a non-issue. In a discrete probability space, the probabilities of all samples already have a finite sum. The sum of any subset of those samples is then also guaranteed to have a finite sum. This is true for all discrete probability spaces and doesn't need to be proved all over again for every exercise.

This is not an issue of convergence, though, like you make it seem. At worst, the issue is wether the event in question is in fact an event at all. But since we are in a discrete probability space (the samples are basically all the positive natural numbers (there's a winner after $n$ tries) and $\infty$ (no winner ever)), where all sets of samples are valid events. Any and all probabilities of valid events are guaranteed to exist.

I disagree with this being a duplicate of the linked question. While the answer to this question follows from those in the "duplicate", this question is more elementary, and can have helpful answers which are easier to grasp than the answers to the duplicate.

To be honest, I don't really get your proof. What is $n_1$? And are you saying that any permutation is just a cycle?

@Shaun: it is, but that doesn't help in the case of $n\in\{1,2\}$

The identity is the empty product of transpositions, so it's a product of 0 transpositions.

Then maybe it's a language thing. The phrase "either ... or" translates to in my language is very specifically xor. If it's less clear in English, then some way to express it would be fine, I think.

Just say either or. In general, don't overdo it with the formal notations. "Every integer is either even or odd" carries the same meaning as $\forall n\in\mathbb Z((2\mid n)\operatorname{xor}\neg(2\mid n))$. But which one is easier to parse? It's my opinion that natural language is often better, or at least just as good as formal notation. Only use formalities when they actually aid understanding.

What 1/0 is in a specific space is not a matter of proof, but of definition. We have to define how we extend the multiplicative inverse to a different space, and then there's nothing to prove. However, we can prove the following: There is no continuous function $f:\mathbb R\to\overline{\mathbb R}$, where $\overline{\mathbb R}$ is the affinely extended real line, such that $f(x)=\frac{1}{x}$ for all $\in\mathbb R\backslash\{0\}$.

Ah, I see. Then no, in the affinely extended line, $\frac10=+\infty$ would be a bad choice, since this would make 1/x discontinuous. It only makes good sense in the projective extension.

I'm not quite sure what you mean by affinely extended. Anyway, it's the projective extension I've been talking about. There is only one single point at infinity there, so no need to specify $+\infty$, since there is no $-\infty$ in this context.

Yes, we can do that.

Not really. But if you want to search for yourself, you should look for projective geometry. The extended real line is also known as the projective real line.

Then that would be a reasonable definition to make, yes.

The extended number line is not a field, meaning that it doesn't play nicely with algebra. It does play nicely with geometry, though. And division on the number line shouldn't be interpreted as the inverse operation to multiplication, but as a geometric operation: the extended number line can be visualized as a circle, with 0 on the bottom, $\infty$ at the top, 1 to r right and -1 to the left. $\frac1x$ should be interpreted as mirroring each point at the axis going through 1 and -1. Not as the multiplicative inverse of $x$.

We can introduce a point $\infty$ and say $\tan(\frac\pi2)=\infty$. With the right notion of distance on $\mathbb R\cup\{\infty\}$, this extended version of $\tan$ is even continuous. And we can even extend $\tan$ to the complex plane and $\mathbb R\cup\{\infty\}$ to $\mathbb C\cup\{\infty\}$ to get a holomorphic function. So your idea is very reasonable. It's just not usually done in lower level courses.

The precise way to say it would be that the continuous extension of $x\mapsto\frac1x$ as a function to the extended real line maps $0\mapsto\infty$. This continuous extension has so many good properties that it would in a way be reasonable to just say $\frac{1}{0}=\infty$, but the issue is that we are also often considering a different kind of extension of the real line, which also contains $-\infty$. And with this different extension, the map from above can't be extended continuously. Due to this, it's better to be very context sensitive when considering such extensions.

Cutting to the chase: I would refrain from saying $\frac{1}{0}=\infty$, unless the context of the extended real line is crystal clear.