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OK so if I flip a coin until I get tail, is never getting tail an outcome? Should it be in the sample space? I think yes (with probability 0) but seems weird. $\text{ Thank you very much.}$

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Yes, it's an outcome. It occurs, as you say, with probability $0$.

To go all measure-theoretic: a $\sigma$-algebra is closed under countable intersections. A measurable space is a set with a $\sigma$-algebra on it; a measure space is a measurable space together with a measure. A probability space is a measure space where the set has measure $1$. The observable outcomes lie in the $\sigma$-algebra; since we may observe "we haven't thrown a head by time $n$" for each $n$, and since $\sigma$-algebras are closed under countable intersections, we must have that "we never throw a head" lies in the $\sigma$-algebra.

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    So probability $=0$ doesn't mean it's impossible, right? Like, if I choose a number randomely (uniformely) in $[0,1]$, then for all $0\leq x\leq1$ choosing $x$ is an outcome, but *every* outcome has probability $0$ ? But if I say probability of getting $x$ in $[0,1/2]$ then it's $1/2$ ? Is this correct? Thank you. – 2017-01-12
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    @NeedForHelp yes, that is all correct. – 2017-01-12
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    @Mark S. Thank you for confirmation. Similarly if probability of event is $=1$, then it does *not* mean that it will always happen in practice? – 2017-01-12
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    @NeedForHelp In real life, probabilities are *never* $0$ or $1$. It's always possible that the coin which you're tossing will be plucked out of the air by a passing eagle, and eaten. But yes, a probability-$1$ event might not occur. Indeed, let $A_x$ be the event "I don't pick the number $x$" when I'm picking uniformly at random from $[0,1]$. Then *some* $A_x$ must not happen, since I must pick *some* $x$; even though they all have probability $1$. – 2017-01-12
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    From Bayes's theorem, by the way, if I assign a probability of $1$ to something, that means I can never, ever legally stop believing it, no matter what evidence is presented to me. Bayes's theorem will never let me update away from $1$. If you're willing to make bets based on your levels of belief, and you think $A$ has probability $1$, then you should be happy to bet literally everything you have and will ever have on $A$. Is there anything you're so sure of? Could there ever be anything you're so sure of? – 2017-01-12
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    (I assert "probabilities are never $0$ or $1$" as if it were fact. Obviously I can't coherently believe that the probability that probabilities are never $0$ or $1$ is $1$.) – 2017-01-12
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    @Patrick Stevens But in mathematics being happy to bet everything you have is not good enough? Probabilistic proof of the problem $3x+1$ gives probability of $1$ that result is true but still problem remains unsolved? – 2017-01-12
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    It… doesn't give probability of $1$. I don't know where you got that from. If I started with a prior of $\frac{1}{2}$ chance that the $3x+1$ problem's answer was "every sequence reaches $1$", then each number we find whose sequence terminates at $1$ increases my confidence, but never so much that my confidence hits $1$. Even if it were proved, my confidence would not be $1$: I could be hallucinating all this, or every mathematician has just missed a glaring error in the proof (which happens!), or all kinds of things. – 2017-01-12
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    @Patrick Stevens See [here](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.607.577&rep=rep1&type=pdf). In the introduction they say "It allows us to show that, within the constraints of a formally defined probabilistic model, the conjecture is true almost surely (a.s.), that is with probability one." – 2017-01-12
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    In that context, I think they mean "for almost every integer, it holds", not that "the result has probability 1 of being true". Just as almost every real number is irrational, rather than "with probability 1, all reals are irrational". – 2017-01-12