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I read a comment under this question:

There are plenty of events that can occur that have zero probability.

This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering

  1. if zero probability and impossibility mean the same?
  2. if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?

Thanks and regards!

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    $\begingroup$ @Vicrobot: No. Probability can be exactly $0$ for possible events, e.g. here and here. $\endgroup$ Sep 2, 2019 at 8:22
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    $\begingroup$ @Vicrobot: There's nothing to admit. If you multiply both sides of $\frac{1}{\infty}=0$ by $\infty$, you get $undefined = undefined$. With your logic, you could prove that a point must have a non-zero area since a square is the sum of many points. $\endgroup$ Sep 2, 2019 at 13:22
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    $\begingroup$ @Vicrobot: Once again, the probability of picking $1$, $\sqrt{2} $ or $\pi$ from any real number between $0$ and $10$ must be exactly $0$. The probability is well defined, it must be at least 0 and it must also be smaller than $\frac{1}{n}$ for any $n$. It means it must be $0$ and not just "near zero" or "very small". It might feel weird but it's true, and you'll accept it someday ;). $\endgroup$ Sep 3, 2019 at 13:19
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    $\begingroup$ @Vicrobot I read the chat. It would help if you use reason rigorously with specific, well-defined language, because human intuitions about real numbers are often wrong. Not just your intuitions, mine as well. For example, picking a random real number uniformly from $[0,1]$, the probability of getting a rational number is exactly $0$. Understanding this surprising result requires understanding the concept of a measure zero subset. The main topic is the Lebesgue measure. Have fun :) $\endgroup$ Oct 11, 2019 at 13:12
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    $\begingroup$ I didn't say anything about " $1/ \infty$ ". Read about Lebesgue measure ... it will give you a better understanding of how there can be events with probability $0$ that are possible. $\endgroup$ Oct 11, 2019 at 14:39

8 Answers 8

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Two Schools

I think the crux of the matter is what probability actually is:

  • The Bayesian view - probabilities are measures of (personal) confidence or belief, so it's quite obvious why an event with probability zero is not the same thing as an impossible event. But perhaps this isn't such a satisfactory answer.
  • The frequentist view - probabilities are the asymptotic frequency of events as the number of independent trials tends to infinity. Here again wee see that something that happens with probability zero is not the same as something impossible; it's just something that happens so infrequently that the numerator in $\dfrac{\text{occurences}}{\text{trials}}$ is dominated by the denominator.

Technically Speaking

Putting aside such philosophical matters, there's also a technical matter to be discussed here. Under the usual measure-theoretic formulation of probability theory, we have a sample space $\Omega$ and a family $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ of events (measurable subsets of $\Omega$), and the probability of an event $A \in \mathcal{F}$ is its measure $\mathbb{P}(A)$. There is nothing in the axioms of measure theory which say that a non-empty set must have a non-zero measure; and if we interpret $\mathcal{F}$ as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability.

Example

To give a concrete example, consider a random variable $X$ which is uniformly distributed on the interval $[0, 1]$. Although $\mathbb{P}[X \in (a, b)] = b - a$ for all $(a, b) \subset [0, 1]$, the axioms of probability force us to conclude that $\mathbb{P}[X = x] = 0$ for any individual $x \in [0, 1]$: for if $\mathbb{P}[X = x] = \varepsilon > 0$, because $X$ is uniformly distributed, by additivity of the probabilities of disjoint events, we'd be forced to conclude that $[0, 1]$ contains at most $\frac{1}{\varepsilon}$ (a finite number!) points, which is absurd.

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  • $\begingroup$ Thanks! "if we interpret F as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability", (1) do you mean that the empty set is a possible event, and not an impossible one? (2) Are impossible events F-measurable? $\endgroup$
    – Tim
    May 24, 2011 at 20:12
  • $\begingroup$ @Tim: I'd say $\emptyset$ is a possible event: it's possible nothing happens. The idea that $\emptyset$ is impossible comes from the fact that an impossible event is outside the scope of the model, whether it be because the "corresponding" sample point is not in $\Omega$ (whatever that means) or because the combination of sample points in question is not measurable. In either case it's meaningless to ask about the probability of such an event because it's undefined. $\endgroup$
    – Zhen Lin
    May 24, 2011 at 20:54
  • $\begingroup$ I like the first paragraph, it defines what exactly P(A) = 0 is. $\endgroup$
    – SmallChess
    Aug 14, 2014 at 2:12
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    $\begingroup$ But $\epsilon = 0$ is also absurd, because then the probability that you pick any number at all is $0$. $\endgroup$
    – Fax
    Sep 13, 2019 at 14:21
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Zero probability isn't impossibility. If you were to choose a random number from the real line, 1 has zero probability of being chosen, but still it's possible to choose 1.

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    $\begingroup$ More information here: en.wikipedia.org/wiki/Almost_surely $\endgroup$ May 24, 2011 at 19:41
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    $\begingroup$ One could argue that such a random choice of a real doesn't have zero probability, but only infinitesimal probability. Thus, it looks like you have zero probability, when you don't. $\endgroup$ Jun 13, 2011 at 13:42
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    $\begingroup$ No, Doug. Fernando is correct. Any finite number divided by an infinite number is 0, not an "infinitesimal probability". Dealing with infinities is a difficult concept to grasp, sort of like Einstein's relativity. $\endgroup$
    – lkessler
    Jul 30, 2014 at 16:56
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    $\begingroup$ @lkessler What, 1/infinity being an infinitesimal is like the definition of an infinitesimal. en.wikipedia.org/wiki/Infinitesimal $\endgroup$
    – semicolon
    Jun 21, 2017 at 19:30
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    $\begingroup$ @YasirSadiq What we're dealing with here is not $1/\infty$, but $\lim_{n\to\infty}(1/n)$, which is precisely $0$(at this point, I seriously suggest you to look up for the epsilon-delta definition of limits). semicolon's point is not valid in the set of standard reals, but in the set of hyperreals/surreals. $\endgroup$
    – Manan
    Aug 12, 2020 at 13:45
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Adding to what others have already mentioned. There is also this notion of plausible event. I am not sure if this is standard. But in the book "Measure Theory and Probability" by Malcolm Ritchie Adams and V. Guillemin, a plausible event is defined as an event which corresponds to a Borel set.

Hence, my understanding of the three words is as follows:

If we take the probability space $(X,\mathscr{F},\mu )$,

An event $A \subseteq X$ is impossible if $A = \emptyset$

An event $A \subseteq X$ is implausible if $A \notin \mathscr{F}$

An event $A \subseteq X$ is improbable if $\mu^*(A) = 0$

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    $\begingroup$ An implausible event doesn't satisfy our intuitive notion of "improbable", since if an event is implausible then so is its complement. $\endgroup$ May 24, 2011 at 20:48
  • $\begingroup$ @Yuval: True. On a slightly similar note, I am not sure intuitively if we would want $\mu(A) = 0$ or $\mu^*(A)$ for an improbable event. $\endgroup$
    – user17762
    May 24, 2011 at 20:59
  • $\begingroup$ Thanks! Just wonder what the star means in $\mu^*(A) = 0$? $\endgroup$
    – Tim
    May 24, 2011 at 20:59
  • $\begingroup$ @Tim: By $\mu^*$, I mean the outer measure $\endgroup$
    – user17762
    May 24, 2011 at 21:01
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    $\begingroup$ @Tim presumably because the outer measure can be applied to all set, not just sets $A\in \mathcal{F}$. $\endgroup$ May 24, 2011 at 21:26
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Mathematicians generally formalize probability using the notion of a probability space and measure theory. In this formalism it is possible for an event to have probability $0$ without being the empty event. Perhaps the simplest "realistic" (and I use the word loosely) example of such an event is the event of flipping only heads infinitely many times. This event has probability $0$, but it is not empty, which is what one might call a formal definition of "impossible."

The underlying probability space is the set of possible ways to flip a coin infinitely many times. An example of an impossible event here is that you flip, say, cat. The coin has only a heads side and a tails side; it doesn't have a cat side, so flipping cat is impossible.

(Whether this formalism says anything reasonable about the real world is debatable. In practice, events of sufficiently small probability are already impossible. The above is just a statement about a certain mathematical formalism that has proven to be useful in certain contexts. In mathematics, we want to prove statements about some class of objects. Sometimes we can prove that the statement holds with probability $1$, but this does not imply that it holds for all objects, and since we actually care about all objects this distinction really does need to be made in mathematics.)

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    $\begingroup$ "In practice, events of sufficiently small probability are already impossible." Not true! You give me a non-zero probability, however small, and I can generate an event on my laptop which has an even smaller probability of occurring. I just have to generate enough random bits. For this, I only need time proportional to the logarithm of the reciprocal of your probability. (Of course you could make this reciprocal logarithm really big, but I would still win for all reasonable definitions of "really big".) $\endgroup$
    – TonyK
    May 24, 2011 at 19:50
  • $\begingroup$ That feels like a cop-out. Unless you can predict what bits you're going to generate in advance... in practice, you should ignore any event which is less likely than, say, winning the lottery ten times (given that you don't currently play the lottery and you believe yourself to be a rational actor). $\endgroup$ May 24, 2011 at 19:52
  • $\begingroup$ Thanks! Concerning my second question, do you mean that probability theory represents impossibility with the empty set, not a non-empty set with probability zero? Is this how probability theory distinguishes zero probability and impossibility? $\endgroup$
    – Tim
    May 24, 2011 at 19:54
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    $\begingroup$ @Qiaochu: Perhaps I haven't explained myself clearly enough. Let me try again. (i) In an uncountable probability space, events of probability zero can happen. No argument there. (ii) In a countable probability space, events of arbitrarily small probability can happen. Agreed? (iii) I can easily program my laptop to generate (pseudo-)random events of arbitrarily small probability. $\endgroup$
    – TonyK
    May 24, 2011 at 20:10
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    $\begingroup$ @Doug: that's an interesting question. I suppose one could write down a generalization of probability measures that are allowed to take value in the nonstandard reals and perhaps such a theory would allow such things. $\endgroup$ Jun 13, 2011 at 13:58
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Let $A$ be an event, $\Pr$ be the probability measure.

$A$ has zero probability if $\Pr(A) = 0$.

$A$ is impossible if $A=\emptyset$.

Impossibility implies zero probability, but the reverse is false. Consider the real line $\mathbb{R}$; if you randomly select a number $x$, the probability that $x=0$ is $0$, but this is not impossible. In fact, the probability that $x$ belongs to some countable set, e.g $\mathbb{Q}$, is also $0$.

From a purely mathematical point of view, impossibility is simply a stronger statement, so impossibility cannot be described by probability measure. However, another way of thinking might shed some light. That is, if the probability that something exists has probability greater than $0$, then it exists. This notion has been used for some mathematical arguments.

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  • $\begingroup$ Thanks! Having said that A is impossible if A is the empty set, how come that impossibility cannot be described by probability measure? Isn't A being the empty set how probability measure describes impossibility? $\endgroup$
    – Tim
    May 24, 2011 at 20:02
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    $\begingroup$ What I mean is that $\Pr(A) = 0$ does not imply $A=\emptyset$, i.e. knowing probability measure $=0$ does not help you figure out if the set is empty or not. $\endgroup$
    – Hawii
    May 24, 2011 at 20:27
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    $\begingroup$ I like the definition of "impossibility cannot be described by probability measure." $\endgroup$
    – SmallChess
    Aug 14, 2014 at 2:15
  • $\begingroup$ So I guess this means the notion of impossibility is external to the 'probabilistic way of thinking' (as it's described here) since whether an event is empty depends on the particular representation (i.e., probability space) used? $\endgroup$ Jan 8, 2016 at 4:05
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Probability theory is an abstract subject, which is not limited to the real world. In cases where it is limited to the real world, an event of zero probability will not occur. But the abstract underpinning of the real-world cases allows for the occurrence of zero-probability events; when you translate these abstract events into events that are physically detectable, their probabilities become non-zero.

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  • $\begingroup$ Thanks! Do you mean that probability theory may not perfectly model the real world, which is the reason that an event assumed to have zero probability in probability theoretical model can have non-zero probability in the real world and therefore can occur in the real world? $\endgroup$
    – Tim
    May 24, 2011 at 19:58
  • $\begingroup$ @Tim: the culprit isn't probability theory in general (here). It's the specific information that you have. If your information is incomplete, your assessment of the probability of various events is necessarily going to be incomplete. Your assessment of the probability of an event is (if you subscribe to the appropriate philosophy of probability) a statement about your mind, not about the world. $\endgroup$ May 24, 2011 at 20:17
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    $\begingroup$ Another way to think about it is that, in the real world, you only have finite probability spaces. The is no "real" way to pick a random number uniformly in $[0,1]$, only to pick successively closer approximations of a real number in $[0,1]$ so that the limit is uniform. Then probability on infinite spaces is about limits of probabilities on finite spaces. This turns out to be useful in the real world because we are dealing with finite spaces so large that the probabilities behave very close to their limits. $\endgroup$ May 24, 2011 at 20:39
  • $\begingroup$ @Tim: I mean that probability theory can model the real world (as far as we know), but that it can do more if we want it to. $\endgroup$
    – TonyK
    May 27, 2011 at 11:07
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I believe the root of this confusion is all because Math(in its nature) is in general(including probabilities) always more serious towards its concepts, than we -the humans- are. And this seriousness is exactly what brings Math its clarificationability(the power of turning vague into non-vague) and simplificationability(the power of turning complicated into simple), two of its core ultimate goals!

An instance of this amount of seriousness: When Math says "probability" it really means "probability"! I.e, mathematical probabilities are ALWAYS about "probable" expectations, and so even its 1 is not certainty, and even its 0 is not impossibility! Rather indeed, its 1 is absolute probability and its 0 is absolute improbability!

It's just an intentional neglection of all the odd possibilities. For instance, when I throw a die, it is possible that I will get a 7...! Why?! Maybe someone has added a dot to the 6 side! But we neglect that possibility on purpose for the sake of simplification, because it's an odd(read improbable) possibility.

Another Example: Yes, it is possible (even absolutely possible!) that the exact number 3 might occur in the continuous interval [1, 4], but it is absolutely improbable(notice how you can't disagree), thus the P=0. (Notice: Computers are never really continuous, so don't even try doing it programmatically, it won't be a counterexample).

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I don't know if this is correct, but this is how I made sense of it, I think this a very intuitive explanation.

Imagine a car moving (from left to right) in a straight line with constant speed of 100MPH, and at x=0 the breaks of the car are engaged. Naturally the car will start to decelerate until it comes to rest (speed = 0 MPH).

The car will certainly come to rest, but the question is:

At what distance D [meters] (with respect to x=0) will the car come to rest? (This value D is not known apriori and can vary from one experiment to the next by a myriad of factors, so we can ask probability questions about the variable D).

Since distance is modeled by real numbers, the variable D is continuous. If we ask: what is P(D=6.354), the probability that it comes to rest at specifically D = 6.354 meters?

then, since the probability distribution of D is continuous, the answer is P(D=6.354)=0. This probability is zero.

Notice that there is nothing special about the number "6.354". That is, for any real number r the probability that D is exactly r is zero, i.e. P(D=r) = 0. This is simply a consequence of how continuous probability distributions are defined.

Regardless, we know that the car will certainly come to rest and that this happens at some specific distance d (where d is a real number). So even though P(D=d) = 0, it does not mean that coming to rest at D=d was impossible; clearly it was not because the car does come to rest at d.

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