# Does 0% chance mean impossible? [duplicate]

Suppose we pick a random real number between 0 and 1 and call it $x$. There are $2^{\aleph_0}$ possible values, so the chance of picking any specific number (such as $x$) in that range is 0. But in the end, we did manage to pick $x$, despite its probability of 0.

Does this mean that a 0% chance is actually possible, or is there some flaw in this logic?

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## marked as duplicate by hardmath, egreg, N. S., T. Bongers, Christopher A. WongJul 30 '14 at 17:16

No, the probability really is zero. And no, probability zero does not mean impossible. – mixedmath Jul 30 '14 at 16:09
@snulty: What do you mean "plays well"? Probability theory is essentially measure theory. The axiom of choice can be used to show there are non-measurable sets, therefore it can be used to show that not every subset of $[0,1]$ can be assigned a probability, if your distribution is "reasonably nice". – Asaf Karagila Jul 30 '14 at 17:33

# Math

Yes, this is the canonical example of a non-trivial event with $0$ probability.

Probability is a measure, and it is quite common for non-empty sets to have $0$ measure (such sets might be dense and uncountable &c).

Conversely, an event happens almost surely (sometimes abbreviated as a.s.) if it happens with probability $1$. Note the qualifier almost! E.g., if you pick a random number in $[0;1]$, it will be almost surely an irrational, moreover, a transcendental number (because their complements - rationals and algebraic numbers - are countable and thus have zero measure). This does not mean that you cannot possibly pick $1/2$.

# Philosophy

## Bayesian

If you view probability as a subjective measure of likelihood that a certain event will occur, then, obviously, you cannot believe that one number in $[0;1]$ is more likely than another one; which means that each individual number has to be assigned probability of $0$.

## Frequentist

If you view probability as the limit of frequency, then a random sequence in $[0;1]$ will probably contain no duplicates, so, as the number of trials goes to $\infty$, the number of successes (i.e., occurrences of the specific number) will be $0$ or $1$, so the probability will be $0$.

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How come picking a random number almost surely results in an irrational/transcendental one? Is it because the rationals are countable and irrationals are not? – Ryan Jul 30 '14 at 16:44
@Ryan: And therefore there are infinitely more irrational numbers, and therefore you are infinitely more likely to pick one. – Mooing Duck Jul 30 '14 at 16:45
@MooingDuck Thanks! – Ryan Jul 30 '14 at 16:46

You should look up sets of measure zero and the complementary notion of almost everywhere.

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There are an infinite number of real numbers between 0 and 1. That means the chance of any of them being picked is zero. However, one of them will surely be picked. That is, one out of an infinite number of them.

Therefore, the chance that you select in advance the one that will be picked is zero.

And once you've selected it, or selected 1,000 random numbers, or selected a googolplex random numbers, you have still selected just 0% of the possible numbers you can select.

Dealing with infinite spaces can lead to somewhat illogical sounding conclusions, but the following is your answer:

Each individual rational or irrational number is possible to be selected, but with 0% chance.

Impossible means there is a 0% chance. But a 0% chance does not mean impossible.

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