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In probability theory, one says that an event happens almost surely (sometimes abbreviated as a. s.) if it happens with probability one. In other words, the set of possible exceptions may be non-empty, but it has probability zero.

In probability experiments on a finite sample space, there is often no difference between almost surely and surely. However, the distinction becomes important when the sample space is an infinite set, because an infinite set can have non-empty subsets of probability zero.

An example of the use of this concept is the strong version of the law of large numbers.

Almost never describes the opposite of almost surely: an event that happens with probability zero happens almost never.

## Definition

In a probability space $(\Omega, \mathcal{F}, P)$, an event $E \in \mathcal{F}$ happens almost surely if

$P(E) \; = \; 1$.

It happens almost never if

$P(E) \; = \; 0$.

## Examples

### Interval

For a uniform distribution on the interval $[0,1] \subset \mathbb R$, the probability of randomly picking exactly a certain number $x \in [0,1]$ is 0, although this event is not impossible. Correspondingly, the probability of picking any number except $x$ is 1, but this event will not necessarily occur.

### Daily Kit Box

Consider the case where the Tech card of a is revealed. As the card can only be Advanced or Mid-Tech, the corresponding sample space is $\{A, M\}$, where the event $\{A\}$ occurs if the card is Advanced Tech, and $\{M\}$ if it is Mid-Tech. The probability of getting Advanced Tech is $p = 0.3$ from which it follows that the complementary event, not getting Advanced Tech (getting Mid-Tech), has the probability $1 - p = 0.7$.

In an experiment where an infinite amount of Tech cards from a Daily Kit Box is revealed repeatedly, any infinite sequence of Advanced and Mid-Tech is a possible outcome of the experiment. However, any particular infinite sequence of Advanced and Mid-Tech has probability zero of being the exact outcome of the (infinite) experiment.

To see why, note that the probability of getting only Advanced Techs with 2 repetitions is

$p \cdot p = p^2 = 0.3^2 = 0.09$.

The probability of getting 3 Advanced Techs with 3 repetitions is

$p \cdot p \cdot p = p^3 = 0.027$,

and that of getting $n$ Advanced Techs with $n$ repetitions is $p^n$. As the probability gets smaller the higher the number of repetitions is, letting $n$ tend to $\infty$ yields zero:

$\lim_{n\to\infty}0.3^n = 0$

Note that the result is the same no matter the value of $p$, so long as we constrain $p$ to be greater than 0, and less than 1. In particular, the event "the sequence contains at least one $M$" happens almost surely (i. e., with probability 1).

However, if instead of an infinite number of repetitions we stop revealing Tech cards after some finite time, say a million turns, then the all-Advanced Tech sequence has non-zero probability. The all-Advanced Tech sequence has probability $p^{1,000,000}\neq 0$, while the probability of getting at least one Mid-Tech is $1 - p^{1,000,000}$ and the event is no longer almost sure.