To say that the sequence $Y_n$ (not necessarily independent) converges almost surely or almost everywhere or with probability 1 or strongly towards $X$ means that
$$P(\limu{ n }{ \infty } Y_n = c) = 1 $$
This means that the values $Y_n$ approach the value of $c$,
Let $X_1, X_2, … $ be a sequence of i.i.d. random variables with mean $\mu $. Then, the sequence of sample means $M_n = (X_1, _2, …, _n) / n$ converges to $\mu $ with probability $1 $.
Or we can simply say that
$$M_n \asto \mu $$
Let $X_1, X_2, … $ be a sequence of i.i.d. random variables that are uniformly distributed in $[0, 1] $, and let $Y_n = \min{ X_1, X_2, …, X_n } $. We wish to show that $Y_n $ converges to $0 $ with probability $1$.
Observe that $Y_{n + 1} \leq Y_n$ for all $n $.
Since this sequence is bounded below by zero, it must have a limit, which we denote by $Y$.
Fix $\epsilon > 0 $. We have $Y \geq \epsilon$ if and only if $X_i \geq \epsilon$ for all $i $.
$$P(Y \geq \epsilon) = P(X_1 \geq \epsilon … X_n \geq \epsilon) = (1 - \epsilon)^n$$
Since this is true for all $n $. We must have
$$P(Y \geq \epsilon) \leq \limu{ n }{ \infty }(1 - \epsilon)^n = 0 $$
This shows that $P(Y \geq \epsilon) = 0 $, $\forall \epsilon > 0$.
We conclude that $P(Y > 0) = 0 $ which implies that $P(Y = 0) = 1 $. Since $Y $ is the limit of $Y_n$, we see that $Y_n \asto 1$.
$$Y \asto c \implies Y \ipto c $$
Consider a discrete-time arrival process.
The set of times is partitioned into consecutive intervals of the form $I_k = \set{ 2^k, 2^k +1, …, 2^{k+1} - 1 }$. Note that the length of $I_k $ is $2^k $, which increases with $k $. During each interval $I_k $, there is exactly one arrival, and all times within an interval are equally likely. The arrival times within different intervals are assumed to be independent. Let us define $Y_n = 1 $ if there is an arrival at time $n $, and $Y_n = 0 $ if there is no arrival.
We have $P(Y_n \neq 0) = \frac{ 1 }{ 2^k } $, if $n \in I_k $. Note that as $n $ increases, it belongs to intervals $I_k $ with increasingly large indices $k $. Consequently,
$$\limu{ n }{ \infty } P(Y_n \neq 0) = \limu{ k }{ \infty } \frac{ 1 }{ 2^k } = 0$$
i.e.
$$Y_n \ipto 0$$
However, when we carry out the experiment, the total number of arrivals is infinite (one arrival during each interval $I_k$). Therefore, $Y_n$ is unity for infinitely many values of $n$, the event $\set{ \limu{ n }{ \infty } Y_n = 0}$ has zero probability, and we do not have convergence with probability 1.
At any given time, there is only a, small, diminishing with $n$, probability of a substantial deviation from $0$, which implies convergence in probability.
On the other hand, given enough time, a substantial deviation from 0 is certain to occur and for this reason, we do not have convergence with probability 1.