What does it really mean when the probability of multiple independent events sum to 1? More than 1? For example, say I roll a fair six-sided die six times, and ask what the likelihood is that I will see a result of "6" among the six outcomes. Each roll, ostensibly, has a 1/6 probability of giving me a "6", and future rolls are not dependent on past rolls. Summing the probability of the events, 1/6+1/6+...+1/6, or, in this case, (1/6)*6, I calculate a probability of 1. But in real life, I know that it is perfectly possible to roll a fair six-sided die six times and not see a "6" among the outcomes. Going a step further, I can roll a fair six-sided die nine times, and ask what the likelihood of seeing a "6" among the outcomes is, for a probability of (1/6)*9 or 1.5, but I can ALSO roll a fair six-sided die nine times in real life and never see a "6". So am I missing something? Does it have to do with the word "fair"? Am I using the wrong formula? Is probability just an inherently flawed description of reality that calls for some subjectivity? Something else?
 A: *

*Here's another angle:
Since the die rolls are independent (i.e., the outcome of one doesn't
affect the probabilities of the outcomes of any other), the probability that the first '4' is obtained on the $\left(n+1\right)^\text{th}$ roll is
$\left(\frac56\right)^n\left(\frac16\right).$
(So the chance of getting a '4' only on the $7^\text{th}$ roll is
$6\%.$)
Thus, there will always be a nonzero chance of still not getting a
'4' however many die rolls.


*Addressing the question directly:
The event of obtaining a '4' on the $p^\text{th}$ roll is not
mutually exclusive of the event of obtaining a '4' on the
$q^\text{th}$ roll, because they can both†
occur. So, the probability of obtaining at least one '4' on the
$p^\text{th}$ and $q^\text{th}$ rolls is smaller than
$\frac16+\frac16.$
   † regardless of whether the rolls
are happening concurrently or in succession 
As the number of rolls $n$ increases, the number of such common
outcomes among the rolls increases at an increasing rate, i.e.,
the probability of getting at least one '4' increases at a
decreasing rate. The desired probability is always smaller than $\frac n6$ and $1.$
A: Your fundamental problem lies in the "Fundamental Principles of Counting".
Though it's intuitive, I'll explicitly state it here as well:

If an event can occur in $m$ different ways, and another event can occur in $n$ different ways, then the total number of occurrences of the events is $m × n$.
The fundamental principle of counting only works when the choice that is to be made are independent of each other.
If one possibility depends on another, then a simple multiplication does not work.

In your case, as you have identified, "future rolls are not dependent on past rolls". So, we can simply multiply (not add) to get the required result.
Examples:

*

*What is the probability of getting 6 on all the 6 dices rolled?


For each roll, getting "$6$" has a probability of $1/6$. 
So, the probability of getting $6$ on all of them at the same time/consecutively is:
P(E)=$\frac16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16=\frac1{6^6}$.



*What is the probability of getting 6 on at least 1 of the dices rolled?


$P(E)=1-P(E')$ where $P(E')$ is the event of not getting $6$ on any dice.
$P(E')=\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56$



*What is the probability of getting a sum of 6?


It can happen only if we get $1$ on every dice (similar to example 1). 
So, the probability of getting $1$ on all of them at the same time/consecutively is:
P(E)=$\frac16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16=\frac1{6^6}$.

