Crisis in my understanding of probability If I were to roll a die, what would ​be the probability of getting $2$? Certainly it would be $\dfrac 16$ (because there are $6$ numbers and sample space contains 6 numbers)
 But I think we can look at it another way. 
We need $2$ right? and from $1-6$ there is only one number which is $2$ i.e. $2$. If I were to roll there would essentially be two types of numbers ie. one which is $2$ and one which is not ie $1,3,4,5,6$. So there are two basic outcomes.
P(E) = number $2$÷total outcomes ie $2$.
P(E)= $\dfrac 12$
So why is the probability $\dfrac 16$(when it is clear that the outcome we desire, can either happen or not happen.) Shouldn't the probability be $\dfrac 12$ instead, why not?
The probability is $\frac{1}{6}$ because there are $6$ elements in the sample space that are all equally likely.

If we take a sample space of $2$ and not $2$, then there are $2$ elements in the sample space, but they are not equally likely.

The classic example is that you don't have a $50\%$ chance of winning the lottery because the two options are "win" and "don't win".
P(E) = number 22÷total outcomes 


The problem in the construction you have created is that the selection part (choose a side out of 6, the set {1,2,3,4,5,6}) doesn't match the total outcomes part (choose from the set {2, not 2}).