If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?

If you roll two fair six-sided dice, what is the probability that the sum is $4$ or higher?

The answer is $\frac{33}{36}$ or $\frac{11}{12}$. I understand how to arrive at this answer. What I don't understand is why the answer isn't $\frac{9}{11}$? When summing the results after rolling two fair six sided dice, there are $11$ equally possible outcomes: $2$ through $12$. Two of these outcomes are below four, meaning $9$ are greater than or equal to four which is how I arrived at $\frac{9}{11}$. Can someone help explain why that is wrong?

Answer

It is wrong because it is not $11$ equally possible outcome.

There is exactly $1$ way to get the sum to be $2$. ($1+1=2$)

but there is more than one way to get $3$. ($1+2=3, 2+1=3$)