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?

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 $3$. ($1+2=3, 2+1=3$)

As pointed out by others, the possible sums of the dice don't all have an equal probability of showing up. To see this, you can write it all out:

Die 1 | Die 2 | result
  1       1       2
  1       2       3
  1       3       4
  ...
  6       5      11
  6       6      12

When you look at the resulting table, there are 36 combinations. 1 of those is a 2 (ie you have a 1 in 36 chance of getting a 2 from D1 + D2), 2 of those are '3' etc.

Now it's easy to see how to get the chance of getting a 4 or higher.

$\frac9{11}$ is wrong precisely because it assumes the probabilities of getting each sum are equal. Here they are not: there's only one way to roll a sum of two (the snake eyes of gambling jargon) but six ways to roll a seven.

if the correct answer is $ \frac {33}{36}$ it means there are 36 possible results.. You should try to write them down..It will be more clear. You just used equiprobability in a case where there isn't equiprobability..$P(1)=0 \quad P(2) =1/36,\quad P(3)=2/36 ,\quad P(4)=3/36,$

$ P(5)=4/36 \quad P(6)=5/36 \quad P(7)=6/36 \quad P(8)=5/36 \quad $

$ P(9)=4/36 \quad P(10)=3/36 \quad P(11)=2/36 \quad P(12)=1/36$

And the answer is just: $$P=1-P(1)-P(2)-P(3)=\frac {33}{36}$$