TITLE: If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?
QUESTION [14 upvotes]: 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?

REPLY [4 votes]: Your probabilities are wrong. For each value of die 1, there are 6 possible values of die 2, the available scores for a roll of two dice are: (7 is the most likely score)
2,3,4,5,6,7

  3,4,5,6,7,8

    4,5,6,7,8,9

      5,6,7,8,9,10

        6,7,8,9,10,11

          7,8,9,10,11,12

So, out of the 36 total possible outcomes, 33 of them have a value of 4 or higher, or 33:36, or 11:12