Sum of two dice without enumerating over a sample space Given two fair dice are rolled,
I want to calculate the probability of their sum being >= 7.
I've calculated 1 - p(sum <= 6) by enumerating over the events of interest in the sample space of 36 to get a probability of 7/12.
Is there method or formula to arrive at this same probability without having to enumerate over the sample space?
 A: How about this:
Let the numbers on the two dice be $X_1$ and $X_2,$
each uniformly distributed on $\{1,2,3,4,5,6\}$
and independent of each other.
Let $Y = 7 - X_2.$ Then $Y$ is
uniformly distributed on $\{1,2,3,4,5,6\}$ and independent of $X_1.$
Observe that 
$$X_1 - Y + 7 =  X_1 - (7 - X_2) + 7 = X_1 + X_2.$$
Therefore
$$P(X_1 + X_2 \geq 7) = P(X_1 - Y + 7 \geq 7) = P(X_1 \geq Y).$$
By symmetry, since $X_1$ and $Y$ have the same distribution,
$P(X_1 > Y) = P(X_1 < Y),$
and by the law of total probability,
$$P(X_1 = Y) + 2P(X_1 < Y) = P(X_1 = Y) + P(X_1 < Y) + P(X_1 > Y) = 1.$$
But since $X_1$ and $Y$ are independent, $P(X_1 = Y)$ is just the probability that $Y$ comes out to some preselected value, that is,
$$P(X_1 = Y) = \frac16.$$
Using a little algebra on the equations found so far, you can show that
$$P(X_1 > Y) = \frac12\left(1 - \frac16\right) = \frac{5}{12}.$$
Then
$$P(X_1 + X_2 \geq 7) = P(X_1 \geq Y) = P(X_1 > Y) + P(X_1 = Y)
= \frac{5}{12} + \frac16 = \frac{7}{12}.$$
This seems like a lot of work to avoid counting, but there you are.
A: Is this enumeration?
$$P(X_1+X_2\leq 6)=\sum_{i=1}^6P(X_1+X_2\leq6\mid X_2=i)P(X_2=i)=$$
$$=\frac16\sum_{i=1}^6P(X_1\leq6-i)=\frac1{36}(5+4+3+2+1)=\frac{15}{36}.$$
And
$$1-\frac{15}{36}=\frac7{12}.$$
