What is the probability that the sum of three thrown fair dice is $\geq13$?
I did the brute force method and wrote down all tupels. But I am not sure if it is right:
I got $\frac{53}{216}$.
Is it correct?
Maybe there is a more elegant way to get the probability but I don't see it.
On
It is the same as the combinations that $4$ dice sum to $19.$ This is only because there are 6 values in the $13-18$ range for three dice. So, each of these possibilities matches with a potential outcome on a theoretical 4th die.
${18\choose 3} - {4\choose 1}{12\choose 3} + {4\choose 2}{6\choose 3}$
Or, even nicer - the combinations of 4 dice that sum to 19 is the same as the combinations of 4 dice that sum to 9.
${8\choose 3} = 56$
All possible sums are $3,4,5,...,18$
By symmetry, the cases for $18$ is same as $3$, $17$ is same as $4$, and so on, so we need to consider only $3$ to $10$.
As a tool, we know the number of solution to $a_1+a_2+a_3=n$ where $a_i$s are all positive integer is ${n-1 \choose 2}$ as we can imagine there are $n$ balls and we put $2$ sticks in between the balls in order to separate them into three separate groups.
Now back to the sums:
$S=3,18: {2\choose 2}=1$
$S=4,17: {3\choose 2}=3$
$S=5,16: {4\choose 2}=6$
$S=6,15: {5\choose 2}=10$
$S=7,14: {6\choose 2}=15$
$S=8,13: {7\choose 2}=21$. Note since $a_1+a_2+a_3=8$ can have no positive solution where $a_i>6$, so we can just use the formula up to this point.
$S=9,12: {8\choose 2}-{3\choose 1}{1\choose 1}=25$. Note here we need to minus the number of solution where we fix one of $a_i$ to be $7$, and for example when $a_3=7$ the number of solutions for $a_1+a_2=9-7$.
$S=10,11: {9\choose 2}-{3\choose 1}{2\choose 1}-{3\choose 1}{1\choose 1}=27$.
Now we count the possibiliy of $S$ at least $13$:
$${1+3+6+10+15+21\over 216}={56\over 216}$$