What is the probability of a sum of $6$ with three fair dice?

Question

When a single fair die is tossed, the probability of a "sum" of $6$ is $1/6$. When two fair dice are tossed, the probability of a sum of $6$ is $5/36$. What is the probability of a sum of $6$ when three fair dice are tossed?

Answer 1

If $p_1(x)$ is the probability of rolling an "x" on one die,

and $p_2(x)$ is the probability of rolling a sum equal to $x$ on two dice.

$p_3(x) = \sum_\limits {n=1}^6 p_1(n)p_2(x-n)$

$p_3(6) = $$\sum_\limits {n=1}^6 p_1(n)p_2(x-n)$

$p_2(x < 2) = 0$ and $p(x) = \frac {1}{6}$ for $1 \le x \le 6$

$p_3(6) = \frac 16 \sum_\limits {n=2}^5 p_2(n) = \frac 16(\frac 1{36} + \frac 2{36} + \frac 3{36} + \frac {4}{36})$

Answer 2

HINT: The number of all possible results is $6^3$. How many vectors $(d_1,d_2,d_3)$ where $d_i$ is the value on $i$-th dice, and $d_1+d_2+d_3=6$ have we got?

Answer 3

Hint. We have that $6=1+1+4=1+2+3=2+2+2$. Can you take it from here?