Originally Posted by

**Plato** You did not take in account the possible permutations of the dice.

For example: $X=2$ the case $(1,1,1,2)$ can occur in four ways.

All of $(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1)$ must be counted for $X=2$

Consider rolling just two dice in the usual game where we add both together.

$\begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{( 1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2 ,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3, 4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4 )}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4) }&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)} &{(6,5)}&{(6,6)}\end{array}$

To get the sum $7$ we count the pairs $(1,6),(1,6),(2,5),(5,2),(3,4),(4,3)$ to get $\mathscr{P}(S=7)=\dfrac{6}{36}$.