Question

Throwing a dice 3 times, where X the combination of total values of 13. What is all possible combinations of value that X can take

Answer 1

A dice is rolled 3 times. A roll can take values $1,2,3,4,5,6$. We need the sum of 3 rolls to be $13$. We enumerate the outcomes that give a sum of 13 for the 3 dies. Let the outcome of the rolls be $X_1,X_2,X_3$ .

Then with $X_1=1$ requires $X_2+X_3=12\Rightarrow X_2=6,X_3=6$ , 1 combination

Then with $X_1=2$ requires $X_2+X_3=11\Rightarrow X_2=6,X_3=5 ;X_2=5,X_3=6$ 2 combinations

Then with $X_1=3$ requires $X_2+X_3=10\Rightarrow X_2=6,X_3=4 ;X_2=5,X_3=5;X_2=4,X_3=6$ 3 combinations

Then with $X_1=4$ requires $X_2+X_3=9\Rightarrow X_2=6,X_3=3 ;X_2=5,X_3=4;X_2=4,X_3=5;X_2=4,X_3=5;X_2=3,X_3=6$ 8 combinations

Then with $X_1=5$ requires $X_2+X_3=8\Rightarrow X_2=6,X_3=2;X_2=5,X_3=3;X_2=4,X_3=4;X_2=5,X_3=3;X_2=2,X_3=6$ 5 combinations

Then with $X_1=6$ requires $X_2+X_3=7\Rightarrow X_2=6,X_3=1; X_2=5,X_3=2; X_2=4,X_3=3; X_2=3,X_4=3; X_2=2,X_3=5; X_2=1,X_3=6$ 6 combinations

Thus the total combinations of values X can take is $1+2+3+8+6={\color{Blue} 22}$ .