Question

$A$ and $B$ throw with $3$ dice : if A throws $8$, what is $B^{\prime }s$ chance of throwing a higher number ?

Answer 1

We would find the cases where B throws a number less than or equal to 8 and then we subtract it from 216.

Number of cases when sum of 3 is on the dice $=(1,1,1)=1$

Number of cases when sum of 4 is on the dice $=(1,1,2);(1,2,1);(2,1,1)=1+2=3$

Number of cases when sum of 5 is on the dice $=(1,2,2);(2,1,2);(2,2,1);(3,1,1);(1,1,3);(1,3,1)=3+3=6$

Number of cases when sum of 6 is on the dice $=6+4=10$

Number of cases when sum of 7 is on the dice $=10+5=15$

Number of cases when sum of 8 is on the dice $=15+6=21$

Hence, total no of cases of B throing higher is $216-(1+3+6+10+15+21)=160$

So probability is $\frac{160}{216}=\frac{20}{27}$