Question

Show that the chances of throwing six with $4,3,$ or $2$ dice respectively are as $1:6:18$.

Answer 1

With 4 dice,

- The number of possible combination to get six are $10\{1113,1122,1311,3111,2211,2112,1131,1221,2121,1212\}$

- Total number of possible combinations $=6^4$

- $\therefore$ The probability $P_4=\frac{10}{6^4}$

With 3 dice,

- The number of possible combination to get six are $10\{114,141,411,123,132,312,213,321,231,222\}$

- Total number of possible combinations $=6^3$

$\therefore$ The probability $P_3=\frac{10}{6^3}$

With 2 dice,

- The number of possible combination to get six are $5\{51,42,33,24,15\}$

- Total number of possible combinations $=6^2$

$\therefore$ The probability $P_2=\frac{5}{6^3}$

$\therefore P_4:P_3:P_2=\frac{10}{6^4}:=\frac{10}{6^3}:\frac{5}{6^3}=1:6:18$