Henry Ernest Dudeney/Modern Puzzles/205 - The Three Dice/Solution

by : $205$

 * The Three Dice

Solution
Mason's chances of winning were $1$ in $6$.

If Jackson selected $8$ and $14$, his chances were the same.

Proof
To make $7$, you need to make the throws:
 * $1 \ 1 \ 5$
 * $1 \ 2 \ 4$
 * $1 \ 3 \ 3$
 * $1 \ 4 \ 2$
 * $1 \ 5 \ 1$
 * $2 \ 1 \ 4$
 * $2 \ 2 \ 3$
 * $2 \ 3 \ 2$
 * $2 \ 4 \ 1$
 * $3 \ 1 \ 3$
 * $3 \ 2 \ 2$
 * $3 \ 3 \ 1$
 * $4 \ 1 \ 2$
 * $4 \ 2 \ 1$
 * $5 \ 1 \ 1$

a total of $15$ throws.

To make $13$, you need to make the throws:
 * $1 \ 6 \ 6$
 * $2 \ 5 \ 6$
 * $2 \ 6 \ 5$
 * $3 \ 4 \ 6$
 * $3 \ 5 \ 5$
 * $3 \ 6 \ 4$
 * $4 \ 3 \ 6$
 * $4 \ 4 \ 5$
 * $4 \ 5 \ 4$
 * $4 \ 6 \ 3$
 * $5 \ 2 \ 6$
 * $5 \ 3 \ 5$
 * $5 \ 4 \ 4$
 * $5 \ 5 \ 3$
 * $5 \ 6 \ 2$
 * $6 \ 6 \ 1$
 * $6 \ 5 \ 2$
 * $6 \ 4 \ 3$
 * $6 \ 3 \ 4$
 * $6 \ 2 \ 5$
 * $6 \ 6 \ 1$

a total of $21$ throws.

Hence Mason wins with a total of $36$ throws.

For each of throw $a \ b \ c$ there exists the complementary throw $\paren {7 - a} \ \paren {7 - b} \ \paren {7 - c}$ which makes a total of $21 - \paren {a + b + c}$.

Hence there is an equal chance of making $21 - \paren {a + b + c}$ as there is of making $a + b + c$.

Hence Jackson's numbers are $21 - 13$ and $21 - 7$, giving $8$ and $14$.