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

Modern Puzzles by Henry Ernest Dudeney: $205$

The Three Dice

Mason and Jackson were playing with three dice.

The player won whenever the numbers thrown added up to one of the two numbers he selected at the beginning of the game.

As a matter of fact, Mason selected $7$ and $13$, and one of his winning throws was $6$, $4$, $3$.

What were his chances of winning a throw?

And what two other numbers should Jackson have selected for his own throws to make his chances of winning exactly equal?

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 $14$.

$\blacksquare$

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $205$. -- The Three Dice
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $474$. The Three Dice