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