Question

On a tiny remote island where the death sentence still exists a man can be granted mercy after receiving the death sentence
He is given 18 red balls and 6 green balls
He must divide them between 3 boxes with at least one ball in each
He is blindfolded and must choose a box at random and then a single ball from within the box
He only receives mercy if a red ball is chosen
Find the probability that he recieves mercy when he distributes the balls in the most favourable mannor

Answer 1

Task:

On a tiny remote island where the death sentence still exists a man can be granted mercy after receiving the death sentence

He is given 18 red balls and 6 green balls

He must divide them between 3 boxes with at least one ball in each

He is blindfolded and must choose a box at random and then a single ball from within the box. He only receives mercy if a red ball is chosen

Find the probability that he receives mercy when he distributes the balls in the most favorable manner.

Solution:

Suppose that a man placed $r_1, r_2$ and $r_3$ red balls in the $1^{\text{st}}$, $2^{\text{nd}}$ and $3^{\text{rd}}$ boxes respectively. And $g_1, g_2, g_3$ green balls in the $1^{\text{st}}$, $2^{\text{nd}}$ and $3^{\text{rd}}$ boxes respectively ($g_i, r_i$ are non-negative whole numbers). We have that

(1) $r_1 + r_2 + r_3 = 18$,

(2) $g_1 + g_2 + g_3 = 6$.

(3) $r_1 + g_2 > 0, i = 1, 2, 3$.

Denote a valid distribution as $d$, $d = d(r_1, r_2, r_3, g_1, g_2, g_3)$ is defined by $r_1, r_2, r_3, g_1, g_2, g_3$, for which (1)-(3) are right. The probability that a man receives mercy is $P(d) = \frac{1}{3} \frac{r_1}{r_1 + g_1} + \frac{1}{3} \frac{r_2}{r_2 + g_2} + \frac{1}{3} \frac{r_3}{r_3 + g_3} = \frac{1}{3} \left( \frac{r_1}{r_1 + g_1} + \frac{r_2}{r_2 + g_2} + \frac{r_3}{r_3 + g_3} \right)$.

Call the balls distribution $d_0$ optimal if $P(d_0) = \max_d P(d) = P_0$.

Then

1) $P_0 \geq \frac{10}{11}$. Proof: $P_0 \geq P(d(16, 1, 1, 6, 0, 0)) = \frac{1}{3} \left( \frac{16}{16 + 6} + \frac{1}{1 + 0} + \frac{1}{1 + 0} \right) = \frac{1}{3} \left( \frac{8}{11} + 2 \right) = \frac{130}{311} = \frac{10}{11}$.

2) $r_1, r_2, r_3 \geq 1$, because otherwise $P(d_0) \leq \frac{1}{3} (1 + 1 + 0) = \frac{2}{3} \leq \frac{10}{11}$.

3) For $1 < a \leq b$, $1 < c \leq d$, $\frac{a}{b} + \frac{c}{d} \leq \frac{a + c - 1}{b + d - 1} + 1$.

Proof: inequality is equivalent to $\frac{a}{b} (b + d - 1) + \frac{c}{d} (b + d - 1) \leq a + b + c + d - 2$

$$\begin{array}{l} a + a \frac{d}{b} - \frac{a}{b} + c + c \frac{b}{d} - \frac{c}{d} \leq a + b + c + d - 2 \\ \frac{a}{b} (d - 1) + \frac{c}{d} (b - 1) \leq b + d - 2 \\ \end{array}$$

But $a \leq b$ and $c \leq d$, therefore, $\frac{a}{b} (d - 1) + \frac{c}{d} (b - 1) \leq d - 1 + b - 1 = b + d - 2$.

Therefore, $\frac{a}{b} (d - 1) + \frac{c}{d} (b - 1) \leq b + d - 2$ is right and $\frac{a}{b} + \frac{c}{d} \leq \frac{a + c - 1}{b + d - 1} + 1$ is right.

4) Using 3): $P(d) = \frac{1}{3} \left( \frac{r_1}{r_1 + g_1} + \frac{r_2}{r_2 + g_2} + \frac{r_3}{r_3 + g_3} \right) \leq \frac{1}{3} \left( \frac{r_1}{r_1 + g_1} + \frac{r_2 + r_3 - 1}{r_2 + g_2 + r_3 + g_3 - 1} + 1 \right) \leq$

$$\frac{1}{3} \left( \frac{r_1 + r_2 + r_3 - 2}{r_1 + g_1 + r_2 + g_2 + r_3 + g_3 - 2} + 2 \right) = \frac{1}{3} \left( \frac{16}{16 + 6} + 2 \right) = \frac{10}{11}.$$

5) 1) and 4) $\Rightarrow P_0 = \frac{10}{11}$.

Answer: $\frac{10}{11}$.