The Puzzle of the Stranded Man

The man's best strategy is to place one white marble in the first two boxes and then the remaining marbles in the third box. This guarantees that he'll escape the island if one of the first two boxes are selected. If the man selects one of the boxes with only one white marble in it, then the probability that the man draws a white marble when selecting one of those boxes is 1. The remaining box will have 21 white marbles and 7 black marbles, so when that box is selected, the probability that the man draws a white marble from that box is $\frac{3}{4}$ . Now we sum the probabilities that he draws a white marble over the boxes:

$(\frac{1}{3})(1) + (\frac{1}{3})(1) + (\frac{1}{3})(\frac{3}{4}) = \frac{11}{12}$

Therefore, the probability that the man escapes the island is $\boxed{\displaystyle\frac{11}{12}}$

This is an interesting optimization problem! For simplicity, let $b := 7,\:w := 23$ be the total number of black and white marbles, respectively, and define for $i=1;\:2;\:3$ :

$\begin{aligned} b_i,\:w_i:&&&\text{number of black and white marbles in chest }i\\ B_i:&&&\text{event, that we choose chest }i\text{ with probability }P(B_i)=\frac{1}{3}\\ E: &&& \text{event, that we escape the island} \end{aligned}$

We calculate the probability to escape the island using Bayes' Theorem. The black and white marbles satisfy two boundary conditions: $\begin{aligned} P(E)&=\sum_{i=1}^3P(E|B_i)P(B_i) = \frac{1}{3}\sum_{i=1}^3P(E|B_i),&&&&&b_1+b_2+b_3&=b,&&&w_1+w_2+w_3&=w \end{aligned}$

Before we start optimizing, consider two edge cases first:

• Case 1: One chest does not contain any white marbles. If we pick that chest, we stay on the island for sure. The other two chests let us escape with a probability less than or equal to one: $P(E)\leq\frac{1}{3}(0+1+1)=\frac{2}{3}$

• Case 2: Two chests contain exactly one white marble. If we pick either of those, we escape the island for sure. The remaining marbles determine the probability of the third chest: $P(E)=\frac{1}{3}\left( 1+1+\frac{w-2}{w-2+b} \right)=\frac{2}{3}+\frac{1}{3}\cdot\frac{w-2}{w-2+b}=\frac{2}{3}+\frac{21}{3\cdot 28}=\frac{11}{12}$ Notice the first case has less probability - it is clearly non-optimal, so every chest must contain (at least) one white marble!

We want to show that case 2 is indeed optimal. We prove: "Any optimal solution has exactly one chest, that may contain something other than one white marble". To begin with, we examine only one pair of chests and leave the rest alone. By symmetry, we can work with chest 1 and 2: $\begin{aligned} K:=P(E|B_1)+P(E|B_2)&=\frac{1}{3}\left( \frac{w_1}{w_1+b_1}+\frac{w_2}{w_2+b_2} \right),&&&b_1+b_2&=:b'\geq 2,& w_1+w_2&=:w'\geq 2 \end{aligned}$ Let's redistribute the marbles of only these two chests, so that one contains exactly one white marble and the other contains the rest. To shorten notation, we define $x_i:=w_i+b_i\geq 1$ with $\red{(*)}\:x_1+x_2=w'+b'$ . Compare the probabilities before and after redistribution: $\begin{aligned} w_1&=1,\:b_1=0:&&&&&K'&:=P(E|B_1)+P(E|B_2)=\frac{1}{3}\left(1+\frac{w'-1}{w'-1+b'}\right)\underset{\red{(*)}}{=}\frac{2(x_1+x_2)-b'-2}{3(x_1+x_2-1)}\\[2em] &&&&&&K'-K&=K'-\frac{1}{3}\left( \frac{w_1}{x_1}+\frac{w_2}{x_2} \right)=\ldots=\frac{ b_1x_2(x_2-1)+b_2x_1(x_1-1) }{3x_1x_2(x_1+x_2-1)}\geq0 \end{aligned}$ Note that $K'\geq K$ - we have increased our probability by creating a chest with exactly one white marble! We can repeat this step until all but the last chest contain exactly one white marble. After a finite number of steps, we end up with case 2, and our probability is now greater than or equal to the one we began with! We conclude, that case 2 is optimal: $P(E)\leq\boxed{\frac{11}{12}}$

Rem.: I phrased the optimization step a bit more general than necessary - it should work for any number of chests less than or equal to the number of white marbles!

Rem.: It is important that we have only a finite number of steps in our optimization algorithm for the proof to work. Sadly, I cannot recall why (or if) a countable number of steps might lead to problems. If someone wants to elaborate, I'm more than interested in an opinion!