1. ## Probability

Imagine that Kay is a prisoner sentenced to death. The Emperor offers Kay 50 black marbles, 50 white marbles and 2 empty bowls. He then says, “Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove ONE marble. If the marble is WHITE you will live, but if the marble is BLACK… you will die.”

How should Kay divide the marbles to maximise his chance of survival?

Note: Probability of him surviving must be >0.5

2. Hello, Punch!

This is a classic (very old) puzzle.

Imagine that Kay is a prisoner sentenced to death.
The Emperor offers Kay 50 black marbles, 50 white marbles and 2 empty bowls.
He then says, “Divide these 100 marbles into these 2 bowls.
You can divide them any way you like as long as you use all the marbles.
Then I will blindfold you and mix the bowls around.
You then can choose one bowl and remove ONE marble.
If the marble is WHITE you will live, but if the marble is BLACK. you will die.”

How should Kay divide the marbles to maximise his chance of survival?
Spoiler:

He should place ONE white marble in bowl A and the rest in bowl B.

. . . . . $\displaystyle \boxed{\begin{array}{c}\text{Bowl A} \\ \text{1 white} \\ \\ \end{array}} \qquad \boxed{\begin{array}{c}\text{Bowl B} \\ \text{49 white} \\ \text{50 black} \end{array}}$

$\displaystyle P(\text{bowl A}) \:=\:\frac{1}{2},\quad P(\text{white}) \:=\:1$
. . Hence: .$\displaystyle P(\text{white from bowl A}) \;=\;\frac{1}{2}\cdot1 \:=\:\frac{1}{2}$

$\displaystyle P(\text{bowl B}) \:=\:\frac{1}{2},\quad P(\text{white}) \:=\:\frac{49}{99}$
. . Hence: .$\displaystyle P(\text{white from bowl B}) \:=\:\frac{1}{2}\cdot\frac{49}{99}$

Therefore: .$\displaystyle P(\text{white}) \;=\;\frac{1}{2} + \frac{49}{198} \;=\;\frac{74}{99}$