Charlie and Mr. Z. are playing the following game:
- Mr. Z. brings $10$ notes of 100£ and $10$ notes of 20£.
- Charlie distributes the $20$ notes into $2$ jars, in a way he chooses.
- Mr. Z. chooses one of the jars randomly.
- Charlie chooses one note from the chosen jar randomly.
- Charlie receives the chosen note and pays Mr. Z. 60£. (In other words, Charlie can either gain 40£ or lose 40£).
How should Charlie distribute the notes into the jars in order to maximize his chances of winning?
My first intuitive thought was that he should distribute the notes evenly, but I found out that it isn't the answer and didn't know how to prove that.
One straight-forward approach came to my mind suggests to search for the maxima of the following function, which represents the probability of Charlie's winning:
for $a$=number of 100£ notes in the first jar, $b$=number of 20£ notes in the first jar, we get:
$$\mathbb{P}\left(Winning\right)=\frac{1}{2}\left(\frac{a}{a+b}+\frac{\left(10-a\right)}{\left(10-a\right)+\left(10-b\right)}\right)$$
but it doesn't involve much probability considerations (and honestly I'm not sure I recall the technique for that).
I would be glad to hear your suggestions to the problem. (It should be related to the subject of conditional probability.)
Charlie should put one $100$ pound note in one jar and all the other notes in the other jar. By doing so, he guarantees that he will win half the time, while still winning nearly half the time if Mr. Z picks the other jar. Using your formula shows that the probability of Charlie winning with this choice is $$\Pr(\text{winning}) = \frac{1}{2}\left(\frac{1}{1 + 0} + \frac{9}{9 + 10}\right) = \frac{1}{2}\left(1 + \frac{9}{19}\right) = \frac{1}{2} \cdot \frac{28}{19} = \frac{14}{19}$$