What is the probability that he escapes the island? A man is stranded on an island. A benevolent genie presents three boxes, 23 white marbles, and 7 black marbles and instructs the man, "You may distribute the marbles into the boxes any way you see fit, but you must use all of the marbles. Once you finish, you will choose a box at random and then choose a marble from the box at random. If the marble is whiten, then I will help you escape from this place.
Assuming the man distributes the marbles in his best interest, what is the probability that he escapes the island?
Apart from the first two cases, the probability that the man draws a white marble from the box is $3/4$
$\left(\frac{1}{3}\right)(1)+\left(\frac{1}{3}\right)(1)+\left(\frac{1}{3}\right)\left(\frac{3}{4}\right)=\frac{11}{12}$
This might solve my problem but my brain still keeps bugging about the outcome.
The main question arises: Why this choice is optimal? I started to think why we are analyzing the  chance of escaping than not escaping.
When I tried to solve this, while splitting up the marbles into multiple boxes: again its not optimal.
How can I find an optimal way of explanation to this now?
 A: First off, we know there has to be at least one white stone in each, or you would cap out at $\frac 2 3$ which is less than what you have.
We can limit what we look at by just considering how many boxes have black stones in them.
The first case is putting all the bad apples in one box.  Obviously in that case the  best thing to do is to only put a single white in the other 2 and use the remaining 21 to dilute the influence of the 7, which is what you achieved above.
If we have 2 cases with black stones, then we only need 1 white in the third box,   so we have 22 white to distribute amongst the two remaining, mixed with 7 black stones.
So, your black stones could be 6 and 1, 5 and 2,  or 4 and 3.  Obviously no single box can have as big a loss potential as $\frac 1 4$,  so at a floor we need to triple the white stones:
Box 1:  6 black and 18 white
Box 2:  1 black and 3 white
This gives each of these the same $\frac 1 4$ as before,  totaling double the loss potential!   We only have a single white stone we can add left to bring one of them slightly below $\frac 1 4$,   so we have a loss potential of $\frac 1 12$ from the better worst box,  and something nonzero from the better box,  thus it is a worst scenario
Similarly with the other configurations, you need 3* as many white as black,  so 21 white, 7 black...only one box can be improved a tiny bit
Now lets say you have black stones in all 3 boxes.   Once again,  the very worst box  we can allow as a loss $\frac 1 4$,  so we need to use 21 white to match the 7 black...leaving us only 2 white stones left to improve the odds.  THe best we ccan do is improve the odds of 1 or 2 of the boxes,  leaving the third at a $\frac 1 4$ failure rate, and the other two at nonzero failure rates,  thus all are worse than the single black box case.
