Q. You are to divide 9 balls - 4 green and 5 red into 2 boxes. You pick a box at random and a ball from it at random. If it is red, you DIEEEE!, if it is green (I will let) you live. How would you maximize your odds of living?

Let`s call the 2 boxes, boxA and boxB

the maximize odds for living is

boxA = 1 Green (100%)
BoxB = 3 Green and 5 Red (3/8)

as you have 50% to get boxA and 50% to get boxB, the odds to get a Green ball is:

(1 + 3/8)*1/2 = 11/16 = 68,75%

All other odds are lower because:

if you increase green balls in a boxA - it`ll continue at 100%, but will lower picking green ball at boxB odds
if you put only red balls in a boxA - your odd will be in max 50%
if you make a mist green and red - your odd will be in max 50% too (2green/1red and 2green/4red)