Quote from: Jeremycoin on December 26, 2015, 10:35:01 AM

Finally I can figure it out Grin

If I can put the ball by myself, I will put 5 red + 3 green in the first box and left the 1 green ball in the second box. So whenever I pick the box, I know that the box with just 1 ball on it is have the green ball.

The answer is correct. (4 mBTC to Jeremycoin)

Original question:
You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 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 do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

Answer to my variant: (Best Explanation) (3 mBTC to fumblingperch)

Quote from: fumblingperch on December 27, 2015, 02:27:17 AM

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)

.