- 29 Dec '04 13:231 red marble in jar 1 and the rest in jar 2 gives about 3/4 chance...74/99 to be precise

put x red marbles in jar 1 along with y black ones.

With chance 1/2 you pick a marble from jar 1, with chance x/x+y this is a red marble.

With chance 1/2 you pick a marble from jar 2, with chance 50-x/100-x-y this is a red marble.

1/2*(x/x+y + 50-x/100-x-y) = ugly

I have not the time to write ugly out, to differentiate to x and y and to see of they happen to have a common zero. - 29 Dec '04 14:15

The statement of your puzzle is incomplete. What , wherefrom and how do you pick out has to be specified.*Originally posted by kcams***this is an easy one,**

you have 2 jars & 100 marbles, 50 red and 50 blue.

you have to place the marbles into the jars such that you maximise the likelihood that you pick out a red marble.

i can get almost 3/4 or 75% likelihood. - 30 Dec '04 04:00

Place the red marbles into both jars. Give the blue marbles to the kid down the street. 100%.*Originally posted by kcams***this is an easy one,**

you have 2 jars & 100 marbles, 50 red and 50 blue.

you have to place the marbles into the jars such that you maximise the likelihood that you pick out a red marble.

i can get almost 3/4 or 75% likelihood.

Place 25 blue marbles into each jar. Carefully, without disturbing the jars, place 25 red marbles on top of the blue marbles. 100%. - 02 Jan '05 15:28There's a 50% chance you'll pick the jar with the red marble. If you pick this jar, you'll definitely get a red marble. (1/2)

If you don't pick that jar, then you'll have a 49/100 chance of getting a red marble.

Therefore the chance of getting a red marble is 1/2 + 1/2 * 49/100 = 100/200 + 49/200 = 149/200. - 12 Jan '05 19:23

*cough* In my days we were taught to simplify expressions like that.*Originally posted by ranjan sinha***It was a case of error due to hurried thinking and typing. Coolwarrior has already posted the correct expression. Thankx.**

Correct me if i'm wrong but it's not 74.99%

My calculation say it's 74/99 = 74.74 % approximately.