riddle: lots of marbles...
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From: Santa Cruz
riddle: lots of marbles...
PLEASE DO NOT LOOK UP THE ANSWERS ON THE INTERNET AND POST THEM HERE. If you want to look up the answers that's fine. The answers are out there, they are not that hard to find. Some people like to try and solve the riddle before giving up and looking up the answers. This is for those people. Enjoy...
you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
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From: Pottsville
what if you just put one red marble in the one jar and the other 99 marbles in the other jar...i don't feel like crunching the numbers, but i think that would be near 75% chance of drawing a red.
i finally got a chance to answer!!! i'm sure that someone will beat me with a better answer though.
i finally got a chance to answer!!! i'm sure that someone will beat me with a better answer though.
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From: Oshkosh
Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that.... what am i missing?
But i'm sure there's more to it than that.... what am i missing?
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From: Pottsville
[QUOTE]Originally posted by Hegulator
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
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From: Irvine
[QUOTE]Originally posted by Hegulator
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
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From: San Diego, CA
Originally posted by lehighwiz
i considered that, but the randomness of the draw will eliminate that as an option i think.
allright, i did the math:
((1/1)*100)/2 + ((49/99)*100)/2 = 74.74%
i considered that, but the randomness of the draw will eliminate that as an option i think.
allright, i did the math:
((1/1)*100)/2 + ((49/99)*100)/2 = 74.74%
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From: Santa Cruz
[QUOTE]Originally posted by Hegulator
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
[B]Assuming a standard-shaped jar and assuming you just grab a random marble out of the top of each jar, just put 25 blue marbles in the bottom of each jar and 25 red marbles on the top of each jar.
But i'm sure there's more to it than that....
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From: Santa Cruz
Originally posted by lehighwiz
i considered that, but the randomness of the draw will eliminate that as an option i think.
allright, i did the math:
((1/1)*100)/2 + ((49/99)*100)/2 = 74.74%
i considered that, but the randomness of the draw will eliminate that as an option i think.
allright, i did the math:
((1/1)*100)/2 + ((49/99)*100)/2 = 74.74%
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