Problem of the Day
A new programming or logic puzzle every Mon-Fri

Marble Grab

You have two bags, one with 50 red marbles and one with 50 blue marbles. Your goal is to grab a red marble from each bag. How can you rearrange the marbles to maximize your chances of grabbing a red marble from each?

Enjoy your weekend!

Permalink: http://problemotd.com/problem/marble-grab/

Comments:

  • Pak - 10 years, 6 months ago

    I can't seem to get better odds than 1/4, at first thought.

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  • QMS - 10 years, 6 months ago

    If you put only one red marble in one bag, and the rest in the other bag, you get a probability of 49/99 which is slightly less than 1/2. Any possible improvement?

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  • Pak - 10 years, 6 months ago

    Your calculation is wrong. 1/50 * 49/50 = approximately 1.96% success.

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  • QMS - 10 years, 6 months ago

    What I mean is only one marble in one bag, so 100% chance to pick up a red marble. And all the remaining marbles in the other bag so 49/99 to pick up a red marble.

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  • Daniel - 10 years, 6 months ago

    Assuming you aren't going to shuffle the bag, I'd pour the marbles in so the red ones are on top, and pick one from the top of the pile.

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  • Anonymous - 10 years, 6 months ago

    Or just pour out the blue marbles and split the red ones between bags.

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  • Ethan - 10 years, 6 months ago

    Mix them up evenly and it's exactly a 50% chance. I think this question kind of misrepresents this one: http://goo.gl/Pmpba

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  • Ethan - 10 years, 6 months ago

    I retract my previous statement. I think QMS has it.

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  • Max Burstein - 10 years, 6 months ago

    I believe this is the best solution as well. Though pouring all the red marbles on top as Daniel mentioned isn't a bad idea either.

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  • tehmagik - 10 years, 6 months ago

    put 1 red marble in 1 bag, put the other 49 in w/ the 50 blue marbles...100% chance from bag 1, 49/99 from bag 2...almost 3/4 chance of drawing 2 red marbles now

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  • Anonymous - 10 years, 6 months ago

    I think the math below is wrong on a lot of the assumptions.

    Having one bag with 1 Red Marble and one bag with 49 Red and 50 Blue yields the chance of.

    1/1 * 49/99 = 1 * 49/99 =  49.49%
    

    This is I believe the best solution because you effectively eliminate your first pick and give yourself the greatest chance of success on the second pick. Any other way you arrange the marbles will lower the second bags chances at no reward to the first.

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