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abelniak1-Visitor
1-Visitor
March 5, 2012
Question

Probability brainteaser

  • March 5, 2012
  • 4 replies
  • 6577 views

There's an interesting probability post over on reddit (jars, balls, choose at random, maximize the odds of choosing something specific). One of the comments is "Someone with the appropriate software can probably plug all this stuff in and get a maximum value." Any takers?

http://www.reddit.com/r/math/comments/qiesr/interesting_question_on_probability_about_sorting/

QUESTION:

You have 3 jars, 20 red balls, 100 blue balls. You need to place all the balls into the jars such that when you blindly pick one ball out of a random jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a ball out of that jar) you can place the balls however you like, but each ball must be in a jar. All balls are identical other than color. Picking will be truly random, so no ingenious strategy will change the mathematical outcome.

1) What is the best to maximize the chances to pick a red ball?

2) What is the probability of picking a red ball in your answer?

4 replies

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AlanStevens19-Tanzanite
19-Tanzanite
March 6, 2012

Ok. Brute force and ignorance works well here! See attached.

Alan

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RichardJ19-Tanzanite
19-Tanzanite
March 6, 2012

Once you have pickd a jar, what do you do if it's empty?

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AlanStevens19-Tanzanite
19-Tanzanite
March 6, 2012

My understanding of the question is that you only get one pick.

Alan

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RichardJ19-Tanzanite
19-Tanzanite
March 6, 2012

It was meant as a facetious question I could put all the blue balls in one jar and all the red ones in another. That doesn't maximize the probability of picking a red ball, but it does bring up the question of what happens if I pick the empty jar.

M
MichaelH1-Visitor
1-Visitor
March 29, 2012

Place a red ball in jar 1, and another in jar 2. In the third jar place the remaining 18 red and 100 blue balls.

Therefore, the probability of picking a red ball with one pick is 2/3+18/3*118=71.8%.

H
hilbert1-Visitor
1-Visitor
March 29, 2012

MichaelH has posted the correct answer, yet the label at the top says "Not Answered"

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RichardJ19-Tanzanite
19-Tanzanite
March 29, 2012

MichaelH has posted the correct answer,

As did Alan Stevens, back on Mar 6th.

A
abelniak1-VisitorAuthor
1-Visitor
March 30, 2012

Hey everyone. Thanks for the submissions! Hope you had fun!

(I just marked this as 'answered'. Note that I listed it as a question to generate chatter. I wasn't moderating it for *the correct* answer, per se - just for disucssion.)

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