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Probability brainteaser

abelniak
1-Newbie

Probability brainteaser

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?

15 REPLIES 15
AlanStevens
15-Moonstone
(To:abelniak)

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

Alan

RichardJ
19-Tanzanite
(To:abelniak)

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

AlanStevens
15-Moonstone
(To:RichardJ)

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

Alan

RichardJ
19-Tanzanite
(To:AlanStevens)

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.

AlanStevens
15-Moonstone
(To:RichardJ)

Yes, I guess the wording of the question can be taken to imply that you must pick out an actual ball. However, if you just ask what is the probability of picking out a red ball,you could allow an empty jar; and then the probability of picking a red ball out of it is zero.

Alan

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%.

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

RichardJ
19-Tanzanite
(To:hilbert)

MichaelH has posted the correct answer,

As did Alan Stevens, back on Mar 6th.

In my defense, I did not open that worksheet until after I posted my answer.

I don't even know why this thread was titled "brainteaser". It wasn't.

The "brute force" approach reminds me of determining "pi" by drawing a circle in a square and randomly shooting at it. Yeah, it works.

RichardJ
19-Tanzanite
(To:MichaelH)

In my defense, I did not open that worksheet until after I posted my answer.

Sorry. I didn't mean to imply that you got your answer from Alan's worksheet and then reposted that answer. Actually, I tihnk that is extremely unlikely. None of your other posts have given any indication that you would do something like that.

My comment was only directed at how long the answer has been posted without "Correct Answer" being assigned.

No offense taken. My response was written in mock indignation style with tongue planted firmly in cheek.

This reminds me of when I was in physics grad school at UCSD and the professor walked out during the middle of a final exam. He announced loudly "I'll be outside having a smoke. If any of you feels the need to cheat now, I strongly suggest you find another career."

Of course, he was right.

AlanStevens
15-Moonstone
(To:MichaelH)

My ballsinjars worksheet does, in fact, contain the 'sensible' solution MichaelH proposed in his earlier post, in addition to the 'brute force and ignorance' method (which, by the way, is an exhaustive evaluation approach rather than a monte-carlo one of the random shooting at a circle in square type). The 'sensible' calculation seems fairly obvious, but an interesting follow-up question might be "Can you prove this is the maximum value?" The exhaustive evaluation does this, but it does so in what seems to me to be an ugly way. Is there a more elegant proof?

Alan

To see how others have discussed this, and the original source, head over to http://www.reddit.com/r/math/comments/qiesr/interesting_question_on_probability_about_sorting/

Sorry if the title was misleading. It was merely posted as a way to pass the (mathematical) time. Probabilty, guessing, estimation, and algebraic expressions, in the form of interrogative wordproblems, are often considered brainsteasers on the Internet.

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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