Best Answer CaptainEd, 19 December 2012 - 07:01 PM

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Guest Message by DevFuse

Started by bonanova, Oct 24 2012 07:45 PM

Best Answer CaptainEd, 19 December 2012 - 07:01 PM

Spoiler for Observation about OP--Break once, break both pieces, discard one

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19 replies to this topic

Posted 26 October 2012 - 10:14 PM

Thank you, bonanova. I've been feeling dumb and behaving lazy. You've prodded me, so I have an answer for the first of the two.

Spoiler for Two random breaks

Posted 27 October 2012 - 04:07 AM

Thank you, bonanova. I've been feeling dumb and behaving lazy. You've prodded me, so I have an answer for the first of the two.

Spoiler for Two random breaks

Good one. Yes and yes.

Your approach should serve well for the second case as well.

*Vidi vici veni.*

Posted 29 October 2012 - 03:45 PM

Spoiler for Step 2 of my catching up to Bonanova's carrot--breaking the stick and then breaking it again

So now I have to figure out the OP--randomly break the stick, then randomly break each piece. Discard one piece. Next time...

Posted 01 November 2012 - 05:24 AM

You are in the right ball park, but a little low.Spoiler for Step 2 of my catching up to Bonanova's carrot--breaking the stick and then breaking it again

So now I have to figure out the OP--randomly break the stick, then randomly break each piece. Discard one piece. Next time...

Ask where the longer stick must be broken given the first break was at f < .5.

Then average the probability of that second break for 0 < f < .5.

*Vidi vici veni.*

Posted 09 November 2012 - 05:24 PM

I have run two simulations, each over 1,000,000 runs.

The first simulates the scenario where one stick is randomly broken and then one of the resulting two sticks is randomly choosen and then randomly broken.

The second scenario simulates where the stick is randomly broken and each of the two resulting sticks are then randomly broken and then one of the resulting four sticks is randomly choosen and thrown away:

Scenario 1: Probability of Sucessful Triangle: 0.19378

Scenario 2: Probability of Sucessful Triangle: 0.19371

Assuming my simulations are correct, it practically makes no difference which scenario is choosen

The first simulates the scenario where one stick is randomly broken and then one of the resulting two sticks is randomly choosen and then randomly broken.

The second scenario simulates where the stick is randomly broken and each of the two resulting sticks are then randomly broken and then one of the resulting four sticks is randomly choosen and thrown away:

Scenario 1: Probability of Sucessful Triangle: 0.19378

Scenario 2: Probability of Sucessful Triangle: 0.19371

Assuming my simulations are correct, it practically makes no difference which scenario is choosen

**Edited by brifri238, 09 November 2012 - 05:33 PM.**

Posted 19 November 2012 - 07:47 PM

Some day, when I look back on this, I'll laugh. But right now, my ability to do word problems is not very satisfying. I believe I did exactly what you advised, I clearly graphed some points randomly generated according to this approach, I saw a graph in which the probability is exactly 1/3, yet my Excel spreadsheet continues to show numbers larger than 1/3, as you said (and presumably, as brifri238 says). Thanks in advance for my education in mathematical thinking!You are in the right ball park, but a little low.

Ask where the longer stick must be broken given the first break was at f < .5.

Then average the probability of that second break for 0 < f < .5.

Posted 20 November 2012 - 05:38 PM

Spoiler for Break once, pick a piece, break again

**Edited by CaptainEd, 20 November 2012 - 05:40 PM.**

Posted 22 November 2012 - 05:58 AM

You got it.

Kudos for hanging in with it.

The missing factor 2 comes when you recognize the integration range is 1/2.

You have to divide by that to get the average over that range.

I didn't notice till now that**grifri238**'s result is a little high..

I got:

*After a first break at point a <.5. the second break must be in the central region of width a within the larger [1-a] piece. If you average a/(1-a) over all values of a from 0 to .5 you get*

-1 + 2log2 = -1 + 1.386294361... = 0.386294361... Half that is 0.1931471806...

I have not analyzed the variant mentioned in OP.

I'm waiting for the 'aha!' Moment that reduces it to this case.. http://brainden.com/..._DIR#/smile.png

Kudos for hanging in with it.

The missing factor 2 comes when you recognize the integration range is 1/2.

You have to divide by that to get the average over that range.

I didn't notice till now that

I got:

-1 + 2log2 = -1 + 1.386294361... = 0.386294361... Half that is 0.1931471806...

I have not analyzed the variant mentioned in OP.

I'm waiting for the 'aha!' Moment that reduces it to this case.. http://brainden.com/..._DIR#/smile.png

*Vidi vici veni.*

Posted 19 December 2012 - 07:01 PM Best Answer

Spoiler for Observation about OP--Break once, break both pieces, discard one

Posted 19 December 2012 - 08:43 PM

Spoiler for Observation about OP--Break once, break both pieces, discard one

Very nice Cap'n!

*Vidi vici veni.*

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