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The "aha!" problems 7. Urning the balls


Best Answer m00li, 06 May 2014 - 01:38 AM

Spoiler for no aha here

continuing...

Spoiler for continuing
Go to the full post


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19 replies to this topic

#1 bonanova

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Posted 01 May 2014 - 12:13 PM

An urn contains b black balls and w white balls.
Balls are removed singly and randomly until all the balls of one color are removed.
What is the expected number of balls that remain in the urn?


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 m00li

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Posted 03 May 2014 - 10:41 AM

Spoiler for no aha here

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#3 bonanova

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Posted 03 May 2014 - 12:26 PM

Heroic work.

Can you (aha) enumerate the equally likely cases more simply?


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#4 DeGe

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Posted 04 May 2014 - 09:12 PM

Spoiler for Aha?


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#5 bonanova

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Posted 05 May 2014 - 01:22 AM

Spoiler for Aha?

 

Spoiler for Good thinking


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#6 plasmid

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Posted 05 May 2014 - 03:35 PM

Well, the aha! solution that I had in mind didn't work.
In case it's helpful to anyone:
Spoiler for results of simulations

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#7 k-man

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Posted 05 May 2014 - 07:33 PM

Spoiler for not quite the "aha"

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#8 m00li

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Posted 06 May 2014 - 01:38 AM   Best Answer

Spoiler for no aha here

continuing...

Spoiler for continuing

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#9 bonanova

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Posted 06 May 2014 - 11:26 AM

 

Spoiler for no aha here

continuing...
Spoiler for continuing

That's the right formula, and it fits plasmid's simulation numbers and k-man's insights.
I'm marking it solved.
If the simplicity of the expression suggests a words-only derivation, I'll hold off saying anything more.


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- Bertrand Russell

#10 bonanova

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Posted 07 May 2014 - 12:19 AM

Spoiler for Aha hint


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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