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Million Dollar Urns


Best Answer superprismatic, 07 October 2013 - 03:03 PM

Spoiler for what I would pay to play

Go to the full post


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16 replies to this topic

#11 plasmid

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Posted 10 October 2013 - 07:31 AM

This is a problem where it's actually advantageous NOT to apply "without loss of generality".
Spoiler for This is an extension of what DeGe was starting to notice, and is a proof of his answer

Edited by plasmid, 10 October 2013 - 07:39 AM.

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#12 bonanova

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Posted 10 October 2013 - 11:45 AM

WOLOG isn't needed in my previous explanation. Rather,

 

Coin 3 goes into one of the urns. Call that urn X.

Now x/y = 2 and for Coin 4, p(X)/p(Y) = 2 which on average maintains <x>/<y> = 2,

making min = 333 333. And, if you could add frational coins, that is the correct result.

 

But...

Spoiler for Simulations don't lie


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#13 superprismatic

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Posted 10 October 2013 - 03:35 PM

Spoiler for DeGe made a slight mistake at the tail end of his computation


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#14 DeGe

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Posted 11 October 2013 - 12:53 PM

Spoiler for DeGe made a slight mistake at the tail end of his computation

I stand corrected


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#15 bonanova

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Posted 11 October 2013 - 08:35 PM

OK someone help me out on this. If the cases are 1 ... N/2, and there are TWO cases for all of them but N/2, for which there is only ONE case, then why isn't the average slightly LESS than N/4?

And since this is my 4999th post, does that mean the next one has to be profound in some way? I'm in trouble. :~}
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#16 superprismatic

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Posted 11 October 2013 - 10:18 PM

OK someone help me out on this. If the cases are 1 ... N/2, and there are TWO cases for all of them but N/2, for which there is only ONE case, then why isn't the average slightly LESS than N/4?

And since this is my 4999th post, does that mean the next one has to be profound in some way? I'm in trouble. :~}

I'm not sure how to answer this, but consider that the average of the first 2n integers is (2n+1)/2, not n.  It is larger by 1/2.


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#17 bonanova

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Posted 12 October 2013 - 02:18 AM

 


OK someone help me out on this. If the cases are 1 ... N/2, and there are TWO cases for all of them but N/2, for which there is only ONE case, then why isn't the average slightly LESS than N/4?

And since this is my 4999th post, does that mean the next one has to be profound in some way? I'm in trouble. :~}

I'm not sure how to answer this, but consider that the average of the first 2n integers is (2n+1)/2, not n.  It is larger by 1/2.
 
Yes.
You'd need to include zero to make it n.
My bad.
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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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