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

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Guest Message by DevFuse

Started by BMAD, Oct 04 2013 07:20 PM

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

Spoiler for what I would pay to play

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16 replies to this topic

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.**

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*/

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

But...

Spoiler for Simulations don't lie

- Bertrand Russell

Posted 10 October 2013 - 03:35 PM

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

Posted 11 October 2013 - 12:53 PM

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

I stand corrected

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. :~}

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

- Bertrand Russell

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.

Posted 12 October 2013 - 02:18 AM

Yes.

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

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. :~}notn. It is larger by 1/2.

You'd need to include zero to make it n.

My bad.

- Bertrand Russell

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