A lucid analysis. BUT,
If we consider the median = 1/2 infinity case, we end up with 3 "equally
probable" cases.
a) both number below median
b) both numbers above median
c) one below and one above median

alternatively we could get 4 "equally probable" cases
1) A below B below
2) A below B above
3) A above B below
4) A above B above

I'm still unable to see how we get a "better than 50%" edge by knowing the
2nd number.

The "normal" distribution would not apply to random numbers - which are
evenly distributed ie. "flat".

Sarbajit

On Thu, Jun 9, 2011 at 5:46 PM, ERIC P. CHARLES <[email protected]> wrote:

> Ok, I'm a bad person for not reading the cited paper, but I was thinking
> about problem late last night. I keep thinking that we need to make
> assumptions about the distribution (regarding bounds and shape), but then I
> can't figure out a combination of assumptions that really seems necessary.
> This is because any distribution has a median (even if it is an incalculable
> median, like 1/2 infinity). Using that as the key:
>
> Given two randomly generated numbers, odds are that one of them is above
> the median, the other is below the median. We need two numbers, so that we
> can tell which one is which. If we restrict ourselves to making a guess
> relative to the first number (because that's what I think Russ was saying),
> then when the first number is the smaller one, we guess that it is below the
> median (and hence the third number has more that a 50% chance of being above
> it). Reverse if the first number is the larger one.
>
> Of course, sometimes we are wrong, and both random numbers are on the same
> side of the median... but on average we are still better off guessing in
> this manner. If we know the shape of the distribution, it should be pretty
> easy to calculate the advantage. For example, if the distribution is normal,
> the smaller score will (on average) be one standard deviation below the
> mean, and hence 84% of the distribution will be above it.
>
> Eric
>
>
> On Wed, Jun 8, 2011 11:10 PM, *Russ Abbott <[email protected]>* wrote:
>
> It doesn't establish the range. All that's really necessary is that there
> be a non-zero probability that the second number falls between the first and
> the third. On those occasions when it does you will have the right answer.
> On all others you will be right 50% of the time.  I saw it in a reprint of 
> this
> paper<http://www.americanscientist.org/issues/issue.aspx?id=5783&y=0&no=&content=true&page=2&css=print>.
> Look for David Blackwell.
>
>  What I like about this phenomenon is that it feels like action at a
> (mathematical) distance -- similar to the Monte Hall problem in which
> showing the content of one door makes it better to switch choices. (If you
> don't know this problem, it's worth looking up, e.g., 
> here<http://www.askamathematician.com/?p=787>
> .)
>
> *-- Russ Abbott*
> *_____________________________________________*
> ***  Professor, Computer Science*
> *  California State University, Los Angeles*
>
> *  Google voice: 747-*999-5105
> *  blog: *http://russabbott.blogspot.com/
>   vita:  http://sites.google.com/site/russabbott/
> *_____________________________________________*
>
>
>
> On Wed, Jun 8, 2011 at 6:52 PM, Sarbajit Roy 
> <[email protected]<#13074553f57f0e8b_>
> > wrote:
>
>> How does knowing the second number establish the range ? Is there any work
>> on this.
>>
>> Sarbajit
>>
>>
>> On Thu, Jun 9, 2011 at 1:15 AM, Russ Abbott 
>> <[email protected]<#13074553f57f0e8b_>
>> > wrote:
>>
>>>  Russell Standish has the right idea.  If you knew the range, say the
>>> first number is higher/lower than the third depending
>>> on whether the first numbers is greater than or less than the middle of the
>>> range. Since you don't know the range, the second random number is used
>>> instead.  Say higher/lower depending on whether the first number is
>>> higher/lower than the second.
>>>
>>>  Also, think about it this way. If the middle number is either greater
>>> than or less than both the first and the third, you have a 50% chance of
>>> being right. If it's between the first and the third, the strategy described
>>> will always be right. Presumably there is a non-zero probability that the
>>> second number will be between the first and the third. Therefore one has a
>>> greater than 50% chance of being right.
>>>
>>> *-- Russ *
>>>
>>>
>>> On Wed, Jun 8, 2011 at 10:06 AM, Owen Densmore 
>>> <[email protected]<#13074553f57f0e8b_>
>>> > wrote:
>>>
>>>> Do you have a pointer to an explanation?
>>>>
>>>>        -- Owen
>>>>
>>>> On Jun 8, 2011, at 12:11 AM, Russ Abbott wrote:
>>>>
>>>> > Although this isn't new, I just came across it (perhaps again) and was
>>>> so enchanted that I wanted to share it.
>>>> >
>>>> > Generate but don't look at three random numbers. (Have someone ensure
>>>> that they are distinct. There is no constraint on the range.) Look at the
>>>> first two. You are now able to guess with a better than 50% chance of being
>>>> right whether the first number is larger than the unseen third.
>>>> >
>>>> > I like this almost as much as the Monte Hall problem.
>>>> >
>>>> > -- Russ
>>>> >
>>>>   > ============================================================
>>>> > FRIAM Applied Complexity Group listserv
>>>> > Meets Fridays 9a-11:30 at cafe at St. John's College
>>>> > lectures, archives, unsubscribe, maps at http://www.friam.org
>>>>
>>>>
>>>> ============================================================
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>>>> Meets Fridays 9a-11:30 at cafe at St. John's College
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>>>>
>>>
>>>
>>> ============================================================
>>> FRIAM Applied Complexity Group listserv
>>> Meets Fridays 9a-11:30 at cafe at St. John's College
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>>>
>>
>>
>  ============================================================
> FRIAM Applied Complexity Group listserv
> Meets Fridays 9a-11:30 at cafe at St. John's College
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>
> Eric Charles
>
> Professional Student and
> Assistant Professor of Psychology
> Penn State University
> Altoona, PA 16601
>
>
>
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