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]> 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]> 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]>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 >>> >>> >>> ============================================================ >>> 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 >>> >> >> >> ============================================================ >> 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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