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 <russ.abb...@gmail.com> 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 <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., ><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 <<#>> 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 <<#>> 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 <<#>> 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> > > > > > > > > > > > > > ============================================================ >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 > Eric Charles Professional Student and Assistant Professor of Psychology Penn State University Altoona, PA 16601
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