a) The assumption was that there is no constraint on the range. b) Knowing 2 numbers (or even a hundred) tells us nothing about the range/boundary for the 3rd (or the 101st).
c) So the only thing I can say is that if the 3 numbers are disclosed to the guesser in ascending order, the probability that the 3rd number is greater than the 1st approaches or equals 1 (making it well over 50%). Sarbajit On Thu, Jun 9, 2011 at 7:36 PM, <[email protected]> wrote: > The first number partitions the distribution. Unless the areas on > either side of the partition are equal, there is a greater than 50 > percent chance that the second number will be drawn from the larger > partition. Assuming that the three numbers are independent and > identically distributed, the probability of drawing the third number > from the larger partition is the same as the probability of drawing > the second number from the larger partition. Basically, the second > number determines whether the third number will be larger or smaller > than the first. > > > Shawn > > On Thu, Jun 9, 2011 at 9:09 AM, ERIC P. CHARLES <[email protected]> wrote: > > Sarbajit, > > Great point, but let me make it a bit more complicated. Possibilities > marked > > with a "+" indicate situations in which we will have a probabilistic > > advantage in our guessing, possibilities marked with a "-" indicate > > situations in which we will have a probabilistic disadvantage in our > > guessing: > > 1) A below B below > > 1a) and A below B + > > 1b) and B below A - > > 2) A below B above + > > 3) A above B below + > > 4) A above B above > > 4a) and A above B + > > 4b) and B above A - > > > > Eric > > > > P.S. The case of a single bounded distribution is definitely the hardest > for > > me to think about, a double bounded or unbounded distribution seems much > > more intuitive. Also, the restriction to guess relative to A makes it > harder > > for me to think about. Imagine instead that all we did was guess that the > > third number would be above the smallest of the first two. > > > > On Thu, Jun 9, 2011 08:35 AM, Sarbajit Roy <[email protected]> wrote: > > > > 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. 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.) > >> > >> -- 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 > >>> > >> > >> ============================================================ > >> 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 > >> > >> > > > > Eric Charles > > > > Professional Student and > > Assistant Professor of Psychology > > Penn State University > > Altoona, PA 16601 > > > > > > > > ============================================================ > > 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
