Errr.... guys? You might want to check the paper<http://www.americanscientist.org/issues/issue.aspx?id=5783&y=0&no=&content=true&page=2&css=print>. It isn't three random numbers. It's two numbers written by a human opponent and a random number. A somewhat different scenario...
—R On Thu, Jun 9, 2011 at 12:39 PM, ERIC P. CHARLES <[email protected]> wrote: > 1) Constraint on the range is irrelevant, it is just a distraction. > > 2) Knowing the actual range or the boundaries is irrelevant, it is only a > distraction. -- heck, even knowing the shape of the distribution and the > actual value of the numbers is a distraction. > > 3) All that matters is that half the time the first number will be below > the median and half the time it will be above the median (by definition of > the median). As Shawn points out (providing what seems like a cleaner > presentation than the one I offered), the second number only serves to help > you guess whether the first number is above or below the median, and no > matter what distribution you have it will on average provide at least some > (Shannon-style) information. > > Surely guessing in this manner will often get you the wrong answer, but all > that was asserted was that you would, on average guess at a better than > chance level. As long as there are more than two numbers in the distribution > from which the random numbers are drawn, and the three numbers are known to > be distinct, it should be true - hence the reducability to the Monte Hall > problem. > > Eric > > P.S. For the Monte Hall reduction, we can do it with simple ordinal > judgements: > Step 1, you are given a door (it does not matter how good a prize is behind > that door) > Step 2, you are told that a second door has a better prize behind it. > Step 3, you are asked if you want to take the prize behind a third door, or > stick with the first door. > > Because we know that what is behind the three doors is different, there are > two possibilities: > 1. You were shown what was behind door 2, because that was the only door > with a prize better than door 1. > 2. You were shown what was behind door 2, because both door 2 and 3 have > better prizes than door 1, and the guy running the show flipped a coin. > > Option 2 is twice as likely as option 1, hence you are ever so slightly > better off switching (on average). > > If we move to a continuous distribution, we can still pretend that it is a > discontinuous-ordinal distribution, so the same logic works. If I tell you > that a second random number was larger than the first... you are better off > guessing that a third random number will also be better than the first. The > actual shape of the distribution will determine how big an advantage you > receive, but there will be an advantage no matter what. > > P.P.S. Thanks Shawn for your explanation, and Russ for mentioning dear old > Monte. > > > On Thu, Jun 9, 2011 01:52 PM, *Sarbajit Roy <[email protected]>* wrote: > > 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]<#13076d5fa992f1a1_13075b4419e4519c_> > > 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]<#13076d5fa992f1a1_13075b4419e4519c_>> >> 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]<#13076d5fa992f1a1_13075b4419e4519c_>> >> 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]<#13076d5fa992f1a1_13075b4419e4519c_>> >> 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]<#13076d5fa992f1a1_13075b4419e4519c_>> >> 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]<#13076d5fa992f1a1_13075b4419e4519c_>> >> 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]<#13076d5fa992f1a1_13075b4419e4519c_> >> > >> >>> 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]<#13076d5fa992f1a1_13075b4419e4519c_> >> > >> >>>> 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 > > 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
