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, <<#>> 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 <<#>> 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 <<#>> 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 <<#>> 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 <<#>> 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 <<#>> 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 > >>> > >>> > >> > >> 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
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