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 e...@psu.edu 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 sroy...@gmail.com 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 e...@psu.edu 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 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
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 sroy...@gmail.com 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 russ.abb...@gmail.com
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
