Re: $1000 to the first person who can solve this problem

On Feb 17, 11:36 am, crimefreesoftw...@xxxxxxxxx wrote:
In a previous post many people helped us in articulating our math
problem and we were given some possible solutions. One solution that
is VERY close to what we need was the Secretary Problem. Thanks again
to those who helped us clarify/articulate our math problem and point
us in the right direction.

Our problem is almost perfectly described in what is known as the
"Secretary Problem."

Please refer to this link for a detailed explanation:http://en.wikipedia.org/wiki/Secretary_problem

The following is an outline of our specific problem:

Sequence of Events:

    1)   You receive Original Numbers one at a time.

    2)   You will eventually receive a total of 10 Original Numbers
(there are 10 Original Numbers to a Set)

    3)   Each time you receive an Original Number, you must decide
immediately (“on the fly”) whether THIS Original Number will be the
Lowest of all 10 Original Numbers in this Set, even though you haven’t
seen all 10 Original Numbers (yet).

    4)   Once you Select an Original Number, it becomes your Selected
Number for This Set, and you skip over/ignore the remaining Original
Numbers in the Set of 10 and you prepare to look at the next Set.

What We Need:

      A formula/method/algorithm that will determine the Selected
Number (on the fly) that it thinks will be the lowest of the Set of 10
Original Numbers.  Repeat this process of Selecting one number for
each subsequent Set of 10 Original Numbers.

Calculations to be kept:

    1)   Average of all of the Original Numbers (every Original Number
in every Set).

    2)   Average of all of the Selected Numbers (each Selected Number
in every Set).

Goal Desired:

       To have the Average of all of the Selected Numbers be lower
than the Average of all of the Original Numbers by 1 point or more.

Purpose:

      To use the (lower) Selected Average as a preferred subset of
numbers as well as to obtain better/lower Average results with much
less workload required.

Current Method We Are Trying:

      “Secretary Problem” formula/method seems to do exactly what we
need, but doesn’t perform as well with our set of Original Numbers.
Since the Secretary Problem assumes equally distributed data, we are
speculating that our Original Numbers must not be equally distributed.

Conclusion:

Perhaps some alteration of the Secretary Problem can be made to
achieve the Goal Desired.

Sample Data to test with:

A large sample of Original Numbers can be seen here:

http://www.crimefreesoftware.com/DataSet1c_8600datapoints.csv

This is a collection of 860 sets of 10 Original Numbers, shown in one
large series of 8,600 Original Numbers.
We have a number of large collections of data sets that can be used to
back-test your formulas. If you want more data, just let us know and
we will provide links to it.

This is a very important problem that we need help solving.  If you
can help provide the solution that achieves the goals outlined above
using the datasets we provide, we are happy to pay you $1000 for your
successful solution. Any help is appreciated.

Thank you,

Dan
crimefreesoftw...@xxxxxxxxx

Looking at your data it's very cyclical in nature. So while the trend
is ascending then the minimum of the set will most likely be first
number. If the trend is descending then the minimum will most likely
be the last number of the set. Using your dataset I knocked up a very
crude algorithm in a spread*** that waits until the 5th number of a
set appears and compares it to the average of the first four numbers.
If n5>avg(n1:n4) then trend=up so choose n1. If n5<avg(n1:n4) then
trend=down so choose n10. Using this method my average of the selected
numbers was 13467 compared to the average of all original numbers
which was 13470.

I'm not sure if this is exactly what you're after since my
"algorithm" (crude as it is) needs to wait for 5 numbers in a set to
appear before it can make a "decision" about the trend. I think what
you're after is an algorithm that can make a decision on EVERY number?

Anyway, hope that helps.
.

Quantcast