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Matrix Completion Cycle Part II

Topic closed. 3 replies. Last post 10 years ago by Thoth.

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Thoth's avatar - binary
Findlay, Ohio
United States
Member #4855
May 28, 2004
400 Posts
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Posted: December 9, 2006, 11:52 pm - IP Logged

Matrix Completion Cycle Part II

 

The graphs I inserted into in part one of this subject(http://www.lotterypost.com/thread/146429) measured the Pick 3 in the following amounts of consecutive games:

 

693
1,386
2,079
2,772
3,465
4,158
4,851
5,544
6,237
6,930
7,623
8,316
9,009
9,702

I have a good reason for measuring the games in those amounts, which I will disclose shortly.  But first, since many players like to track the games in nice even intervals, I will break the completion rates down into successive spans of 500 consecutive games.  The chart below shows how many of the 1,000 straight combinations should be drawn within the amount of consecutive games that are listed in the first column.  The actual expected amount of different combinations is shown along with its total percentage of the 1,000 combo matrix.  Looking at the chart, you can see that only 777 different combinations should be drawn within 1,500 consecutive games.  So much for the fallacy of expecting each combo to hit about once in every 1,000 games!  Additionally, if one can truly expect only 777 different combos to fall within 1,500 games, then one must also anticipate that 223 combos will remain to be drawn!  If ever you are looking at a historical listing of the previous results and you see a large group of numbers that have yet to hit, just remember this: in the same way that probability governs the amount of unique combos that should be drawn in a given period of time, it will also dictate the amount of combos that should not be drawn!  So, whenever you discover a group of "never-hit" numbers, just remember that they're really not the anomalous freaks of nature in the game of randomness, but rather they are negative side of the expected...always expect the unexpected, those numbers are essentially supposed to be there.

 

  

 

The graph below illustrates the expected amount of "hitters" and "non-hitters" for the given amounts of consecutive games.  Four different states are listed on the graph.  The top half of the graph illustrates how closely each state follows the calculated expected amounts to be drawn and the bottom half illustrates how closely each state follows the calculated expected amounts to NOT be drawn.  It doesn't matter what state you put on the graph, they will all follow suit accordingly and with only negligible fluctuations. 

 

 

On this particular graph, the Tri State Pick 3 (NH, ME, VT) shows the largest deviation from the expected path.  What actually happened here was that between games 2,000 and 3,000, more unique combos were drawn than were expected.  This is actually a bonus for anyone who plays from the group of non-hitters.  Even though the deviation appears large on the graph, the difference is actually pretty small.  Here are the amounts of different combinations drawn next to the calculated expected amount:

 

Games

Expected

Actual

2000

865

882

2500

918

936

3000

950

960

3500

970

975

 

Next, you will notice how the paths of each line all meet and cross at the 500 different combinations mark!  This is extremely important.  The red zone on the graph represents the probability median.  This zone is the halfway point where 50% (or 500) of the total 1000 combination matrix becomes drawn.  It goes without saying; this is also the point where the other 50% (or 500) will not be drawn.  You'll notice that the red zone doesn't actually end on the 500th consecutive draw.  It actually lands on game 693.  Talk about making 1+1=3, well here you go folks.  It takes 693 games for 500 different combinations to be drawn in straight form.  This means that within 693 games that there are about 193 numbers that are repeats, which are numbers that have already previously hit at least once.

I made these graphs to demonstrate how probability actually dictates the flow of randomness within the game.  The information was presented as a projected timeline to show just how long it will take for each of the 1000 combinations to be drawn.  The results are quite accurate and speak for themselves.  Of course, the same rules technically apply to the game no matter where you start the measurement.  I started at the very first drawing ever held in each of the states I tested.  In all reality though, you can start the measurements anywhere in any games history and witness fittingly accurate results.  Take any series of 1000 consecutive games and you will have close to 632 of the 1000 different numbers drawn.  Take any series of 1500 games and you should always have close to 777 of the 1000 different numbers drawn.  Last but not least, take any span of 693 games and you will see just about 50% (or 500) of the 1000 different Pick 3 combinations drawn!

 

At the beginning of this post I mentioned that I had listed the amounts of consecutive games differently in part one of the topic.  What I did was list the spans of consecutive games in 693 game increments.  The reason for this is to demonstrate that Pick 3 can be observed to operate much like the classic coin-toss game!  You can apply the listed odds of 1 in 1000 to any individual Pick 3 combination and all the while assume and/or choose to believe that it should hit about once in every 1000 games, but according to the laws of probability this simply isn't practical.  Perhaps there is another way to see it...how about we explore the odds from the outer reaches of random, lets call it the probability time continuum...E=MC^2...oops wrong formula LOL...how about M=1-(1-.001)^693

 

The Pick 3 odds from this domain would look something like this:

 

.50
693

 

That's half a chance over the course of 693 games.  In all reality, these "odds" hold true in every Pick 3 game and are easily observable with the right tools.  In the same way that it took 693 games to see the first 500 different numbers drawn in each of the states I tested, every combination by itself has about a 50% chance of being drawn in any consecutive span of 693 games.  Think of it this way: since 500 different combos are drawn during any span of 693 games, any one of them could easily be any of the 1000 that are available.  In essence, the performance of each of the 1,000 combos parallels the repeated flipping of a coin.  Each flip is a measure of 693 games (I call this measure a Median Run). The result of each flip is either heads or tails for each specific combination that is tested throughout the run, which signifies either a hit or no-hit for the combo. 

With that in mind, any string of outcomes can also be seen as a random stream of ones and zeros...thus giving us a binary breakdown of the randomness In Pick 3!  This will be covered in much greater detail in Part 3.

~Probability=Odds in Motion~

    JADELottery's avatar - MeAtWork 03.PNG
    The Quantum Master
    West Concord, MN
    United States
    Member #21
    December 7, 2001
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    Posted: December 9, 2006, 11:55 pm - IP Logged

    Totally absorbed and understood.

    Presented 'AS IS' and for Entertainment Purposes Only.
    Any gain or loss is your responsibility.
    Use at your own risk.

    Order is a Subset of Chaos
    Knowledge is Beyond Belief
    Wisdom is Not Censored
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    Jehocifer

      Amazing Grace's avatar - lion
      rainbow lake
      Canada
      Member #25177
      November 2, 2005
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      Posted: December 10, 2006, 12:20 am - IP Logged

      Well this is good news , it confirms my land mine system will work, just need to tweek a little more,

      Thanks for the Info

        Thoth's avatar - binary
        Findlay, Ohio
        United States
        Member #4855
        May 28, 2004
        400 Posts
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        Posted: December 10, 2006, 1:16 am - IP Logged

        Well this is good news , it confirms my land mine system will work, just need to tweek a little more,

        Thanks for the Info

        Your welcome AG and thanks Jade,

        Heres a little something extra for the mind to ponder:

         

        This is the same graph only this time I added three-digit combinations made from the decimal values of Pi.  The leading digit three is not included.  All combos were made from the digits to the right of the decimal in the sequence that they appear in, so in 3.141529653.....the first combo is 141.  The second is 592 and the third combo is 653, etc.  The last of the 1000 possible three-digit combos drawn was 580 at what would be the equivalant of 6,076 consecutive games!   

        ~Probability=Odds in Motion~