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Pick 3 Matrix Completion Cycle

Topic closed. 10 replies. Last post 9 years ago by Thoth.

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Findlay, Ohio
United States
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May 28, 2004
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 Posted: December 9, 2006, 2:35 am - IP Logged

Have you ever put together a historical listing of all the combos drawn in your states Pick 3 game and wondered why there are so many combinations that have never hit?  Have you ever noticed how some combos hit with a much higher frequency than others?  Have you ever wondered just how long it will take for all of the 1,000 pick 3 straights to be drawn in your states game?  The answers to these questions can be found within the laws of probability.

Believe it or not, the laws of probability actually dictate the long term performance of the game.   The projected long term outcome for EVERY Pick 3 game IS highly predictable.  In essence, every Pick 3 game in the country is identical with respects to how the randomness within each game operates.  There is a very real and distinct timeline that all Pick 3 games follow.  I like to call this timeline the Matrix Completion Cycle.  Many trackers and game analyzers prefer to gather statistical data and derive approximated probabilities or expected results from them.  I like to do just the opposite...first I calculate the probability, then I watch the stats verify them like magic.

So, how long will it take for every Pick 3 number to be drawn in your state?  Heres a formula that dictates the answer: log(1-MP)/log(1-p).  This is saliu's formula with the "DC" replaced with "MP".  The formula doesnt really change any but the logic in it does slightly.  The MP is used to denote the "Matix Percentage"...which is any percentage of the matrix entered as a decimal.  So 50% is .5 and 75% is .75 etc.  Now, since the game is random and nothing can truly be predicted with absolute certainty, do not enter 100% (1.00)!  You can, however, enter 99.99% (.9999)...or as many nines as you like.  The "P" in the formula is the decimal probability for each straight combo in the matrix.  In Pick 3, each straight combo represents 1 in 1,000, so the P is 1/1000, which equals 1÷1000 = .001.

Simply put, this formula will tell you how many consecutive games must take place before you can expect to see a specific percentage of the matrix drawn!  As an example, if you wanted to know how many games it takes to get to the point where 750 of the 1000 possible straights are drawn, you would run the formula as: log(1-.75)/log(1-.001).  The formula then spits out the number 1385.601.  Which means 1385.601 consecutive games or 1386 when rounded.  The number of trials this formula spits out  for any percentage entered into it is extremely accurate!  Take a look at the first 1386 games in your state and count how many DIFFERENT Pick 3 straight were drawn.  It will be very, very close to 750!

I labeled this post as the Matrix Completion Cycle.  The word "completion" can be paralelled to the term "100%", which I just said to never use in the formula, so I should probably clarify what i mean here little better.  The matrix is completed when all 1,000 combos are drawn.  To project all 1,000 combos would imply 100% of the matrix.  Since we can't honestly use 100%, the next best thing will have to do, which in this case is the 99.9% mark.  Of the 1,000 combos in the matrix, 99.9% equates to 999 combos.  The number of games given by the formula is accurate, but is better to treat the figures as close approximations.  Most state will not measure up perfectly...meaning that the actual amount of different numbers drawn will normally be slightly over or slightly under the calculated expected result, but I assure you, they will be very close!  Since there is a minor fluctuation in the actual results, many times the one last combo (or last few combos) will hit before or soon after the 99.9% mark...which is 6,905 consecutive games.  It's usually those last one or two combos that many see as the great anomalous stinkers that seemingly defy the odds.

The first of the three graphs below illustrates the expected percent of matrix to be drawn within a givin number of games.  Notice how the line crosses the 50% mark at 693 games.  It only take 693 games to have very close to 500 of the 1,000 pick 3 combinations drawn.  The odd part is, you usually wont see the the last of the other 500 non-hitters until sometime between games 6,237 and 7,623, which is the 99.81% to 99.95% range!  This is the range where most states Pick 3 matrices are completed!

Rather than using percents, the next chart shows the actual amount of combos drawn within the listed consecutive number of games.  I used the entire histories of Indiana, West Virginia and Georgia.  Notice how closely each state follows the expected result!.  Note: the ending points on the graph indicate that all three states completed their cycles on game 6237.  This is not really the case, all three states had the their last combo drawn between games 5544 and 6237, but this would have requred alot more work to illustrate on the graph.  Other than that, the data points are all valid and accurate.

The last graph shows the amounts as percentages.  Here again, it's amazing how closely the six listed states follow the calculated and expected result!  Each of the states follows so closely that it becomes difficult to discern one line from another.  If we added the three states from the graph above, they would follow along just as well.

For each of the nine states, here are the actual drawing# that their cycle or matrix was completed in as well as the combo the hit:

 STATE GAME COMBO OH 8118 157 PA 7623 214 NJ 6092 453 MD 6371 720 IL 5883 698 MI 8044 486 WV 6132 516 GA 5712 848 IN 5973 063

There is much more to discuss regarding this data, I'll post more before the weekends through.

~Probability=Odds in Motion~

The Quantum Master
West Concord, MN
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 Posted: December 9, 2006, 10:02 am - IP Logged

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AL
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January 5, 2006
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 Posted: December 9, 2006, 10:26 am - IP Logged

thanks looking forward to seeing it

CA
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December 10, 2003
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 Posted: December 9, 2006, 11:52 am - IP Logged

It's interesting to note that all progressions were completed in these states between 4096 and 8192 draws.

Computers, as almost anyone knows, only work with two numbers - 0 and 1. Either a switch is off (0) or on (1). So, if any type of repeating cycle is to be found, it will occur on the cusp of a direct multiple of 2. With regard to the number of drawings mentioned here, the next possible multiple of 2 is 8,192 - 213. Check Drawing #1 and Drawing #8193. If the drawn numbers are the same the RNG used for the draw may be repeating itself. Proof would be to check Drawing #2 and Drawing #8194. If they are the same as well it is possible that the repeating cycle of the RNG has been found.

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

=^.^=

Findlay, Ohio
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 Posted: December 10, 2006, 1:31 am - IP Logged

It's interesting to note that all progressions were completed in these states between 4096 and 8192 draws.

Computers, as almost anyone knows, only work with two numbers - 0 and 1. Either a switch is off (0) or on (1). So, if any type of repeating cycle is to be found, it will occur on the cusp of a direct multiple of 2. With regard to the number of drawings mentioned here, the next possible multiple of 2 is 8,192 - 213. Check Drawing #1 and Drawing #8193. If the drawn numbers are the same the RNG used for the draw may be repeating itself. Proof would be to check Drawing #2 and Drawing #8194. If they are the same as well it is possible that the repeating cycle of the RNG has been found.

Interesting idea, but we could never get so so lucky!

~Probability=Odds in Motion~

Switching between Fairfax, VA and Belgium
Belgium
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 Posted: December 12, 2006, 7:18 pm - IP Logged

I would have to do some serious search for the code, it has been a long time,
but i remember one number (amount of draws before every straight number came up, on average of all pick3 games in the US and Canada) popping up when it concerns the "turn of the pick3 game",
and that number was around 7800 draws.

lasas3

An onion a day keeps everyone away!!!

Redford/MI
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 Posted: December 12, 2006, 8:04 pm - IP Logged

Thanks for posting this information.

It was very informative & I appreciate the fact that you thought to share it here with all of us at the post.

L ttaL   T

NY
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October 16, 2005
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 Posted: December 13, 2006, 5:31 pm - IP Logged

Thanks for the good info. Now how about the same info for a pick 4 game?

For those who don't read between the lines, I'll point out that for Georgia. the state that drew the final number in the least number of games, it still took 7.82 years, assuming they had  2 draws a day for the full 7.82 years. Remember that the next time somebody says a number is due because it hasn't come up for 3 or 4 or 5 years.

Findlay, Ohio
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 Posted: December 13, 2006, 8:30 pm - IP Logged

Thanks for the good info. Now how about the same info for a pick 4 game?

For those who don't read between the lines, I'll point out that for Georgia. the state that drew the final number in the least number of games, it still took 7.82 years, assuming they had  2 draws a day for the full 7.82 years. Remember that the next time somebody says a number is due because it hasn't come up for 3 or 4 or 5 years.

...I have more info that better explains how this cycle works.  I hope to post it later on tonight or in the next day or so. I haven't done too much work with the Pick 4 but I have done enough to know that the median of 6,932 games is accurate.  With that in mind, the Pick 4 should closely follow these amounts:

 ConsecutiveGames Percent ofMatrix Drawn Numbers DrawnFrom Matrix Remaining to be drawn 6,932 50.00% 5000 5000 13,864 75.00% 7500 2500 20,796 87.50% 8750 1250 27,728 93.75% 9375 625 34,660 96.88% 9688 312 41,592 98.44% 9844 156 48,524 99.22% 9922 78 55,456 99.61% 9961 39 62,388 99.80% 9980 20 69,320 99.90% 9990 10 76,252 99.95% 9995 5 83,184 99.98% 9998 2 90,116 99.99% 9999 1 97,048 99.99% 9999 1 103,980 100.00% 10000 0

In a Pick 4 game you could expect the last one or two combos to hit between games 76,252 and 83,184.  Though Im sure there will be cases where it ends between 55,456 and 62,388.

The expected amount of different numbers drawn from the 10,000 combination matrix in exactly 10,000 games is 63.21% or 6321 combos.

~Probability=Odds in Motion~

NY
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 Posted: December 14, 2006, 1:59 pm - IP Logged

Thanks, again.

And again, I'll read between the lines for those who think a number is overdue because other numbers have come up several times, while some numbers haven't come up at all. If the final number were drawn at game 55,456 it would mean that the other numbers had all been draw an average of 5.55 times. Since that distribution wouldn't be uniform we would expect that a lot of numbers had only been drawn once, while a similar number should have been drawn 11 times, and some would have been drawn 15 to 20 times.

I don't expect most people to understand the more complicated math that calculates these results, but here's a really simple way of looking at it.  Once 95% of the possible numbers have been drawn, in each subsequent drawing there's a 95% chance that the number drawn will be one that has already been drawn previously. A lot of people believe that a number is more likely to come up because it is "overdue" but the reality is that  those overdue numbers are less likely to be drawn than one of the numbers that isn't "overdue." Of course that has nothing to do with random events evening out in the long run. It's simply because once 95% of the numbers have been drawn there are 19 numbers that aren't overdue for each number that is.

I'll also point out that the percentage of the possible combinations that are actually played can be calculated the same way.  With about 176 million possible combination for MM, selling 176 million tickets would mean that about 63% of the combinations had been played if those combinations were chosen randomly, and that would mean that with 176 million tickets sold there would only be a 63% chance of having a winner.

Findlay, Ohio
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 Posted: December 14, 2006, 9:08 pm - IP Logged

KY FLOYD

"I'll also point out that the percentage of the possible combinations that are actually played can be calculated the same way.  With about 176 million possible combination for MM, selling 176 million tickets would mean that about 63% of the combinations had been played if those combinations were chosen randomly, and that would mean that with 176 million tickets sold there would only be a 63% chance of having a winner."

Quite true!  It would be fun to go blow \$693.00 on Pick 3 quick-picks to see how just close to 500 the amount of unique tickets would be.  Random numbers are usually distributed in this way, provided that each number has the same probability of occurring...whether they be the actual drawing results or the random tickets themselves shouldnt matter.  Im not sure if this is referred to as hypergeometric or binomial distribution?

Once 95% of the possible numbers have been drawn, in each subsequent drawing there's a 95% chance that the number drawn will be one that has already been drawn previously.

You are the first person to come right out and mention the "repeat factor"...lol, one less post for me to do!  That is what's directly responsible for the way the matrix breaks down.

~Probability=Odds in Motion~

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