Matrix Decomposition by Half-Life
In part one and part two I demonstrated how the 1,000 number Pick 3 matrix accurately follows a calculated and expected completion cycle. For any Pick 3 game, the largest deviation between the actual performance of the game and the calculated expected results usually occurs at the ending point of the cycle. This ending point is where the last one or two of the 1,000 different Pick combos is actually drawn.
On the graphs in part one and two, each of the states (shown below) follow the expected path very closely. The numbers beside each state is the game number where the last of the 1,000 combinations was drawn for each state. As you can see, Georgia completed its cycle pretty early in comparison to Ohio's 8,118 games. While the difference in the amount of games it took for each of the states to complete its cycle seems huge, it is actually expected to occur in this way.
When you get down to the nitty-gritty of what's really going on behind the scenes in the randomness that's controlling the Pick 3, it only makes perfect sense to see the various states ending at different draw counts. I have mentioned several times now that within any span of 693 consecutive games, that very close to 50% or 500 of the 1,000 Pick 3 combinations will be drawn. This can be any span of 693 consecutive games pulled from anywhere in any states historical listing of previous results. It doesn't have to be just the first 693 games in your states history. You can pick any random spot on the list and begin the 693 game measurement and the 50% rule will apply.
Everything Empirical: I'm never really content to just chuck out mathematical facts without offering up some supporting data. My data for this matter will illustrate that the 50% rule is pretty darn accurate...or as least as accurate as one can expect when measuring the performance of random outcomes. I should mention that the rule is not a perfect 50% most the time, but rather a very tight fluctuation that ranges between 48% and 52% in at least 99.9% of all cases. So, as a rule of thumb, you can always count on between 480 and 520 different combos being drawn during any span of 693 games.
The graph below shows 111 separate span measurements, 21 of which are from a Pick 3 style three-digit groupings of the digits of Pi (p)! The other 90 are from the following states: MD, DE, IL, PA, and OH. I measured the three-digit groupings of Pi digits in the order they occur in after the decimal point. This simulates an additional Pick 3 game and also shows that even Pi Digits aren't exempt from the laws of probability. One curious aspect of Pi is that the randomness in its three-digit sequences appears to be a little more perfect than your typical pick 3 game, but that's a whole different topic. At first appearance the graph kind of makes the performance of each line look radical and highly erratic, but look at the numbers of the Y-Axis on the left side of the graph, they are:
+30 +20 +10 500 -10 -20 -30.
The line goes up or down to meet the amount of combinations drawn for every span of 693 games. This is usually slightly above or slightly below 500. Occasionally, the line will land right on 500 exact.
The next graph illustrates the same principal as the first graph, but with even more states shown at once. There is a larger cluster of lines on the left side of the graph than on the right because several of the states only had eight, nine, or ten consecutive spans in their history. Only the states with the most spans will extend onto the right side of the graph. These eleven states add an additional 109 separate span measurements to the analysis.
Since it's established that very close to 500 of the 1,000 different Pick 3 straights will be drawn during the first 693 drawings of any states Pick 3, why then does it usually take a good 6,000 or more games for the remaining 500 combinations to finally be drawn?
Matrix Decomposition by 50/50 or "Half-Life"
At game 693 there were close to 500 different combos drawn from the matrix. This obviously left us with close to 500 straights that didn't hit. These are the combos that need to be drawn in order to complete the matrix. Just as 500 different numbers came up during the first 693 game span, so too will 500 different numbers be drawn during the second 693 game span. With this in mind, how many of this new group of 500 different combinations will be combinations from the original group of 500 that DID NOT hit during the first span? The answer is simple...because each particular Pick 3 straight on its own has a 50% probability of hitting at any time during any measured span, then its only fair to say that approximately 50% of the 500 non-hitters from the first span will hit during their 50% chance in the second span. Obviously, 50% of 500 is 250. So by the completion of two median spans, which is game 1,386 (693 x 2), we should have very close to 750 different numbers drawn from the matrix. This leaves us 250 more to wait on. Not coincidentally, the formula 1-(1-.001)^1386 arrives at .75, which is 75%...which is the percentage of the matrix that 750 combinations equals.
As for the 250 straights that have yet to be drawn, we can expect 50% (125) of those to be drawn during the third consecutive span, which gives us a total of 875 of the 1,000 straights drawn from the matrix, leaving us with 125 still remaining. Three spans equals 2079 games...1-(1-.001)^2079 = .875...or 87.5%, which is 875 of 1000. Of that 125 remaining, 50% or 62.5 (63) will hit during the fourth span, which ends at 2772 consecutive games. This leaves us with around 63 remaining.
This 50/50 split off continues as follows:
- 31.5 (32) in span five ending at 3465 games
- 15.75 (16) in span six ending at 4158 games
- 7.875 (8) in span seven ending at 4851 games
- 3.93 (4) in span eight ending at 5544 games
- 1.96 (2) in span nine ending at 6237 games
- .98(1) in span ten ending at 6930 games
And finally, during span 11 (ending at 7623 games) that one last combo should be drawn. Many times, this last combo is drawn in span eight, nine or ten. There is just enough fluctuation in the percentages that allow for this to happen. It just the same way, this one last combo (or perhaps two) can also make it out as far as span 12 and 13. Let a quarter represent the last remaining straight beginning on the first draw of span 11. You flip the quarter and it's in the air spinning. This midair spinning action represents the next series of 693 games. When this series of games is over, the quarter lands and its either heads or tails. If it lands on heads, the number is drawn. If it lands on tails, the number is not drawn. We all know how the game of coin tossing works. We could toss 2 or 3 tails in a row pretty easily...thus stretching the matrix completion time out even farther. The same analogy goes for the four combos remaining after span eight. All four can easily hit during span nine. This is the equivalent of throwing up 4 separate quarters and having them all land on heads (1 in 16)...it's quite possible, it's just not as probable as two of the numbers hitting and two of them not hitting (6 in16). Then again, three of the four could hit (4 in 16) or only one of the four could hit (4 in 16). The last option is none of the four hitting at all, which is 1 in 16.
The next two graphs show how closely the original 500 non-hitters (from the first 693 games) follow the expected 50/50 hit or "decay" rate. Don't you just love the terms "half-life" and "decay" rate. They give the game an ill-sense of radioactivity LOL!
In both of the graphs above, the highest numbers at the top (the 500's) are the amount of combinations not drawn within the first span of 693 games. All subsequent numbers are the amounts of combinations drawn from that original "500" over the course of consecutive 693 game spans.
~Probability=Odds in Motion~