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Not sure where to put this question (D3)...Prev TopicNext Topic
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... so I put it here.
I'm also not sure the game works with spatial/temporal concepts like "first" and "second" and "third," but for the sake of clarity, I'll just go with those in order to ask….
I believe of every number the Daily 3 selects—say, for example, 123—one of those numbers would be the First number the D3 selects. Then it picks a second and third number. Then, it mixes them all up, to produce the result.
Let's say for argument's sake, it's 2: that was the First number selected. Then, it chose (somehow) 1 and 3, second and third. Then it mixed them all up to produce 123.
Though the 2 in my example is the middle number, it was the First number chosen.
I'm not sure about the second and third numbers. But I believe the D3 game chooses a different First number, every single time it plays, from the last selected First number. If it chose 2 as it's First, it will not choose 2 for First the next time.
Sure, you can get the same result. You can get, 123 and the next time, 231. But that only means either the 1 or 3 was chosen First, on the second play.
And so, the question:
Would this even help somehow in creating strategies? Or would it be pretty much useless?
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... so I put it here.
I'm also not sure the game works with spatial/temporal concepts like "first" and "second" and "third," but for the sake of clarity, I'll just go with those in order to ask….
I believe of every number the Daily 3 selects—say, for example, 123—one of those numbers would be the First number the D3 selects. Then it picks a second and third number. Then, it mixes them all up, to produce the result.
Let's say for argument's sake, it's 2: that was the First number selected. Then, it chose (somehow) 1 and 3, second and third. Then it mixed them all up to produce 123.
Though the 2 in my example is the middle number, it was the First number chosen.
I'm not sure about the second and third numbers. But I believe the D3 game chooses a different First number, every single time it plays, from the last selected First number. If it chose 2 as it's First, it will not choose 2 for First the next time.
Sure, you can get the same result. You can get, 123 and the next time, 231. But that only means either the 1 or 3 was chosen First, on the second play.
And so, the question:
Would this even help somehow in creating strategies? Or would it be pretty much useless?
I'm probably here unless I'm not.
Dreaming would be a perfectly useless function if it's only purpose was to entertain. -
Thanks for the thoughtful answer, Garyo. I want to expand on what I'm talking about, with some examples maybe. But I wonder if you could elaborate on what you mean by...
Normal distribution shows a digit repeating back to another position 10% +/-2%. (made the +/-2% up since it fluctuates in Texas between 11.9 and 13% at various times). If it is less than 5% in your state, it would be HUGE advantage.
Not sure I understand this. Can you explain?
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Best guess is the original programming for this is on a 5 1/4 floppy collecting duct in some Goodwill. Likely in Somalia, Tanzania, or sold to the Russians as some super secret spyware. Hope the KGB is enjoying figuring out what it does.
This is a makeshift of the original. Pretty easy to follow. N1-N3 is the digits drawn, Sum, SUp is a positive or negative number showing which direction the sum is going, RP1-RP3 shows the repeat position. Nbr1 - Nbr3 shows the actual digit that repeated. I coded them in yellow light green and orange to help.
Jump to the 435 draw and see there was 1 repeat in the first position and it was the number 4. in 974 there was 1 repeat in the third position and it was 4. In 219 the 9 goes from the first position to the third.
The 6 repeat from 566 to 679 poses an interesting question. Do we treat that as two digits repeating? Or just one? I chose to ssay a single digit repeated.
The repeating 7 in 797 is where I truly dislike this method of counting repeats. Although it is true one digit did repeat twice, it was still only one digit. And yet programming insist THERE WAS TWO REPEATS.
If I total all the double hits AND TRIPLE HITS using this method it will be about 25% to 27% hit ratio per position. That would a bit high by the 70% standard we see quoted, but that's the facts.
That repeat hit ratio would go down quite a bit if we only count repeats from single to singles.
Pre-emptive Strike: Yes, I did selectively choose this part of the chart to prove my point. There are probably another 1000 places I could have chosen but this one is close to the top. And I'm lazy like that.
Thoughts?
G
I'm probably here unless I'm not.
Dreaming would be a perfectly useless function if it's only purpose was to entertain. -
(HEADNOTE: Gary, I was nearly done with writing this as your post appeared, so forgive me if it doesn't directly address your further detailed and illuminating re-response. I hope this provides some response back, but will have to think about yours first before adequately replying. Meanwhile, I'll just finish this, and hope this works to some satisfaction….)
It's highly unlikely what I'm ruminating about hasn't been discussed on this board before—I'm so new here, it's just impossible to go over every previous post for however many decades.
But what I'm wondering about, is the nature of the D3 game itself (specifically, the California D3, which does use an algorithm; and, here, I'm only going to talk about singles plays, not doubles).
I posit that the game picks a First number. Then it picks a Second number, and a Third number.
I posit that the game always picks a different First number. Always, and without exception. But the order of the game's picking isn't revealed in the actual result (e.g., result 234 might mean the game chose the 2 as First, or the 3, or the 4: you just don't know, though).
But to expand: it seems evident the game goes ahead and picks different Second numbers, and Third numbers, all the time. So let's just assume for the moment, for the sake of argument, that it does.
So, you can easily still get a repeat number: 234 can reappear as 324, with the Firsts, Seconds, and Thirds mixed up. Proof this is probably the case? The fact that you occasionally do get full repeat numbers like this, but for the most part (?—does it ever happen?) you never get three repeats or more.
If Firsts, Seconds, and Thirds never repeat, then one can go on to say—converting the game's choices now spatially—you will never get a repeated digit from one level of play to the next.
And so, taking the exactly last two California D3 plays from today
F S T
0 2 9 E (b)
5 9 7 D (a)What we see visually could be the order of Firsts, Seconds, and Thirds: F(a)(b) = 5/0, S(a)(b) = 9/2, T(a)(b) = 7/9. One could mix these up a few different ways, too; but I'm saying what it would not be is F(a)(b) v S(a)(b) v T(a)(b) ≠ 9/9. (I hope I'm using my symbols right there!)
How would this help? Well, if you knew that digit position X of a play couldn't (by definition) be beside, below, nor above itself, that could very well help to strategize[sic.] play.
(It all becomes like a grand Sudoku game—surely that comparison's been made in the past?)
Taking the plays above on face value for the sake of an example (i.e., assuming they are presented visually as well as in reality as First, Second, Third digit placements—this would be perilous in actual strategies, because the true identity of F/S/T placements is unclear), then we would know…
Above F(b), i.e., F(c), you couldn't have another digit 0. This would affect the choice of S(c), because it would be dependent upon what digit appears in F(c) - but we already know, that S(c) can't be digit 2. And so on to T(c), which we already know can't be digit 9.
One would simply have to rearrange the play into correct order F/S/T... but how does one do this? I think there are ways. I even think the pool for each set, F/S/T - beyond its own natural selection (i.e., by definition, they can't all be digits 0-9 at once) - are limited by the current play/s of the game. The more you limit the pools, then filter them through their possible positioning... the more you shrink the possible 120 boxes of play down to viable, playable sets.
... er, right?
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Going back 20 years in the Texas history this is the result of the same three digits hitting back to back. The red shows where the digits hit in exact order. Alternating green and yellow is to save eye strain. (Mostly my eye strain)
The blue is the only place where the same three digits hit back to back to back.
The orange and rose block is just an oddity. 277 hits. 277 hits again. And after one draw says, "Let's do it again!"
27 times in 6522 draws is .00414 of the total. And with the exception of the sums 3 and 4, it happens in the "fatter" part of the sum chart, where most of the combinations are found.
In Texas where the drawings are by ball machine in front of a live audience, it would be a bit hard to put the balls in the order you want without someone seeing you. Not so with an RNG.
27 times in 20 years? I'd be more inclined to bet against it happening than betting it would. And if I did bet, I'd at least scramble the numbers for better odds.
G
I'm probably here unless I'm not.
Dreaming would be a perfectly useless function if it's only purpose was to entertain. -
This is all good stuff, Garyo, and it can be used to support what I'm saying. The 137 triple repeat can easily be reorganized so that each digit doesn't touch above/below/beside: just switch the 1 and 3 in the topmost blue line. And although these three would then represent non-touching digits, it doesn't mean it was the original order drawn (or, wouldn't mean it, in a computerized version of the game).
Maybe it doesn't matter what order they came up? If one can rearrange these digits any way afterward, logically, one can do it before hand; or at least create box scenarios that can then be played.
I don't think the doubles work this way. Me, I've always divided the P3 and P4 games into two games in one: there's the singles game, with its own rules; then, the doubles/triples game, with its own rules. Both games are going on within one Game, and both are interdependent… and both have slightly different rules governing them. I don't believe the rules governing digit placement necessarily apply in doubles.
But I also don't think the doubles "break up" ongoing singles games, and vice-versa. That is, if you had say, 123; then say, 889, a double; if a single appeared next, it's not like the double never appeared—it simply works its own placements off the 889, rather than having to adhere to the rules that would apply from the previous single 123.
Again, I have no proof of these things… this is just what I've observed, over detailed study. But I think they're pretty certain, from my vantage point.
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It would be quite different boxing a map of plays out, according to this thesis.
Traditional box maps I've seen, use letters for digits. A=0, B=1, etc. They end up being quite orderly: ABC, ABD, ABE, etc.
But for this map, you'd need a wholly different symbol system: either multiple letters, or some other kind of denotation. Because each slot would require up to three different symbols to map its digits.
You could do this, say: A, a, a. These would represent 0 in digit space First (A), digit space Second (a), and digit space third (a), respectively.
Let's say the play 012 came up (we're assuming in this example the correct temporal/spatial placement of play's digits), and you were working out the follow up singles play. Digit space First could only contain digits that run B-J (1-9, since 0 can't be played); digit space Second would run a, c-j (0, 2-9, since 1 can't be played); and digit space Third would run a, b, d-j (0, 1, 3-9, since 2 can't be played). An end result - say, 461 - would look like this then: Dfb.
You'd have an end map of mixed up and missing letters; but it would represent all available plays, and should only add up to 84 boxes.
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Thanks for the kind words! Credit to where it is due. Married a geeky mathematician who couldn't open a pickle jar, but could do finances in her head faster than most use a calculator. Brilliant woman.
Somewhere on this LP site is a chart you should get your hands on. Likely in the Pick 3 encyclopedia's blog, Win D, if he still has a blog up. That chart shows how to cover every combination of one digit boxed for $42.50(?), or all the singles for $27(?). Some people refer to it as the "If You Know One Digit" system made famous by Steve Player.
Playing singles and using filters you can get it down to about $13. Been there. Done that. Using the "elimination" technique, 6 hit, so it won't hit again, I played.
&(&!@@#!!!! 6 hit two nights in a row! Can't happen three times.
%&^&@#@#!@@!!! 6 hit three nights in a row. But.......
%!&&&@@@@^^%$&#^#@!!!!!!!! NO WAY FOUR NIGHTS IN A ROW?
Ended up spending over $60 to win $40.
Still take a look at it. There is a lot of GOOD, SOLID INFO on Win D's blog.
But back to the original point of a number repeating 10% of the time. Went ahead and grabbed the CA Pick 3 drawings. Made some changes in the code, added another dozen lines to track repeats by position. Labelled Rep1 thru Rep 9.
Rep1 is the first digit repeating back to the first position. Rep2 First digit repeats to the second position. Rep3 first to third. Rep 4 second digit to first position. Rep5 second to second position, Rep6 second digit to third position. Rep7 third to first, Rep8 third to second, Rep 9 third to third.
You see there is only a one in ten chance of a given number repeating back in any position. Ten digits equals a one in ten chance of any digit falling in any position.
Certainly different than the bloated RP percentages. And a tad more realistic IMHO.
How does this affect your thinking?
Couple of other thoughts, interesting side notes if you will.....
1) The second digit (1245 total) has not repeated back to the first position in 34 draws. Might be worth looking at for a possible play.
2) I love working with numbers and I appreciate the thanks, but how about sending the winning MegaMillions instead? Don't have to buy anything, just send the numbers and I'll do the rest. TIA
G
I'm probably here unless I'm not.
Dreaming would be a perfectly useless function if it's only purpose was to entertain. -
(HEADNOTE: Gary, I was nearly done with writing this as your post appeared, so forgive me if it doesn't directly address your further detailed and illuminating re-response. I hope this provides some response back, but will have to think about yours first before adequately replying. Meanwhile, I'll just finish this, and hope this works to some satisfaction….)
It's highly unlikely what I'm ruminating about hasn't been discussed on this board before—I'm so new here, it's just impossible to go over every previous post for however many decades.
But what I'm wondering about, is the nature of the D3 game itself (specifically, the California D3, which does use an algorithm; and, here, I'm only going to talk about singles plays, not doubles).
I posit that the game picks a First number. Then it picks a Second number, and a Third number.
I posit that the game always picks a different First number. Always, and without exception. But the order of the game's picking isn't revealed in the actual result (e.g., result 234 might mean the game chose the 2 as First, or the 3, or the 4: you just don't know, though).
But to expand: it seems evident the game goes ahead and picks different Second numbers, and Third numbers, all the time. So let's just assume for the moment, for the sake of argument, that it does.
So, you can easily still get a repeat number: 234 can reappear as 324, with the Firsts, Seconds, and Thirds mixed up. Proof this is probably the case? The fact that you occasionally do get full repeat numbers like this, but for the most part (?—does it ever happen?) you never get three repeats or more.
If Firsts, Seconds, and Thirds never repeat, then one can go on to say—converting the game's choices now spatially—you will never get a repeated digit from one level of play to the next.
And so, taking the exactly last two California D3 plays from today
F S T
0 2 9 E (b)
5 9 7 D (a)What we see visually could be the order of Firsts, Seconds, and Thirds: F(a)(b) = 5/0, S(a)(b) = 9/2, T(a)(b) = 7/9. One could mix these up a few different ways, too; but I'm saying what it would not be is F(a)(b) v S(a)(b) v T(a)(b) ≠ 9/9. (I hope I'm using my symbols right there!)
How would this help? Well, if you knew that digit position X of a play couldn't (by definition) be beside, below, nor above itself, that could very well help to strategize[sic.] play.
(It all becomes like a grand Sudoku game—surely that comparison's been made in the past?)
Taking the plays above on face value for the sake of an example (i.e., assuming they are presented visually as well as in reality as First, Second, Third digit placements—this would be perilous in actual strategies, because the true identity of F/S/T placements is unclear), then we would know…
Above F(b), i.e., F(c), you couldn't have another digit 0. This would affect the choice of S(c), because it would be dependent upon what digit appears in F(c) - but we already know, that S(c) can't be digit 2. And so on to T(c), which we already know can't be digit 9.
One would simply have to rearrange the play into correct order F/S/T... but how does one do this? I think there are ways. I even think the pool for each set, F/S/T - beyond its own natural selection (i.e., by definition, they can't all be digits 0-9 at once) - are limited by the current play/s of the game. The more you limit the pools, then filter them through their possible positioning... the more you shrink the possible 120 boxes of play down to viable, playable sets.
... er, right?
-
I'm probably here unless I'm not.
Dreaming would be a perfectly useless function if it's only purpose was to entertain. -
-
Garyo - I assume you mean (?) that because there's a 10 chance, statistically shown, of any digit returning to its position, that my "theory" is incorrect, because I'm claiming an other-than-that system.
Remember: just because one part of an algorithm might be discovered, doesn't necessarily mean all parts of the algorithm are discovered. There could be added "rules" that ensure a basic output, over time, of 10% per digit placement. In fact, I'm sure there's such a system in place.
Someone smarter than me (go outside and throw a rock) can do the math that works out: 120 singles boxes X 730 potential plays per year X however many years this game has been going on. You'd get a figure somewhere in the 1000s… and yet, THREE IN A ROW is more or less unheard of?… I bet the odds don't reflect that reality. And if that's so, I must be at least partially correct (i.e., that the game doesn't just randomly figure out three digits with no other considerations).
Adobea78 - I think your system seems about the best one around, at least for identifying straights. But I remain confused, mostly because I'm just not anywhere near your league of intelligence/knowledge… where does that "filter" and "base" come from? Can you link me to the basics of your system? It seems to be everywhere here on the LP, but not simply in one place. Or am I missing it?
