Kentucky United States
Member #32,651
February 14, 2006
10,317 Posts
Offline
Quote: Originally posted by Pick3Guy on Oct 28, 2014
Well, it depends on whether it's straight or box. So, if you're boxing consecutive digits then it's expected to ,on average, hit once in every 16 drawings. But, if it's straight then it's ,on average, once in every 100 drawings.
Anyway, we can't really speak about when something is due. This is the same thing as saying that we know when it's going to come out. But, it may go well passed the range of 17 to 23 drawings without hitting . In this case, we can say that the game is more ripe for consecutives.
I probably shouldn't use the word "due" and said the mathematical probability is once every 22 drawings. The window of play is no more than 8 drawings unless a betting strategy for extended play and still show a profit is created. The idea is for the group to hopefully perform within probability and only played when the timing is right.
Consecutive numbers hit more than probability in my state in the last 3 years (about 1000 drawings). For play a skip chart could be used to determine how many times they hit within a range. Not saying it's easy to do, but a few bucks could be made with a little work.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by Stack47 on Oct 2, 2014
Looks like your every-day-garden-size-variety-self-proclaimed lottery expert to me.
"Ahh, the old days when everything was new. Some very talented folks use to drop by back then. Strong willed and just full of beans and brains too they were. LOL"
I've read lots of lottery related articles on the Net and never saw anything resembling a "Holy Grail" of pick-3 systems, but I never expected to find it either. However if there is such a thing, I'll never find it reading articles about the evils of gambling by someone like our local con-man who believes there are more than two people on LP giving him any credibility.
There is nothing wrong with pointing out the probabilities and the huge odds against, but to make their case they usually have to ignore other probabilities and statistics that are deciding factors in making the bet. At the end of the day, it comes down to the concept of making simple choice on what to bet and how much to bet on it; something the people on the dark side can't comprehend.
Well, there ,now, is such a thing as a "holy grail" for the Pick 3. But, unfortunately, you'd never find it ,as a whole, reading any articles ,or otherwise, anywhere. Although, you may ,and can, find articles that ,more or less, touch upon the concepts and methodology involved.
Simi Valley, CA United States
Member #156,933
July 4, 2014
823 Posts
Offline
Quote: Originally posted by Pick3Guy on Oct 28, 2014
So, where does the law of averages say that ,at some point, the subset will hit within five plays of it already hitting? But, you're right that the proposed strategy doesn't really work? Why not? That's because it's more involved than that. You'd need to take the patterns into account too.
Of course it should! Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)
If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.
If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.
Or a third.
Or a fourth.
Or a fifth.
Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.
And how many do you need, really... if the pay-outs work in your favor?...
But the CA algorithm prevents this. Or, it sure as hell seems to.
You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!
We have no dreams at all, or interesting ones. We should learn to be awake the same way—not at all, or in an interesting manner. -- Friedrich Nietzsche
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by Tialuvslotto on Oct 27, 2014
"Well, of course, there are groups that do hit more often than others."
Would you care to expand on this statement? It sounds like something that should be discussed on LP!
Well, as we know, the numbers can be broken down into groups of different sizes. So, groups containing more numbers can be expected to hit more often. For example, the group of singles hits more often than that of doubles. Likewise, doubles hit much more often than triples. Anyway, the list goes on. But, I'm sure we all know this already.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by PeerGynt on Oct 27, 2014
When to play, of course, is always dicey - if one knew exactly when to play, one could do quite well; and even the phrase "when to play" assumes one knows what to play - no one needs to know when to play losing numbers!
If the computer-run D3 games didn't work off algorithms, one could just select any set of numbers, and start tracking them in a very orderly, dull fashion, and consistently win. Everyone could win with relatively little effort. The whole reason the algorithm is there, is to avoid just such playing - no matter how tiny or how large you slice and dice the numbers into groups, those groups don't play according to a consistency inherent to purely random selection processes.
But do notice how the game seems to have an eerie consistency to it, and a dependable one at that. Take pairs: track some random pairs. Note how some'll come real quick, and how others will be out for long periods of time... maybe even multiple long periods of time... but then... always... eventually... a short amount of time.
So you get, just talking about pairs now: Pairs that are out short periods of time, we'll say, <7 plays. And then you get pairs that are out vast amounts of time... a few times in a row, sometimes... but then finally, they, too, are out <7 plays.
Think about that for a minute. There's something in there worth thinking about.
Well, it seems you're underestimating the difficulty involved in a game that's not using algorithms. But, why do you believe that it's not so hard to beat a purely random game? If that's true, then there wouldn't be so many people losing with mechanical ball drawings.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by grwurston on Oct 28, 2014
Yes, they are. They are favored to appear as the lowest number in the set. 85 out of the 120 P3 boxed singles have 0, or 1, or 2, as the lowest number. On the same token, the 7 or 8 or 9 will appear as the highest number in the set on 85 out of the 120 single boxed draws, so in that aspect you are correct.
The point I was trying to make was that if you know that 0, or 1, or 2, will be the lowest number 71% of the time, why would you not use a combo containing one or two of them? You are just putting the odds in your favor.
On the P4, it is 175 out of the 210 boxed singles, or 83%.
Well, of course, any of digits 0,1,2 may ,or will, be the lowest to appear when any of them do come out. But, they don't have any advantage over any other set of 3 digits in terms of odds. For example, the digits 3,4,5 will occur about as often as 0,1,2 will occur.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by Stack47 on Oct 29, 2014
I probably shouldn't use the word "due" and said the mathematical probability is once every 22 drawings. The window of play is no more than 8 drawings unless a betting strategy for extended play and still show a profit is created. The idea is for the group to hopefully perform within probability and only played when the timing is right.
Consecutive numbers hit more than probability in my state in the last 3 years (about 1000 drawings). For play a skip chart could be used to determine how many times they hit within a range. Not saying it's easy to do, but a few bucks could be made with a little work.
So, how are you coming up with 22 drawings using 10 consecutives? Perhaps, you're referring to consecutive pairs? Also, Kentucky has two drawings each day and only one on Sunday. So, it only takes about a year and a half to reach 1000 drawings.
Anyway, there will be times when a group hits more ,or less, often than it's expected to. But, counting how many times a group hits will not be much help. So, suppose you determine that consecutives hit less often than what would be expected in a given year. The same thing could happen the following year too. What's more, you'd still have to know when to play.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by PeerGynt on Oct 29, 2014
Of course it should! Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)
If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.
If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.
Or a third.
Or a fourth.
Or a fifth.
Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.
And how many do you need, really... if the pay-outs work in your favor?...
But the CA algorithm prevents this. Or, it sure as hell seems to.
You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!
Well, first of all, there are 210 numbers when boxing. Also, triples can't be boxed. But, the law of averages says nothing about how many plays. This is just what you're saying.
Texas United States
Member #150,790
December 31, 2013
824 Posts
Offline
Quote: Originally posted by Pick3Guy on Oct 29, 2014
Well, first of all, there are 210 numbers when boxing. Also, triples can't be boxed. But, the law of averages says nothing about how many plays. This is just what you're saying.
I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.
This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws. The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.
And you can similarly define a playing window for almost any situation the game can throw at you.
"There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."
United States
Member #116,339
September 8, 2011
5,096 Posts
Offline
Quote: Originally posted by PeerGynt on Oct 29, 2014
Of course it should! Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)
If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.
If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.
Or a third.
Or a fourth.
Or a fifth.
Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.
And how many do you need, really... if the pay-outs work in your favor?...
But the CA algorithm prevents this. Or, it sure as hell seems to.
You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!
Use this link for your sub groups> inference is personal
CA>
Drawing Date
Pick 3
Pick 4
Midday
Evening
Midday
Evening
Tue, Oct 28, 2014
4-7-9
1-2-6
9-6-5-1
Mon, Oct 27, 2014
2-5-1
8-6-0
7-1-9-8
Sun, Oct 26, 2014
4-6-6
6-7-7
2-0-2-6
Sat, Oct 25, 2014
1-3-2
9-8-3
7-7-6-5
Fri, Oct 24, 2014
7-8-1
9-4-2
6-0-4-5
Thu, Oct 23, 2014
5-6-8
0-5-3
3-3-1-3
Wed, Oct 22, 2014
8-7-0
2-5-3
2-1-8-3
Tue, Oct 21, 2014
3-3-8
1-9-6
5-4-8-8
Mon, Oct 20, 2014
6-2-6
8-8-0
6-7-6-4
Sun, Oct 19, 2014
3-2-0
4-7-3
5-1-5-7
draw 320> sequential digit difference (Triggers)>1-2 (ie 3-1 and 2-0)
T1> 1,3,8,0 T2> 2,4,7,9
Grouping for prediction interval points>A/ 0-1-2-3-4, B/ 7-8-9
wheel each group or form pairs from each group for box and straight hits>
Inference> A draw set targets a section of the pool(Subsetting), so you remain focus, keep things simple!, we're not
not bothered any parametric percentile , we're only concerned with the subset each moment--lottery is random!
Grouping> 2-3-4-5 6-7-8-9 , form triads with each group and add your x
6764>1-1-2>T1>1380,T2>2479, Groups>A 0-1-2-3-4, B/7-8-9> form triads from each group
NB> The ideal of hitting the next draw is delusional, is not even cost effective, if your subset has >50% degree of confidence(by observation), then why not let time run it course(Time frame).Look at across the groups, your x is right there!
Crested Butte, CO United States
Member #69,862
January 18, 2009
1,428 Posts
Offline
Quote: Originally posted by Tialuvslotto on Oct 29, 2014
I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.
This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws. The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.
And you can similarly define a playing window for almost any situation the game can throw at you.
I don't agree with 90% its much lower more like 70%.
Grouping> 2-3-4-5 6-7-8-9 , form triads with each group and add your x
6764>1-1-2>T1>1380,T2>2479, Groups>A 0-1-2-3-4, B/7-8-9> form triads from each group
NB> The ideal of hitting the next draw is delusional, is not even cost effective, if your subset has >50% degree of confidence(by observation), then why not let time run it course(Time frame).Look at across the groups, your x is right there!
CAL>Fantasy 5>Matrix 5/39, Set N=39> see link for Triggers to form subsets(groups)
United States
Member #90,247
April 24, 2010
13,962 Posts
Offline
Quote: Originally posted by Pick3Guy on Oct 29, 2014
Well, of course, any of digits 0,1,2 may ,or will, be the lowest to appear when any of them do come out. But, they don't have any advantage over any other set of 3 digits in terms of odds. For example, the digits 3,4,5 will occur about as often as 0,1,2 will occur.
If you know the lowest number. Pick 3. 120 Single Boxed Numbers. Arranged By Root Sums.
United States
Member #155,987
June 5, 2014
497 Posts
Offline
Quote: Originally posted by Tialuvslotto on Oct 29, 2014
I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.
This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws. The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.
And you can similarly define a playing window for almost any situation the game can throw at you.
So, exactly how, or where, are you getting 90%? But, admittedly, a logical basis was just found ,within my system, for Win D's double trap. Even so, I still need to gather empirical data to support this. This would require the existence of certain, significant patterns.