United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 18, 2016, 4:12 pm - IP Logged

Quote: Originally posted by JADELottery on January 18, 2016

so everyone can see this in a different way than the graph, we highlighted the 6, 7, and 8 clockwise only moves for column A up to the moment 6, 7, 8 returned, shown below.

keep in mind, 6, 7 and 8 clockwise moves should happen about once every 3.333... draws.

scroll down and discover the missing moves.

"keep in mind, 6, 7 and 8 clockwise moves should happen about once every 3.333... draws."

Now what is the probability that they won't occur on any single draw? 70%s

What is the probability that they won't occur for 42 straight draws?

1 in (1/(.7 ** 42) = 1 in 3,205,400

But, as I pointed out before, you could use any set of 3 "clockwise moves" and look for a gap and there are 128 of them.

Each of those 128 possibilities has the same 1 in 3,205,400 chance so the chance of any 1 of them having a gap of 42 is 1 in 25,000

There are also three digits, I assume you've looked at all of them, so cut another factor of 3 out because each of those could have randomly generated that set and it just happened to occur in the first digits.

Without even considering how many possible algorithms you've run against the data before seeing something interesting, the odds have no dropped to 1 in 8346

The drawing you are using has over 15,000 samples so it is quite reasonable to expect one or more gaps that is 42 or longer in one of the three digits using any combination of 3 numbers in the range of 1-10.

BTW, stating that there is a 1 in 3,205,400 chance doesn't mean that is the odds of what you are seeing happening in the data set you are using,

it is the odds that any sequence of 42 numbers has no "clockwise move" of 6, 7 or 8.

As a previous post points out, if you look long enough with enough different algorithms you will eventually find some unusual/rare occurrence

For example, what are the odds that the exact sequence of "clockwise moves" that you see in the 9 days starting on Apriil 1, 2015 would occur? The odds of that are 1 in 1,000,000,000

Try enough different algorithms, manipulate the data enough different ways and you are certain to find some unusual patterns.

Take, just as an example, that 10x10 matrix that you posted. It's sort of like a Suduko puzzle where all the digits from 1-10 need to be in every row and column.

You've arbitrarily come up with a "clockwise move" formula that arranges them in a particular way but in reality I could make that first row one of over 3,000,000 permutations of the digits 1-10 and build 3,000.000 unique tables (aka a formula for computing a "move"). So I'm convinced the draw has a flaw and to prove it I test each of those 3,000.000 tables and look for a pattern of a gap of 42 or longer in 6,7,8.

Wanna bet I can find one using the 2nd digit of the MN draw?

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 18, 2016, 4:17 pm - IP Logged

Quote: Originally posted by dddwww on January 18, 2016

"keep in mind, 6, 7 and 8 clockwise moves should happen about once every 3.333... draws."

Now what is the probability that they won't occur on any single draw? 70%s

What is the probability that they won't occur for 42 straight draws?

1 in (1/(.7 ** 42) = 1 in 3,205,400

But, as I pointed out before, you could use any set of 3 "clockwise moves" and look for a gap and there are 128 of them.

Each of those 128 possibilities has the same 1 in 3,205,400 chance so the chance of any 1 of them having a gap of 42 is 1 in 25,000

There are also three digits, I assume you've looked at all of them, so cut another factor of 3 out because each of those could have randomly generated that set and it just happened to occur in the first digits.

Without even considering how many possible algorithms you've run against the data before seeing something interesting, the odds have no dropped to 1 in 8346

The drawing you are using has over 15,000 samples so it is quite reasonable to expect one or more gaps that is 42 or longer in one of the three digits using any combination of 3 numbers in the range of 1-10.

BTW, stating that there is a 1 in 3,205,400 chance doesn't mean that is the odds of what you are seeing happening in the data set you are using,

it is the odds that any sequence of 42 numbers has no "clockwise move" of 6, 7 or 8.

As a previous post points out, if you look long enough with enough different algorithms you will eventually find some unusual/rare occurrence

For example, what are the odds that the exact sequence of "clockwise moves" that you see in the 9 days starting on Apriil 1, 2015 would occur? The odds of that are 1 in 1,000,000,000

Try enough different algorithms, manipulate the data enough different ways and you are certain to find some unusual patterns.

Take, just as an example, that 10x10 matrix that you posted. It's sort of like a Suduko puzzle where all the digits from 1-10 need to be in every row and column.

You've arbitrarily come up with a "clockwise move" formula that arranges them in a particular way but in reality I could make that first row one of over 3,000,000 permutations of the digits 1-10 and build 3,000.000 unique tables (aka a formula for computing a "move"). So I'm convinced the draw has a flaw and to prove it I test each of those 3,000.000 tables and look for a pattern of a gap of 42 or longer in 6,7,8.

Wanna bet I can find one using the 2nd digit of the MN draw?

"What is the probability that they won't occur for 42 straight draws?"

actually, the draw difference in the table we showed for 6, 7 and 8 are 48, 52 and 43.

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 Douglas Paul Smallish Jehocifer

United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 18, 2016, 4:22 pm - IP Logged

Now before you say it isn't reasonable to use some scrambled up 10x10 matrix, my answer would be why not? If I was selling a book or system I'd explain the lottery uses some random number generating scrambling algorithm and it has a flaw in it that my matrix exposes.

It wouldn't even be that scrambled, only the rows would be scrambled, the columns would have your nice sequence of numbers. Or I could scramble just the first column and come up with another 3,000,000 tables to try.

Gotta try everything to uncover a flaw in the lotteries random number generator.

I'll be coding this one up tomorrow and have it looking through the 2nd digits to prove there is a flaw there too.

You just missed it because you weren't as thorough as I'll be in looking for the unscrambling key :)

United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 18, 2016, 4:30 pm - IP Logged

Quote: Originally posted by JADELottery on January 18, 2016

"What is the probability that they won't occur for 42 straight draws?"

actually, the draw difference in the table we showed for 6, 7 and 8 are 48, 52 and 43.

You are now quoting individual "clockwise move" gaps (aka draw difference) instead of all three at the same time? What is the relevance?

Those are completely different. The odds that you will get a gap of 52 for a single "clockwise move" value (such as 6) is 1 in 242.

The odds of a gap of 43 is 1 in 93 and for 48 it is 1 in 157

Again, those are not the odds for seeing one in the entire lottery dataset, that's the odds of ANY consecutive sequence of (for example) 52 not having one particular digit in it.

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 18, 2016, 4:37 pm - IP Logged

Quote: Originally posted by dddwww on January 18, 2016

Now before you say it isn't reasonable to use some scrambled up 10x10 matrix, my answer would be why not? If I was selling a book or system I'd explain the lottery uses some random number generating scrambling algorithm and it has a flaw in it that my matrix exposes.

It wouldn't even be that scrambled, only the rows would be scrambled, the columns would have your nice sequence of numbers. Or I could scramble just the first column and come up with another 3,000,000 tables to try.

Gotta try everything to uncover a flaw in the lotteries random number generator.

I'll be coding this one up tomorrow and have it looking through the 2nd digits to prove there is a flaw there too.

You just missed it because you weren't as thorough as I'll be in looking for the unscrambling key :)

well, that is interesting, because we've been studying of your response to our stimulus.

your performance is about what we are expecting.

we do so enjoy your feedback.

we're open for more.

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 Douglas Paul Smallish Jehocifer

United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 18, 2016, 4:41 pm - IP Logged

Quote: Originally posted by JADELottery on January 18, 2016

"Short gaps are uncommon, in 10,000 runs using sequences of 15,200 numbers the shortest maximum gap was 17, the longest was 50."

no, the shortest is 1 and the maximum is unbounded.

I didn't say the shortest possible or the longest possible, I said in my 10,000 runs using 15,200 samples for each run those were the shortest and longest I saw.

And in fact the shortest possible is 0 (the sequence of "clockwise moves" consists entirely of 6,7 and 8's.

The probability of a sequence of 15,200 digits producing "clockwise moves" that only consist of 6,7, and 8 is roughly the same as red coming up on a roulette wheel 15000 times in a row. That is 1 in a number with almost 8000's zeros.

The probabilities of shorter runs drop more quickly than the probability of longer runs.

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 18, 2016, 5:37 pm - IP Logged

Quote: Originally posted by dddwww on January 18, 2016

I didn't say the shortest possible or the longest possible, I said in my 10,000 runs using 15,200 samples for each run those were the shortest and longest I saw.

And in fact the shortest possible is 0 (the sequence of "clockwise moves" consists entirely of 6,7 and 8's.

The probability of a sequence of 15,200 digits producing "clockwise moves" that only consist of 6,7, and 8 is roughly the same as red coming up on a roulette wheel 15000 times in a row. That is 1 in a number with almost 8000's zeros.

The probabilities of shorter runs drop more quickly than the probability of longer runs.

"And in fact the shortest possible is 0 (the sequence of "clockwise moves" consists entirely of 6,7 and 8's."

no, this is wrong.

going from one draw occurrence to the next consecutive draw occurrence is a draw difference of exactly 1.

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 Douglas Paul Smallish Jehocifer

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 18, 2016, 9:43 pm - IP Logged

we'll post more in a bit, but the average rate or draws between occurrences is 35,094,972 for the specific 6, 7, and 8 missed consecutive occurrences of 48, 52 and 43 draws simultaneously.

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 Douglas Paul Smallish Jehocifer

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 18, 2016, 10:26 pm - IP Logged

we like C for some things, but C# is better to work with.

below is Visual C# code we setup a 1,500 iteration run to find number of draws to match the conditions we have observed in the clockwise only move for 6, 7 and 8 at 48, 52 and 43 draws respectively.

the sample run we did gave us an average rate of about 35,094,972 draws to meet those conditions.

__________

using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Windows.Forms;

namespace WindowsFormsApplication1 { public partial class Form1 : Form { public Form1() { InitializeComponent(); }

private void button1_Click(object sender, EventArgs e) { //setup random generation and variables Random rnd = new Random();

int a, b, draw, i, j, ClockwiseOnly; bool ConditionNotFound; int[] ClockwiseLastDraw = new int[11]; int[] ClockwiseDrawDifference = new int[11];

//loop for iterating conditional find for (i = 1; i <= 1500; i++) { //zero out last draw and draw difference for (j = 0; j < 11; j++) { ClockwiseLastDraw[j] = 0; ClockwiseDrawDifference[j] = 0; }

//load random a (ball # 0 - 9) //initialize draw and condition a = (int)(10 * rnd.NextDouble()); draw = 2; ConditionNotFound = true;

//set loop to find number of draws to get the clockwise draw difference //of 48 draw for 6, 52 draws for 7 and 43 draws for 8 do { //load random b (ball # 0 - 9) b = (int)(10 * rnd.NextDouble());

//check if clockwise move had a previous draw occurrence if (ClockwiseLastDraw[ClockwiseOnly] != 0) { //if clockwise move had a previous draw occurrence find the draw difference and save ClockwiseDrawDifference[ClockwiseOnly] = draw - ClockwiseLastDraw[ClockwiseOnly]; }

//save last draw for the specific clockwise move ClockwiseLastDraw[ClockwiseOnly] = draw;

//move b value to a to accommodate new b value a = b;

//increment draw draw++;

//test clockwise draw difference if ((ClockwiseDrawDifference[6] >= 48) && (ClockwiseDrawDifference[7] >= 52) && (ClockwiseDrawDifference[8] >= 43)) { //if condition for clockwise draw difference of 6 is 48 and 7 is 52 and 8 is 43 //then put iteration index and draws to meet condition in text box textBox1.Text = i + "\t" + draw + "\r\n" + textBox1.Text; ConditionNotFound = !ConditionNotFound; } } while (ConditionNotFound); //end loop on condition found to get to next iteration

//update form for running visual observations Update(); } }

United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 19, 2016, 1:48 pm - IP Logged

Quote: Originally posted by JADELottery on January 18, 2016

we'll post more in a bit, but the average rate or draws between occurrences is 35,094,972 for the specific 6, 7, and 8 missed consecutive occurrences of 48, 52 and 43 draws simultaneously.

A perfect example of completely misunderstanding how probability works and obviously those different length "missed consecutive occurances" couldn't possibly have have occurred "simultaneously" as you state.

You weren't looking for a specific set of non-overlapping missed consecutive occurences of 6,7,8

You might as well point out that the probability of two consecutive Powerball drawings having this sequence of numbers:

4 8 19 27 34 10

3515261646

is 1 in 85,381,621,928,990,244

It's meaningless.

You claimed that the gap of 27 in 6,7, and 8 not appearing was evidence of a "flaw".

The odds of ANY particular 48, 52, and 43 sequence of missing/not missing combinations for 6,7,8 will have the SAME EXACT probability of occurrence.

United States Member #171734 January 11, 2016 127 Posts Offline

Posted: January 19, 2016, 2:11 pm - IP Logged

I was correct in thinking that there have been multiple algorithms used in an attempt to predict the MN Daily 3 numbers. I always appreciate the opportunity to test a clearly defined prediction so I went back and looked at the JADE One Hit Wonder predictions from Jan 2010 through April 2010.

The JADE One Hit Wonder uses an unspecified algorithm to produce 3 digits in a specific sequence and the following explanation of how to play those 3 digits is provided:

Alright, now that the One Hit Wonder Chart is here, you might be wondering yourself. What is it and what do I do with it?

'What is it?' is easy. It's a single combo for that day's draw and is derived with the same method used in the JADE Pick 3 Pick 4 Selector program by using the past few draws to come up with the combo. This one combo is special though; each digit position is important in determining what the next real draw is and isn't. This leads to the 'What do I do with it?" question.

Each digit position is designed to be useful in determining the next draw outcome. There are a set of possible matches that can help determine what kind of play you'd like to work with. One is the obvious 'exact' or 'straight' match; this is just the one combo all by itself. The others are when some or none of the positions are matching.

We'll use the Pick 3 as an example. Let's say the One Hit Combo for that day's draw is 1 2 3. Now, there are some things to look for, one is position and the other is the missing position. In the case of 1 2 3, in exact position, either all three the numbers are drawn, any two of the numbers are drawn, any one of the numbers are drawn, or none of the numbers are drawn for that day's draw. This would look something like the following:

1 2 3

X 2 3 - 1 X 3 - 1 2 X

X X 3 - X 2 X - 1 X X

X X X

X is a don't care number, but it's important to understand that X cannot have any of the One Hit Wonder numbers in those positions. If we look at the first position with the 1, either 1 hits or it doesn't. If it doesn't then it must be a number other than 1; they are 0 2 3 4 5 6 7 8 9 for the first position only. You can use the One Hit Wonder Chart to either include a particular position or exclude it.

In each of the above cases, there are a set number of combos that can be found. Below are the don't care possible combos for each don't care case.

1 2 3 - 1 combo

X 2 3 - 9 combos, 1 X 3 - 9 combos, 1 2 X - 9 combos

X X 3 - 81 combos, X 2 X - 81 combos, 1 X X - 81 combos

X X X - 729 combos

What this system does is generate a total of 271 unique combinations that are bet as straight sequence. You are a winner if a digit of the 3 predicted shows up in the exact same position in the drawn number. You can win up to 3 times (one for each digit position) so your maximum win is $1500 if you pay $1/ticket.

Your possible outcomes are:

Lose $271

Win $229

Win $729

Win $1229

Since the MN Lottery pays out at 1/2 the expected odds, each $1 you spend on MN Daily 3 tickets has an expected value of 50 cents (ie you'll lose 1/2 your money on the average bet).

Between 1/1/2010 and 4/30/2010 there were 111 JADE One Hit Wonder predictions for the 111 MN Daily 3 drawings.

Betting the JADE One Hit Wonder number for all 271 combinations (as above) during that period would have resulted in a $16,581 loss. The expected loss would have been $15040.50

Expected loss is 0.5 * (271 * 111)

So the JADE One Hit Wonder produced results that are consistent with what a random selection of predicted numbers would have produced.

Out of the 271*111 = 30,081 tickets purchased there were 27 winners.

During the 111 daily draws, there were 25 days were a single ticket won and 2 days when 2 tickets won.

In order to profit from the MN Daily 3 you have to win at a rate greater than twice the rate that random chance would predict. The JADE One was unable to exceed even what random chance would predict, much less double that.

West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline

Posted: January 19, 2016, 2:26 pm - IP Logged

Quote: Originally posted by dddwww on January 19, 2016

I was correct in thinking that there have been multiple algorithms used in an attempt to predict the MN Daily 3 numbers. I always appreciate the opportunity to test a clearly defined prediction so I went back and looked at the JADE One Hit Wonder predictions from Jan 2010 through April 2010.

The JADE One Hit Wonder uses an unspecified algorithm to produce 3 digits in a specific sequence and the following explanation of how to play those 3 digits is provided:

Alright, now that the One Hit Wonder Chart is here, you might be wondering yourself. What is it and what do I do with it?

'What is it?' is easy. It's a single combo for that day's draw and is derived with the same method used in the JADE Pick 3 Pick 4 Selector program by using the past few draws to come up with the combo. This one combo is special though; each digit position is important in determining what the next real draw is and isn't. This leads to the 'What do I do with it?" question.

Each digit position is designed to be useful in determining the next draw outcome. There are a set of possible matches that can help determine what kind of play you'd like to work with. One is the obvious 'exact' or 'straight' match; this is just the one combo all by itself. The others are when some or none of the positions are matching.

We'll use the Pick 3 as an example. Let's say the One Hit Combo for that day's draw is 1 2 3. Now, there are some things to look for, one is position and the other is the missing position. In the case of 1 2 3, in exact position, either all three the numbers are drawn, any two of the numbers are drawn, any one of the numbers are drawn, or none of the numbers are drawn for that day's draw. This would look something like the following:

1 2 3

X 2 3 - 1 X 3 - 1 2 X

X X 3 - X 2 X - 1 X X

X X X

X is a don't care number, but it's important to understand that X cannot have any of the One Hit Wonder numbers in those positions. If we look at the first position with the 1, either 1 hits or it doesn't. If it doesn't then it must be a number other than 1; they are 0 2 3 4 5 6 7 8 9 for the first position only. You can use the One Hit Wonder Chart to either include a particular position or exclude it.

In each of the above cases, there are a set number of combos that can be found. Below are the don't care possible combos for each don't care case.

1 2 3 - 1 combo

X 2 3 - 9 combos, 1 X 3 - 9 combos, 1 2 X - 9 combos

X X 3 - 81 combos, X 2 X - 81 combos, 1 X X - 81 combos

X X X - 729 combos

What this system does is generate a total of 271 unique combinations that are bet as straight sequence. You are a winner if a digit of the 3 predicted shows up in the exact same position in the drawn number. You can win up to 3 times (one for each digit position) so your maximum win is $1500 if you pay $1/ticket.

Your possible outcomes are:

Lose $271

Win $229

Win $729

Win $1229

Since the MN Lottery pays out at 1/2 the expected odds, each $1 you spend on MN Daily 3 tickets has an expected value of 50 cents (ie you'll lose 1/2 your money on the average bet).

Between 1/1/2010 and 4/30/2010 there were 111 JADE One Hit Wonder predictions for the 111 MN Daily 3 drawings.

Betting the JADE One Hit Wonder number for all 271 combinations (as above) during that period would have resulted in a $16,581 loss. The expected loss would have been $15040.50

Expected loss is 0.5 * (271 * 111)

So the JADE One Hit Wonder produced results that are consistent with what a random selection of predicted numbers would have produced.

Out of the 271*111 = 30,081 tickets purchased there were 27 winners.

During the 111 daily draws, there were 25 days were a single ticket won and 2 days when 2 tickets won.

In order to profit from the MN Daily 3 you have to win at a rate greater than twice the rate that random chance would predict. The JADE One was unable to exceed even what random chance would predict, much less double that.