United States Member #164727 March 12, 2015 2536 Posts Offline

Posted: October 25, 2016, 10:10 pm - IP Logged

Aware of the premise that most of the time at least one number from the previous winner shows in the next drawing has given me an idea that I never thought of until recently, and it was staring me in the face all this time...lol. So simple I overlooked it.

Here are the rules/filers:

A: No pairs from the last TWO draws. For example, 123 has 3 pairs...12-13-23. None of those pairs should be in your selections for the next drawing.

B: No combinations that are all high, or all low numbers.

I used the last winning number midday in Florida today. The winner was 973, so I used the 9, the 7, and the 3.

Starting with the number 9 in 973, I realized there are only a certain amount of pairs hat can be made, thus reducing the cost. And no doubles...... For example...using the 9 example again, you have 09-19-29-39-49-59-69-79-89

So for our first group we use the 09. Our last two winning numbers in Florida were 194 and 973, so no pairs can be included from those two latest winning numbers.

Let's start the list for the first group,

The 9=09-19-29-39-49-59-69-79-89

First the 09....so now we have one of the all possible pairs for the number 9, we must find the third digit. We don't want doubles, all low, all high, all even, all odd, so we use those filters when we make combos out of those pairs.

Okay, our first pair is 09. We need a third digit. Let's start with 0. We can't use 0 because that would be 090, and that's a double, DISQUALIFIED!

Next number to pair with 09 is 1. 091 is also DISQUALIFIED...why? because one of the rules states that no pairs from the previous two winners can be included. Two draws prior, the winner was 194, and there goes the 19.

Next is the 2 paired with the 09. 092 is a qualified combo to keep because it violates no rules whatsoever.

So we have 092. how about 093? Yes. it also qualifies, no rules broken.

092-093...094?...NOPE...why? because the pair in 094...94 was in last night's winner...194

So we have 092 and 093...

Let's move on to 095..."Which btw was tonight's Florida winner which was on my list of predictions). of course 095 was kept for not breaking any of those rules stated above.

096? Yes, no rules broken....092-093-095-096

097?...NOPE, today's mid winner was 973, the 97 pair disqualifies it.

098? That's good to go, and finally 099 which is a no-no because it's a double.

So far for the first group we have 092-093-095-096.

Do this with the other two numbers to get all the numbers.

United States Member #164727 March 12, 2015 2536 Posts Offline

Posted: October 25, 2016, 10:19 pm - IP Logged

Forgot to mention that after doing the 09 workout, you'll also have to do the 19 workout, then the 29-39..etc...for that first group. I know it seems like a lot of numbers, but trust me, once you apply the 4 filters that occur so rarely in general, you'll end up with about 35 combos altogether.

OKC, OK United States Member #120575 December 22, 2011 824 Posts Offline

Posted: October 25, 2016, 11:57 pm - IP Logged

Quote: Originally posted by amber123 on October 25, 2016

Forgot to mention that after doing the 09 workout, you'll also have to do the 19 workout, then the 29-39..etc...for that first group. I know it seems like a lot of numbers, but trust me, once you apply the 4 filters that occur so rarely in general, you'll end up with about 35 combos altogether.

Thanks Amber.

This is what they call "a single digit returned", there is also a "two-digit returned". Her is the general concept:

Example (a,b,c) is previous drawing result: For a single digit returned the next drawing (exclude double and triple) HAS to be in one of the following groups:

1) a&[all pairs of[(0:9)-(a,b,c)] This means: Remove the last drawing digits (a,b,c) from the 10 digits 0 to 9, take the remaining to form all possible pairs and

finally combine those pairs with the digit a.

2) b&[all pairs of (0:9)-(a,b,c)] Apply the same concept as above using digit b

3) c&[all pairs of (0:9)-(a,b,c)] Apply the same concept as above using the digit c

Finally you can apply your preferred filter at your own risk

For two- digit returned: Form the 3 pairs: ab,ac, bc

then

1) ab&[all single digits of(0:9)-(a,b,c)] this means to combine the pair (ab) with each of the remaining 7 single digits

2) ac&[all single digits of (0:9)-(a,b,c)] do the same as above

3) bc&[all single digits of (0:9)-(a,b,c)] .

An example for single digit returned

On 10/23 drawing was 251 next day 10/24 the drawing was 583

Let say on 10/24 you want to play a single digit returned. You would do this:

1) 2&[ all pairs of(0:9)-(2,5,1)] =2&(all pairs of (0,3,4,6,7,8))

2) 5&(all pairs of (0,3,4,6,7,8) the winning number 583 is in this group

3) 1&(all pairs of (0,3,4,6,7,8)

Of course you would apply your preferred filter to reduce the picks.

ORLANDO, FLORIDA United States Member #4924 June 3, 2004 5914 Posts Online

Posted: October 26, 2016, 6:39 am - IP Logged

Quote: Originally posted by pavizlo$ on October 25, 2016

Thanks Amber.

This is what they call "a single digit returned", there is also a "two-digit returned". Her is the general concept:

Example (a,b,c) is previous drawing result: For a single digit returned the next drawing (exclude double and triple) HAS to be in one of the following groups:

1) a&[all pairs of[(0:9)-(a,b,c)] This means: Remove the last drawing digits (a,b,c) from the 10 digits 0 to 9, take the remaining to form all possible pairs and

finally combine those pairs with the digit a.

2) b&[all pairs of (0:9)-(a,b,c)] Apply the same concept as above using digit b

3) c&[all pairs of (0:9)-(a,b,c)] Apply the same concept as above using the digit c

Finally you can apply your preferred filter at your own risk

For two- digit returned: Form the 3 pairs: ab,ac, bc

then

1) ab&[all single digits of(0:9)-(a,b,c)] this means to combine the pair (ab) with each of the remaining 7 single digits

2) ac&[all single digits of (0:9)-(a,b,c)] do the same as above

3) bc&[all single digits of (0:9)-(a,b,c)] .

An example for single digit returned

On 10/23 drawing was 251 next day 10/24 the drawing was 583

Let say on 10/24 you want to play a single digit returned. You would do this:

1) 2&[ all pairs of(0:9)-(2,5,1)] =2&(all pairs of (0,3,4,6,7,8))

2) 5&(all pairs of (0,3,4,6,7,8) the winning number 583 is in this group

3) 1&(all pairs of (0,3,4,6,7,8)

Of course you would apply your preferred filter to reduce the picks.

ORLANDO, FLORIDA United States Member #4924 June 3, 2004 5914 Posts Online

Posted: October 26, 2016, 6:43 am - IP Logged

Quote: Originally posted by amber123 on October 25, 2016

Aware of the premise that most of the time at least one number from the previous winner shows in the next drawing has given me an idea that I never thought of until recently, and it was staring me in the face all this time...lol. So simple I overlooked it.

Here are the rules/filers:

A: No pairs from the last TWO draws. For example, 123 has 3 pairs...12-13-23. None of those pairs should be in your selections for the next drawing.

B: No combinations that are all high, or all low numbers.

I used the last winning number midday in Florida today. The winner was 973, so I used the 9, the 7, and the 3.

Starting with the number 9 in 973, I realized there are only a certain amount of pairs hat can be made, thus reducing the cost. And no doubles...... For example...using the 9 example again, you have 09-19-29-39-49-59-69-79-89

So for our first group we use the 09. Our last two winning numbers in Florida were 194 and 973, so no pairs can be included from those two latest winning numbers.

Let's start the list for the first group,

The 9=09-19-29-39-49-59-69-79-89

First the 09....so now we have one of the all possible pairs for the number 9, we must find the third digit. We don't want doubles, all low, all high, all even, all odd, so we use those filters when we make combos out of those pairs.

Okay, our first pair is 09. We need a third digit. Let's start with 0. We can't use 0 because that would be 090, and that's a double, DISQUALIFIED!

Next number to pair with 09 is 1. 091 is also DISQUALIFIED...why? because one of the rules states that no pairs from the previous two winners can be included. Two draws prior, the winner was 194, and there goes the 19.

Next is the 2 paired with the 09. 092 is a qualified combo to keep because it violates no rules whatsoever.

So we have 092. how about 093? Yes. it also qualifies, no rules broken.

092-093...094?...NOPE...why? because the pair in 094...94 was in last night's winner...194

So we have 092 and 093...

Let's move on to 095..."Which btw was tonight's Florida winner which was on my list of predictions). of course 095 was kept for not breaking any of those rules stated above.

096? Yes, no rules broken....092-093-095-096

097?...NOPE, today's mid winner was 973, the 97 pair disqualifies it.

098? That's good to go, and finally 099 which is a no-no because it's a double.

So far for the first group we have 092-093-095-096.

Do this with the other two numbers to get all the numbers.

amber, please clarify the statement below

A: No pairs from the last TWO draws. For example, 123 has 3 pairs...12-13-23. None of those pairs should be in your selections for the next drawing.

if that is true, should there be another set of pairs?

United States Member #164727 March 12, 2015 2536 Posts Offline

Posted: October 26, 2016, 11:28 am - IP Logged

Quote: Originally posted by CARBOB on October 26, 2016

amber, please clarify the statement below

A: No pairs from the last TWO draws. For example, 123 has 3 pairs...12-13-23. None of those pairs should be in your selections for the next drawing.

if that is true, should there be another set of pairs?

No, I think I should clarify.

Yesterday's Florida winning numbers were 973 and 059. 973 pairs=97 93 73.....059=05 09 59

After you make your combos, there should be NO combos on your list that have 97 93 73 05 09 59 in them.

For example, one of my combos at the end of my workout is 327. This combo should be deleted from my list because there's a 73 in it. It's basically another filter.

ORLANDO, FLORIDA United States Member #4924 June 3, 2004 5914 Posts Online

Posted: October 26, 2016, 12:10 pm - IP Logged

Quote: Originally posted by amber123 on October 26, 2016

No, I think I should clarify.

Yesterday's Florida winning numbers were 973 and 059. 973 pairs=97 93 73.....059=05 09 59

After you make your combos, there should be NO combos on your list that have 97 93 73 05 09 59 in them.

For example, one of my combos at the end of my workout is 327. This combo should be deleted from my list because there's a 73 in it. It's basically another filter.

Broken Arrow, OK United States Member #155206 May 11, 2014 1875 Posts Offline

Posted: October 26, 2016, 2:57 pm - IP Logged

Hey Amber, great system, and thanks for sharing. Only one thing that I'm confused about....let's say the last two drawings were 123 and 456. Now, I understand that none of the PAIRS from either of these drawing can be used again in the workout, but do you get your PICKS from both of these drawings or only from the LAST drawing. in other words, would I ONLY use these:

10 thru 19

20 thru 29

30 thru 39

or would I ALSO use 40 thru 49, 50 thru 59, and 60 thru 69 to create picks for the next drawing? When I use BOTH sets, I come up with an average of about 46 combos to play, but only around 35 combos if I ONLY use the first set? Thanks again!

"What lies behind us and what lies before us are tiny matters compared to what lies within us"

Zaperlopopotam Belgium Member #173932 March 26, 2016 979 Posts Offline

Posted: October 26, 2016, 3:17 pm - IP Logged

Quote: Originally posted by amber123 on October 25, 2016

Aware of the premise that most of the time at least one number from the previous winner shows in the next drawing has given me an idea that I never thought of until recently, and it was staring me in the face all this time...lol. So simple I overlooked it.

Here are the rules/filers:

A: No pairs from the last TWO draws. For example, 123 has 3 pairs...12-13-23. None of those pairs should be in your selections for the next drawing.

B: No combinations that are all high, or all low numbers.

I used the last winning number midday in Florida today. The winner was 973, so I used the 9, the 7, and the 3.

Starting with the number 9 in 973, I realized there are only a certain amount of pairs hat can be made, thus reducing the cost. And no doubles...... For example...using the 9 example again, you have 09-19-29-39-49-59-69-79-89

So for our first group we use the 09. Our last two winning numbers in Florida were 194 and 973, so no pairs can be included from those two latest winning numbers.

Let's start the list for the first group,

The 9=09-19-29-39-49-59-69-79-89

First the 09....so now we have one of the all possible pairs for the number 9, we must find the third digit. We don't want doubles, all low, all high, all even, all odd, so we use those filters when we make combos out of those pairs.

Okay, our first pair is 09. We need a third digit. Let's start with 0. We can't use 0 because that would be 090, and that's a double, DISQUALIFIED!

Next number to pair with 09 is 1. 091 is also DISQUALIFIED...why? because one of the rules states that no pairs from the previous two winners can be included. Two draws prior, the winner was 194, and there goes the 19.

Next is the 2 paired with the 09. 092 is a qualified combo to keep because it violates no rules whatsoever.

So we have 092. how about 093? Yes. it also qualifies, no rules broken.

092-093...094?...NOPE...why? because the pair in 094...94 was in last night's winner...194

So we have 092 and 093...

Let's move on to 095..."Which btw was tonight's Florida winner which was on my list of predictions). of course 095 was kept for not breaking any of those rules stated above.

096? Yes, no rules broken....092-093-095-096

097?...NOPE, today's mid winner was 973, the 97 pair disqualifies it.

098? That's good to go, and finally 099 which is a no-no because it's a double.

So far for the first group we have 092-093-095-096.

Do this with the other two numbers to get all the numbers.

a) 123 456 eliminate all doubles from listed. b)c)d) done