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# P-4 for P-3 pairs giver system

Topic closed. 20 replies. Last post 2 years ago by retxx.

 Page 1 of 2

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 7, 2013, 5:50 pm - IP Logged

I was just glancing over the p-4 and p-3 numbers and noticed something that occurs quite regularly.

At least here in florida, If you take the the last evening P-4 winner and look at the gap differences between the numbers, a good portion of the time two of those numbers will show up in the next day's mid or eve P-3.

Example, Florida, Dec. 2012 through January 2013,

Nov. 30 eve-2572   Between 2 and 5 is a gap of 3 numbers, so the first number is 3. Between 5 and 7 is a gap of 2 numbers, so the second number is 2, and between the 7 and and 2 is a gap of 5, so the third number is 5.

So we have 325

Dec. 1 mid-533

Sometimes all three numbers produced by the P-4 show a few days down the line.

More examples below.

Dec. 2 eve-6331=302

Dec. 3 eve-312

Dec. 9 eve-9083=985

Dec. 10 mid-995

Dec. 10 eve-7391=468

Dec. 11 eve-064

Dec. 15 eve-7845=141

Dec. 16 mid-421

Dec. 17 eve-9950=045

Dec. 18 eve-407

Dec. 19 eve-4437=014

Dec. 20 eve-048

Dec. 23 eve-8890=019

Dec. 24 mid-129

Dec. 24 eve-5153=642

Dec. 25 mid-642------------------STRAIGHT HIT !

Dec. 27 eve-0103=113

Dec. 28 mid-134

Dec. 31 eve-7960=236   236

Jan. 1 mid-223         eve-367

Jan. 2 eve-7802=182

Jan. 3 eve-871

Jan. 4 eve-3562=214

Jan. 5 mid-146

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 7, 2013, 5:54 pm - IP Logged

So for tonight's eve we use last night's P-4 winner, 7754=021

I'm going to predict the 21 in the front or back. We'll see.

OHIO
United States
Member #5328
June 30, 2004
1284 Posts
Offline
 Posted: January 7, 2013, 6:47 pm - IP Logged

very nice find.

have a great winning day!

Miss Kitty

United States
Member #14
November 9, 2001
27843 Posts
Offline
 Posted: January 7, 2013, 6:51 pm - IP Logged

Thanks

love to nibble those micey feet.

OHIO
United States
Member #5328
June 30, 2004
1284 Posts
Offline
 Posted: January 7, 2013, 7:01 pm - IP Logged

works in ohio for the most part..nice tool to have!

have a great winning day!

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 7, 2013, 9:57 pm - IP Logged

Thank You All for the nice comments.

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 7, 2013, 10:01 pm - IP Logged

I was just glancing over the p-4 and p-3 numbers and noticed something that occurs quite regularly.

At least here in florida, If you take the the last evening P-4 winner and look at the gap differences between the numbers, a good portion of the time two of those numbers will show up in the next day's mid or eve P-3.

Example, Florida, Dec. 2012 through January 2013,

Nov. 30 eve-2572   Between 2 and 5 is a gap of 3 numbers, so the first number is 3. Between 5 and 7 is a gap of 2 numbers, so the second number is 2, and between the 7 and and 2 is a gap of 5, so the third number is 5.

So we have 325

Dec. 1 mid-533

Sometimes all three numbers produced by the P-4 show a few days down the line.

More examples below.

Dec. 2 eve-6331=302

Dec. 3 eve-312

Dec. 9 eve-9083=985

Dec. 10 mid-995

Dec. 10 eve-7391=468

Dec. 11 eve-064

Dec. 15 eve-7845=141

Dec. 16 mid-421

Dec. 17 eve-9950=045

Dec. 18 eve-407

Dec. 19 eve-4437=014

Dec. 20 eve-048

Dec. 23 eve-8890=019

Dec. 24 mid-129

Dec. 24 eve-5153=642

Dec. 25 mid-642------------------STRAIGHT HIT !

Dec. 27 eve-0103=113

Dec. 28 mid-134

Dec. 31 eve-7960=236   236

Jan. 1 mid-223         eve-367

Jan. 2 eve-7802=182

Jan. 3 eve-871

Jan. 4 eve-3562=214

Jan. 5 mid-146

Dec. 24 eve-5153=642

Dec. 25 mid-642------------------STRAIGHT HIT !

This is actually wrong. It should be 442. My bad !

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 7, 2013, 10:15 pm - IP Logged

The winner was 117. No cigar, but there's always tomorrow. It's been two days since a hit, so it's overdue.

United States
Member #137125
December 27, 2012
15 Posts
Offline
 Posted: January 8, 2013, 9:58 am - IP Logged

I was just glancing over the p-4 and p-3 numbers and noticed something that occurs quite regularly.

At least here in florida, If you take the the last evening P-4 winner and look at the gap differences between the numbers, a good portion of the time two of those numbers will show up in the next day's mid or eve P-3.

Example, Florida, Dec. 2012 through January 2013,

Nov. 30 eve-2572   Between 2 and 5 is a gap of 3 numbers, so the first number is 3. Between 5 and 7 is a gap of 2 numbers, so the second number is 2, and between the 7 and and 2 is a gap of 5, so the third number is 5.

So we have 325

Dec. 1 mid-533

Sometimes all three numbers produced by the P-4 show a few days down the line.

More examples below.

Dec. 2 eve-6331=302

Dec. 3 eve-312

Dec. 9 eve-9083=985

Dec. 10 mid-995

Dec. 10 eve-7391=468

Dec. 11 eve-064

Dec. 15 eve-7845=141

Dec. 16 mid-421

Dec. 17 eve-9950=045

Dec. 18 eve-407

Dec. 19 eve-4437=014

Dec. 20 eve-048

Dec. 23 eve-8890=019

Dec. 24 mid-129

Dec. 24 eve-5153=642

Dec. 25 mid-642------------------STRAIGHT HIT !

Dec. 27 eve-0103=113

Dec. 28 mid-134

Dec. 31 eve-7960=236   236

Jan. 1 mid-223         eve-367

Jan. 2 eve-7802=182

Jan. 3 eve-871

Jan. 4 eve-3562=214

Jan. 5 mid-146

INSTANT NEXT NUMBERS FOR P4 GAMES

EX: CA JAN 1 2013 NUMBERS 4539

4539 PLUS MIRROR

4539=9084

USING TOTAL IN P4 GAMES 9999 SUBTRACT BOTH THE WINNING AND MIRROR NUMBERS

9999                                    9999

- 4539                                - 9084

5460                                     0915

NOW NUMBERS THAT ARE REPEAT SHOULD BE OUT

45 OF 4539            90 OF 9084

54 OF 5460             09 OF 0915

NOW I HAD 39 AND 60                     84 AND 15 LEFT

SINCE 39 AND 84 ARE ON TOP OF BOTH 60 AND 15

3 OF 39 SHOULD BE WITH 6 OF 60

9 OF 39 WITH 0 OF 60

8 OF 84 WITH 1 OF 15

4 OF 84 WITH 5 OF 15

NOW I WILL HAD 3-6 9-0 8-1 4-5

SO WHAT I DO IS NOW SUBTRACT THE RIGHT NUMBERS TO THE LEFT NUMBERS

3-6 IS 6 MINUS 3= 3 SINCE 0 CANT SUBTRACT ABY NUMBERS I WILL CARRY 1 TO IT 0 + 1 = 10 NOW 10 MINUS 9= 1

IF ANY NUMBERS OF RIGHT SIDES IS SMALLER THAN LEFT SIDES JUST CARRY 1 TO IT

SO 11 MINUS 8= 3 AND 5 MINUS 4= 1

NOW I HAD THE NUMBERS 3131

SO WHAT I DO IS ADD THAT NUMBERS TO THE WINNING NUMBERS OF 4539 ON JAN 1 2013 OF CA

4539

+ 3131

7670 THIS JUST REGULAR MATH

NOW SINCE I HAD 3131 ALL I NEED IS A PAIR OF 2 NUMBERS AND ADD THEM TOGTHER SO 3 + 1 = 4

THE 7670 PLUS 4

I GET 7674 WHICH IS THE NEXT WINNING NUMBER IN CA

A STRAIGHT AUTOMATC WIN INSTANTLY

FINALLY I BEAT THE LOTTO SYSTEM OF CA

TAKE THAT SYSTEM AND THAT TOO LOL.

I THINK IT MAY WORK IN P3 TOO

I HAVENT TEST IT OUT YET ON P3 GAMES

SO THANKS ALOT

ANY SUGGESTIONS TO MAKE THIS SECRET TECHNIQUE MORE PERFECT IS ACCEPTED

THANKS

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 8, 2013, 10:05 am - IP Logged

INSTANT NEXT NUMBERS FOR P4 GAMES

EX: CA JAN 1 2013 NUMBERS 4539

4539 PLUS MIRROR

4539=9084

USING TOTAL IN P4 GAMES 9999 SUBTRACT BOTH THE WINNING AND MIRROR NUMBERS

9999                                    9999

- 4539                                - 9084

5460                                     0915

NOW NUMBERS THAT ARE REPEAT SHOULD BE OUT

45 OF 4539            90 OF 9084

54 OF 5460             09 OF 0915

NOW I HAD 39 AND 60                     84 AND 15 LEFT

SINCE 39 AND 84 ARE ON TOP OF BOTH 60 AND 15

3 OF 39 SHOULD BE WITH 6 OF 60

9 OF 39 WITH 0 OF 60

8 OF 84 WITH 1 OF 15

4 OF 84 WITH 5 OF 15

NOW I WILL HAD 3-6 9-0 8-1 4-5

SO WHAT I DO IS NOW SUBTRACT THE RIGHT NUMBERS TO THE LEFT NUMBERS

3-6 IS 6 MINUS 3= 3 SINCE 0 CANT SUBTRACT ABY NUMBERS I WILL CARRY 1 TO IT 0 + 1 = 10 NOW 10 MINUS 9= 1

IF ANY NUMBERS OF RIGHT SIDES IS SMALLER THAN LEFT SIDES JUST CARRY 1 TO IT

SO 11 MINUS 8= 3 AND 5 MINUS 4= 1

NOW I HAD THE NUMBERS 3131

SO WHAT I DO IS ADD THAT NUMBERS TO THE WINNING NUMBERS OF 4539 ON JAN 1 2013 OF CA

4539

+ 3131

7670 THIS JUST REGULAR MATH

NOW SINCE I HAD 3131 ALL I NEED IS A PAIR OF 2 NUMBERS AND ADD THEM TOGTHER SO 3 + 1 = 4

THE 7670 PLUS 4

I GET 7674 WHICH IS THE NEXT WINNING NUMBER IN CA

A STRAIGHT AUTOMATC WIN INSTANTLY

FINALLY I BEAT THE LOTTO SYSTEM OF CA

TAKE THAT SYSTEM AND THAT TOO LOL.

I THINK IT MAY WORK IN P3 TOO

I HAVENT TEST IT OUT YET ON P3 GAMES

SO THANKS ALOT

ANY SUGGESTIONS TO MAKE THIS SECRET TECHNIQUE MORE PERFECT IS ACCEPTED

THANKS

Thanks, i will try to absorb the workout but i must first get past the first hill by asking how you got 9084.

Shouldn't it be 9584?

The mirror of 5 is 0, so 5+0=5, no? or am i doing it wrong?

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 8, 2013, 10:38 am - IP Logged

INSTANT NEXT NUMBERS FOR P4 GAMES

EX: CA JAN 1 2013 NUMBERS 4539

4539 PLUS MIRROR

4539=9084

USING TOTAL IN P4 GAMES 9999 SUBTRACT BOTH THE WINNING AND MIRROR NUMBERS

9999                                    9999

- 4539                                - 9084

5460                                     0915

NOW NUMBERS THAT ARE REPEAT SHOULD BE OUT

45 OF 4539            90 OF 9084

54 OF 5460             09 OF 0915

NOW I HAD 39 AND 60                     84 AND 15 LEFT

SINCE 39 AND 84 ARE ON TOP OF BOTH 60 AND 15

3 OF 39 SHOULD BE WITH 6 OF 60

9 OF 39 WITH 0 OF 60

8 OF 84 WITH 1 OF 15

4 OF 84 WITH 5 OF 15

NOW I WILL HAD 3-6 9-0 8-1 4-5

SO WHAT I DO IS NOW SUBTRACT THE RIGHT NUMBERS TO THE LEFT NUMBERS

3-6 IS 6 MINUS 3= 3 SINCE 0 CANT SUBTRACT ABY NUMBERS I WILL CARRY 1 TO IT 0 + 1 = 10 NOW 10 MINUS 9= 1

IF ANY NUMBERS OF RIGHT SIDES IS SMALLER THAN LEFT SIDES JUST CARRY 1 TO IT

SO 11 MINUS 8= 3 AND 5 MINUS 4= 1

NOW I HAD THE NUMBERS 3131

SO WHAT I DO IS ADD THAT NUMBERS TO THE WINNING NUMBERS OF 4539 ON JAN 1 2013 OF CA

4539

+ 3131

7670 THIS JUST REGULAR MATH

NOW SINCE I HAD 3131 ALL I NEED IS A PAIR OF 2 NUMBERS AND ADD THEM TOGTHER SO 3 + 1 = 4

THE 7670 PLUS 4

I GET 7674 WHICH IS THE NEXT WINNING NUMBER IN CA

A STRAIGHT AUTOMATC WIN INSTANTLY

FINALLY I BEAT THE LOTTO SYSTEM OF CA

TAKE THAT SYSTEM AND THAT TOO LOL.

I THINK IT MAY WORK IN P3 TOO

I HAVENT TEST IT OUT YET ON P3 GAMES

SO THANKS ALOT

ANY SUGGESTIONS TO MAKE THIS SECRET TECHNIQUE MORE PERFECT IS ACCEPTED

THANKS

I also ran into another problem. Using Florida Jan. 1 as the starting point eve-4194

4914+ mirror=3373

9999 minus 4914=5085

9999 minus 3373=6626

Now here's where I run into a snag. You say to eliminate any repeating pairs, so let's start with 4914. I don't see any repeating pairs, so how would I conitnue? Unless you count 49 Twice, like 4914     4914

And the same problem with 3373, 5085, and 6626. I can't continue until I get a resolution for the first problem with the mirror, and then this one with the pairs.

I'll wait for your response. Thanks.

United States
Member #137433
January 3, 2013
469 Posts
Offline
 Posted: January 8, 2013, 10:44 am - IP Logged

Works in GA.  Thankts

United States
Member #137125
December 27, 2012
15 Posts
Offline
 Posted: January 8, 2013, 12:25 pm - IP Logged

Thanks, i will try to absorb the workout but i must first get past the first hill by asking how you got 9084.

Shouldn't it be 9584?

The mirror of 5 is 0, so 5+0=5, no? or am i doing it wrong?

WHAT I MEANT BY 4539 PLUS MIRROR

IS THAT I MIRROR THE NUMBER 4539

TO ITS MIRROR NUMBER

9084

United States
Member #128795
June 2, 2012
4857 Posts
Online
 Posted: January 8, 2013, 12:32 pm - IP Logged

WHAT I MEANT BY 4539 PLUS MIRROR

IS THAT I MIRROR THE NUMBER 4539

TO ITS MIRROR NUMBER

9084

Got it, thanks. Now to the second problem of not finding any duplicate pairs. Thanks.

United States
Member #137125
December 27, 2012
15 Posts
Offline
 Posted: January 8, 2013, 1:08 pm - IP Logged

I also ran into another problem. Using Florida Jan. 1 as the starting point eve-4194

4914+ mirror=3373

9999 minus 4914=5085

9999 minus 3373=6626

Now here's where I run into a snag. You say to eliminate any repeating pairs, so let's start with 4914. I don't see any repeating pairs, so how would I conitnue? Unless you count 49 Twice, like 4914     4914

And the same problem with 3373, 5085, and 6626. I can't continue until I get a resolution for the first problem with the mirror, and then this one with the pairs.

I'll wait for your response. Thanks.

OK I SEE

WELL FOR ME

IT SEEMS TO WORK IN CA LOTTERY

BUT I NOT SURE YET ON THE REST OF THE PREVIOUS NUMBERS CUX I ONLY HAD THIS IDEA SINCE YESTERDAY ONLY

AND HAD TESTED IT OUT ON THE JAN 1 P4 GAME "ONCE" ONLY IN CA

IT SEEMS TO WORK FOR THIS SINGLE DIGITS ON JAN 1 ONLY

RIGHT NOW I TRYING TO TEST OUT TO SEE IF I CAN GET AT LEAST 3 OUT OF 4 NEXT WINNING NUMBERS

AND DO SAME METHOD TO SEE IF I CAN GET THE NEXT 1 SINGLE WINNING NUMBER TO BE WITH THE 3 NEXT WINNING NUMBERS

THATS WHAT I AM GOING TO DO AT LEAST NOW.

WELL WHAT I DID WAS USING 9999 MINUS 4539 AND GET 5460 IN HERE

9999

-4539

5460

NOTICED: SINCE THERE IS THE PAIR OF 45 IN 4539 AND 54 OF 5460.

U TAKE THEM OUT FROM 4 TO 4 AND 5 TO 5 CRISS CROSS TAKE OUTS

I GUESS U CAN  SAY IT THE BIG     X        TAKE OUTS

BECUX IT HAD TWO OF THE SAME NUMBERS WHICH IS 4 AND 5

WHICH LEFT ONLY THE 3 AND 9 OF 4539 ON TOP OF RIGHT SIDES

NEXT TO THE NUMBER 4 AND 5 OF 4539 WHICH WAS TAKEN OUT BY CRISS CROSS

THE BOTTOM RIGHT SIDES WITH ONLY THE 6 AND 0 OF 5460

THE SAME IS FOR THE MIRROR NUMBERS AS WELL

WITH 8 AND 4 OF 9084 ON TOP RIGHT SIDES AND 1 AND 5 OF BOTTOM RIGHT SIDES

9999                                9999

- 4539                            - 9084

54 60                               09 15

SO WHAT I DID NOW IS GROUP THEM ALL TOGETHER LIKE THIS:

3 TO 6

9 TO 0

8 TO 1

4 TO 5

NOW NEXT I SUBTRACT NUMBERS FROM RIGHT SIDES TO LEFT

6 MINUS 3 = 3

SINCE 0 CANT SUBTRACT ANY NUMBERS AT ALL I JUST CARRY THE ONE TO THE ZERO MAKING IT A TEN 10

SO 10 MINUS 9 = 1

IF ANY NUMBERS FROM RIGHT SIDES IS LESS THAN LEFT SIDES NUMBERS JUST CARRY 1 TO IT

SO IT CAN SUBTRACT THE LEFT SIDES

NOW NEXT IS 1 MINUS 8 SO CARRY 1 TO IT

1 NOW IS 11 MINUS 8= 3

5 MINUS 4 = 1

SO I GET THE NUMBERS 3131

SE 3131 AND ADDED TO THE WINNING NUMBERS OF 4539

U GET 7670

THEN USE THE 3131 AGAIN BUT THIS TIME ONLY A PAIR SO I GET

31, ADD THEM I GET 4

SO I ADD 4 TO THE RESULT OF 7670 TO BE

7674 THE NEXT STRAIGHT WINNING NUMBERS FOR JAN 2 2013 IN CALIFORNIA.

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