NOW : 2/3 MEANS 2 OUT OF 3 DIGITS HITS 80% WHEN PLAYED ALL;
MY SUGGESTION:
DOES ANY ONE HAS AN IDEA HOW TO CREATE A SYSTEM WITH LESS COMBINATIONS TO MAKE IT PROFITABLE? USING THIS SET OF TEN NUMBERS?
THANKS ALL
There are several different ways to create group of 3 digit combos using all 10 digits that has a 2 out of 3 guarantee and I believe it's possible to get the same guarantee using less than 10 combos. Since 72% of all combos have 3 different digits and over time will be drawn about 72% time, combined with the 2 matches from some of the doubles the other 28% percent of the drawings, your 80% is pretty accurate.
Most pick-3 games have no payoff for matching 2 digits so we have to find another method for practical play. There is a method using all 10 digits that arranges them into 10 combos with the same 2 out of 3 guarantee where the player can decide which order to place the 10 digits that is practical for play.
Create an ordered list of the 10 digits using any method placing in order of most favored digits to least favored digits. It might look something like this:
4-7-0-3-6-9-2-8-5-1
A-B-C-D-E-F-G-H-I-J
The 10 combos are:
A-B-C 407
A-D-E 436
A-F-G 492
A-H-I 481
B-D-F 039
B-E-G 062
B-H-J 081
C-D-G 732
C-E-F 769
C-I-J 751
Exchange the letters with the digits and you'll have 10 combos with match a 2 guarantee and if the digits are placed in the correct order, match all 3. 407-436-492-481-039-062-081-732-769-751 will have a boxed 2 digit match in all 720 combos with 3 different digits.
This would be a boxed hit but a player could get creative and get straight hits by setting the digit order by using the 3 digits they felt were most likely to be in first position for A, B, and C. The digits G and J are always in the third position, the digits F twice , and C, E, and I once. It depends on how the ordered list of 10 digits is choosen, but it's possible to match all 3 digits straight everytime 3 different digits are drawn.
Getting an $8 for $1 payoff by playing 10 boxes doesn't sound very exciting, but by finding a way to arrange the ordered list to get two boxed hits our of every 8 drawings could be very profitable by raising the bet.
Germany
Member #62,825
July 10, 2008
5 Posts
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Quote: Originally posted by Stack47 on Jul 15, 2008
There are several different ways to create group of 3 digit combos using all 10 digits that has a 2 out of 3 guarantee and I believe it's possible to get the same guarantee using less than 10 combos. Since 72% of all combos have 3 different digits and over time will be drawn about 72% time, combined with the 2 matches from some of the doubles the other 28% percent of the drawings, your 80% is pretty accurate.
Most pick-3 games have no payoff for matching 2 digits so we have to find another method for practical play. There is a method using all 10 digits that arranges them into 10 combos with the same 2 out of 3 guarantee where the player can decide which order to place the 10 digits that is practical for play.
Create an ordered list of the 10 digits using any method placing in order of most favored digits to least favored digits. It might look something like this:
4-7-0-3-6-9-2-8-5-1
A-B-C-D-E-F-G-H-I-J
The 10 combos are:
A-B-C 407
A-D-E 436
A-F-G 492
A-H-I 481
B-D-F 039
B-E-G 062
B-H-J 081
C-D-G 732
C-E-F 769
C-I-J 751
Exchange the letters with the digits and you'll have 10 combos with match a 2 guarantee and if the digits are placed in the correct order, match all 3. 407-436-492-481-039-062-081-732-769-751 will have a boxed 2 digit match in all 720 combos with 3 different digits.
This would be a boxed hit but a player could get creative and get straight hits by setting the digit order by using the 3 digits they felt were most likely to be in first position for A, B, and C. The digits G and J are always in the third position, the digits F twice , and C, E, and I once. It depends on how the ordered list of 10 digits is choosen, but it's possible to match all 3 digits straight everytime 3 different digits are drawn.
Getting an $8 for $1 payoff by playing 10 boxes doesn't sound very exciting, but by finding a way to arrange the ordered list to get two boxed hits our of every 8 drawings could be very profitable by raising the bet.
are 91.6% of a 2if3of10 lines cover or wheel which as Stack47 pointed out can be closed to 100% in 8 lines.
When you say 2 of the digits are correct 80% of the time, I wonder how often they are consecutive 8-9 vs. non consecutive 7-9. You could make a system out of consecutive pairs or a pair and a third digit.
If we drop two numbers and optimize the rest we get . . .
120, 189, 289, 347, 356, 456, 567, 890
Close as we can get it and still keep the cover 100%
You will have at least two digits correct in one number every time three different digits are drawn. Because only 8 possible digits remain to take the place of the third digit necessary to a win, your odds become 1 in 8 of a box win when all different digits are drawn.
are 91.6% of a 2if3of10 lines cover or wheel which as Stack47 pointed out can be closed to 100% in 8 lines.
When you say 2 of the digits are correct 80% of the time, I wonder how often they are consecutive 8-9 vs. non consecutive 7-9. You could make a system out of consecutive pairs or a pair and a third digit.
If we drop two numbers and optimize the rest we get . . .
120, 189, 289, 347, 356, 456, 567, 890
Close as we can get it and still keep the cover 100%
You will have at least two digits correct in one number every time three different digits are drawn. Because only 8 possible digits remain to take the place of the third digit necessary to a win, your odds become 1 in 8 of a box win when all different digits are drawn.
BobP
First Bob P, I am eternally grateful to you. your stuff on wheeling and lottery systems is THE BEST on the net.
I have all the pleasure to read you again and again,and let me thank you again for replying to my post.
just to tell,take it for fun, if i hit big, i won´t get you only abox of beer, as you mention in your previous posts, you deserve more.....
best wishes from BERLIN Germany, you see you have admirers from All the world.