winning system discussion

[grayth]

Hey everyone,

I've been crunching same as everyone else and I'm trying to nail down a stratedgy....and thought i would run it by you guys....

and see what we come up with..

From the data I've seen for ontario pick3 there is a 50% chance that any 1 of the last 3 drawn numbers will show up in the next draw. This is true historically speaking. Also I ran a test and found that there is a 75% chance that any 1 number drawn from the last draw will show up within 2 draws.

The problem lies in that you still have a 30% chance to guess the right number out of the three drawn numbers....when i ran a test and just choose the first digit i lost more time than won.

But if you chose 2 numbers out of the 3 you would have a 66% chance you chose correctly and if you ran the combinations for it for no more than 2 draws 75% of the time on average you'll win....

That being the case and there being 1000 combinations would that mean that each number only has 100 possible combinations...I don't think so cause as i study the sequential numbers....there is 100 combos with 0 digit in number 1 spot and then even more for the second spot etc for each number 1-9 so does that mean there are 1000 combo's for each numer to fall in any slot?

I trying to work out a way to lessen the possible combos given this occurance fact and see if there is a way to bring it down to 100 combos or so...

questions...theories...flames etc to this strat..?

Troy

 

[LT]

Hi grayth ... welcome to the forums

Here are some thoughts re the use of the last three numbers .. which one to pick?

First some background info:

Pick3 numbers with 3 unique digits are drawn appox 72% of the time.
Doubles are drawn approx 27% of the time.
Triples are drawn approx 1% of the time.

Pick3 numbers with 2 even and 1 odd digit or 1 even and 2 odd
are drawn aprox 71% of the time.

Pick3 boxed numbers with a sum between 12 and 15 make up more
than half the numbers and therefore are drawn most often.

While "cold" concepts do not work with a 649 type lottery. "dueness" is very applicable to P3 and 4 predictions.

Find which sums between 12 and 15 are "due" (the longest out). Pick a couple of sum sets.

Eliminate all double and triple combos.

Eliminate all odd and all even combos.

Now look for sets that contain the numbers from the last draw and play those sets.

If you wanted to really tune it in find the coldest number (out of all three positions) between 0-9 and play only those sets as described above but the combos must now have one of the past three numbers as well as the overall longest out number and the sum must be a due sum between 12 and 15.

That should make for some winners

LT

 

[charles2]

Assuming Boxed picks not straight and not including double or triple numbers then you have to buy this many tickets for this amount of numbers to cover:

to cover 4 numbers, say,0-1-2-3 = need to buy 4 tickets combinations ***** best bets since winnings are usually between about $100 to $60 *****

to cover 5 numbers, say,0-1-2-3-4 = need to buy 10 tickets combinations ***** best bets since winnings are usually between about $100 to $60 *****

to cover 6 numbers, say,0-1-2-3-4-5 = need to buy 20 tickets combinations ***** best bets since winnings are usually between about $100 to $60 *****

to cover 7 numbers, say,0-1-2-3-4-5-6 = need to buy 35 tickets combinations ***** best bets since winnings are usually between about $100 to $60 *****

to cover 8 numbers, say,0-1-2-3-4-5-6-7 = need to buy 56 tickets combinations ***** best bets since winnings are usually between about $100 to $60 *****

to cover 9 numbers, say,0-1-2-3-4-5-6-7-8 = need to buy 84 tickets combinations

to cover 10 [this is all this numbers and winnings are usually between about $100 to $60 so i don't think it's worth it] numbers, say,0-1-2-3-4-5-6-7-8-9 = need to buy 120 tickets combinations

i like using the 4 or 5 or or 7 number methods and you should win often

 

[grayth]

Thanks for the tips LT I'm building a spreadsheet to track the sum hits and skips...

Charles.....

Do you have a spreadsheet that creates the combos on the fly based on what amount of numbers you want to cover??

Also it would appear you pretty much always play boxed?

I assume because of the lower odds...but also the winnings are lower as well.

Troy

 

[charles2]

Substitute with your numbers you want to play [i hope it isn't to long again )] Good Luck

ya i never play straight cause i never win

Boxed

u'r 4 #'s
4 tickets

1-2-3
1-2-4
1-3-4
2-3-4

u'r 5 #'s
10 tickets

1-2-3
1-2-4
1-2-5
1-3-4
1-3-5
1-4-5
2-3-4
2-3-5
2-4-5
3-4-5

u'r 6 #'s
20 tickets

1-2-3
1-3-4
1-4-5
1-5-6
1-2-4
2-3-4
2-4-5
2-5-6
2-6-1
2-3-5
3-4-5
3-5-6
3-5-1
3-6-4
3-6-2
2-5-1
4-5-6
4-6-1
4-6-2
6-1-3

u'r 7 #'s
35 tickets

1-2-3
1-2-4
1-2-5
1-2-6
1-2-7
1-3-4
1-3-5
1-3-6
1-3-7
1-4-5
1-4-6
1-4-7
1-5-6
1-5-7
1-6-7
2-3-4
2-3-5
2-3-6
2-3-7
2-4-5
2-4-6
2-4-7
2-5-6
2-5-7
2-6-7
3-4-5
3-4-6
3-4-7
3-5-6
3-5-7
3-6-7
4-5-6
4-5-7
4-6-7
5-6-7

u'r 8 #'s
56 tickets

1-2-3
1-2-4
1-2-5
1-2-6
1-2-7
1-2-8
1-3-4
1-3-5
1-3-6
1-3-7
1-3-8
1-4-5
1-4-6
1-4-7
1-4-8
1-5-6
1-5-7
1-5-8
1-6-7
1-6-8
1-7-8
2-3-4
2-3-5
2-3-6
2-3-7
2-3-8
2-4-5
2-4-6
2-4-7
2-4-8
2-5-6
2-5-7
2-5-8
2-6-7
2-6-8
2-7-8
3-4-5
3-4-6
3-4-7
3-4-8
3-5-6
3-5-7
3-5-8
3-6-7
3-6-8
3-7-8
4-5-6
4-5-7
4-5-8
4-6-7
4-6-8
4-7-8
5-6-7
5-6-8
5-7-8
6-7-8

u'r 9 #'s
84 tickets

1-2-3
1-2-4
1-2-5
1-2-6
1-2-7
1-2-8
1-2-9
1-3-4
1-3-5
1-3-6
1-3-7
1-3-8
1-3-9
1-4-5
1-4-6
1-4-7
1-4-8
1-4-9
1-5-6
1-5-7
1-5-8
1-5-9
1-6-7
1-6-8
1-6-9
1-7-8
1-7-9
1-8-9
2-3-4
2-3-5
2-3-6
2-3-7
2-3-8
2-3-9
2-4-5
2-4-6
2-4-7
2-4-8
2-4-9
2-5-6
2-5-7
2-5-8
2-5-9
2-6-7
2-6-8
2-6-9
2-7-8
2-7-9
2-8-9
3-4-5
3-4-6
3-4-7
3-4-8
3-4-9
3-5-6
3-5-7
3-5-8
3-5-9
3-6-7
3-6-8
3-6-9
3-7-8
3-7-9
3-8-9
4-5-6
4-5-7
4-5-8
4-5-9
4-6-7
4-6-8
4-6-9
4-7-8
4-7-9
4-8-9
5-6-7
5-6-8
5-6-9
5-7-8
5-7-9
5-8-9
6-7-8
6-7-9
6-8-9
7-8-9

u'r 10 #'s
120 tickets

1-2-3
1-2-4
1-2-5
1-2-6
1-2-7
1-2-8
1-2-9
1-2-10
1-3-4
1-3-5
1-3-6
1-3-7
1-3-8
1-3-9
1-3-10
1-4-5
1-4-6
1-4-7
1-4-8
1-4-9
1-4-10
1-5-6
1-5-7
1-5-8
1-5-9
1-5-10
1-6-7
1-6-8
1-6-9
1-6-10
1-7-8
1-7-9
1-7-10
1-8-9
1-8-10
1-9-10
2-3-4
2-3-5
2-3-6
2-3-7
2-3-8
2-3-9
2-3-10
2-4-5
2-4-6
2-4-7
2-4-8
2-4-9
2-4-10
2-5-6
2-5-7
2-5-8
2-5-9
2-5-10
2-6-7
2-6-8
2-6-9
2-6-10
2-7-8
2-7-9
2-7-10
2-8-9
2-8-10
2-9-10
3-4-5
3-4-6
3-4-7
3-4-8
3-4-9
3-4-10
3-5-6
3-5-7
3-5-8
3-5-9
3-5-10
3-6-7
3-6-8
3-6-9
3-6-10
3-7-8
3-7-9
3-7-10
3-8-9
3-8-10
3-9-10
4-5-6
4-5-7
4-5-8
4-5-9
4-5-10
4-6-7
4-6-8
4-6-9
4-6-10
4-7-8
4-7-9
4-7-10
4-8-9
4-8-10
4-9-10
5-6-7
5-6-8
5-6-9
5-6-10
5-7-8
5-7-9
5-7-10
5-8-9
5-8-10
5-9-10
6-7-8
6-7-9
6-7-10
6-8-9
6-8-10
6-9-10
7-8-9
7-8-10
7-9-10
8-9-10

 

[Brad]

LT,

Pick3 boxed numbers with a sum between 12 and 15 make up more than half the numbers and therefore are drawn most often.

Maybe I'm not reading this correctly but I have it as 220 total possible boxed combos, out of those there are 60 combos with a sum of 12-15. If all doubles and triples are removed, one's left with 40 with a sum of 12-15 from a total of 120 combos . Can you explain further

 

[LT]

Hey Brad and all,

Re the system ... I made an error in the percentage ratio - the number of boxed bets between the sum ranges of 12 to 15 are not 1/2 the numbers they are 1/3 of the numbers. Didn't have my brain pluged in right or my glasses on straight when I was looking at only the top of my chart

I have posted a pick 3 boxed sums chart in the free lotto section. Dowmload and it is EZ to see the combos.

Here is what should be correct.

There are a total of 220 overall boxed combinations.
The sums of 12 to 15 provide 60 combos in total minus 20 with doubles and triples giving us 40 "good bets" Just as Brad has calculated.

The sum ranges on either side of the distribution ie 16 to 27 and 0 to 11 give us 80 boxed combos each and when the doubles and triples are taken out they yeild 40 combos each so the 40 "good bets" are only one third of the potential sets.

Now lets take those overall best bets and do the repeated number theory system that grayth is talking about.

I'll use the example of the Ontario P3. Last night's winner (April 1/03) was 567.

I haven't had time to analyse the "due sum" so lets just look at all combos in the common sum ranges of 12 to 15 and next I'm selecting all combos with just the 5 or the 6 or the 7 hoping that one of the numbers will repeat.

That gives me these bets:
069, 078, 159, 168, 258, 059, 068, 158, 347, 058, 346, 147, 237, 246, 345

Now I always play for more than just the immediate next draw, when something is due, so lets see how these 15 combos work over the next 4 draws. 4 draws for a total of $60 because the win for a boxed combo is approx $60 +

Let's see if there are any winners

 

[Brad]

Thanks LT,

now I must confess that I can't add (without a calculator) . My statement "...one's left with 40 with a sum of 12-15 from a total of 122 combos ..." should read 120 combos..." to correspond with your figures

I'll correct my previous post .

 

[LT]

I can't add without a calculator either
Feel good .. U were just a little off re the number .. I was WAY off re the percentage

I was just doing "what if" tests on my P3 database and there are some wins when one plays the 3 numbers from one date through a tight range of sums for a term of 4 to 5 draws.
I'll investigate further

 

[charles2]

OK, this is what I come up with for the Ontario Pick 3 stats, BTW that 567 win yesterday was obviously a lot of people's fav numbers, 5-6-7 hey cause there were an unusually a lot of winners hence the low payout of $66.

This is my total sums for the ontario P3, so I'm using a sum of between 9 to 18 *************, Boxed and no doubles or triples and no hot cold or due numbers, I would personally be hitting on the 5 or 6 or 7 numbers cause it seems to represent best value for the money.

total pick 3 sums =
2 = 27
3 = 26
9 = 1
11 = 25
12 = 24
17 = 2
21 = 3
28 = 5
34 = 4
36 = 23
47 = 22
51 = 6
61 = 21
71 = 7
71 = 20
82 = 19
93 = 8
108 = 10 *************
115 = 16 *************
118 = 9 *************
126 = 18 *************
128 = 12 *************
140 = 13 *************
143 = 14 *************
145 = 15 *************
150 = 17 *************
153 = 11 *************


Boxed for April 2 2003

4 numbers used = 4 tickets reduced to 2 = 0-4-5 + 2-4-5

5 numbers used = 10 tickets reduced to 5 = 0-4-5 + 3-4-5 + 2-3-4 + 2-4-5 + 2-3-5

6 numbers used = 20 tickets reduced to 12 = 3-4-5 + 4-5-6 + 0-4-5 + 0-5-6 + 0-3-6 + 3-5-6 + 2-3-6 + 3-4-6 + 2-3-5 + 2-5-6 + 2-4-5 + 2-3-4

7 numbers used = 35 tickets reduced to 20 = 0-4-5 + 3-4-8 + 3-4-5 + 3-4-6 + 2-3-4 + 4-5-8 + 4-5-6 + 2-4-5 + 0-3-8 + 0-3-6 + 0-5-8 + 0-5-6 + 3-5-8 + 3-6-8 + 2-3-8 + 3-5-6 + 2-3-5 + 2-3-6 + 2-5-8 + 2-5-6

8 numbers used = 56 tickets reduced to 34 = 0-4-9 + 0-4-5 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 2-3-4 + 4-5-8 + 4-5-9 + 2-4-9 + 4-5-6 + 2-4-5 + 0-3-8 + 0-3-9 + 0-3-6 + 0-8-9 + 0-5-8 + 0-5-9 + 0-6-9 + 0-2-9 + 0-5-6 + 3-5-8 + 3-6-8 + 2-3-8 + 3-5-9 + 3-6-9 + 2-3-9 + 3-5-6 + 2-3-5 + 2-3-6 + 2-5-8 + 2-5-9 + 2-6-9 + 2-5-6

9 numbers used = 84 tickets reduced to 53 = 0-4-9 + 0-4-5 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 2-3-4 + 4-5-8 + 1-4-8 + 4-5-9 + 2-4-9 + 1-4-9 + 4-5-6 + 2-4-5 + 1-4-5 + 1-4-6 + 0-3-8 + 0-3-9 + 0-3-6 + 0-8-9 + 0-5-8 + 0-1-8 + 0-5-9 + 0-6-9 + 0-2-9 + 0-1-9 + 0-5-6 + 3-5-8 + 3-6-8 + 2-3-8 + 1-3-8 + 3-5-9 + 3-6-9 + 2-3-9 + 1-3-9 + 3-5-6 + 2-3-5 + 1-3-5 + 2-3-6 + 1-3-6 + 1-8-9 + 2-5-8 + 1-5-8 + 1-6-8 + 1-2-8 + 2-5-9 + 1-5-9 + 2-6-9 + 1-6-9 + 1-2-9 + 2-5-6 + 1-5-6 + 1-2-6

10 numbers used = 120 tickets reduced to 72 = 0-4-9 + 0-4-5 + 0-4-7 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 3-4-7 + 2-3-4 + 4-5-8 + 1-4-8 + 4-5-9 + 2-4-9 + 1-4-9 + 4-5-6 + 4-5-7 + 2-4-5 + 1-4-5 + 4-6-7 + 1-4-6 + 2-4-7 + 1-4-7 + 0-3-8 + 0-3-9 + 0-3-6 + 0-3-7 + 0-8-9 + 0-5-8 + 0-7-8 + 0-1-8 + 0-5-9 + 0-6-9 + 0-7-9 + 0-2-9 + 0-1-9 + 0-5-6 + 0-5-7 + 0-6-7 + 0-2-7 + 3-5-8 + 3-6-8 + 3-7-8 + 2-3-8 + 1-3-8 + 3-6-9 + 2-3-9 + 3-5-6 + 2-3-5 + 3-6-7 + 2-3-6 + 1-3-6 + 2-3-7 + 1-8-9 + 2-5-8 + 1-5-8 + 1-6-8 + 2-7-8 + 1-7-8 + + 1-2-8 + 2-5-9 + 2-6-9 + 1-6-9 + 2-7-9 + 1-2-9 + 5-6-7 + 2-5-6 + 1-5-6 + 2-5-7 + 2-6-7 + 1-6-7 + 1-2-6 + 1-2-7


Boxed for April 2 2003 using hot Numbers so no 1's or 2's
HOT NUMBERS FOR LAST 5 DRAWS
4 = 4
2 = 0
2 = 3
2 = 5
2 = 8
1 = 6
1 = 7
1 = 9
0 = 1
0 = 2

Boxed for April 2 2003 using hot Numbers so no 1's or 2's

4 numbers used = 4 tickets reduced to 1 = 0-4-5

5 numbers used = 10 tickets reduced to 2 = 0-4-5 + 3-4-5

6 numbers used = 20 tickets reduced to 7 = 3-4-5 + 4-5-6 + 0-4-5 + 0-5-6 + 0-3-6 + 3-5-6 + 3-4-6

7 numbers used = 35 tickets reduced to 13 = 0-4-5 + 3-4-8 + 3-4-5 + 3-4-6 + 4-5-8 + 4-5-6 + 0-3-8 + 0-3-6 + 0-5-8 + 0-5-6 + 3-5-8 + 3-6-8 + 3-5-6

8 numbers used = 56 tickets reduced to 22 = 0-4-9 + 0-4-5 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 4-5-8 + 4-5-9 + 4-5-6 + 0-3-8 + 0-3-9 + 0-3-6 + 0-8-9 + 0-5-8 + 0-5-9 + 0-6-9 +0-5-6 + 3-5-8 + 3-6-8 + 3-5-9 + 3-6-9 + 3-5-6

9 numbers used = 84 tickets reduced to 22 = 0-4-9 + 0-4-5 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 4-5-8 + 4-5-9 + 4-5-6 + 0-3-8 + 0-3-9 + 0-3-6 + 0-8-9 + 0-5-8 + 0-5-9 + 0-6-9 + 0-5-6 + 3-5-8 + 3-6-8 + 3-5-9 + 3-6-9 + 3-5-6

10 numbers used = 120 tickets reduced to 33 = 0-4-9 + 0-4-5 + 0-4-7 + 3-4-8 + 3-4-9 + 3-4-5 + 3-4-6 + 3-4-7 + 4-5-8 + 4-5-9 + 4-5-6 + 4-5-7 + 4-6-7 + 0-3-8 + 0-3-9 + 0-3-6 + 0-3-7 + 0-8-9 + 0-5-8 + 0-7-8 + 0-5-9 + 0-6-9 + 0-7-9 + 0-5-6 + 0-5-7 + 0-6-7 + 3-5-8 + 3-6-8 + 3-7-8 + 3-6-9 + 3-5-6 + 3-6-7 + 5-6-7

 

[grayth]

thanks everyone

Thanks, LT and Charles,


hey charles congrats on your win the other day...I've taken eveyones suggestions as I try and build a strat and have tested out a possibility using a little bit from the repeating number theory, some ideas from LT on sum values and from charles his box wheels....

I ran a test on myself for the past 5 weeks of data in my database by cutting it out then attempting to pick my numbers wheel them and then run my fictitious tickets for 3 days...if I had no winners...then part of my strat was to sub in no more than 2 numbers into the wheel then run for the rest of the remaining 5 days...as I decided to go in 5 day intervals...

the results were pretty good on a test run again these were all boxed runs.

First 5 days I had 3 winners
next 5 days 2 winners
next 5 days another 2 winners
4th set of 5 days no winners
5 set only had 3 draws as that brought me up to todays draw 1 winner...

total of 8 wins over 23 days

cost in wheeling 6 numbers 460.00
winnings aprox average 80=640.00

nets out a profit...

question now...I've been trying to make the 6 numbers in 20 tickets wheel more efficient to help make it a more cost effective tool to include in my strat...

I found a 6 numbers in 13 ticket wheel but it's not balanced and when i ran the same test using it it netted out less wins...so I think the only thing to do is try LT suggestion but instead run a more loose strat and dump any sums off the wheeled numbers under 9 and over 16...this should help kill 1 or 2 tickets in a 6 number balanced wheel...

unless anyone else has a way to net out 15 tickets in 6 numbers and still have it fairly balanced? Or any new suggestions here?

Tomorrow I'm running a real try and will see what happens this week with my numbers...

Troy