Using sums is a very effective way to win in the pick3. The most common group of sums are the 12-15 group. To arrive at a sum just add all three digits together. If you take away doubles you have only 10 numbers to work in each of these 4 groups. In the month of Jan. for NY there were 16 hits from the 12-15 group. Not bad considering there only 40 possible non-match numbers from these sums. Go back the past couple weeks in your state and look for the hottest sum and most due sum from the 12-15 group, and play the numbers in it. For example in NY the sum 13 is the most due to show, while the 15 has been the hottest this year. Each time a number hits in the 12-15 group mark it down, this way you can see whats hot, and overdue. Here is a list of the 12-15 sums, minus the doubles.

Yeah, very good stuff Blackapple...I came to a very similar conclusion, which is why I started a new thread to explore this idea in more detail with a separate thread and keep this topic geared more towards the skip research, Maybe you can add some more insight to the following thread:

http://www.lotterypost.com/thread/150343/766102?q=WSN1, thanks for your input.....

I would like to look at applying the sum group with overdue max skip and benford application and see how to best maximize the two concepts together....

Findlay, Ohio United States Member #4855 May 28, 2004 400 Posts Offline

Posted: February 7, 2007, 10:07 pm - IP Logged

I thought about tracking the performance of the skips of combos according to root sums and seeing how they hold up to Benfords Law (as I'm sure they will). This skips for straight numbers that outline Benfords Law can be broken down into 9 groups of 1,111 possibilities each (at least 99.9999% of the time or more anyways).

When you break down the Pick 3 combos into root sums you will have 10 groups: 9 of them (sums 1 through 9) will consist of 111 combos each and the 10th (sum 0) will only have one. The 10th group can be be ignored as it only represents 1 combo out of 1000. If the law applies, every sum should have around 33+% of its hits come at either one game later (a repeat) or 10 through 19 games later. I havent taken the time to test this yet but I'd bet money thats how it will be.

Using the standard sums...0 through 27... makes things more difficult because the different sum groups contain different amounts of numbers which makes certain groups much more probable and frequent than others. You would have to have different scales to track the different sums on.

Anywhere & Everywhere United States Member #10713 January 23, 2005 290 Posts Offline

Posted: February 7, 2007, 10:44 pm - IP Logged

Quote: Originally posted by Thoth on February 7, 2007

I thought about tracking the performance of the skips of combos according to root sums and seeing how they hold up to Benfords Law (as I'm sure they will). This skips for straight numbers that outline Benfords Law can be broken down into 9 groups of 1,111 possibilities each (at least 99.9999% of the time or more anyways).

When you break down the Pick 3 combos into root sums you will have 10 groups: 9 of them (sums 1 through 9) will consist of 111 combos each and the 10th (sum 0) will only have one. The 10th group can be be ignored as it only represents 1 combo out of 1000. If the law applies, every sum should have around 33+% of its hits come at either one game later (a repeat) or 10 through 19 games later. I havent taken the time to test this yet but I'd bet money thats how it will be.

Using the standard sums...0 through 27... makes things more difficult because the different sum groups contain different amounts of numbers which makes certain groups much more probable and frequent than others. You would have to have different scales to track the different sums on.

Thoth, very interesting observation. I agree that the sums would be a lil more difficult to track using the current model, due to the variances within the individual sums. Let's took at testing your theory on the root sums, i think it has significance

Anywhere & Everywhere United States Member #10713 January 23, 2005 290 Posts Offline

Posted: February 9, 2007, 1:34 am - IP Logged

Quote: Originally posted by Thoth on February 5, 2007

I like waiting for one of the "Benford numbers" to hit...then gamble on the 55% chance that another one will hit within the next two games. Of course other factors have to be included to narrow the picks in that group down

Thoth, how about we up-the-anty for straight hits with fewer combos by using the Benford skips for each 0-9 digits by position, to get a smaller core groups of most likely digits by position, and use the benford skips with root sums 0-9. You would have only a few numbers to play, with greater success rate

Anywhere & Everywhere United States Member #10713 January 23, 2005 290 Posts Offline

Posted: February 9, 2007, 3:08 am - IP Logged

I just completed a brief but interesting test I went back to TN eve draw 9/20/06 = 480, and located the digits by each position that matched the Benford Skip Model created by Thoth, the following explains my breakthrough:

Positional Benford Digits to wheel: POS1= 0,4,5,9 POS2= 7,8,9 POS3= 0,2,5,9 {4x3x4}= 48 st. combos If Benford LDR filter is applied: Wheel LDR: 2,5,9 final st. combos = 13#s to play. 9/21/06 draw results=966, we got the POS1 digit correct, however we missed pos2 and pos3 and the LDR

Day 2, use the same wheel to get all 3 digits by position correct or eliminate #9 from the POS1 POS1= 0,4,5,9 POS2= 7,8,9 POS3= 0,2,5,9 {4x3x4}= 48#s st. combos/36#s Wheel Benford LDR after last draw: 1,5,8,9: final st. combos = 18#s/13#s to play 9/22/06 draw result=099

Do you see the winning straight in final selections from these filters and 18#s / 13#s or 48#s/36#s with one filter

Findlay, Ohio United States Member #4855 May 28, 2004 400 Posts Offline

Posted: February 10, 2007, 11:10 am - IP Logged

It's interesting to see someone applying it as a filter to the digits. I havent really had time to do much testing lately, I'll try to post some more info this coming week.

Normally, thier should be three or four digits in each position that meet the criteria. This would gove you anywhere from 27 to 48 straights to play most the time. The odds of a "BL digit" hitting in all three positions at the same time to produce a win is going to be somewhere between 27/1000 (1 in 37 reduced odds) to 48/1000 (1 in 21 reduced odds) most of the time.

Of course if you look at the benford group of straights (there is usually around 320 to 340 at any given time) and you can correctly guess the root sum, that should also cut the pick down to around 37... (330÷9) straights. Almost every way of looking at it is going to give you between 27 and 48 numbers to play...I havent found a way to reduce the picks below 27 yet.

Anywhere & Everywhere United States Member #10713 January 23, 2005 290 Posts Offline

Posted: February 10, 2007, 2:18 pm - IP Logged

Quote: Originally posted by Thoth on February 10, 2007

It's interesting to see someone applying it as a filter to the digits. I havent really had time to do much testing lately, I'll try to post some more info this coming week.

Normally, thier should be three or four digits in each position that meet the criteria. This would gove you anywhere from 27 to 48 straights to play most the time. The odds of a "BL digit" hitting in all three positions at the same time to produce a win is going to be somewhere between 27/1000 (1 in 37 reduced odds) to 48/1000 (1 in 21 reduced odds) most of the time.

Of course if you look at the benford group of straights (there is usually around 320 to 340 at any given time) and you can correctly guess the root sum, that should also cut the pick down to around 37... (330÷9) straights. Almost every way of looking at it is going to give you between 27 and 48 numbers to play...I havent found a way to reduce the picks below 27 yet.

Wow, you are correct. I did a quick test to validate. I would be interested to find out what additional information you come up with after additional testing. Based upon the information given to this point, Im thinking a clever way execute the BL#s, after a hit, with about 2-4 Rootsums that fall within the BL skip parameter, this will give you about 75 - 150 #s to play for a straight hit. Im not sure what the reduced odds factor is to hit within 1-2 draws, but I think it pretty significant. Because, at stated before, BL after one hit = 55% chance within 2 draws, then 2-4 in 21 or 37 odds with root sums. If we can isolate the max skip with this concept, play only during the highest probability setup, and apply the progression where needed, I think this will be a formadible strategy.

Also, one can apply the BL#s digits positionally or any position for a straight. Cool stuff.

Thoth, this is giving me another inspiration as I found with the 12-15 sums, I see a hugh odds advantage factor when all odds and evens hit. I will send you a PM, maybe you help me identify the odds factor here.

Anywhere & Everywhere United States Member #10713 January 23, 2005 290 Posts Offline

Posted: February 10, 2007, 6:20 pm - IP Logged

I also noticed that 25% of the game is ALL HIGH & LOW and of that about 60% of the next draw comes back as a non ALL and 40% comes back as ALL in either direction. Also, about 92% of these ALL HIGH & LOWS will produce non all within 2 draws. That is 750 st. combinations, now if apply BL in of one of two ways we get the following options:

Play the BL#s which are a mix after ALL high/low from the 1000 poll

or

Wait til a BL ALL high or low combo hit, then play the BL# mix group to hit

I dont how many #s that is but I would assume about 75% of your 330 BL#s =250#s to play for a st. hit. I think that field of numbers is still a little high, however, it statiscally it gives you a great odds advantage. Thoth maybe you or someone can add to this odd stats and or other solid filters, that work.