Athens Greece Member #84743 January 3, 2010 3 Posts Offline

Posted: May 20, 2010, 9:40 am - IP Logged

Hello,

I'm new in the forums here and i'm not sure if the issue of my topic, belongs here...

Anyway... I would like to know, what's the purpose of the skips. Lets suppose that we see a skip chart... We see a lot of numbers... What should we take under consideration in order to find the right number to play?

num. skip 1 skip 2 skip 3 skip 4 skip 5 etc

03 12 1 22 8 5

what should i look or calculate from this point?

I'd appreciate it very much, if you'll help me out in this problem.

Pennsylvania United States Member #2218 September 1, 2003 5387 Posts Offline

Posted: May 21, 2010, 12:26 pm - IP Logged

Quote: Originally posted by chrismor on May 20, 2010

Hello,

I'm new in the forums here and i'm not sure if the issue of my topic, belongs here...

Anyway... I would like to know, what's the purpose of the skips. Lets suppose that we see a skip chart... We see a lot of numbers... What should we take under consideration in order to find the right number to play?

num. skip 1 skip 2 skip 3 skip 4 skip 5 etc

03 12 1 22 8 5

what should i look or calculate from this point?

I'd appreciate it very much, if you'll help me out in this problem.

Thank you in advance!

If you are playing a Pick 3 game where you have balls 0 through 9, what you could do is take each Position 1, Position 2, and Position 3.

Total number of past draws = 126 (If draw is everday would equal 18 weeks)

Look at Position 1 balls 0 through 9

1. Calculate the number of hits for each digit from 126 draws.

2. For example it the Digit 0 hit 14 times in the last 126 draws you would have a ratio of 14:126 and a probability of 9:00 (126/14) and a ratio of 1:9

3. Calculate the "Current Skip #" from the past 126 drawinsgs (how many draws back Digit 0 hit). For example if Digit 0 hit 32 draws ago in Position 1 the skip would be 32

4. Calculate the "Highest/Maximum Skip #" from the past 126 drawings. For example if the Digit 0 hit 14 times you would look at the 14 skips in the last 126 draws. Take the highest skip number.

Current = 32 Previous order of skips = 21,5,5,2,1,15,11,5,3,13,1,1,8,1

5. Calculate the "Median Value" for Digit #1. So for example the 14 skip values were the following:

21,5,5,2,1,15,11,5,3,13,1,1,8,1

The Median would calculate to a Median Value of "2"

So what are we going to take from the calculations above.

The image above was taken from Pennsylvania Pick 3 Evenng Pick 3 drawings.

I highlighted the Digit 4 in Position 1. You can see that on average the Digit 4 should hit a total of 12.6 times in 126 drawings.

The current hits from the past 126 drawings totals 8 times.

The next column you see it is below the average 12.6 by -4.

In the next column it has a skip of 27 meaning the last draw that the digit 4 appeared was Saturday April 24, 2010 (4-6-8)

The ratio is 8:126 with a probability of 15.75 (126/8) with a ratio of 1:15 and a Median of 8.50

The Median of 8.50 was taken from the skips from the 8 previous skips: 1,22,4,10,6,16,8,9

So the Digit 4 would be a possible play for tonight based on the fact it has a Median Value of 8.50 and a Current Skip of 27, which is over 3 times the Medain Value 8.5. The Average number of times the Digit 4 should appear is 12.6 times over the past 126 draws and it only hit a total of 8 times, -4 less than the average.

Pennsylvania United States Member #2218 September 1, 2003 5387 Posts Offline

Posted: May 26, 2010, 4:05 pm - IP Logged

Quote: Originally posted by jbgmarker on May 26, 2010

Hello winsumloosesum,

Not to pick on you, but the Median you list for the numbers:

21,5,5,2,1,15,11,5,3,13,1,1,8,1 as 2 is not right.

It is 5.

I only mention this in case you are using a function that needs to be changed to give you better accuracy in your work.

Regards,

Jerry

Jerry you are correct. I originally calculated the digit 0 (zero) with a median of 5. Then looking at the all the digits for position 1 I changed to digit 4 as a possible play for the next draw.

United States Member #73037 April 3, 2009 147 Posts Offline

Posted: May 28, 2010, 11:54 am - IP Logged

Quote: Originally posted by winsumloosesum on May 21, 2010

If you are playing a Pick 3 game where you have balls 0 through 9, what you could do is take each Position 1, Position 2, and Position 3.

Total number of past draws = 126 (If draw is everday would equal 18 weeks)

Look at Position 1 balls 0 through 9

1. Calculate the number of hits for each digit from 126 draws.

2. For example it the Digit 0 hit 14 times in the last 126 draws you would have a ratio of 14:126 and a probability of 9:00 (126/14) and a ratio of 1:9

3. Calculate the "Current Skip #" from the past 126 drawinsgs (how many draws back Digit 0 hit). For example if Digit 0 hit 32 draws ago in Position 1 the skip would be 32

4. Calculate the "Highest/Maximum Skip #" from the past 126 drawings. For example if the Digit 0 hit 14 times you would look at the 14 skips in the last 126 draws. Take the highest skip number.

Current = 32 Previous order of skips = 21,5,5,2,1,15,11,5,3,13,1,1,8,1

5. Calculate the "Median Value" for Digit #1. So for example the 14 skip values were the following:

21,5,5,2,1,15,11,5,3,13,1,1,8,1

The Median would calculate to a Median Value of "2"

So what are we going to take from the calculations above.

The image above was taken from Pennsylvania Pick 3 Evenng Pick 3 drawings.

I highlighted the Digit 4 in Position 1. You can see that on average the Digit 4 should hit a total of 12.6 times in 126 drawings.

The current hits from the past 126 drawings totals 8 times.

The next column you see it is below the average 12.6 by -4.

In the next column it has a skip of 27 meaning the last draw that the digit 4 appeared was Saturday April 24, 2010 (4-6-8)

The ratio is 8:126 with a probability of 15.75 (126/8) with a ratio of 1:15 and a Median of 8.50

The Median of 8.50 was taken from the skips from the 8 previous skips: 1,22,4,10,6,16,8,9

So the Digit 4 would be a possible play for tonight based on the fact it has a Median Value of 8.50 and a Current Skip of 27, which is over 3 times the Medain Value 8.5. The Average number of times the Digit 4 should appear is 12.6 times over the past 126 draws and it only hit a total of 8 times, -4 less than the average.

Why did you choose 126 for your calculations? Why not use "prime" numbers? ie: 13, 29, 61, 127, etc.?

I think this is what I was asking about when I couldn't access your file. (My computer crashed approx. 3 weeks ago, I'm waiting for repairs to be completed. I'm using a borrowed computer while waiting.)

If you would, please explain the decision to use the 126 number. I've got more questions, but they can wait until I'm back up and running again...

Thank you,

KnuckleHead

The only DUMB question is the one question you DID NOT ask...

United States Member #69530 January 11, 2009 7803 Posts Offline

Posted: June 4, 2010, 4:54 pm - IP Logged

Quote: Originally posted by winsumloosesum on May 26, 2010

Jerry you are correct. I originally calculated the digit 0 (zero) with a median of 5. Then looking at the all the digits for position 1 I changed to digit 4 as a possible play for the next draw.

Thanks for spotting this.

Winsumloosesum?

Have you ever decided to move away from this type of statistical approach and try something else/new like short term trends?

You ever stop to think that pre-draws are the main reason those skips last for sooo long?

An absence of data also means an absence of statistical accuracy.....