Stone Mountain*Georgia United States Member #828 November 2, 2002 10491 Posts Offline

Posted: September 23, 2005, 8:38 pm - IP Logged

Well .....that hurt! Just spent 1 hour on a large post and it just went POOF! Went to print and it would not print. Service unavailable.

Oh well.... had no idea it would be so large. Went through one at a time and marked each one of the wins for the last year. They were counted Wins if you always wheeled the 7 digits that did not play in the number the draw before. It was hard.

Anyway.... ended up with 92 of them out of265 Six way numbers in the last 365 draws. During the year it looked like there were only 4 or so times where it would not work for extended period of time. Extended means 6 to 10 draws where there were no wins. If someone else wants to copy and paste the list and bold in the hits ..... go for it.

So... Of the times the 265 singles hit..... the 7 digit wheel hit 92 times or 35%real world playing single to single. Formula should be very close to that.

Anyone else want to go through each one of these and copy paste the wins for the 7 digit wheel ....JUMP in... LOL

Stone Mountain*Georgia United States Member #828 November 2, 2002 10491 Posts Offline

Posted: September 23, 2005, 9:23 pm - IP Logged

I guess ....bottom line so far ,,,for now... is this. The hit rate for a full 365 draw year including the doubles on this real world example ended up with 92 hits.

That means without any regard to SKIPS OR DOUBLES during the year and playing every darn day like a robot..... regardless of the flow,,,, you would expect the 7 digit wheel to win..... around 1/4 th. the time. This is playing from single to single number. Always playing the wheel based on the (no show) digits in the last 6 way... not the doubles. Playing that wheel until the next single hit.

So ...its almost like the doubles hit rate ...of about 27% or around 100 a yr. in a 365 game. You wouldn't play doubles everyday ofcourse.... you count the days between hits etc. etc. Same thing here.

Still no idea how to put it into a formula for Pick three....forget Pick 4.

This is good enough for gov. work..... for now.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

CA United States Member #2987 December 10, 2003 832 Posts Offline

Posted: September 24, 2005, 12:19 am - IP Logged

WIN D -

The formula would have to be calculated on a 271/729 base, as any given number appears 271 times in all one thousand 3-number combinations. Fortunately for calculation purposes, 729 is the cube of 9.

There'd be a million possibilities - one of a thousand numbers followed by another one of a thousand numbers. You'd have to include the doubles and triples in the building process.

I'm a bit busy right now but if I get some spare time, sounds interesting.

gl

j

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

Wisconsin United States Member #1303 March 27, 2003 1508 Posts Offline

Posted: September 24, 2005, 4:10 pm - IP Logged

Quote: Originally posted by WIN D on September 23, 2005

Math wise .....how often will we get a no digit return from the previous draw? Pick 3 and Pick 4.

If you don't show your work I'll have to count the entire answer wrong. LOL

Thanks

WIN D,

Somewhere on LP in the past, I did a research for Pick 3 in a number of states as to how often they do get a one-digit return (just the opposite from what yo uask, but the math answer to what you want would be easy to calculate knowing when a digit does return) -- as I remember, a digit returns approximately (and consistently) about 47-48% of the time. Therefore, it is yet another of those 50/50 indicators....not much help. Half the time you'll be correct. Not good enough.

Stone Mountain*Georgia United States Member #828 November 2, 2002 10491 Posts Offline

Posted: September 24, 2005, 4:51 pm - IP Logged

But Badger..... doubles are even worse return rate of only 27% and people still play doubles.

Anyway.... that question of how often a digit does return.... The thing that throws that off is the times it goes over board. The times2 digit return. That happens about every 9 draws on average and will mess up a person playing for a one digit return.

It all comes down to the fact that one can't play any system every day but must look for a special set of circumstances before we pull the trigger. We have to aim. Although shooting from the hip is fun ...it will leave you twisting in the wind eventually...LOL The best system will always end up as some form of running between the rain drops.

Anyway...real world is one thing. Thank Goodness for Johnph77 and his abilities to push out a formula on these. I can't wait. Thanks John.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

CA United States Member #2987 December 10, 2003 832 Posts Offline

Posted: September 28, 2005, 7:19 pm - IP Logged

WIN D -

This one was so easy once the method came to me I'm almost embarrassed to admit it. It wasn't done by formulas but rather by programming.

Triples:

As inferred by my previous post, if a triple is the first three-digit number, there are 729 possibilities that the next number drawn will not have a duplicate of the first draw. Note that 729 is the cube of 9.

Doubles:

I took the numbers 0 and 1 for an example. Of the 1,000 possible numbers, 488 have the numbers 0 and/or 1 in them, leaving 512 possibilities without duplication of any numbers in the first draw. Note that 512 is the cube of 8.

All different:

I took the numbers 0, 1 and 2 for an example. Of the 1,000 possible numbers, 657 have the numbers 0, 1 and/or 2 in them, leaving 343 possibilities without duplication of any numbers in the first draw. Note that 343 is the cube of 7.

These are the only three possibilities of occurance in the Pick 3 game, and the percentage of possible non-duplication for each example is 72.9%, 51.2% and 34.3%, respectively.

But if you're looking for the overall possibility of non-duplication regardless of what is contained in the first draw, this is the way it's worked out. There are 10 possible triples with 729 possibilities each for a total of 7,290 non-duplicative possibilities. There are 270 doubles with 512 possibilities each for a total of 138,240 non-duplicative possibilities. There are 720 three digit numbers where all three digits differ with 343 possibilites each for a total of 246,960 non-duplicative possibilities. Totaling these leaves 392,490 possibilities out of 1,000,000 total possibilities where a second draw will not contain any number in the in the first draw, a percentage of 39.249%.

Hope this answers the question.

gl

John

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

Stone Mountain*Georgia United States Member #828 November 2, 2002 10491 Posts Offline

Posted: October 3, 2005, 10:40 am - IP Logged

Thank you John. Your special talents are appreciated here so very much. It will take some time for me to organize all the implications. There will be several.

39.249 Always suspected ...but never confirmed without your help. It will be nice to build on a rock rather than sand for a change.

This may be of interest to those that follow the Pick 4 game. The example that follows is from the Georgia midday game. This list contains just the single numbers ABCD. There were 167 singles during the last year. The ABCD numbers that did notcontain any digits returning from the previous ABCD are noted in BOLD print below.

Of the 167 singles listed only 9 did not contain any digits from the previous 24 way number. It would seem this result could only be useful in the reverse. I did this with my glasses but .....

I am surprised at this example. I really expected a much higher figure. Finding the (real world) return rate on at least ONE digit shocked me. Looks like the formula will end up at 95% . Almost like hunting triples in the Pick 3 game.... Very rare. It would be interesting to see the formula for at least a 2 digit return on ABCD's

The main consideration this player takes away from this example is ..ofcourse *** Always include at least one digit from the previous a- b- c -d number. Unless...you want to lose over 94.6% of the time. or even better

" Always include at least one digit from the previous A-B-C-D number and you will be right 94.6 % of the time. "

mid-Ohio United States Member #9 March 24, 2001 19813 Posts Offline

Posted: October 3, 2005, 12:27 pm - IP Logged

Quote: Originally posted by WIN D on September 23, 2005

Math wise .....how often will we get a no digit return from the previous draw? Pick 3 and Pick 4.

If you don't show your work I'll have to count the entire answer wrong. LOL

Thanks

There are 1000 possible combinations of 3's. If you remove 3 numbers then there 343 possible combinations of 3's, so there would be 343/1000 odds that a combination coming up today would not have one of the numbers in the previous drawing if the previous drawing had 3 different numbers.

However if the last number was a double then there would 512 combinations of 3's that wouldn't contain those two numbers, so there would 512/1000 odds that a combination coming up today would not have one of the two numbers in the previous drawing.

However if the last number was a triple then there would 729 combinations of 3's that wouldn't contain that one number, so there would 729/1000 odds that a combination coming up today would not have the number in the previous drawing.

It all depends on the amount of different numbers in the last drawing.

Note:The pick4 is figured in a similar way

* you don't need to buy more tickets, just buy a winning ticket *