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HOT DIGIT ?? or something more?

Topic closed. 129 replies. Last post 2 months ago by Jayhawk58.

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Posted: September 1, 2016, 2:11 pm - IP Logged

Jayhawk-

Just to be sure your numbers are clear:

"60 occurrences of consecutive pairs and 36 occurrences of mirror pairs" Are all of those exclusive or do some overlap. If some overlap, as one of your previous examples did, then you might need to adjust the 96 out of 124 draws stat.

In regards to one of the consecutive pair digits showing in either 2 draws or 3 draws, are you numbers for 3 draws inclusive of your numbers for 2 draws? What I mean is, when you say one of next 2 draws 42 times, one of the next 3 draws 49 times, does that mean the third day showed a digit 7 times? or 49 times? I'm assuming 7 times for the total of 49.

If I am reading these right, then over 3 draws out for consecutive pairs you are showing about 82% of the time in your sample. For mirror pairs about 83%. The expected rate using random 2 digits instead would be about 87%. So for whatever its worth, the sample you are evaluating does not appear to suggest that using this method improves your chances of finding a digit over random. Doesn't mean it won't work, just means this particular sample doesn't provide evidence of it.

There is overlap of pairs. Some show as consecutive and mirror pairs. As last night's draw was. 501...consecutive is the 01 and the mirror would be the 05.

However I approach them as individual and would play as such. The numbers are indicative of the draw the result was in. Third draw was 7 occurrences. Obviously the best way to play this is the next 2 draws. 

I looked at the random as hitting in two draws and I think that would be hard to figure. For instance if the draw was 368. And you chose the 8 to repeat in the next draw. If it didn't would you stay with it? Looking at draw history, it doesn't appear it would play out. So choose another digit? If so, you are now comparing apples and oranges.

8/12 EVE 139
8/13 MID 583

8/13 EVE 409

8/14 MID 120

8/14 EVE 400


Draw history above: in the 8/12 EVE draw you have 139. So if you picked the digit 1 to go after 2 draws with you would get nothing. If you picked the 3 digit, you would hit the next draw. If you picked the 9, you would hit the second draw. But would you stay with that digit or move on?  What would make you pick the 3 digit? In this case the 3 digit was out 9 draws. Not exactly a hot digit but could be described as due. The third draw shown is a consecutive and a mirror. How to play? In my opinion consecutive pairs  tend to return the digit stronger than mirror pairs. But if you played both you would get the 0 in the next draw. And the 4 was in the next. 

So again, how do you determine what the random digit to select would be?

 Rock Chalk Jayhawk

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    Posted: September 1, 2016, 2:24 pm - IP Logged

    There is overlap of pairs. Some show as consecutive and mirror pairs. As last night's draw was. 501...consecutive is the 01 and the mirror would be the 05.

    However I approach them as individual and would play as such. The numbers are indicative of the draw the result was in. Third draw was 7 occurrences. Obviously the best way to play this is the next 2 draws. 

    I looked at the random as hitting in two draws and I think that would be hard to figure. For instance if the draw was 368. And you chose the 8 to repeat in the next draw. If it didn't would you stay with it? Looking at draw history, it doesn't appear it would play out. So choose another digit? If so, you are now comparing apples and oranges.

    8/12 EVE 139
    8/13 MID 583

    8/13 EVE 409

    8/14 MID 120

    8/14 EVE 400


    Draw history above: in the 8/12 EVE draw you have 139. So if you picked the digit 1 to go after 2 draws with you would get nothing. If you picked the 3 digit, you would hit the next draw. If you picked the 9, you would hit the second draw. But would you stay with that digit or move on?  What would make you pick the 3 digit? In this case the 3 digit was out 9 draws. Not exactly a hot digit but could be described as due. The third draw shown is a consecutive and a mirror. How to play? In my opinion consecutive pairs  tend to return the digit stronger than mirror pairs. But if you played both you would get the 0 in the next draw. And the 4 was in the next. 

    So again, how do you determine what the random digit to select would be?

    How do you determine what the random digit to select would be? Randomly. :)

    It is not difficult to determine the expected results using random over 2 draws. Pick any 2 distinct digits. In the next draw, in each of the three places of the pick 3, there is a 20% chance that one of those two digits will show. To calculate for the 3 places, you have to use the negative side, so for each place there is an 80% chance that neither of the digits will show. For 3 places, that comes out to .8^3, which equals .512, or a 51.2% chance that neither of the digits will show in any of the 3 places. Convert that to positive, and you get 48.8% chance at least one of the digits will show at least once. Do the same thing with 2 draws, and it comes out to about 73.8% chance that at least one of the digits will show at least once.

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      Posted: September 1, 2016, 2:26 pm - IP Logged

      There is overlap of pairs. Some show as consecutive and mirror pairs. As last night's draw was. 501...consecutive is the 01 and the mirror would be the 05.

      However I approach them as individual and would play as such. The numbers are indicative of the draw the result was in. Third draw was 7 occurrences. Obviously the best way to play this is the next 2 draws. 

      I looked at the random as hitting in two draws and I think that would be hard to figure. For instance if the draw was 368. And you chose the 8 to repeat in the next draw. If it didn't would you stay with it? Looking at draw history, it doesn't appear it would play out. So choose another digit? If so, you are now comparing apples and oranges.

      8/12 EVE 139
      8/13 MID 583

      8/13 EVE 409

      8/14 MID 120

      8/14 EVE 400


      Draw history above: in the 8/12 EVE draw you have 139. So if you picked the digit 1 to go after 2 draws with you would get nothing. If you picked the 3 digit, you would hit the next draw. If you picked the 9, you would hit the second draw. But would you stay with that digit or move on?  What would make you pick the 3 digit? In this case the 3 digit was out 9 draws. Not exactly a hot digit but could be described as due. The third draw shown is a consecutive and a mirror. How to play? In my opinion consecutive pairs  tend to return the digit stronger than mirror pairs. But if you played both you would get the 0 in the next draw. And the 4 was in the next. 

      So again, how do you determine what the random digit to select would be?

      I also think you have to remember that when you say "if you picked the 3 digit, you would hit next draw", that is only true if you also picked the 5 and the 8 digits, unless you were playing all of the potential combos with a 3.

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        Posted: September 1, 2016, 3:34 pm - IP Logged

        I also think you have to remember that when you say "if you picked the 3 digit, you would hit next draw", that is only true if you also picked the 5 and the 8 digits, unless you were playing all of the potential combos with a 3.

        Playing all potential combos of 3 would have to be the only way to go. Of course, that's after  doing a 3-4 rundown, eliminating combo's to a playable point.

         

        Today's MID-009.

        So this is the way it looks since start of the post. Still doubting? Of course I filtered it out. Actually this time I didn't ever have it in. I went to the next 3-4 rundown and didn't consult yesterday's. Gave it up too soon. It was there. This has been an issue for me. I think I'm going to start including all pairs from 3 3-4 run downs and then eliminate out. If I had left it in....19 combo's. would have doubled up on the doubles. Why? Two combo's with 0 and consecutives will get you a double. Look it up......

        Sep 01  MID-009

        Aug 31  MID-862   EVE-501

        Aug 30  MID-209   EVE-887

        Aug 29  MID-646   EVE-352

         Rock Chalk Jayhawk

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          Posted: September 1, 2016, 4:53 pm - IP Logged

          Playing all potential combos of 3 would have to be the only way to go. Of course, that's after  doing a 3-4 rundown, eliminating combo's to a playable point.

           

          Today's MID-009.

          So this is the way it looks since start of the post. Still doubting? Of course I filtered it out. Actually this time I didn't ever have it in. I went to the next 3-4 rundown and didn't consult yesterday's. Gave it up too soon. It was there. This has been an issue for me. I think I'm going to start including all pairs from 3 3-4 run downs and then eliminate out. If I had left it in....19 combo's. would have doubled up on the doubles. Why? Two combo's with 0 and consecutives will get you a double. Look it up......

          Sep 01  MID-009

          Aug 31  MID-862   EVE-501

          Aug 30  MID-209   EVE-887

          Aug 29  MID-646   EVE-352

          I guess you would have to put me in the "still doubting" camp, although I think I would better describe myself as "not convinced". Reason being that the only evidence you've put forward doesn't show your system coming out any better than one would expect with random selections. Small sample size so far though so I don't know how much meaning to put into it.

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            Posted: September 1, 2016, 6:31 pm - IP Logged

            Back to basics, the odds of any pair for each UNIQUE draw is 1/100, this odds remains constant, any draw history will not change the odds.

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              Posted: September 1, 2016, 7:05 pm - IP Logged

              Back to basics, the odds of any pair for each UNIQUE draw is 1/100, this odds remains constant, any draw history will not change the odds.

              1/100 would be the probability of a straight  pair in a pick 2 game, the probability of a pair in pick 3 boxed would be better but I think it might be a little complicated to actually calculate. I'll see what I can do and post it later.

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                Posted: September 1, 2016, 7:27 pm - IP Logged

                I guess you would have to put me in the "still doubting" camp, although I think I would better describe myself as "not convinced". Reason being that the only evidence you've put forward doesn't show your system coming out any better than one would expect with random selections. Small sample size so far though so I don't know how much meaning to put into it.

                How large a sample would you like?

                 Rock Chalk Jayhawk

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                  Posted: September 1, 2016, 7:31 pm - IP Logged

                  1/100 would be the probability of a straight  pair in a pick 2 game, the probability of a pair in pick 3 boxed would be better but I think it might be a little complicated to actually calculate. I'll see what I can do and post it later.

                  I believe the probability of any distinct pair showing in any order, any position in a single pick 3 draw would be 6%, or 6/100, as I believe there are 60 different combinations out of the possible 1000 that would include a distinct pair.

                  Distinct Pair AB, X=any of digits 0-9

                  Possible combinations:

                  A B X = 10 possible

                  A X B = 10 possible

                  B A X = 10 possible

                  B X A = 10 possible

                  X A B = 10 possible

                  X B A = 10 possible

                  60 total.

                  Sorry Jayhawk, I don't think the probability of a pair has much to do with the topic of your thread.

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                    Posted: September 1, 2016, 7:35 pm - IP Logged

                    Back to basics, the odds of any pair for each UNIQUE draw is 1/100, this odds remains constant, any draw history will not change the odds.

                    I'm not trying to prove any odds changing concerning pairs. I am more interested in what happens after the pair. I'm not even trying to predict pairs. 

                    What I am trying to show is the "probability" of a digit occurring after the pair. If it is 50/50, and I am recognising that any digit can occur but these happen more, or enough, to be predictable. That is a indicator I can use.

                    "So the totals become Random.......55     55 draws that are based on random, not knowing what could be there.

                                Consecutive or mirror.......47      47 draws that are based on this happened, so this could and has shown to happen."

                    I think that is something I can use. I know it will not occur all the time. 

                     Rock Chalk Jayhawk

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                      Posted: September 1, 2016, 7:37 pm - IP Logged

                      How large a sample would you like?

                      To be considered evidence that a consecutive pair increases the chance that one of those digit shows in two or three draws? Maybe 500 or 1000 draws, with a positive (or hit if you want to call it that) rate decently higher than what would be expected by random.

                      My opinion only. Just was thinking what it would take for me to really look at increasing my play when I saw the trigger. maybe 10000 draws to significantly increase my play. Maybe I'm too picky, but I figure if it works over the past 100 draws but doesn't work over the past 1000 draws then I'm not very confident in it.

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                        Posted: September 1, 2016, 7:37 pm - IP Logged

                        I believe the probability of any distinct pair showing in any order, any position in a single pick 3 draw would be 6%, or 6/100, as I believe there are 60 different combinations out of the possible 1000 that would include a distinct pair.

                        Distinct Pair AB, X=any of digits 0-9

                        Possible combinations:

                        A B X = 10 possible

                        A X B = 10 possible

                        B A X = 10 possible

                        B X A = 10 possible

                        X A B = 10 possible

                        X B A = 10 possible

                        60 total.

                        Sorry Jayhawk, I don't think the probability of a pair has much to do with the topic of your thread.

                        It's ok. It happens.

                        I would like to commend you Wisconsin.

                        Our dialog and exchange of information has been very pleasant compared to some of our past discussions. This is the LP I remember.

                         Rock Chalk Jayhawk

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                          Posted: September 2, 2016, 4:33 am - IP Logged

                          I'm not trying to prove any odds changing concerning pairs. I am more interested in what happens after the pair. I'm not even trying to predict pairs. 

                          What I am trying to show is the "probability" of a digit occurring after the pair. If it is 50/50, and I am recognising that any digit can occur but these happen more, or enough, to be predictable. That is a indicator I can use.

                          "So the totals become Random.......55     55 draws that are based on random, not knowing what could be there.

                                      Consecutive or mirror.......47      47 draws that are based on this happened, so this could and has shown to happen."

                          I think that is something I can use. I know it will not occur all the time. 

                          Using those numbers tells me that it's close enough to be be used as a 50/50 filter.

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                            Posted: September 2, 2016, 7:15 am - IP Logged

                            It's ok. It happens.

                            I would like to commend you Wisconsin.

                            Our dialog and exchange of information has been very pleasant compared to some of our past discussions. This is the LP I remember.

                            I Agree! This thread has been very interesting and informative. 

                            "There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

                            ~Robert A. Heinlein

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                              Posted: September 2, 2016, 8:35 am - IP Logged

                              Using those numbers tells me that it's close enough to be be used as a 50/50 filter.

                              50/50 filters can be effective. For those not familiar, a few links.

                              www.lotterypost.com/thread/164643

                              www.lotterypost.com/thread/280727

                              www.lotterypost.com/thread/293875

                              www.lotterypost.com/thread/245804

                               Rock Chalk Jayhawk