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How many 3 digit combos can be made from 6 digits (no pairs) PLEASE HELP !!

Topic closed. 20 replies. Last post 3 years ago by Stack47.

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Warwick, Rhode Island
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Posted: July 20, 2013, 9:06 pm - IP Logged

Hello All.

I'm a newbie here, actually I once was a member of LP long ago, and now I'm back into the lottery frenzy.

Anyhow, a question for you math wizards. How many 3 digit (Straight) combos can you get out of 6 different digits (no pairs or triad)

I worked on tonights number, using a mirror of yesterdays number, along with the Lottery Bible table.

Here's what I had. Mirror was 983.

983 - 448 - 624 -600 - 846

938 - 174 - 839 - 396 - 015

389 - 648 - 884 - 426 - 662

398 - 770 - 916 - 225 - 461

839 - 693 - 938 - 471 - 617

893 - 275 - 411 - 720 - 966

Kept the 406721 from the doubles, played 25 random picks from these 6 digits.

Number drawn was016

I had 401, 406 701, 706 out of the 25 picks (box).

I'm just curious how many possible combos there were.  Eventually I'll play them as a .50 combo, it's just now I'm getting back into it, and need to tweak my system.

Back about 6 years ago I had hit a 4 digit Str/box combo for $2604 ($2,500 + $104) using simular tactics, so I'm just testing the waters right now.

Thank you in advance, for any time and help you give.

P.S. Still waiting for my new Lottery Bible to come in.

Thanks All US Flag GOD BLESS AMERICA.


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    Posted: July 20, 2013, 9:26 pm - IP Logged

    120

    6c3=20.  X 6 ways =120

      jimjwright's avatar - Yellow 3.png
      Park City, UT
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      Posted: July 20, 2013, 9:29 pm - IP Logged

      If you are asking for straight combos where position matters with no duplicates you would have 6 choices for 1st digit, 5 choices for 2nd digit, and 4 choices for 3rd digit.

      So that would be 6*5*4=120 straight combos.

      If you were playing boxed  then it would be C(n, r) = n! / r!(n-r)! or in your case (6,3) =  6! / (3! * 3!) = 20 boxed combos.

      Jimmy

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        Warwick, Rhode Island
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        Posted: July 20, 2013, 9:35 pm - IP Logged

        Thank you for the speedy reply cash4ninja.

        Is that a difficult formula to understand/explain ?  Just in case I run into a 5 or possible 7 digit situation.

        I also need to research on this site, as to where I can input 6 digits, with all 3 digit results, such as a wheel.

        Once again.

        Thank you.

          RJOh's avatar - chipmunk
          mid-Ohio
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          Posted: July 20, 2013, 9:35 pm - IP Logged

          Hello All.

          I'm a newbie here, actually I once was a member of LP long ago, and now I'm back into the lottery frenzy.

          Anyhow, a question for you math wizards. How many 3 digit (Straight) combos can you get out of 6 different digits (no pairs or triad)

          I worked on tonights number, using a mirror of yesterdays number, along with the Lottery Bible table.

          Here's what I had. Mirror was 983.

          983 - 448 - 624 -600 - 846

          938 - 174 - 839 - 396 - 015

          389 - 648 - 884 - 426 - 662

          398 - 770 - 916 - 225 - 461

          839 - 693 - 938 - 471 - 617

          893 - 275 - 411 - 720 - 966

          Kept the 406721 from the doubles, played 25 random picks from these 6 digits.

          Number drawn was016

          I had 401, 406 701, 706 out of the 25 picks (box).

          I'm just curious how many possible combos there were.  Eventually I'll play them as a .50 combo, it's just now I'm getting back into it, and need to tweak my system.

          Back about 6 years ago I had hit a 4 digit Str/box combo for $2604 ($2,500 + $104) using simular tactics, so I'm just testing the waters right now.

          Thank you in advance, for any time and help you give.

          P.S. Still waiting for my new Lottery Bible to come in.

          Thanks All US Flag GOD BLESS AMERICA.

          Once a member, always a member unless you get the boot.  What was your name before?

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

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            Warwick, Rhode Island
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            Posted: July 20, 2013, 9:41 pm - IP Logged

            I beleive it was buff4. but I can't say for sure. I joined as a new member again, and chose a different name as I assumed my account verification was under an old email address, and would only be a hassle to track down. But I am positive that it was here on Lottery Post. his is by far the best lottery site on the net.

            Thank you for your interest.

            P.S. No, I wasn't booted LOL.

              NEVNEE47's avatar - DiscoBallGlowing

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              Posted: July 20, 2013, 9:43 pm - IP Logged

              now that is math...i am curious to know when to use the numbers...e.g 930 will bring 953 when is my big question?

                NEVNEE47's avatar - DiscoBallGlowing

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                Posted: July 20, 2013, 9:46 pm - IP Logged

                Hello All.

                I'm a newbie here, actually I once was a member of LP long ago, and now I'm back into the lottery frenzy.

                Anyhow, a question for you math wizards. How many 3 digit (Straight) combos can you get out of 6 different digits (no pairs or triad)

                I worked on tonights number, using a mirror of yesterdays number, along with the Lottery Bible table.

                Here's what I had. Mirror was 983.

                983 - 448 - 624 -600 - 846

                938 - 174 - 839 - 396 - 015

                389 - 648 - 884 - 426 - 662

                398 - 770 - 916 - 225 - 461

                839 - 693 - 938 - 471 - 617

                893 - 275 - 411 - 720 - 966

                Kept the 406721 from the doubles, played 25 random picks from these 6 digits.

                Number drawn was016

                I had 401, 406 701, 706 out of the 25 picks (box).

                I'm just curious how many possible combos there were.  Eventually I'll play them as a .50 combo, it's just now I'm getting back into it, and need to tweak my system.

                Back about 6 years ago I had hit a 4 digit Str/box combo for $2604 ($2,500 + $104) using simular tactics, so I'm just testing the waters right now.

                Thank you in advance, for any time and help you give.

                P.S. Still waiting for my new Lottery Bible to come in.

                Thanks All US Flag GOD BLESS AMERICA.

                i am tryn to understand...

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                  Warwick, Rhode Island
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                  Posted: July 20, 2013, 9:51 pm - IP Logged

                  Jimmy, I started reading your reply and it was making total sense, that was until you got into the calculus type equation, and totally lost me,

                  But thank you just the same for your time and info.

                  I will remember the 6 for the first, 5 for the second and 4th for the 3rd.

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                    Warwick, Rhode Island
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                    Posted: July 20, 2013, 10:03 pm - IP Logged

                    Darn.... 120 combos in this case would not have been bad.

                    It would have cost me $60 (.50 per set) to win $250 (str8).

                    Would love to do that a few time a week.... but from doing research from Jan 1 to now, I find that the "Mirror" of the previous day only works between 9-14 times per month.

                    Hard work and research (read read read) will get me back on course.

                    Thanks all.

                    US FlagGod Bless.

                      jimjwright's avatar - Yellow 3.png
                      Park City, UT
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                      Posted: July 20, 2013, 10:05 pm - IP Logged

                      Jimmy, I started reading your reply and it was making total sense, that was until you got into the calculus type equation, and totally lost me,

                      But thank you just the same for your time and info.

                      I will remember the 6 for the first, 5 for the second and 4th for the 3rd.

                      If you want to understand the math then google Permutations and Combinations as they are applicable to lottery and how combinations are formed.

                      Permutations without repetition would be straight combinations.  Combinations without repetition would be boxed combos.  mathisfun is a good site that explains the concepts fairly clearly.

                      If you are just looking for straight combos with no repetition for Pick 3 then the formula for n digits to play is (n) * (n-1) * (n-2).

                      So lets say you want to play only 5 digits.  Then n=5 so the forumula would be 5 * 4 * 3 = 60. 

                      Jimmy

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                        Posted: July 20, 2013, 10:11 pm - IP Logged

                        You rule Jimmy !!

                        I will surely check out that site. Math was always my favorite subject, and now it's my grandsons (8 yr old) whom I adopted when he was 1.

                        This site rocks.....

                        jbuff.

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                          Posted: July 20, 2013, 10:34 pm - IP Logged

                          They say you learn something new every day. Thanks to Jimmy, I have now put Mathisfun in my favorites, learned the difference between a combination and a Permutation, as well as factorial function.

                          Never stop learning.

                          Many thanks.

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                            Posted: July 21, 2013, 2:16 am - IP Logged

                            I've been researching here for 14 hours, trying to learn more, but it's 2 AM and almost time to call it a night.

                            This is a question mostly for Jimmy who kindly responded earlier in the post. My question is this.

                            You had said that if I wanted to wheel 6 digits (Permutation) or no doubles, that I would use this formula.

                            6 in the first, 5 in the second, and 4 in the 3rd. But what I don't understand is, if I were looking to cover all 6 digits straight, how could that happen if I was only playing 4 of the 6 in the 3rd spot ?

                            I'm not questioning your knowledge, and do thank you for your input, I just don't understand how those 4 digits in the 3rd row would cover all 6 numbers that I have.

                            I almost feel as if it should be 6 in the first, 6 in the second and 6 in the third, or 216 sets.

                            Thank you once more.

                            Good night all.

                              jimjwright's avatar - Yellow 3.png
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                              Posted: July 21, 2013, 3:18 am - IP Logged

                              I've been researching here for 14 hours, trying to learn more, but it's 2 AM and almost time to call it a night.

                              This is a question mostly for Jimmy who kindly responded earlier in the post. My question is this.

                              You had said that if I wanted to wheel 6 digits (Permutation) or no doubles, that I would use this formula.

                              6 in the first, 5 in the second, and 4 in the 3rd. But what I don't understand is, if I were looking to cover all 6 digits straight, how could that happen if I was only playing 4 of the 6 in the 3rd spot ?

                              I'm not questioning your knowledge, and do thank you for your input, I just don't understand how those 4 digits in the 3rd row would cover all 6 numbers that I have.

                              I almost feel as if it should be 6 in the first, 6 in the second and 6 in the third, or 216 sets.

                              Thank you once more.

                              Good night all.

                              You originally said that no digits repeat (ie. no doubles or triples) if that is not true then yes it would be 6 * 6 * 6 = 216.

                              Lets use a simple example of only wanting to play the 4 digits 2, 3, 4, 5.  The formula would n * (n-1) * (n-2).  So we would have 4 * 3 * 2 = 24.

                              Lets manually do the combos:

                               1.  2 3 4
                                2.  2 3 5
                                3.  2 4 3
                                4.  2 4 5
                                5.  2 5 3
                                6.  2 5 4
                                7.  3 2 4
                                8.  3 2 5
                                9.  3 4 2
                              10. 3 4 5
                              11. 3 5 2
                              12. 3 5 4
                              13. 4 2 3
                              14. 4 2 5
                              15. 4 3 2
                              16. 4 3 5
                              17. 4 5 2
                              18. 4 5 3
                              19. 5 2 3
                              20. 5 2 4
                              21. 5 3 2
                              22. 5 3 4
                              23. 5 4 2
                              24. 5 4 3

                              You can see from first 6 lines once you commit digit 2 for position 1 then you have 3 choices for position 2.  Once you commit digit 3 for position 2 then you have 2 choices for position 3.

                              The more formal formula would be n! / (n-r)! where you would specify n and r would always be 3 for Pick3.

                              For your original example it would be 6! / (6-3)! = 6! / 3! = 6 * 5 * 4 = 120.

                              For my example it would be 4! / (4-3)! = 4! / 1! = 24

                              Jimmy