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Trying to calculate the number of combinations with criteria

Topic closed. 14 replies. Last post 5 years ago by nickbrownsfan.

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Posted: April 11, 2012, 7:05 am - IP Logged

I know that in a 6/45 lottery, there are 8,145,060

Lets say you split the 45 numbers into two groups, with group A having the numbers 1-30 and group B 31-45. You make a prediction that 5 of the 6 numbers will be drawn from group A and 1 of the 6 will be drawn from group B. If your prediction is correct, your probability of winning has certainly increased, but has the size of combinations changed? If so, what is the formula to calculate this change.

Would this be any different if the numbers in group A and group B overlapped.

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    cleveland ohio
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    Posted: April 11, 2012, 9:48 am - IP Logged

    not sure if this is what your looking for but you would take the amount of chances and multiply them together.

    In your example for your first choice you have 30 numbers to choice from

    your next choice would have 29 (you've already selected 1) etc

    So you would have 30*29*28*27*26 next you need to factor out all the doubles because each number is unique so you have 6 choices on where to place the first number then 5 choices on the 2nd number ect all the way to 1.

    You then divide the first set by the 2nd set to get your total combinations.

    30*29*28*27*26/5*4*3*2*1 = 142506

    you have a 1 in 15 chance at the last number so multiply them together 142506*15= 2137590

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      Posted: April 11, 2012, 10:32 am - IP Logged

      If your maths is correct. Does that mean that by me making a correct prediction, I have reduced my odds from 8mil to 2mil, while still playing all 45 balls.

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        cleveland ohio
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        Posted: April 13, 2012, 12:35 am - IP Logged

        Well before someone comes in here and says that the odds do not change what ever numbers you play are, Yes given the peramiters you have set your odds have lowered by a considerable amount.

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          bgonçalves
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          Posted: April 13, 2012, 9:10 am - IP Logged

          Well before someone comes in here and says that the odds do not change what ever numbers you play are, Yes given the peramiters you have set your odds have lowered by a considerable amount.

          Hello, falling, the best division is in 4 parts, because, when billions of simulations of the groups will have zero or one number and one will have at least two (there may be more but at least two numbers m always 100% of the sweepstakes, you can check ok, ie you can see the drive (a number) stop and trios of cad 4 groups and combine such a hot group trio, will combine with another group pair of cold, and another number from another corner.

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            Posted: April 13, 2012, 10:51 pm - IP Logged

            I don't understand what you're trying to say

              RL-RANDOMLOGIC's avatar - usafce

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              Posted: April 18, 2012, 4:35 pm - IP Logged

              ff

              Conditional odds.  Setting these types of perimeters is a great way to play. Even if your

              assumptions are incorrect because your tickets still carry the same odds as a QP.  If on

              the other hand you are correct then you have a much better chance of winning something.

              Removing one ball from a 5-59 matrix reduces the pool by 424,270 sets.  If you can remove

              one digit from a drawing successfully, say digit 7 for instance, you reduce the pool by 2,036,701

              sets for the above matrix and any set you play still has as good of odds as any QP regardless

              of what you choose to remove.   A pick-5 QP removes all the numbers but 5 and many swear by

              them, go figure!!  If I remove 20 numbers from the pool and play the rest I am considered an

              idiot by others who spout QP's as being superior somehow.   This appears to me to be stupidity

              gone to seed.  Don't let the odds slow you down, play on.

              Welcome to LP

              RL

              Working on my Ph.D.  "University of hard Knocks"

              I will consider the opinion that my winnings are a product of chance if you are willing to consider

              they are not.  Many great discoveries come while searching for something else

              USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

                US Flag Trump / 2016 & 2020  

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                Posted: April 19, 2012, 4:44 am - IP Logged

                How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

                  RJOh's avatar - chipmunk
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                  Posted: April 19, 2012, 5:09 am - IP Logged

                  I know that in a 6/45 lottery, there are 8,145,060

                  Lets say you split the 45 numbers into two groups, with group A having the numbers 1-30 and group B 31-45. You make a prediction that 5 of the 6 numbers will be drawn from group A and 1 of the 6 will be drawn from group B. If your prediction is correct, your probability of winning has certainly increased, but has the size of combinations changed? If so, what is the formula to calculate this change.

                  Would this be any different if the numbers in group A and group B overlapped.

                  You would calculate the odds like you do for PM and MM which use two groups of numbers.

                  5 of 30 = 142,506  x 1 of 15 =15   or 15 x 142,506 = 2,137,590

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

                    RJOh's avatar - chipmunk
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                    Posted: April 19, 2012, 5:13 am - IP Logged

                    How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

                    Are you planing to use the numbers 1-15 twice since both groups cover 1-15?

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

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                      Posted: April 19, 2012, 5:53 am - IP Logged

                      Are you planing to use the numbers 1-15 twice since both groups cover 1-15?

                      Yes. For my first 3 numbers, I am just picking randomly from 1 to 45. For my last 3 numbers, I am picking randomly from 1 to 15.

                        RJOh's avatar - chipmunk
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                        Posted: April 19, 2012, 11:36 am - IP Logged

                        Yes. For my first 3 numbers, I am just picking randomly from 1 to 45. For my last 3 numbers, I am picking randomly from 1 to 15.

                        Sounds like you're back where you started picking 6 numbers from a pool of 45.  Choosing at least three of them from the lower 15 doesn't change the way the number of possible combinations are calculated.

                         * you don't need to buy more tickets, just buy a winning ticket * 
                           
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                          Posted: April 19, 2012, 12:28 pm - IP Logged

                          Sounds like you're back where you started picking 6 numbers from a pool of 45.  Choosing at least three of them from the lower 15 doesn't change the way the number of possible combinations are calculated.

                          I'm pretty sure it does. To illustrate. In a 2 of 6 game, i get 15 possible combinations. If I were to choose one from 1-6 and the other from 1-3, I effectively eliminated the combinations, 4/5, 4/6 & 5/6, bring the number of possible combinations down to 12. I'm trying to figure out the formula to calculate this.

                            RJOh's avatar - chipmunk
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                            Posted: April 19, 2012, 6:54 pm - IP Logged

                            I'm pretty sure it does. To illustrate. In a 2 of 6 game, i get 15 possible combinations. If I were to choose one from 1-6 and the other from 1-3, I effectively eliminated the combinations, 4/5, 4/6 & 5/6, bring the number of possible combinations down to 12. I'm trying to figure out the formula to calculate this.

                            When you figure out the formula, post it and we will learn something new.  Good luck.

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

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                              cleveland ohio
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                              Posted: April 22, 2012, 1:47 pm - IP Logged

                              How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

                              There is 2 ways to look at this the first way 3 of 45 14190 * 3 of 15 455 with leaves 6456450 total combinations but Im not sure that is totally correct because you can still have 2 overlapping numbers. That would be more for looking for 3 mega balls or something being able to pick the number 1 in each set sooo... I am going to say you need to pick the 3 numbers from 1-15 first the combinations there stay the same.

                              15*14*13/3*2*1= 455   Now for the 2nd set you have already placed 3 numbers in the 1-15 area which leaves you picking 3 of 42 (total possible numbers left to pick from)

                              42*41*40/3*2*1=11480

                              11480*455=5223400