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What are the odds

Topic closed. 193 replies. Last post 2 years ago by sandnan.

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RL-RANDOMLOGIC's avatar - usafce

United States
Member #59354
March 13, 2008
3962 Posts
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Posted: April 28, 2014, 9:49 am - IP Logged

Are the lines you play giving you the best bang for your dollar?

Note!
The cNr calculations are done using the standard windows calculator set to scientific mode. I have gone over this several times to reduce errors but some may still exist. My left hand types faster than my right which causes hard to find errors in the number values. With that being said, here we go.     

I will keep this simple as possible so that everyone regardless of you math level will be able to understand it. I will show why optimizing our sets will increase our wins by increasing the number of overall chances. Line optimizing is a very valuable tool for anyone who plays more than one line per game.

Jade posted something similar to this a year or so back and I think this gives a little more insight into the question he posed at that time. If I remember correctly he asked which was a better bet, playing 1 line for 10 games or playing 10 lines in a single game.  I will show how that playing 1 line for 10 games has better overall chances of winning a prize than playing 10 lines for a single game. This is not the reason for this post but it helps to understand it but  first a little math.

The figures here apply to a pick 5 of 39 game but hold true for any double digits lottery game you just need to plug in the correct variabels. .

Below are a few problems for which the nCr function can be used.  These are all lottery related and cover various aspects of calculating combinations.

Below are a couple methods one could use to calculate the number of times any one number appears within a 5-39 matrix.

Option #1. 575757*5/39 = 73815

Option #2  (39-1)=38!/(5!*34!)= 501942 and 575757-501942=73815

The first option is very simple but what if we need to calculate the number of times any 2 numbers appear together. If we multiply 73,815 by 2 we get 73815*2=147624 which is incorrect.

The correct answer is 7,770.   

To calculates pairs (39-2)= 37!/(3!*34!)=7770
To calculates trays (39-3)= 36!/(2!*34!)=630
To calculates quads (39-4)= 35!/(1!*34!)=35

Here I show how to calculate the odds using the nCr function and some simple math.

To do this we first need to know a couple things. Again for this example I will use a 5-39 but if working on a different matrix, just plug in your variables. The list below shows the number of combinations of 2, 3, 4 etc.. within pick 5 and 6 lottery games.

To find these values using the nCr function follow the example -> 5!/(2!,3!)=10 which shows the total 2 number combos within a set of 5 numbers.

combinations of 2 in 5 = 10
combinations of 3 in 5 = 10
combinations of 4 in 5 = 5
combinations of 5 in 5 = 1

combinations of 2 in 6 = 15
combinations of 3 in 6 = 20
combinations of 4 in 6 = 15
combinations of 5 in 6 = 6
combinations of 6 in 6 = 1     
   
This list can be easily memorized which will cut down on the number of calculations needed to for most of the problems here.

First we will calculate the odds for matching 2 of 5 numbers.  To do this we have to plug in the correct variables. Look at this example. 34!/(3!*31!)=5984  Notice the (n) value is 34.  This is gotten by subtracting 5 from 39.  The next variable (r) is 3. To get this value we subtract the prize number match from the total numbers in single line, 5-2=3, The third variable is gotten by subtracting (r) from (n), 34-3=31

Here are the three variables for calculating 2,3,4 of 5 numbers matches.

2of5 are #1, 39-5=(34), #2=5-2=(3) and #3=34-3=(31)

3of5 are #1, 39-5=(34), #2=5-3=(2) and #3=34-2=(32)

4of5 are #1, 39-5=(34), #2=5-4=(1) and #3=34-1=(33)
 
34!/(3!*31!)=5984 
34!/(2!*32!)=561 
34!/(1!*33!)=34   

Next we multiply the results by the number of combinations within the a set of 5 numbers.
combinations of 2 in 5 = 10
combinations of 3 in 5 = 10
combinations of 4 in 5 = 5
combinations of 5 in 5 = 1

34!/(3!*31!)=5984*10 =59840 
34!/(2!*32!)= 561*10 = 5610
34!/(1!*33!)=  34* 5 =  170
5 of 5          1* 1 =    1               
___________________________
Total                =65621

This gives us the number of lines within the entire matrix that if drawn will win a prize.

To calculate the individual odds for matching a prize we divide the total number of sets within the matrix by the total of each possible number match

Odds of matching 2 numbers = 575757/59840=1 in 9.62     
Odds of matching 3 numbers = 575757/5610 =1 in 102.63
Odds of matching 4 numbers = 575757/170  =1 in 3386.80
Odds of matching 5 numbers = 575757/1    =1 in 575757

To calculate the overall odds we add all the possible prizes then divide the total lines by this amount. 575757/65821=8.77

The overall odds are 1 in 8.77 for winning a prize for a 5-39 game.  This applies only to single line plays regardless of the numbers played. What this means is that there are 65,621 possible lines that if any one is drawn we will win a prize.  I know this is a bit redundant but we need to understand it.

We now know exactly what our overall odds of winning a prize when playing one ticket but what if we play more than one line.   

Now that we have calculated the total chances one set has for a single drawing we can calculate the total chances for playing one line for 10 games. 65621*10=656210. Playing one line each game gives us the best cost/chance ratio we can get. The CC ratio for a one line play is calculated by dividing the cost by the total chances. $1.00/65621=$0.000015239.

You may be a bit confused by this so let's dig a little deeper.  What if we play 2 lines? Lets look at the two examples below and I will introduce what I call "prize overlap lines", POL for short. 

Player #1
01-02-03-04-05
06-07-08-09-10

Player #2
01-02-03-04-05
01-02-03-04-06

Above we have two players that show the extreme ranges for betting on the lottery. Player #1 plays 10 different numbers on his/her 2 lines and player #2 plays 6 total numbers. There are no repeating numbers in the lines of player #1 but player 2's lines share 4 numbers. 

To calculate the number of POL's I use a algorithm that won't be posted here but will use the results gotten from it. 

I need to back up a bit and explain what POL's are.  Remember above where I calculated the number of lines within a matrix that if drawn could win a prize, these totals are based on a single play. When we play more that one line the overall chances will change depending on the numbers played.       

A PLO is any line in the matrix that shares a win with more than one line played.

Example.
Let's say that the set 01-04-08-10-24 is drawn. Both lines of player #1 match 2 numbers,  01-04 and 08-10.  The set drawn is a POL, "Prize Overlap Line"

Player #1's share 3,100 lines/POL's.
Player #2's share 43,797 lines/POL's.

How does this effect the overall chances?

We can see from above what the overall chances are for a single bet game but when playing more than one line we have to make adjustments when we calculate the overall.

Player #1's first line has 65621 chances of winning a prize and so does the second but some of the lines are POL's.

01-02-03-04-05 = 65621 total chances
06-07-08-09-10 = 65621 total chances

There are 3100 lines within the matrix that share prizes with both lines so 65621+65621=131242-3100=128142
This is the overall chances these two lines have of winning a prize in the next draw and 57575/128142=4.493

Now lets look at the overall chances for player #2. These lines both have the same overall chances as player #1 but the POL's are much higher. 65621+65621=131242-43797=87445 so overall chances are 575757/87445=6.584.

We can see from this that player #1 will on average win a prize for every $4.39 he bets while player #2 will win one prize for every $6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

The CC ratio gives us the cost per chance so $2.00/128142=.000015608 vs $2.00/87445=.000022872.

Playing one line per game will always have the lowest cost to chance ratio.  The down side is that our one line will never give better odds than 1 in 8.77.

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  

    CARBOB's avatar - FL LOTTERY_LOGO.png
    ORLANDO, FLORIDA
    United States
    Member #4924
    June 3, 2004
    5893 Posts
    Offline
    Posted: April 28, 2014, 9:59 am - IP Logged

    Are the lines you play giving you the best bang for your dollar?

    Note!
    The cNr calculations are done using the standard windows calculator set to scientific mode. I have gone over this several times to reduce errors but some may still exist. My left hand types faster than my right which causes hard to find errors in the number values. With that being said, here we go.     

    I will keep this simple as possible so that everyone regardless of you math level will be able to understand it. I will show why optimizing our sets will increase our wins by increasing the number of overall chances. Line optimizing is a very valuable tool for anyone who plays more than one line per game.

    Jade posted something similar to this a year or so back and I think this gives a little more insight into the question he posed at that time. If I remember correctly he asked which was a better bet, playing 1 line for 10 games or playing 10 lines in a single game.  I will show how that playing 1 line for 10 games has better overall chances of winning a prize than playing 10 lines for a single game. This is not the reason for this post but it helps to understand it but  first a little math.

    The figures here apply to a pick 5 of 39 game but hold true for any double digits lottery game you just need to plug in the correct variabels. .

    Below are a few problems for which the nCr function can be used.  These are all lottery related and cover various aspects of calculating combinations.

    Below are a couple methods one could use to calculate the number of times any one number appears within a 5-39 matrix.

    Option #1. 575757*5/39 = 73815

    Option #2  (39-1)=38!/(5!*34!)= 501942 and 575757-501942=73815

    The first option is very simple but what if we need to calculate the number of times any 2 numbers appear together. If we multiply 73,815 by 2 we get 73815*2=147624 which is incorrect.

    The correct answer is 7,770.   

    To calculates pairs (39-2)= 37!/(3!*34!)=7770
    To calculates trays (39-3)= 36!/(2!*34!)=630
    To calculates quads (39-4)= 35!/(1!*34!)=35

    Here I show how to calculate the odds using the nCr function and some simple math.

    To do this we first need to know a couple things. Again for this example I will use a 5-39 but if working on a different matrix, just plug in your variables. The list below shows the number of combinations of 2, 3, 4 etc.. within pick 5 and 6 lottery games.

    To find these values using the nCr function follow the example -> 5!/(2!,3!)=10 which shows the total 2 number combos within a set of 5 numbers.

    combinations of 2 in 5 = 10
    combinations of 3 in 5 = 10
    combinations of 4 in 5 = 5
    combinations of 5 in 5 = 1

    combinations of 2 in 6 = 15
    combinations of 3 in 6 = 20
    combinations of 4 in 6 = 15
    combinations of 5 in 6 = 6
    combinations of 6 in 6 = 1     
       
    This list can be easily memorized which will cut down on the number of calculations needed to for most of the problems here.

    First we will calculate the odds for matching 2 of 5 numbers.  To do this we have to plug in the correct variables. Look at this example. 34!/(3!*31!)=5984  Notice the (n) value is 34.  This is gotten by subtracting 5 from 39.  The next variable (r) is 3. To get this value we subtract the prize number match from the total numbers in single line, 5-2=3, The third variable is gotten by subtracting (r) from (n), 34-3=31

    Here are the three variables for calculating 2,3,4 of 5 numbers matches.

    2of5 are #1, 39-5=(34), #2=5-2=(3) and #3=34-3=(31)

    3of5 are #1, 39-5=(34), #2=5-3=(2) and #3=34-2=(32)

    4of5 are #1, 39-5=(34), #2=5-4=(1) and #3=34-1=(33)
     
    34!/(3!*31!)=5984 
    34!/(2!*32!)=561 
    34!/(1!*33!)=34   

    Next we multiply the results by the number of combinations within the a set of 5 numbers.
    combinations of 2 in 5 = 10
    combinations of 3 in 5 = 10
    combinations of 4 in 5 = 5
    combinations of 5 in 5 = 1

    34!/(3!*31!)=5984*10 =59840 
    34!/(2!*32!)= 561*10 = 5610
    34!/(1!*33!)=  34* 5 =  170
    5 of 5          1* 1 =    1               
    ___________________________
    Total                =65621

    This gives us the number of lines within the entire matrix that if drawn will win a prize.

    To calculate the individual odds for matching a prize we divide the total number of sets within the matrix by the total of each possible number match

    Odds of matching 2 numbers = 575757/59840=1 in 9.62     
    Odds of matching 3 numbers = 575757/5610 =1 in 102.63
    Odds of matching 4 numbers = 575757/170  =1 in 3386.80
    Odds of matching 5 numbers = 575757/1    =1 in 575757

    To calculate the overall odds we add all the possible prizes then divide the total lines by this amount. 575757/65821=8.77

    The overall odds are 1 in 8.77 for winning a prize for a 5-39 game.  This applies only to single line plays regardless of the numbers played. What this means is that there are 65,621 possible lines that if any one is drawn we will win a prize.  I know this is a bit redundant but we need to understand it.

    We now know exactly what our overall odds of winning a prize when playing one ticket but what if we play more than one line.   

    Now that we have calculated the total chances one set has for a single drawing we can calculate the total chances for playing one line for 10 games. 65621*10=656210. Playing one line each game gives us the best cost/chance ratio we can get. The CC ratio for a one line play is calculated by dividing the cost by the total chances. $1.00/65621=$0.000015239.

    You may be a bit confused by this so let's dig a little deeper.  What if we play 2 lines? Lets look at the two examples below and I will introduce what I call "prize overlap lines", POL for short. 

    Player #1
    01-02-03-04-05
    06-07-08-09-10

    Player #2
    01-02-03-04-05
    01-02-03-04-06

    Above we have two players that show the extreme ranges for betting on the lottery. Player #1 plays 10 different numbers on his/her 2 lines and player #2 plays 6 total numbers. There are no repeating numbers in the lines of player #1 but player 2's lines share 4 numbers. 

    To calculate the number of POL's I use a algorithm that won't be posted here but will use the results gotten from it. 

    I need to back up a bit and explain what POL's are.  Remember above where I calculated the number of lines within a matrix that if drawn could win a prize, these totals are based on a single play. When we play more that one line the overall chances will change depending on the numbers played.       

    A PLO is any line in the matrix that shares a win with more than one line played.

    Example.
    Let's say that the set 01-04-08-10-24 is drawn. Both lines of player #1 match 2 numbers,  01-04 and 08-10.  The set drawn is a POL, "Prize Overlap Line"

    Player #1's share 3,100 lines/POL's.
    Player #2's share 43,797 lines/POL's.

    How does this effect the overall chances?

    We can see from above what the overall chances are for a single bet game but when playing more than one line we have to make adjustments when we calculate the overall.

    Player #1's first line has 65621 chances of winning a prize and so does the second but some of the lines are POL's.

    01-02-03-04-05 = 65621 total chances
    06-07-08-09-10 = 65621 total chances

    There are 3100 lines within the matrix that share prizes with both lines so 65621+65621=131242-3100=128142
    This is the overall chances these two lines have of winning a prize in the next draw and 57575/128142=4.493

    Now lets look at the overall chances for player #2. These lines both have the same overall chances as player #1 but the POL's are much higher. 65621+65621=131242-43797=87445 so overall chances are 575757/87445=6.584.

    We can see from this that player #1 will on average win a prize for every $4.39 he bets while player #2 will win one prize for every $6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

    The CC ratio gives us the cost per chance so $2.00/128142=.000015608 vs $2.00/87445=.000022872.

    Playing one line per game will always have the lowest cost to chance ratio.  The down side is that our one line will never give better odds than 1 in 8.77.

    RL

    RL,

    Thank you for doing this, very informative. I have a question, in a 5/36 game how many lines would I have to play to ensure each digit is covered at least once. If this takes you away from the direction you are going or headed, forget it.

      RL-RANDOMLOGIC's avatar - usafce

      United States
      Member #59354
      March 13, 2008
      3962 Posts
      Offline
      Posted: April 28, 2014, 10:38 am - IP Logged

      Carbob

      A round figure would be 36/5=7.2.

      Also note that the statement below is incorrect.

      We can see from this that player #1 will on average win a prize for every $4.39 he bets while player #2 will win one prize for every $6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

      The dollar value is doubled because we are playing two lines. You can't drop

      below the overall odds for a dollar bet. The first part proves this is true. 

      Each line we purchase will never have the same overall chances as the previous

      line because of overlaps.  Our chances can however be increased but it comes

      at a price. I should have said for each combined bet and left the dollar sign

      off. Sorry for the mistake, it was due to a edit mistake. The statement was to

      show the differences in the number of chances based on the numbers we put into

      play.

      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  

        RL-RANDOMLOGIC's avatar - usafce

        United States
        Member #59354
        March 13, 2008
        3962 Posts
        Offline
        Posted: April 28, 2014, 10:57 am - IP Logged

        Bob

        I reread your question and I am working on a digit version of this that shows why I play digits.  We don't have

        to cover all the digits the lines in the matrix take care of that.  If we put 5 digits into play, lets say 1-2-3-5-7 then

        we are playing 19 lines in a 5-36. 

        01-02-03-05-07

        11-12-13-15-17

        21-22-23-25-27

        31-32-33-35

        This is over half the numbers in the matrix but let's say that we only play sets that have all 5 digits in each line.

        For my 5-39 I can only build 3 lines before I have to reuse a number.  I will get into detail about this and show

        why it works.  Five digits, 3-base + 2 non-base will reduce the matrix to less than 6000 lines. If I say that each

        non-base digit can only show once in a set then it drops to 3000 and change.  When I first read your reply I

        missed the word digit and thought you were talking numbers, sorry again for yet another mistake.

         

        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  

          RL-RANDOMLOGIC's avatar - usafce

          United States
          Member #59354
          March 13, 2008
          3962 Posts
          Offline
          Posted: April 28, 2014, 11:36 am - IP Logged

          In a nutshell.

          Unless a person can predict the exact numbers then the worst possible bet one could make is to

          play many lines using the same numbers for each set.  The overall chances would not increase at

          all.  The first ticket we buy has the best CC ratio as every additional ticket we purchase drops in

          value.   When I play I don't consider my bet in lines played but the number of overall chances my

          tickets give me.  If I can pay $10.00 for  250,000 chances or $10.00 for 145,000 chances, which one

          is the better bet.  It's not a $1.00 per ticket bet it's a $10.00 fee for 250,000 chances.   

          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  


            United States
            Member #93947
            July 10, 2010
            2180 Posts
            Offline
            Posted: April 28, 2014, 1:40 pm - IP Logged

            RL-RANDOMLOGIC,

            COST PER CHANCE

            "We can see from this that player #1 will on average win a prize for every $4.39 he bets while player #2 will win one prize for every $6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

            The CC ratio gives us the cost per chance so $2.00/128142=.000015608 vs $2.00/87445=.000022872."

            I haven't checked your math but your results are consistent with what is widely known, and that is that the "DISTRIBUTION" of your winnings can be manipulated by choosing betting patterns like those you suggest.  I discussed this in 2010.

            https://www.lotterypost.com/thread/218174/1728520

            Unfortunately, "COST PER CHANCE" is NOT "EXPECTED VALUE," which is really what "winning $" is all about.  BTW, are you sure JadeLottery showed there was any difference over the long haul in the 10/1 vs 1/10 question?

            See my signature line.

            --Jimmy4164

              RL-RANDOMLOGIC's avatar - usafce

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              Member #59354
              March 13, 2008
              3962 Posts
              Offline
              Posted: April 28, 2014, 4:47 pm - IP Logged

              Jimbo

              I was wondering if you could stay away, I guess not.  I also did not make any reference to what Jade's post said

              one way or the other.  As far as I remember he just posted the question for us to ponder, I could be wrong about

              that as I don't remember visiting the post but a few times.   I did pose a very similar question in your last topic but

              you never replied.  Maybe you caught on to my little play on math, maybe not.  However I must say I am glad to see

              you post without using the word innumeracy.  The first statement above was retracted as being incorrect and the $

              sign should have not been used.  The figures are for a 2 line bet without regard to the price of the tickets.  The CC

              ratio I believe to be correct but my haste in getting this up has already yielded a few mistakes.       

              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  

                RL-RANDOMLOGIC's avatar - usafce

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                Member #59354
                March 13, 2008
                3962 Posts
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                Posted: April 29, 2014, 3:19 pm - IP Logged

                Part 2

                First let me say that I will be posting this in several parts due to time restraints.  I have about 10
                projects I am working on and to stay on tract I decided to devote no more than one hour a day to
                lottery.  This includes all my lottery related stuff so it may be a few days between post.

                Now I will talk a little about the digits and why I use them.  All of us are familiar with the base 10
                numerical system.  Using digits is really nothing more than going back to the basics we were taught
                in grade school.  The number 37 represents 3*10 + 7*1 so working with digits is really nothing new.

                Back in the early days of computers, PC's were slow, I mean really slow. Back when I first started
                witting lottery stuff I was limited to using brute force to prove certain calculations which could
                take days or even weeks.  I started looking for shortcuts to prove/solve problems even though I had
                never even heard of a algorithm.  My first algorithm reduced over a 500 lines of code to less than 5
                and took a few minutes to solve a problem that before took days.  I would never have been able to
                figure out how to write that algorithm without reducing numbers to digits and have been using them
                since that time. 

                Digits and the lottery.
                The digits show at a greater frequency then the numbers they represent. The greater the frequency the
                easier it is to approximate when something will show.  I can't predict the numbers but I can make a
                few good approximations of when certain digits will show.  Digit tracking provides more information
                that can be used to analyzing data.  Not all digits have equal overall chances of showing in the next
                draw. Digits alone are not enough to increase your winnings but they are a good place to start.

                System players mostly look for biases which in turn gives them reasons to choose one value over another.
                Without a bias we are left with using methods that are not mathematical. IMHO there are two types of
                tools that can be used for picking numbers. 

                Predictors and system.
                Predictors attempt to predict the best numbers to play with little or no user input and reliey on historic
                data analysis. Systems provide the user with many data types which should show biases for one value over
                another. Predictors make these choices for you systems don't.

                Most here know my stance on the lottery and random but I will go over it again.  A lottery has a finite
                number of combinations which can be drawn.  Each number has a equal chance of being drawn but over time
                the sets or lines drawn will mimic the overall matrix in population and distribution.  This applies not
                only to the numbers but almost any observation we can make. The only random attribute of a lottery is
                the order in which the numbers are drawn.  Analysis of secondary data such as digits can show us a bias
                that is not a product of random but a product of the matrix.  The digit system is really just a lottery
                profiling tool very similar to those used in law enforcement and other disciplines.       

                Below is a list of the last 10 drawings for the MO. cash 5-39 game

                DAY  DATE       NUMBERS        Total Digits   Digits
                =================================================================
                SUN  04/27/14   04 09 11 18 20      6         1-2-4-8-9-0
                SAT  04/26/14   19 20 26 27 33      7         1-2-3-6-7-9-0
                FRI  04/25/14   05 19 21 29 35      5         1-2-3-5-9
                THU  04/24/14   03 23 33 36 39      4         2-3-6-9
                WED  04/23/14   07 10 11 37 38      5         1-3-7-8-0
                TUE  04/22/14   04 11 32 38 39      6         1-2-3-4-8-9
                MON  04/21/14   01 02 07 30 35      6         1-2-3-5-7-0
                SUN  04/20/14   02 03 05 27 30      5         2-3-5-7-0
                SAT  04/19/14   05 16 32 36 39      6         1-2-3-5-6-9
                FRI  04/18/14   01 11 25 30 31      5         1-2-3-5-0


                The digit system divides the digits into two groups called [Base=1-2-3] and [Non-base=4-5-6-7-8-9-0]
                Base digits are digits that can take up either the left or right position within a number as in 32.

                Eighty percent of the last 10 games have 5 or 6 total digits. There is one game with 7 and one with 4.
                Fifty percent of the draws all share the digits 1-2-3 and fourty percent share digits 1-2-3-9.  Digit
                (0) and (9) appear in sixty percent overall.  The digit (0) is not counted when used to pad a single
                digit number as in the number 01.

                I will keep this simple so that it can be done using pen and paper.

                A digit setup is done by first deciding how many digits we want to use in each set. For this example
                I will will be using 6.  The six digits I choose are 1-2-3 + 5-7-9.  Every set I build from these 6
                digits must contain at least one of each digit.

                These six digits repersent the numbers

                01-02-03-05-07-09
                11-12-13-15-17-19
                21-22-23-25-27-29
                31-32-33-35-37-39

                So we have 24 numbers.  There are 42,504 combinations of 5 numbers in 24 but if we say that every set
                must contain at least 1 of every digit then the total combinations are reduced to 5421 a reduction of
                37,083 lines.  If we say that the non-base digits can only be used once then the total combinations
                are reduced to 3033. 

                This is however way to many sets to play so how do we select which numbers to actually play.  Lets say
                that the set drawn is 01-07-23-25-39

                In the 3033 lines there are

                680 that match 2of5
                182 that match 3of5
                 24 that match 4of5
                  1 that match 5of5
                ____________________
                total  = 887 overall chances for winning a prize.

                3033 / 887 = 3.42  overall odds of winning.

                Above is a best case but what if digit 9 does not show and the set 03-22-27-33-35 is drawn.  This set
                has 4 digits 2-3-5-7.

                620 that match 2of5
                112 that match 3of5
                  3 that match 4of5
                  0 that match 5of5
                ____________________
                total = 735 overall chances of winning a prize

                3033 / 735 = 4.13 overall odds of winning a prize.


                Now lets do one more but this time we will add another digit to the set drawn and again leave out digit
                9. Set drawn = 07-15-24-33-38.  We know that the max prize will be a 3of5 because 2 of the numbers 24
                and 38 are not in our list of 24 numbers above and the digits are 1-2-3-4-5-7-8 = 7

                310 that match 2of5
                 29 that match 3of5
                  0 that match 4of5
                  0 that match 5of5
                _____________________
                total = 349 overall chances of winning a prize

                3033 / 349 = 8.69


                Remember the overall odds are 1 in 8.77 overall for winning a prize for this game and all 3 of the simulations
                here fall below that number.  If playing more than one line and use the information posted at the start of this
                topic then you can see that playing digits can work out very good even when the actual draw takes another
                direction.  As with any method we do need to make so many correct selections but we don't have to be spot on.

                In part 3 I will dig a little deeper and show how to select the numbers to play from the list.

                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  

                  CARBOB's avatar - FL LOTTERY_LOGO.png
                  ORLANDO, FLORIDA
                  United States
                  Member #4924
                  June 3, 2004
                  5893 Posts
                  Offline
                  Posted: April 29, 2014, 4:32 pm - IP Logged

                  RL, I'm with you, thank you.


                    United States
                    Member #93947
                    July 10, 2010
                    2180 Posts
                    Offline
                    Posted: April 30, 2014, 1:54 am - IP Logged

                    Jimbo

                    I was wondering if you could stay away, I guess not.  I also did not make any reference to what Jade's post said

                    one way or the other.  As far as I remember he just posted the question for us to ponder, I could be wrong about

                    that as I don't remember visiting the post but a few times.   I did pose a very similar question in your last topic but

                    you never replied.  Maybe you caught on to my little play on math, maybe not.  However I must say I am glad to see

                    you post without using the word innumeracy.  The first statement above was retracted as being incorrect and the $

                    sign should have not been used.  The figures are for a 2 line bet without regard to the price of the tickets.  The CC

                    ratio I believe to be correct but my haste in getting this up has already yielded a few mistakes.       

                    RL

                    RL-RANDOMLOGIC,

                    You said,  "I also did not make any reference to what Jade's post said one way or the other.  As far as I remember he just posted the question for us to ponder, I could be wrong about that as I don't remember visiting the post but a few times."


                    His simulation supported the theoretical answer in that it resulted in no statistically significant difference between "1 Play for 10 Draws" vs  "10 Plays for 1 Draw."

                    https://www.lotterypost.com/thread/229884

                    Your Cost Per Chance [of winning some
                    thing] is NOT the EXPECTED VALUE of a Play, or set of plays.  Consequently, your methods will produce more HITS, but in the long run, your WINNINGS will be the same as with Quick Picks or any other selection method you choose.  A stack of $1 and $2 dollar winning tickets doesn't buy you much, especially when you consider what you spent to accumulate them.  If your goal is to impress the lottery outlet clerks with lots of small wins, you're on the right track! You are not the only player who believes that more 1 and 2 number matches increases your chances of matching 3, 4, and 5/6.  You are wrong.

                    Your saving grace is that your methods will definitely not decrease your winnings. Smile

                    --Jimmy4164

                      JADELottery's avatar - MeAtWork 03.PNG
                      The Quantum Master
                      West Concord, MN
                      United States
                      Member #21
                      December 7, 2001
                      3675 Posts
                      Offline
                      Posted: April 30, 2014, 11:45 am - IP Logged

                      Are the lines you play giving you the best bang for your dollar?

                      Note!
                      The cNr calculations are done using the standard windows calculator set to scientific mode. I have gone over this several times to reduce errors but some may still exist. My left hand types faster than my right which causes hard to find errors in the number values. With that being said, here we go.     

                      I will keep this simple as possible so that everyone regardless of you math level will be able to understand it. I will show why optimizing our sets will increase our wins by increasing the number of overall chances. Line optimizing is a very valuable tool for anyone who plays more than one line per game.

                      Jade posted something similar to this a year or so back and I think this gives a little more insight into the question he posed at that time. If I remember correctly he asked which was a better bet, playing 1 line for 10 games or playing 10 lines in a single game.  I will show how that playing 1 line for 10 games has better overall chances of winning a prize than playing 10 lines for a single game. This is not the reason for this post but it helps to understand it but  first a little math.

                      The figures here apply to a pick 5 of 39 game but hold true for any double digits lottery game you just need to plug in the correct variabels. .

                      Below are a few problems for which the nCr function can be used.  These are all lottery related and cover various aspects of calculating combinations.

                      Below are a couple methods one could use to calculate the number of times any one number appears within a 5-39 matrix.

                      Option #1. 575757*5/39 = 73815

                      Option #2  (39-1)=38!/(5!*34!)= 501942 and 575757-501942=73815

                      The first option is very simple but what if we need to calculate the number of times any 2 numbers appear together. If we multiply 73,815 by 2 we get 73815*2=147624 which is incorrect.

                      The correct answer is 7,770.   

                      To calculates pairs (39-2)= 37!/(3!*34!)=7770
                      To calculates trays (39-3)= 36!/(2!*34!)=630
                      To calculates quads (39-4)= 35!/(1!*34!)=35

                      Here I show how to calculate the odds using the nCr function and some simple math.

                      To do this we first need to know a couple things. Again for this example I will use a 5-39 but if working on a different matrix, just plug in your variables. The list below shows the number of combinations of 2, 3, 4 etc.. within pick 5 and 6 lottery games.

                      To find these values using the nCr function follow the example -> 5!/(2!,3!)=10 which shows the total 2 number combos within a set of 5 numbers.

                      combinations of 2 in 5 = 10
                      combinations of 3 in 5 = 10
                      combinations of 4 in 5 = 5
                      combinations of 5 in 5 = 1

                      combinations of 2 in 6 = 15
                      combinations of 3 in 6 = 20
                      combinations of 4 in 6 = 15
                      combinations of 5 in 6 = 6
                      combinations of 6 in 6 = 1     
                         
                      This list can be easily memorized which will cut down on the number of calculations needed to for most of the problems here.

                      First we will calculate the odds for matching 2 of 5 numbers.  To do this we have to plug in the correct variables. Look at this example. 34!/(3!*31!)=5984  Notice the (n) value is 34.  This is gotten by subtracting 5 from 39.  The next variable (r) is 3. To get this value we subtract the prize number match from the total numbers in single line, 5-2=3, The third variable is gotten by subtracting (r) from (n), 34-3=31

                      Here are the three variables for calculating 2,3,4 of 5 numbers matches.

                      2of5 are #1, 39-5=(34), #2=5-2=(3) and #3=34-3=(31)

                      3of5 are #1, 39-5=(34), #2=5-3=(2) and #3=34-2=(32)

                      4of5 are #1, 39-5=(34), #2=5-4=(1) and #3=34-1=(33)
                       
                      34!/(3!*31!)=5984 
                      34!/(2!*32!)=561 
                      34!/(1!*33!)=34   

                      Next we multiply the results by the number of combinations within the a set of 5 numbers.
                      combinations of 2 in 5 = 10
                      combinations of 3 in 5 = 10
                      combinations of 4 in 5 = 5
                      combinations of 5 in 5 = 1

                      34!/(3!*31!)=5984*10 =59840 
                      34!/(2!*32!)= 561*10 = 5610
                      34!/(1!*33!)=  34* 5 =  170
                      5 of 5          1* 1 =    1               
                      ___________________________
                      Total                =65621

                      This gives us the number of lines within the entire matrix that if drawn will win a prize.

                      To calculate the individual odds for matching a prize we divide the total number of sets within the matrix by the total of each possible number match

                      Odds of matching 2 numbers = 575757/59840=1 in 9.62     
                      Odds of matching 3 numbers = 575757/5610 =1 in 102.63
                      Odds of matching 4 numbers = 575757/170  =1 in 3386.80
                      Odds of matching 5 numbers = 575757/1    =1 in 575757

                      To calculate the overall odds we add all the possible prizes then divide the total lines by this amount. 575757/65821=8.77

                      The overall odds are 1 in 8.77 for winning a prize for a 5-39 game.  This applies only to single line plays regardless of the numbers played. What this means is that there are 65,621 possible lines that if any one is drawn we will win a prize.  I know this is a bit redundant but we need to understand it.

                      We now know exactly what our overall odds of winning a prize when playing one ticket but what if we play more than one line.   

                      Now that we have calculated the total chances one set has for a single drawing we can calculate the total chances for playing one line for 10 games. 65621*10=656210. Playing one line each game gives us the best cost/chance ratio we can get. The CC ratio for a one line play is calculated by dividing the cost by the total chances. $1.00/65621=$0.000015239.

                      You may be a bit confused by this so let's dig a little deeper.  What if we play 2 lines? Lets look at the two examples below and I will introduce what I call "prize overlap lines", POL for short. 

                      Player #1
                      01-02-03-04-05
                      06-07-08-09-10

                      Player #2
                      01-02-03-04-05
                      01-02-03-04-06

                      Above we have two players that show the extreme ranges for betting on the lottery. Player #1 plays 10 different numbers on his/her 2 lines and player #2 plays 6 total numbers. There are no repeating numbers in the lines of player #1 but player 2's lines share 4 numbers. 

                      To calculate the number of POL's I use a algorithm that won't be posted here but will use the results gotten from it. 

                      I need to back up a bit and explain what POL's are.  Remember above where I calculated the number of lines within a matrix that if drawn could win a prize, these totals are based on a single play. When we play more that one line the overall chances will change depending on the numbers played.       

                      A PLO is any line in the matrix that shares a win with more than one line played.

                      Example.
                      Let's say that the set 01-04-08-10-24 is drawn. Both lines of player #1 match 2 numbers,  01-04 and 08-10.  The set drawn is a POL, "Prize Overlap Line"

                      Player #1's share 3,100 lines/POL's.
                      Player #2's share 43,797 lines/POL's.

                      How does this effect the overall chances?

                      We can see from above what the overall chances are for a single bet game but when playing more than one line we have to make adjustments when we calculate the overall.

                      Player #1's first line has 65621 chances of winning a prize and so does the second but some of the lines are POL's.

                      01-02-03-04-05 = 65621 total chances
                      06-07-08-09-10 = 65621 total chances

                      There are 3100 lines within the matrix that share prizes with both lines so 65621+65621=131242-3100=128142
                      This is the overall chances these two lines have of winning a prize in the next draw and 57575/128142=4.493

                      Now lets look at the overall chances for player #2. These lines both have the same overall chances as player #1 but the POL's are much higher. 65621+65621=131242-43797=87445 so overall chances are 575757/87445=6.584.

                      We can see from this that player #1 will on average win a prize for every $4.39 he bets while player #2 will win one prize for every $6.58. For my 5-39 a match-2 breaks even and 3 and up makes money.

                      The CC ratio gives us the cost per chance so $2.00/128142=.000015608 vs $2.00/87445=.000022872.

                      Playing one line per game will always have the lowest cost to chance ratio.  The down side is that our one line will never give better odds than 1 in 8.77.

                      RL

                      It was If you had 10 bucks to bet... back in 2011.

                      We keep our posted topics in our favorites, http://www.lotterypost.com/member/1170/favorites

                      Presented 'AS IS' and for Entertainment Purposes Only.
                      Any gain or loss is your responsibility.
                      Use at your own risk.

                      Order is a Subset of Chaos
                      Knowledge is Beyond Belief
                      Wisdom is Not Censored
                      Douglas Paul Smallish
                      Jehocifer

                        JADELottery's avatar - MeAtWork 03.PNG
                        The Quantum Master
                        West Concord, MN
                        United States
                        Member #21
                        December 7, 2001
                        3675 Posts
                        Offline
                        Posted: April 30, 2014, 12:36 pm - IP Logged

                        This what we be usin'

                         

                        Factorial - n! = n · (n -1) · (n - 2) · ... · 3 · 2 · 1 and 0! = 1

                        Permutation - P(n, r) = n! / (n - r)!

                        Combination - C(n, r) = P(n, r) / r!

                        Probability of Win - Wp(n, r, w) = (C(r, w) · C(n - r, r - w)) / C(n, r)

                        Odds to Win - Wo(n, r, w) is 1 in (C(n, r) - C(r, w) · C(n - r, r - w)) / (C(r, w) · C(n - r, r - w))

                        n - total numbers in lottery pool
                        r - pick size from the pool of numbers
                        w - winning drawn numbers in a played combo

                        Presented 'AS IS' and for Entertainment Purposes Only.
                        Any gain or loss is your responsibility.
                        Use at your own risk.

                        Order is a Subset of Chaos
                        Knowledge is Beyond Belief
                        Wisdom is Not Censored
                        Douglas Paul Smallish
                        Jehocifer

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                          United States
                          Member #154821
                          April 30, 2014
                          2 Posts
                          Offline
                          Posted: April 30, 2014, 3:21 pm - IP Logged

                          RL

                          It is spring and I am getting the lottery fever too.

                          Thanks for the information you have posted so far.

                            SergeM's avatar - slow icon.png
                            Economy class
                            Belgium
                            Member #123700
                            February 27, 2012
                            4035 Posts
                            Offline
                            Posted: April 30, 2014, 3:24 pm - IP Logged

                              Combinations(39,5) = 39!/[(39-5)!5!] = 39*38*37*36*35/5/4/3/2/1 

                              RL-RANDOMLOGIC's avatar - usafce

                              United States
                              Member #59354
                              March 13, 2008
                              3962 Posts
                              Offline
                              Posted: April 30, 2014, 7:19 pm - IP Logged

                              Jade

                              I tried to format the combo stuff by keystroke using the windows calculator, I figured some might not be fimilar

                              with the nCr function and this would make it easier.  Big thanks for sharing.

                              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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