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quiz pick-4 daily game odds

Topic closed. 48 replies. Last post 12 months ago by RL-RANDOMLOGIC.

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Posted: January 22, 2016, 2:41 pm - IP Logged

What are the odds of matching 1 digit  to the correct position with 4 selections?  10% * 4  40%

What are the odds of matching 2 digits to the correct position with 4 selections?    1% * 3    3%

What are the odds of matching 3 digits to the correct position with 4 selections?     .01% *2  .02%

Not quite.

The first is almost as straightforward as what you did, but you missed one detail. For getting at least 1 of the 4 numbers right you've got a 1 in 10 chance for each of the 4 positions, so 4 * 10% = 40%. To get only 1 of 4 correct you have to get the other 3 wrong. That means you need to multiply the probability of getting at least 1 right by the probability of getting 3 of them wrong. That makes it .4 * .9^3 = .2916, or 29.16%. 1 in 3.43

For getting 2 right it's the same thing, but there 6 ways to get it right, not 3: AB, AC, AD, BC, BD, CD. So getting at least 2 right is .1*.1 *6 = .06. Once again you have to get the others (2 this time)wrong, which is .9*.9 = .81. Thus .81*.06 = .0486, or 4.86%. 1 in 20.58

To get at least 3 numbers is .1^3 = .001 (that's .1%, not .01%), and there are 4 ways to have 3 right (ABC, ABD, ACD, BCD). That gives us 4*.001 = .004. You've only got to get one number wrong, so .9*.004 = .0036, or .36%.  1 in 277.78

That last one is really easy to check against the actual numbers. If the winning number is 1234 there are 10 ways you could have 123?, but one of the 10 would give you all 4. That leaves 9 ways to have the 123 correct and the 4 wrong. The same is true for each of the other numbers. There are 10 ways to have 12?4, and one of them is 1234. Thus 36 of the 10,000 possibilities would give you 3 of the 4, so 36/10,000 = .0036

    JADELottery's avatar - Inky Move-01.gif
    The Quantum Master
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    Posted: January 22, 2016, 4:17 pm - IP Logged

    (C(n, m) · 9(n - m)) / 10n

    n - pick size

    m - match size

    C(n, m) = n! / (m! · (n - m)!)

    n! is the factorial; where n! = n · (n - 1) · (n - 2) · … · 3 · 2 · 1 and 0! = 1

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    Jehocifer

      JADELottery's avatar - Inky Move-01.gif
      The Quantum Master
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      Posted: January 22, 2016, 4:41 pm - IP Logged

      the odds are 1 : (10n / (C(n, m) · 9(n - m))) - 1

       

      9(n - m) is 9 raised to (n - m)

      superscript not high enough visually.

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      Jehocifer

        JADELottery's avatar - Inky Move-01.gif
        The Quantum Master
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        Posted: January 22, 2016, 4:46 pm - IP Logged
        MatchProbabilityOdds
        065.61%1 : 0.52
        129.16%1 : 2.43
        24.86%1 : 19.58
        30.36%1 : 276.78
        40.01%1 : 9999.00

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        Order is a Subset of Chaos
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        Jehocifer

          RL-RANDOMLOGIC's avatar - usafce

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          Posted: January 22, 2016, 11:42 pm - IP Logged

          Jade

          Bingo.  In another post I started playing and posting a few pick-4 games and my hit rates for

          2 positional digits were very high and along the way I also had a few lines with 3 positional

          digits.  I was sure that I was close to hitting a straight which led me to work out the odds.

          Turns out that a 3 digit positional hit is way, way and I do mean way far away from a 4 of 4. 

          A 3 positional hit is not really worth mention.  In 74 lines spread out over around 18 plays I

          had 3 tickets with 3 positional digits.  Average 74/3=24.67.  This is over 10x better than the

          odds would suggest but close to a straight, not at all.  I guess if my attempts were consistant

          then I could expect a hit on around 1000 lines but that's just a guess-ta-ment.   

           

          RL

          P.S.  I had hoped you would hold off a little longer before posting.  LOL

          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

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            RL-RANDOMLOGIC's avatar - usafce

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            Posted: January 22, 2016, 11:51 pm - IP Logged

            KY

            I missed your post so it looks like you beat jade to the punch.  Jade looks to have a typo as it's 277.78

            for a 3 positional match not 276.78

            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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              The Quantum Master
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              Posted: January 23, 2016, 1:20 am - IP Logged

              KY

              I missed your post so it looks like you beat jade to the punch.  Jade looks to have a typo as it's 277.78

              for a 3 positional match not 276.78

              RL

              Those are true odds, the ratio of success to failure, typically reduced to 1 success to failure or 1 : least number of failures.

              If p is the probability of success, then the odds become 1 : (1 / p) - 1

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                The Quantum Master
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                Posted: January 23, 2016, 1:28 am - IP Logged

                In the odds for matching 4 it is 1 : 9999.

                It means for every 1 possible success there are 9999 possible failures.

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                Order is a Subset of Chaos
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                Jehocifer

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                  The Quantum Master
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                  Posted: January 23, 2016, 1:34 am - IP Logged

                  The match 0 was written 1 : 0.52 just to be consistent with 1 : least number of failures.

                  It could have been written 2 : 1.04, or 2 successes for approximately 1.04 failures.

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                  Order is a Subset of Chaos
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                    RL-RANDOMLOGIC's avatar - usafce

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                    Posted: January 23, 2016, 5:34 am - IP Logged

                    Jade

                    I built a calculator to get my values and the way it works is to first calculate the number of lines

                    needed to cover the number of matches then it divides, 10,000/36=277.778.  In your post above

                    your odds for a matching-3 positional digits is listed as 1 in 276.778.  I figured it was just a typo.

                    I don't understand the calculation for matching 0 digits as I get 10,000/6,561 = 1 in 1.524.     

                    My calculations to cover each match from 0 to 4

                    0 = 6,561

                    1 = 2916

                    2 = 486

                    3 = 36

                    4 = 1

                    OK, it just hit me, thanks for the correction.

                    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: January 23, 2016, 8:56 am - IP Logged

                      I started playing pick-4 a while back and while thinking about the odds I came to the conclusion

                      that it's not as easy as one might think.  Lets say that we match 3 of the 4 digits to the correct

                      position, what are the odds for doing this.  One might think matching 3 digits would be 1 in 1000

                      but this is not true unless you only select 3 of the 4 digits places to play.

                       

                      Math odds quiz.  Straight only

                      What are the odds of matching 1 digit  to the correct position with 4 selections?

                      What are the odds of matching 2 digits to the correct position with 4 selections?

                      What are the odds of matching 3 digits to the correct position with 4 selections?

                      RL

                      Maybe you should  rephrase  your question:

                      Math odds quiz.  Straight only

                      How many picks  for each position one digit(504 or 1000)

                      How many picks for a Front Pair(100) or any pair(???) 

                      How many picks for Front Triad(10) or any triad (260??)

                      Probability, hence odds , is  a derivative of the binomial format NPr or NCr. So why not concentrate on this filter for your strategy and cost analysis. If you're very good with posed scenario, then start testing even with the large picks.Often times, large picks is 'put off' for  some folks(scared of cost), but if 504 picks can earn me 5k within 5 draws, then the consistency should be tested. Note that the number of picks  can be reduced by the size of  N.

                      Good Luck

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                        The Quantum Master
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                        Posted: January 23, 2016, 10:09 am - IP Logged

                        Jade

                        I built a calculator to get my values and the way it works is to first calculate the number of lines

                        needed to cover the number of matches then it divides, 10,000/36=277.778.  In your post above

                        your odds for a matching-3 positional digits is listed as 1 in 276.778.  I figured it was just a typo.

                        I don't understand the calculation for matching 0 digits as I get 10,000/6,561 = 1 in 1.524.     

                        My calculations to cover each match from 0 to 4

                        0 = 6,561

                        1 = 2916

                        2 = 486

                        3 = 36

                        4 = 1

                        OK, it just hit me, thanks for the correction.

                        RL

                        ok, good.

                        the lottery industry odds are the following:

                        1 : (10n / (C(n, m) · 9(n - m)))

                        it's just off by 1 and the reciprocal of the probability, p, or 1 / p.

                        this works for any Daily Game Pick Size lottery: Pick 0, Pick 1, Pick 2, Pick 3, Pick 4, ...

                        where n ≥ 0, m ≥ 0 and m ≤ n.

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                        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

                          RL-RANDOMLOGIC's avatar - usafce

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                          Posted: January 23, 2016, 4:38 pm - IP Logged

                          adobea

                          I did mention straights only just above the odds questions.  I am new to the daily games and as such a bit

                          behind in my math for these games.  When I first considered the odds I made estimates without working

                          them out, turns out, my estimates were way off.  I built a program and was surprised to see that the odds

                          for matching 3 positional digits in a pick-4 game was only 1 in 276.87.  Before this I thought I was really doing

                          well.  The wife then brought up the method we use to select out digits.  I am still trying to come up with the

                          probabilities / odds for the steps.

                          We use a Same/change chart and my steps program to select digits.  There are 3 steps for each of the 4 digit

                          positions for a total of 12 values that have to be set.  We are averaging 9 to 11 of the 12 steps game to game

                          but I am having a hard time figuring out the odds.  The steps values selected are converted to digits so IMHO

                          I should get a better sense of how I am doing by calculating the odds for the steps.  Hitting 11 out of 12 steps

                          seems to be doing better than the odds for a 3 positional hit.  The steps program converts the digits into a

                          3 digit code.

                          The three digit code is a follows.

                          Step-1 breaks the digits into 3 groups,  1=(012), 2=(3456) and step-3=(789)

                          Step-2 is odd or even 1=odd 2=even

                          Step-3 is 1 = low and 2=high.

                          Step codes

                          121=digit 0

                          111=digit 1 

                          122=digit 2

                          211=digit 3

                          221=digit 4

                          212=digit 5

                          222=digit 6

                          311=digit 7

                          321=digit 8

                          312=digit 9

                          codes 112 and 321 are not used.

                          Step-1 sets the group to use. Step-2 says it will be odd or even and step-3 takes the lower or highest remaining value.

                          There are still 10 total choices for each digit selected so the overall odds are are the same but I find it's much easier to

                          select the correct digit by analyzing the codes 1 step at a time.

                           

                          Here is another problem I am working to solve.  What are the odds for hitting 9, 10 11 etc of the 12 step values needed

                          in a pick-4 game?  This is straights only because I am working on the digits by position.  If I hit 11 of the 12 I will only

                          match 3 positional digits but steps wise I had 91.6% correct.  I want to calculate the expected cost and number of plays

                          to hit a straight estimated on my test plays.

                           

                          I also want to figure out is how much advantage if any I am gaining from using the steps vs picking the digits from a pool

                          of 0 to 9.  Pick-4 is a hit-miss game, a 3 of 4 don't count. 

                           

                          Test play data.

                          I have played 18 games and 74 lines total.  My average for 2 positional digit hits is better than 1 to 1 overall and in the

                          18 games played I have had 3ea lines with 3 positional hits hitting 11 of the 12 steps.   My average for the 3 positional

                          hits is 1 in 24.7.

                          From this data I should be able to come up with a good estimate of how long I can expect to play and the cost of winning

                          a p-4 straight. 

                           

                          Here is a scan of one of the 11 of 12 step plays.  You can see that I missed the 4th digit by one step value.  Based on

                          the steps this play came closer than a 3 of 4 hit.  In the 4 position I managed to hit step-1 with odds of 1 in 3 and step-2

                          with odds of 1 in 2.  If looking only at the digits, I hit 3 of 4 for 75% but steps wise I hit 91.6%   What are the odds for

                          hitting 9, 10 11 etc.. of the 12 steps?  Remember that step-1 has odds of 1 in 3 and step-2 and step-3 are 1 in 2. 

                          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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                            Posted: January 23, 2016, 5:03 pm - IP Logged

                            P.S.

                            Playing large is a option as my 2 positional digit hits are better than 1 to 1 per play overall.  To cover

                            a 6-way p-4 pair bet require 486 lines.  MO pays $6K for a $1.00 P-4 str8, the only downside is filling

                            out 486 boards every day or twice a day if mid and evening games are played.

                            Here are a couple pics of my positional wheel and the prize matches.  winning line = 7664

                             

                            P.S. What I mean by the above hit rate is that I play anywhere from 1 to 5 lines per game and I

                            often manage more than one line with 2 positional digit hits and in a few games every line had

                            two positional digits.

                             

                            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: January 23, 2016, 6:37 pm - IP Logged

                              For the odds it is often easier and faster to generate and count them.