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A Modest Proposal

Topic closed. 43 replies. Last post 13 years ago by Colin F.

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Australia
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December 22, 2003
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Posted: May 23, 2004, 12:02 am - IP Logged
Quote: Originally posted by rdarmand on May 22, 2004



Quote: Originally posted by Colin F on May 22, 2004



Let's up the ante a bit. Our new variation of coin tossing requres you to bet on the results of the next 6 tosses. So your ticket will be something like HTHTHH. rdArmand are you going to bet HHHHHH, TTTTTT or something else

Colin





I'm not sure I understand the thrust of your question.





rdarmand

You do. Answer the question.

I'm not a Chess Player but that doesn't mean I can't think ahead a few moves. You've worked out what comes next and that's not agreeable to your play gambit.

I look forward to testing your system.

Regards
Colin

(To other members: watch this guy - he's up to something. Defend your beliefs to the utmost. THIS IS A CALL TO ARMS!!!) 

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    Posted: May 23, 2004, 4:45 am - IP Logged
    Quote: Originally posted by Colin F on May 23, 2004



    Quote: Originally posted by rdarmand on May 22, 2004



    Quote: Originally posted by Colin F on May 22, 2004



    Let's up the ante a bit. Our new variation of coin tossing requres you to bet on the results of the next 6 tosses. So your ticket will be something like HTHTHH. rdArmand are you going to bet HHHHHH, TTTTTT or something else

    Colin





    I'm not sure I understand the thrust of your question.





    rdarmand

    You do. Answer the question.






    Geez, Colin, you don't have to be rude.

    If I say I don't understand, I mean it. The reason I didn't understand is because you seemed to be asking me to bet on the next six flips -- which is completely different from what I was discussing with Win D, which was betting on the next one draw, regardless of what previous draws have been.

    If you don't see the difference, then that's the source of your difficulty.

    Betting on Heads (Tails) for the next six flips is a 2^6 proposition, with 31-to-1 odds against being right. Betting on the next one flip -- even when the last five flips are all Heads (Tails) -- is a 2^1 proposition, with 1-to-1 odds against being right.

    Why would I want to bet on the next six flips? Unless, of course, you give me great odds.

    Where n is equal to the number of tickets bought, the odds of winning in the MegaMillions lottery is given by 1 - [(1 - 0.023399196957)^n]


      Australia
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      Posted: May 23, 2004, 10:40 am - IP Logged

      rdarmand

      That's the reaction I was looking for. Where is Mr Cool now? I never raised the idea of having a bet with you. If you look ahead 6 Draws and pick something other than HHHHHH or TTTTTT then your saying that something other than those is more likely to happen. If it can be that case in the future why is it inapplicable to the past?

      The cause of your sudden frustration and anger which contrasts so strongly with your previous composure is because you now realize you've just been blown out of the water.

      I don't need to quote anyone; what I know I've confirmed. But I thank you for the opportunity to express what I know better.

      Colin

        S.Windsor, CT
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        Posted: May 23, 2004, 11:36 am - IP Logged
        Quote: Originally posted by Colin F on May 22, 2004



        Quote: Originally posted by rdarmand on May 21, 2004



        Probability theory absolutely denies that the notions of hot numbers and the law of averages have any relevance to lottery drawings. As far as statistics is concerned, all systems based on these notions are moonshine.


        That being said &







        rdarmand

        Could you tell me the names of these Probability Books that make direct reference to "Hot" numbers and the "Law of Averages" as applied to Lotto Games.

        I know there are  books which I deliberately have not read by Prof Jones, Gail Howard et al who probabably expouse them.

        In relation to Lotto the two books I purchased and found very usefull were:

        Combinatorics  by V K Balakrishnan  Professor of Mathematics University of Maine

        The Probability Tutoring Book  by Carol Ash  University of Illinois

        No mention in either of those books.

        In any case whatever you do or call it if it produces results better that  random selections you can't scoff at it. I had the pleasure of meeting Sir Edmund Hillary as a school boy and I bet a few people told him he couldn't climb Mt Everest. When he was at the top I bet they were still saying it.

        Colin




          S.Windsor, CT
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          Posted: May 23, 2004, 11:39 am - IP Logged

          To ColinF

          Another book you might read is mention by me under Mathematics. It is "How to Win More" by N,Henze&

          H.Riedwyl.

            Prometheus1's avatar - trace9

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            Posted: May 23, 2004, 11:40 am - IP Logged

             

             Well done Colin!  Simple and easy to understand by all. Jolly good!  That pretty well sums it up. He can't have it both ways. Perhaps, it was just a misunderstanding. We can only hope.

              PROMETHEUS       

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              Posted: May 23, 2004, 12:05 pm - IP Logged
              Quote: Originally posted by Colin F on May 23, 2004



              If it can be that case in the future why is it inapplicable to the past?

              Colin





              Colin,

              I've already explained this, in a post you yourself applauded. Take a look at it again; maybe something will click for you that didn't click before: http://www.lotterypost.com/thread/87083

              You seem to think you've scored some kind of victory because I differentiated between odds when no coin tosses are known, and odds when the previous 5 or 6 tosses are known. The odds are different because they address two different questions. The odds for the future can't be the odds for the past.  Sorry. That's why all predictive systems are so much crapola.

              If you want to understand, go back and read the old post again. If you just want to rant, then rant.

              Where n is equal to the number of tickets bought, the odds of winning in the MegaMillions lottery is given by 1 - [(1 - 0.023399196957)^n]

                Prometheus1's avatar - trace9

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                Posted: May 23, 2004, 12:36 pm - IP Logged

                 http://www.saliu.com/Saliu2.htm 

                 From a mathematicians point of view radarmand. Enjoy. 

                  PROMETHEUS       

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                  Posted: May 23, 2004, 2:12 pm - IP Logged
                  Quote: Originally posted by Prometheus1 on May 23, 2004



                   http://www.saliu.com/Saliu2.htm 

                   From a mathematicians point of view radarmand. Enjoy. 





                  Interesting. It's not a difficult idea.

                  Where a is the likelihood of not winning, b is some sought-after degree of probability of not winning, and x is the number of trials necessary to achieve b:



                  x = (log b)/(log a)


                  EXAMPLE: Your odds of getting the top prize in Pick 3 are 0.001; therefore your odds of not winning the top prize are 0.999. The number of trials necessary to achieve a 50% chance of not winning the top prize are: (log 0.5)/(log 0.999) = -0.301029995664/-0.000434511774 = 692.8. The webpage says 692; perhaps we can attribute that to rounding.

                  EXAMPLE: Your odds of winning any prize in the MegaMillions lottery are 0.023399196957. The number of trials necessary to achieve a 50% chance of not winning any prize are therefore (log 0.5)/(log 0.976600803043) = -0.301029995664/-0.010282922938 = 29.3.

                  Most people would prefer to think in terms of winning, rather than not winning; just subtract your desired odds of winning from 1 and use the difference for your b value.

                  Where n is equal to the number of tickets bought, the odds of winning in the MegaMillions lottery is given by 1 - [(1 - 0.023399196957)^n]

                    Prometheus1's avatar - trace9

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                    Posted: May 23, 2004, 2:43 pm - IP Logged

                     Damn, you have very selective eyes radarmand. The point of showing you that site was to politely warn you against taking that bet.   Win d was going to take you for every penny you had! It was a sucker's bet. If you continue to ignore these helpful hints and selectively overlook helpful instruction your either a fool or just from a dead planet. Good Luck, it's all you have going for you. 

                     Number of TrialsNNecessary For An Event of Probability p to Appear With The Degree of Certainty DC








                    DC


                    ¯

                    p=

                    .90

                    p=

                    .80

                    p=

                    .75

                    p=

                    .66

                    p=

                    1/2

                    p=

                    1/3

                    p=

                    1/4

                    p=

                    1/6

                    p=

                    1/8

                    p=

                    1/10

                    p=

                    1/16

                    p=

                    1/32

                    p=

                    1/64

                    p=

                    1/100

                    p=

                    1/1,000

                    10% - - - - - - - - - 1 1 3 6 10 105
                    25% - - - - - - 1 1 2 3 4 9 18 28 287
                    50% 1 1 1 1 1 1 2 3 6 7 10 21 44 68 692
                    75% 1 1 1 2 2 3 4 7 11 13 21 43 88 137 1,385
                    90% 1 2 2 2 3 5 8 12 17 22 35 72 146 229 2,301
                    95% 1 2 2 3 4 7 10 16 22 29 46 94 190 298 2,994
                    99% 2 3 3 4 7 11 16 25 34 44 71 145 292 458 4,602
                    99.9% 3 4 5 6 10 17 24 37 52 66 107 217 438 687 6,904

                         Let's try to make sense of those numbers. The easiest to understand are the numbers in the column under the heading p=1/2. It analyzes the coin tossing game of chance. There are 2 dvents in the game: heads and tails. Thus, the individual probability for either dvent is p = 1/2. Look at the row 50%: it has the number 1 in it. It means that it takes 1 dvent (coin toss, that is) in order to have a 50-50 chance (or degree of certainty of 50%) that either heads or tails will come out. More explicitly, suppose I bet on heads. My chance is 50% that heads will appear in the 1 st coin toss. The chance or degree of certainty increases to 99.9% that heads will come out within 10 tosses!

                      PROMETHEUS       

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                      Posted: May 23, 2004, 2:55 pm - IP Logged

                      Sorry, Prometheus. It's you who don't understand what you're reading.

                      Where n is equal to the number of tickets bought, the odds of winning in the MegaMillions lottery is given by 1 - [(1 - 0.023399196957)^n]

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                        Posted: May 23, 2004, 9:22 pm - IP Logged

                        Some Ion Saliu influence i see going here Prometheus? I'll admit that he is a clever fellow. Anyways, this subject has already been concluded in the other post "mathematics-probabilty formula of likely dvents" and i've already stated that there is, and always will be 1 in 2 chance of either heads or tails landing. Everyone here seems to be correct in there own way, just misunderstanding oneanother.

                        here's the link to that post  http://www.lotterypost.com/thread/87083

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                          Posted: May 23, 2004, 9:33 pm - IP Logged

                          I would like to add my 2 ce


                            Australia
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                            Posted: May 24, 2004, 3:34 am - IP Logged

                            Intermission

                            As I walk along the Bois de Boulogne
                            With an independent air;
                            You can see the people stare,
                            You can hear them all declare,
                            'Oh he is a millionaire!'
                            You can see them sigh and wink their eye,
                            And then they wink their other eye
                            At the Man who broke the bank at Monte Carlo.

                            Name: Charles Wells
                            Actual Winnings: $200,000
                            Method: Played the Wheel 11 hours a day for 3 days

                            End Intermission

                            Colin