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So what would the "Holy Grail" of Pick 3 playing be?

Topic closed. 268 replies. Last post 3 years ago by Atomic Dog.

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Posted: October 29, 2014, 12:38 am - IP Logged

 Well, it depends on whether it's straight or box. So, if you're boxing consecutive digits then it's expected to ,on average, hit once in every 16 drawings. But, if it's straight then it's ,on average, once in every 100 drawings.

 Anyway, we can't really speak about when something is due. This is the same thing as saying that we know when it's going to come out. But, it may go well passed the range of 17 to 23 drawings without hitting . In this case, we can say that the game is more ripe for consecutives.

I probably shouldn't use the word "due" and said the mathematical probability is once every 22 drawings. The window of play is no more than 8 drawings unless a betting strategy for extended play and still show a profit is created. The idea is for the group to hopefully perform within probability and only played when the timing is right.

Consecutive numbers hit more than probability in my state in the last 3 years (about 1000 drawings). For play a skip chart could be used to determine how many times they hit within a range. Not saying it's easy to do, but a few bucks could be made with a little work.


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    Posted: October 29, 2014, 12:47 am - IP Logged

    Looks like your every-day-garden-size-variety-self-proclaimed lottery expert to me.

    "Ahh, the old days when everything was new. Some very talented folks use to drop by back then.  Strong willed and just full of beans and brains too they were. LOL"

    I've read lots of lottery related articles on the Net and never saw anything resembling a "Holy Grail" of pick-3 systems, but I never expected to find it either. However if there is such a thing, I'll never find it reading articles about the evils of gambling by someone like our local con-man who believes there are more than two people on LP giving him any credibility.

    There is nothing wrong with pointing out the probabilities and the huge odds against, but to make their case they usually have to ignore other probabilities and statistics that are deciding factors in making the bet. At the end of the day, it comes down to the concept of making simple choice on what to bet and how much to bet on it; something the people on the dark side can't comprehend.

     Well, there ,now, is such a thing as a "holy grail" for the Pick 3. But, unfortunately, you'd never find it ,as a whole, reading any articles ,or otherwise, anywhere. Although, you may ,and can, find articles that ,more or less, touch upon the concepts and methodology involved.


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      Posted: October 29, 2014, 1:05 am - IP Logged

      I think these 88 boxes are the closest you can get to the "Holy Grail". Put them into 4 groups

      Mostly High

      056 057 058 059 067 068 069 078 079 089 156 157 158 167 168 169 178 179 189 256 257 259 267 268 269 278 279 289 356 358 359 367 368 378 379 389 457 458 459 467 469 478 479 489

      Mostly Low

      015 016 017 018 019 025 026 027 028 029 035 037 038 039 045 046 047 049 125 126 127 128 129 136 137 138 139 145 146 148 149 235 236 237 238 239 245 247 248 249 346 347 348 349

      Mostly Odd

      013 015 017 019 035 037 039 057 059 079 125 127 129 134 136 138 145 149 156 158 167 169 178 189 235 237 239 257 259 279 347 349 356 358 367 378 389 457 459 479 569 578 589 679

      Mostly Even

      014 016 018 023 025 027 029 034 038 045 047 049 056 058 067 069 078 089 124 126 128 146 148 168 236 238 245 247 249 256 267 269 278 289 346 348 368 458 467 469 478 489 568 689

       Well, the "holy grail" is much more involved than ,simply, grouping numbers. It's ,literally, an entire logical construct of how and when to play.

        PeerGynt's avatar - nw archer.jpg
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        Posted: October 29, 2014, 1:26 am - IP Logged

         So, where does the law of averages say that ,at some point, the subset will hit within five plays of it already hitting? But, you're right that the proposed strategy doesn't really work? Why not? That's because it's more involved than that. You'd need to take the patterns into account too.

        Of course it should!  Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)

        If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.

        If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.

        Or a third.

        Or a fourth.

        Or a fifth.

        Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.

        And how many do you need, really... if the pay-outs work in your favor?...

        But the CA algorithm prevents this.  Or, it sure as hell seems to.

        You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!

        Wink

        We have no dreams at all, or interesting ones. We should learn to be awake the same way—not at all, or in an interesting manner.   -- Friedrich Nietzsche


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          Posted: October 29, 2014, 1:30 am - IP Logged

          "Well, of course, there are groups that do hit more often than others."

          Would you care to expand on this statement?  It sounds like something that should be discussed on LP!

           Well, as we know, the numbers can be broken down into groups of different sizes. So, groups containing more numbers can be expected to hit more often. For example, the group of singles hits more often than that of doubles. Likewise, doubles hit much more often than triples. Anyway, the list goes on. But, I'm sure we all know this already.


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            Posted: October 29, 2014, 2:00 am - IP Logged

            When to play, of course, is always dicey - if one knew exactly when to play, one could do quite well; and even the phrase "when to play" assumes one knows what to play - no one needs to know when to play losing numbers!

            If the computer-run D3 games didn't work off algorithms, one could just select any set of numbers, and start tracking them in a very orderly, dull fashion, and consistently win.  Everyone could win with relatively little effort.  The whole reason the algorithm is there, is to avoid just such playing - no matter how tiny or how large you slice and dice the numbers into groups, those groups don't play according to a consistency inherent to purely random selection processes.

            But do notice how the game seems to have an eerie consistency to it, and a dependable one at that.  Take pairs: track some random pairs.  Note how some'll come real quick, and how others will be out for long periods of time... maybe even multiple long periods of time... but then... always... eventually... a short amount of time.

            So you get, just talking about pairs now: Pairs that are out short periods of time, we'll say, <7 plays.  And then you get pairs that are out vast amounts of time... a few times in a row, sometimes... but then finally, they, too, are out <7 plays.

            Think about that for a minute. There's something in there worth thinking about.  Wink

             Well, it seems you're underestimating the difficulty involved in a game that's not using algorithms. But, why do you believe that it's not so hard to beat a purely random game? If that's true, then there wouldn't be so many people losing with mechanical ball drawings.


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              Posted: October 29, 2014, 2:17 am - IP Logged

              Yes, they are. They are favored to appear as the lowest number in the set. 85 out of the 120 P3 boxed singles have 0, or 1, or 2, as the lowest number. On the same token, the 7 or 8 or 9 will appear as the highest number in the set on 85 out of the 120 single boxed draws, so in that aspect you are correct. 

              The point I was trying to make was that if you know that 0, or 1, or 2, will be the lowest number 71% of the time, why would you not use a combo containing one or two of them? You are just putting the odds in your favor.

              On the P4, it is 175 out of the 210 boxed singles, or 83%.

               Well, of course, any of digits 0,1,2 may ,or will, be the lowest to appear when any of them do come out. But, they don't have any advantage over any other set of 3 digits in terms of odds. For example, the digits 3,4,5 will occur about as often as 0,1,2 will occur.


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                Posted: October 29, 2014, 3:26 am - IP Logged

                I probably shouldn't use the word "due" and said the mathematical probability is once every 22 drawings. The window of play is no more than 8 drawings unless a betting strategy for extended play and still show a profit is created. The idea is for the group to hopefully perform within probability and only played when the timing is right.

                Consecutive numbers hit more than probability in my state in the last 3 years (about 1000 drawings). For play a skip chart could be used to determine how many times they hit within a range. Not saying it's easy to do, but a few bucks could be made with a little work.

                 So, how are you coming up with 22 drawings using 10 consecutives? Perhaps, you're referring to consecutive pairs? Also, Kentucky has two drawings each day and only one on Sunday. So, it only takes about a year and a half to reach 1000 drawings. 

                 Anyway, there will be times when a group hits more ,or less, often than it's expected to. But, counting how many times a group hits will not be much help. So, suppose you determine that consecutives hit less often than what would be expected in a given year. The same thing could happen the following year too. What's more, you'd still have to know when to play.


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                  Posted: October 29, 2014, 3:36 am - IP Logged

                  Of course it should!  Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)

                  If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.

                  If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.

                  Or a third.

                  Or a fourth.

                  Or a fifth.

                  Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.

                  And how many do you need, really... if the pay-outs work in your favor?...

                  But the CA algorithm prevents this.  Or, it sure as hell seems to.

                  You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!

                  Wink

                   Well, first of all, there are 210 numbers when boxing. Also, triples can't be boxed. But, the law of averages says nothing about how many plays. This is just what you're saying.

                    Tialuvslotto's avatar - Jailin
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                    Posted: October 29, 2014, 8:09 am - IP Logged

                     Well, first of all, there are 210 numbers when boxing. Also, triples can't be boxed. But, the law of averages says nothing about how many plays. This is just what you're saying.

                    I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.

                    This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws.  The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.

                    And you can similarly define a playing window for almost any situation the game can throw at you.

                    "There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

                    ~Robert A. Heinlein

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                      Posted: October 29, 2014, 11:56 am - IP Logged

                      Of course it should!  Assuming we're first not talking about a stupidly big set of numbers (like the Mega or Super or Power - fuggedaboutit), but we're talking about the D3, which is either 1000 numbers, or if you're playing boxes, it's only 175 numbers (and that includes ten rarely seen triples!)

                      If you were to set your limit at playing your sub-group - say, five picks - to five plays out, say (i.e., giving it five tries); and then (assuming five losses) stop and wait until the game "resets" - that is, wait until one of your set comes up, and THEN start all over again - the law of averages says, come on, you know this, at some point, maybe after many many attempts, one of those numbers WILL hit, within five plays.

                      If you're so completely crazily unlucky that that one set you chose doesn't for way too long to really keep track of... then surely, it can't happen to a second set.

                      Or a third.

                      Or a fourth.

                      Or a fifth.

                      Now we're just talking a rigged game, if you can't get ONE GROUP out of FIVE to repeat in a system like this.

                      And how many do you need, really... if the pay-outs work in your favor?...

                      But the CA algorithm prevents this.  Or, it sure as hell seems to.

                      You know what though?... there's an easy way to check for sure... actually do it... anyone here can do this... and you can do it for free!

                      Wink

                      Use this link for your sub groups> inference is personal

                      CA>

                      Drawing DatePick 3Pick 4
                      MiddayEveningMiddayEvening
                      Tue, Oct 28, 20144-7-91-2-69-6-5-1
                      Mon, Oct 27, 20142-5-18-6-07-1-9-8
                      Sun, Oct 26, 20144-6-66-7-72-0-2-6
                      Sat, Oct 25, 20141-3-29-8-37-7-6-5
                      Fri, Oct 24, 20147-8-19-4-26-0-4-5
                      Thu, Oct 23, 20145-6-80-5-33-3-1-3
                      Wed, Oct 22, 20148-7-02-5-32-1-8-3
                      Tue, Oct 21, 20143-3-81-9-65-4-8-8
                      Mon, Oct 20, 20146-2-68-8-06-7-6-4
                      Sun, Oct 19, 20143-2-04-7-35-1-5-7

                      draw 320> sequential digit difference (Triggers)>1-2 (ie 3-1 and 2-0)

                      T1> 1,3,8,0      T2> 2,4,7,9

                      Grouping for prediction interval points>A/ 0-1-2-3-4,   B/ 7-8-9 

                      wheel each group  or form pairs from each group for box and straight hits>

                      Inference> A draw set targets a section of the pool(Subsetting), so you remain focus, keep things simple!, we're not

                      not bothered  any parametric percentile , we're only concerned with the subset each moment--lottery is random!

                      draw>473> triggers> 3-4 >T3>3568, T4> 4657,, Grouping A/ 3-4-5    , 6-7-8

                      You form pairs from each group or wheel them> 34x-35x-45x  , 67x-68x-78x

                       

                      P4

                      5175>seq difference(T) 4-6-2>T4>4657,,,T6>6835,   T2>2479

                      Grouping> 2-3-4-5     6-7-8-9   , form triads with each group and add your x

                       

                      6764>1-1-2>T1>1380,T2>2479, Groups>A 0-1-2-3-4,  B/7-8-9>  form triads from each group

                      NB> The ideal of hitting the next draw is delusional, is not even cost effective, if your subset  has >50% degree of confidence(by observation), then why not let time run it course(Time frame).Look at across the groups, your x is right there!

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                        Posted: October 29, 2014, 3:17 pm - IP Logged

                        I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.

                        This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws.  The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.

                        And you can similarly define a playing window for almost any situation the game can throw at you.

                        I don't agree with 90% its much lower more like 70%.

                        Jimmy

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                          Posted: October 29, 2014, 4:41 pm - IP Logged

                          Use this link for your sub groups> inference is personal

                          CA>

                          Drawing DatePick 3Pick 4
                          MiddayEveningMiddayEvening
                          Tue, Oct 28, 20144-7-91-2-69-6-5-1
                          Mon, Oct 27, 20142-5-18-6-07-1-9-8
                          Sun, Oct 26, 20144-6-66-7-72-0-2-6
                          Sat, Oct 25, 20141-3-29-8-37-7-6-5
                          Fri, Oct 24, 20147-8-19-4-26-0-4-5
                          Thu, Oct 23, 20145-6-80-5-33-3-1-3
                          Wed, Oct 22, 20148-7-02-5-32-1-8-3
                          Tue, Oct 21, 20143-3-81-9-65-4-8-8
                          Mon, Oct 20, 20146-2-68-8-06-7-6-4
                          Sun, Oct 19, 20143-2-04-7-35-1-5-7

                          draw 320> sequential digit difference (Triggers)>1-2 (ie 3-1 and 2-0)

                          T1> 1,3,8,0      T2> 2,4,7,9

                          Grouping for prediction interval points>A/ 0-1-2-3-4,   B/ 7-8-9 

                          wheel each group  or form pairs from each group for box and straight hits>

                          Inference> A draw set targets a section of the pool(Subsetting), so you remain focus, keep things simple!, we're not

                          not bothered  any parametric percentile , we're only concerned with the subset each moment--lottery is random!

                          draw>473> triggers> 3-4 >T3>3568, T4> 4657,, Grouping A/ 3-4-5    , 6-7-8

                          You form pairs from each group or wheel them> 34x-35x-45x  , 67x-68x-78x

                           

                          P4

                          5175>seq difference(T) 4-6-2>T4>4657,,,T6>6835,   T2>2479

                          Grouping> 2-3-4-5     6-7-8-9   , form triads with each group and add your x

                           

                          6764>1-1-2>T1>1380,T2>2479, Groups>A 0-1-2-3-4,  B/7-8-9>  form triads from each group

                          NB> The ideal of hitting the next draw is delusional, is not even cost effective, if your subset  has >50% degree of confidence(by observation), then why not let time run it course(Time frame).Look at across the groups, your x is right there!

                          CAL>Fantasy 5>Matrix 5/39,  Set N=39> see link for Triggers to form subsets(groups)

                          Tue, Oct 28, 201408-09-19-20-28?Prize Payouts
                          Mon, Oct 27, 201413-18-22-25-37?Prize Payouts
                          Sun, Oct 26, 201403-08-09-17-31?Prize Payouts
                          Sat, Oct 25, 201410-22-26-29-30?Prize Payouts
                          Fri, Oct 24, 201402-11-22-25-37?Prize Payouts
                          Thu, Oct 23, 201403-08-17-27-29?Prize Payouts
                          Wed, Oct 22, 201406-14-33-34-37?Prize Payouts
                          Tue, Oct 21, 201402-11-13-21-35?Prize Payouts
                          Mon, Oct 20, 201406-12-21-28-35?Prize Payouts
                          Sun, Oct 19, 201409-18-32-37-39?Prize Payouts
                          Sat, Oct 18, 201402-05-25-37-39?Prize Payouts
                          Fri, Oct 17, 201401-08-12-18-27?Prize Payouts
                          Thu, Oct 16, 201411-18-22-27-28?Prize Payouts
                          Wed, Oct 15, 201401-03-19-29-36?Prize Payouts
                          Tue, Oct 14, 201409-14-20-21-34?Prize Payouts
                          Mon, Oct 13, 201409-22-27-29-35?Prize Payouts
                          Sun, Oct 12, 201403-13-20-31-32?Prize Payouts
                          Sat, Oct 11, 201414-16-17-20-33?Prize Payouts
                          Fri, Oct 10, 201401-02-06-19-34?Prize Payouts
                          Thu, Oct 9, 201403-20-21-28-39?Prize Payouts
                          Wed, Oct 8, 201402-18-23-33-39?Prize Payouts
                          Tue, Oct 7, 201401-08-30-34-35?Prize Payouts
                          Mon, Oct 6, 201404-09-10-32-35?Prize Payouts
                          Sun, Oct 5, 201404-12-14-23-24?Prize Payouts
                          Sat, Oct 4, 201405-09-22-32-39?Prize Payouts
                          Fri, Oct 3, 201403-11-23-30-39?Prize Payouts
                          Thu, Oct 2, 201403-15-26-32-34?Prize Payouts
                          Wed, Oct 1, 201402-15-30-35-37?Prize Payouts
                          Tue, Sep 30, 201408-20-24-33-39?Prize Payouts
                          Mon, Sep 29, 201402-03-13-18-38?Prize Payouts
                          Sun, Sep 28, 201413-22-23-25-29?Prize Payouts
                          Sat, Sep 27, 201411-13-16-21-28?Prize Payouts
                          Fri, Sep 26, 201410-21-22-30-36?Prize Payouts
                          Thu, Sep 25, 201401-03-12-27-38?Prize Payouts
                          Wed, Sep 24, 201404-06-16-29-33?Prize Payouts
                          Tue, Sep 23, 201413-14-22-28-33?Prize Payouts
                          Mon, Sep 22, 201403-15-26-30-37?Prize Payouts
                          Sun, Sep 21, 201407-15-17-21-28?Prize Payouts
                          Sat, Sep 20, 201412-14-15-18-20?Prize Payouts
                          Fri, Sep 19, 201405-08-16-29-37?Prize Payouts

                           

                          Draw 5-8-16-29-37   >seq .digit difference >3-8-13-8>Triggers>T3> 3,5,36,38     T8>8,10,31,33     T1>1,3,38,1

                          Grouping(interval points for T's)>A/1-2-3-4-5      B/ 8-9-10       C/31-32-33      D/36-37-38

                          WHEEL ABDCD> 2 numbers from a group gives you 3/5 hits   

                          Draw 12-14-15-18-20> 2-1-3-2>T2>2,4,37,39    T1>1,3,38,1    T3> 3,5,36,38

                          GROUPING(Interval points-range) >A/1-2-3-4-5-6      B/36-37-38-39 , Wheel AB 

                           

                          Draw>7-15-17-21-28> 8-2-4-7>T8> 8,10,31,33   T2>2,4,37,39     T4>4,6,35,37    T7> 7,9,32,34

                          Grouping> A/2-3-4-5   B/6-7-8-9-10     C/31-32-33-34    D/35-36-37-38-39

                           

                           

                          REDUCED WHEEL FOR ABCD is just 28 picks for a month wage for possible 5/5 hit,  3/5 hits will keep you in play during

                          the time frame

                            grwurston's avatar - Lottery-050.jpg
                            Winning is great.
                            bel air maryland
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                            Posted: October 29, 2014, 6:54 pm - IP Logged

                             Well, of course, any of digits 0,1,2 may ,or will, be the lowest to appear when any of them do come out. But, they don't have any advantage over any other set of 3 digits in terms of odds. For example, the digits 3,4,5 will occur about as often as 0,1,2 will occur.

                            If you know the lowest number. Pick 3. 120 Single Boxed Numbers. Arranged By Root Sums.

                            Zero  36 #s

                            RS1 019-028-037-046  RS2 029-038-047-056  RS3 012-039-048-057  RS4 013-049-058-067  RS5 014-023-059-068

                            RS6 015-024-034-079  RS7 016-025-034-079  RS8 017-026-035-089  RS9 018-027-036-045

                            One  28 #s

                            RS1 127-136-145  RS2 128-137-146  RS3 129-128-147-156  RS4 139-148-157  RS5 149-158-167  RS6 123-159-168

                            RS7 124-169-178  RS8 125-134-179  RS9 126-135-189

                            Two  21 #s

                            RS1 235-289  RS2 236-245  RS3 237-246  RS4 238-247-256  RS5 239-248-257  RS6 249-258-267  RS7 259-268

                            RS8 269-278  RS9 234-279

                            Three 15 #s

                            RS1 379  RS2 389  RS3 345  RS4 346  RS5 347-356  RS6 348-357  RS7 349-358-367  RS8 359-368  RS9 369-378

                            Four  10 #s

                            RS1 469-478  RS2 479  RS3 489  RS6 456  RS7 457  RS8 458-467 RS9 459-468

                            Five  6 #s 

                            RS1 568  RS2 569-578  RS3 579  RS4 589  RS9 569

                            Six  3 #s 

                            RS3 678  RS4 679  RS5 689

                            Seven  1 #

                            RS5 789


                              United States
                              Member #155994
                              June 5, 2014
                              497 Posts
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                              Posted: October 29, 2014, 11:49 pm - IP Logged

                              I have to take issue with you on this one -- the law of averages does tell us how many plays, and that information is contained in the probability distribution.

                              This is the basis for Win D's double trap: If you haven't had a double for 7 draws, play doubles for 3 draws.  The probability is 90% of a double falling on skip 8, 9 or 10 in this situation.

                              And you can similarly define a playing window for almost any situation the game can throw at you.

                               So, exactly how, or where, are you getting 90%? But, admittedly, a logical basis was just found ,within my system, for Win D's double trap. Even so, I still need to gather empirical data to support this. This would require the existence of certain, significant patterns.

                                 
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