A winning p-3 strategy(I think)Previous TopicNext Topic

  I would like some feedback on something I came up this while I was sitting  around drinking beer and talking to my dog(she's a wonderfull listener) Let me show you what I think I got

     We know that about 70% of the time a # from last draw will be reapeated in the next draw.ok. If I wheel the #'s so that 1 #is combined with every other #:

 last draw(s.c.) was 179, assuming 1# will repeat gives me the following

   012-013-014-015-016-018-027-029-037-039-047-049-057-059-067-069-078-089-123-124-125-126-128-134-135-136-138-145-146-148-156-158-168-237-239-247-249-257-259-267-269-278-289-347-349-357-359-367-369--378-389-457-459-467-469-478-567-569-578-589-678-689

  If I haven't mistyped something that should give me 63 #'s.

 At 1$ a pop that's 17$ profit if I box em' If I play them all srt it's 122 profit a pop.Of course I wouldn't want to try and play every draw but if I could hit  this even 4or 5 times week I could do pretty good for myself.The whole thing seems so straight foward and simple that a little voice in the back of my head is saying, wait a minute milo,your'e fixin to step into a pile of it.Buuuuut I only see a few way's I could lose and they can be avoided (mostly,lol)

 one,doubles or trips come up,I lose.doubles csn be dodged (mostly)

 two,a # from the previous doesn't repeat.I lose Also can be dodged(mostly)

 three,a reapeating pair from the previous draw could come up,I lose.I came with a system for winning with repeating pairs a few months ago that I thought was gonna work great.Thank God I played it on paper first.it would have ate my lunch and been back for supper,lol.still I'm pretty sure even these can be avoided. If nothin else it only costs 22$ to play for returning pairs if I REALLY think it's coming.I could add them to my wheel but that would eat up my profit and put the whole thing in the red.Course I could play them by thierselve's and,hoooowee!Is this why they call it gambling?lmao

  Anyway I'm throwing this out here for y'all to look at and think about.Any comments ideas,criticisims will be appreciated.Good luck to all.

South Carolina Evening

1 (and only 1) digit repeated 1,371 times out of 2,802 draws.

That is 48.9% of the time.

I don't know how you wheeled them, but if you use:

3 digits from the last draw to get 1 digit out of, you still have 7 digits left for each of the other 2 positions.

So:

3 X 7 X 7 = A wheeler gave me 126 combos, but there will of course be boxed repeats there, as there are only 120 boxed singles.

So, there will be:

It is as you said, 63 boxed singles, but Jack says that they might only come about 49% of the time, so 60 + combinations might still be too many.

Maybe don't play the most overdue boxed digit of the 10 digits, that is filter out combinations which have that one digit, depends how overdue it might be, maybe also filter out some boxed singles that already came out, but check stats on boxed singles before you try that.

Maybe filter out a few boxed pairs from past draw(s), but first check their stats to be sure.

Check stats of Sums, LDRs, Roots, Widths and see if any patterns can be filtered out from them.

63 Boxed singles:

034
035
036
037
038
039
045
046
047
048
049
056
057
058
059
067
068
069
078
079
089
134
135
136
137
138
139
145
146
147
148
149
156
157
158
159
167
168
169
178
179
189
234
235
236
237
238
239
245
246
247
248
249
256
257
258
259
267
268
269
278
279
289

Fernando,

He is using ONE and ONLY one repeat from previous draw.  The 63 is correct for that.

Yes, He is!

The straight 3 x 7 x 7 of course gives more combos, but there are repeats there, once the repeats are deleted or filtered out there are only 63 Boxed Singles!

thx for the info jack but I play both the eve&mid as one contionus game.For the purpose of general discusion I'm not sure it matters what one game in paticular is doing.I understand that differnt games will behave in ways that contradict what the rules of probabilty and common sense tell us they will do.The thing is I belive that if a random system behaves abnormaly(48%vrs70%)then it is smart money to bet that the game will start to behave in a way to return back to random.

  I hope I'm making some sense here. I'm not an accomplished mathemitiction and I have to take it on faith that the idea a coin toss will come up heads or tails is 50/50 is not a provable idea but one accepted as a basic assumption and needing no mathmatical proof.The ideal that one # will repeat from the previous draw 70%+ of the time is one derived from the so-called laws of probability and I have to accept them .

thx for your input jack

I have yet to see ANY state or ANY draw that has 1 and only 1 repeat anywhere near 70%.  You say that it is in the "laws of probability", so please show me how you come up with that.  I use actual data to come up with my percentages.

If you have a state/draw that is more than 49% for 1 and only 1 digit repeat, maybe you could show that data.

Thank you.

thx for the reply lantern.Here is thwheel I'm using.

012(last draw)

34567

89(the rest of the #'s)

034-035-036-037-038-039-045-046-047-048-049-056-057-058-059-067-068-069-078-079-089-134-135-136-137-138--139-145-146-147-148-149-156-157-158-159-167-168-169-178-179-189-234-235-236-237-238-239-245-246-247-248-249-256-257-258-259-267-268-269-278-279-289

  If I was not going to use a paticular # I think it might be simpler to just exclude it and build a wheel w/o that #

I have a couple of spreadsheets that I made up a few months ago for doing 1 and only 1 repeat.  Since someone pointed out that the number that repeats most of the time repeats in a different position, one of my spreadsheets does that for me.

Jack,man you are forcing me to actually go check on something that I have always to as a matter of course.thx for that and damm you for forceing me to do all that extra work lol.

 good luck to you

For GA last night's draw of 651 ... here are the 63 combos with the digits 6, 5, and 1 in different positions than they hit.

062 063 064 067 068 069 102 103 104 107 108 109 213 214 217 218 219 235 245 263 264 267 268 269 275 285 295 314 317 318 319 364 367 368 369 417 418 419 467 468 469 475 485 495 502 503 504 507 508 509 534 537 538 539 589 718 719 768 769 785 795 819 869

Also Jack, I seem to have mistated what I meant.Let me rephrase like this.70% of the time AT LEAST one  #will come up that is a repeat from the prvious draw.To be honest you've got me wondering about that assumption

gl

I'm not sure what "assumption" you are talking about.  Please notice that in ALL of my posts I stated ONE and ONLY one number repeat.  If you are talking about 1, 2, or 3 digits repeating, then 70% is probably correct.  From the very start of this thread we were talking about having ONLY 1 digit repeat.

I started this thread.notice my #3 way to lose

From your very first post:

 last draw(s.c.) was 179, assuming 1# will repeat gives me the following

   012-013-014-015-016-018-027-029-037-039-047-049-057-059-067-069-078-089-123-124-125-126-128-134-135-136-138-145-146-148-156-158-168-237-239-247-249-257-259-267-269-278-289-347-349-357-359-367-369--378-389-457-459-467-469-478-567-569-578-589-678-689

  If I haven't mistyped something that should give me 63 #'s.

Using ONE and ONLY one digit from previous draw gives you 63 combos.  I agree.