A winning p-3 strategy(I think)

Reply #19

milochance — Mon, 09 Nov 2009 23:50:19 GMT

I know I can't put up with too many , on the other hand i'm talking about gambling and how I can increase my chances of going home with the money.I just saw a way where I might be able to do that. A lot of pepole seem to disagree with the the way I figured my chances and that's a good thing.I believe down to my toes that money can be made playing p-3.i'm going to look at my original idea (tomorow'thank you) and see what I think then.I just think that if I do it right (avoid the three losers I po... [ More ]

Reply #18

LANTERN — Mon, 09 Nov 2009 23:32:43 GMT

SC draws the X are repeat digit(s) from the last draw

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When ever we can we try to be High Tech.

Reply #17

Coin Toss — Mon, 09 Nov 2009 23:09:38 GMT

Assuming $1 per play, how many consecutive $63 losses are people willing to put up with, or how many times putting $63 in action and collecting $40?

milochance, you may want to consider:

https://www.lotterypost.com/thread/199654

Reply #16

milochance — Mon, 09 Nov 2009 22:15:23 GMT

Thank you for that info.Just out of curiosity,what happens if you ignore doubles

Reply #15

paurths — Mon, 09 Nov 2009 20:43:26 GMT

Here's some live data, concerning 1 digit repeats:

CT eve: every 2.13 draws

GA mid: every 2.18 draws

GA eve: every 2.09 draws

and so on...

70% is no where near the average...

2 digit returns happen on average every

CT eve: every 7.4 draws

GA mid: every 7.7 draws

GA eve: every 8.1 draws

these averages are valid for all states...

cheers

Ricky

Reply #14

Jack-C — Mon, 09 Nov 2009 17:27:05 GMT

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.... [ More ]

Reply #13

milochance — Mon, 09 Nov 2009 17:23:36 GMT

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

Reply #12

Jack-C — Mon, 09 Nov 2009 17:15:42 GMT

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.

Reply #11

milochance — Mon, 09 Nov 2009 17:12:03 GMT

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

Reply #10

Jack-C — Mon, 09 Nov 2009 17:00:30 GMT

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

Reply #9

milochance — Mon, 09 Nov 2009 16:58:44 GMT

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

Reply #8

Jack-C — Mon, 09 Nov 2009 16:55:00 GMT

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.

Reply #7

milochance — Mon, 09 Nov 2009 16:51:59 GMT

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

Reply #6

Jack-C — Mon, 09 Nov 2009 16:39:56 GMT

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.

Reply #5

milochance — Mon, 09 Nov 2009 16:31:29 GMT

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... [ More ]

Reply #4

LANTERN — Mon, 09 Nov 2009 16:23:58 GMT

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

Reply #3

Jack-C — Mon, 09 Nov 2009 16:20:44 GMT

Fernando,

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

Reply #2

LANTERN — Mon, 09 Nov 2009 16:18:51 GMT

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 th... [ More ]

Reply #1

Jack-C — Mon, 09 Nov 2009 15:22:56 GMT

South Carolina Evening

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

That is 48.9% of the time.

A winning p-3 strategy(I think)

milochance — Mon, 09 Nov 2009 14:42:43 GMT

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... [ More ]