Help JADELottery

Reply #35

Thoth — Sat, 18 Nov 2006 08:22:51 GMT

Actually my sample is not flawed for what I wanted to know. What I wanted to know was the probability of the event itself, which to me is defined as the same boxed combo having been drawn in the specific 17 different states and entities a total of 21 times from October 11th through November 8th. This period of time included 780 individual drawings. I didn't intend my question to get an answer with much meaning or utility as I only asked it becaused I was curious of the chance of it occuring... [ More ]

Reply #34

ayenowitall — Sat, 18 Nov 2006 06:34:20 GMT

Nevermind; I undertsand what you're counting now, but I think your sample is seriously flawed if you're only going to count the drawings in states where the 014 box occurred. If that's what you're doing, why even bother to count the days or drawings when the 014 box didn't show up in those states? I don't understand what meaning or utility can be attached to any statistics derived from the sample you're using.

Reply #33

ayenowitall — Sat, 18 Nov 2006 00:58:49 GMT

Thoth,

What about Arizona, Florida, Idaho, Indiana, Kansas, Louisiana, Maryland, Nebraska, North Carolina, New Jersey, Ohio, Oklahoma, Ontario, South Carolina, Texas, the Tri-State, Virginia, D.C., West Virginia, and Wisconsin?

What source are you using? Check out the Daily Numbers Games page from the Results drop-down menu.

Reply #32

Bryan — Fri, 17 Nov 2006 00:56:01 GMT

I was mistaken. It went 5 days this time. And I played it straight yesterday, but not tonight.

Draw Date State Game Results

Thu, Nov 16, 2006PennsylvaniaDaily Number0-4-1

Reply #31

Thoth — Thu, 16 Nov 2006 17:02:55 GMT

DRAWS:111 10-1110-1210-1310-14

StateKeySunMonTuesWedThursFriSat

CA2(M E) 2222

DE2(M E) 2222

MI2(M E) 2222

MO2(M E) 2222

PA2(M E) 2222

NY2(M E) 2222

IL2 No Sun M 2222

KY2 No Sun M 2222

GA2 No Sun M 2222

TN2 No Sun M 2222

CT2 No S S M 2221

WC1 E 1111

QB1 E 1111

WA1 E 1111

MN1 E 1111

NM1 E No Sun 1111

PR1 E No Sun 1111

DRAWS:18810-1510-1610-1710-1810-1910-2010-21

StateKeySunMonTuesWedTh... [ More ]

Reply #30

Thoth — Thu, 16 Nov 2006 06:05:18 GMT

I noticed the same problem with duplication, I'll post the solution I used to get around it in a few days when I have time.

I'm gonna recheck that 979 draw count, can't believe I made an error that large lol (sure sounds like I did though)...but then again, I was pretty tired the night I added 'em all up.

Reply #29

JADELottery — Wed, 15 Nov 2006 16:41:39 GMT

Thoth,

I'm still working on this. However, I think it may have led me to something more interesting, this may become a new topic on it's own. I can not explain too much now. Thank you for sparking the idea. This may have already been doen by someone else, but I'm not sure. Most of the research I gone through tends to suggest it may not have been tried before. I'll let you know or look for a new topic in the Mathematics Forum. Thanks again.

Reply #28

Bryan — Wed, 15 Nov 2006 07:40:59 GMT

Thanks for checking it out guys. I'm going to continue to play it every 3rd day it misses until it shows the trend is over. Today is the 3rd day. It still hasn't when over the 3rd day. So, it really doesn't have to hit until tomorrow (16th) for it to break the pattern it has been following.

Wish me Luck,

Bryan :)

Thanks Again...

Reply #27

JADELottery — Wed, 15 Nov 2006 06:31:39 GMT

Did a count by one week, it's 376 draws per week, removing one of the Iowa-Illinois repeats. That's 1504 draws for 4 weeks. The draw count of 1596 is correct. Also, I did a year to date check out for all states up to 45 weeks and the 014 hit count is within acceptable limits. There were a total of 16,920 draws; 014 had a .0161764 probabilty of hitting 115 times, that's about 1.61794 % or 1 in 61 chance of hitting 115 times. There's no 014 anomaly, it's probably in a correction/catch-up cycle for... [ More ]

Reply #26

ayenowitall — Wed, 15 Nov 2006 01:42:12 GMT

Select the Daily Numbers Games option from the Results drop-down menu on this site. Count the draws. I think you'll see that my count of 1569 draws is a lot closer to the correct amount than 979.

Reply #25

JADELottery — Tue, 14 Nov 2006 07:09:19 GMT

As x gets larger when s = 10, the value of y drops to near 0. Most average computers don't deal with numbers that have 1E-800's as exponents. It just converts it to the nearest value of 0. Besides, you really want to use this equation for higher values of s in the 100's, 1000's, or larger.

s =

10

x

y

as a %

0

1.580540418

158.0540418

1

0.001032712

0.103271243

2

3.89863E-014

3.89863E-012

3

8.5036E-032

8.5036E-030... [ More ]

Reply #24

jarasan — Tue, 14 Nov 2006 03:54:17 GMT

y = (1 / 0.012 p s) e -((x - 0.006 s)²/(0.012 s)

x - number of hits based on a given sample size

s - number of samples

y - probability that x number of hits will have happen by s number of samples

note: to get percent probability multiply y by 100% Ok jade. lets do a little dimensional analysis.

x= 10

s =10

Pi= 3.14

e=2.73 e E1

y= (1/sqrootof .012 x 3.14 x 10) x e exp ( 10-.006 x10)sqd./ (.012x10)

y = 1/sqrtof .3768 x e exp(99.88/ .12... [ More ]

Reply #23

JADELottery — Mon, 13 Nov 2006 11:37:17 GMT

Thoth,

Just writing to let you know, I'm still working this one out in my mind. Also, I will be gone for a few days on other projects. I'll check now and then to see what's going on, but probably won't have much time to post. Anyway, I want to pass on a few thoughts about the problem you restated. That helped me understand it more. I actually worked on something similar to this a while ago. It had to do with finding the basic permutational forms for Pick 1 through Pick N, N being any quantity... [ More ]

Reply #22

JADELottery — Mon, 13 Nov 2006 07:30:53 GMT

I see a possible problem in you logic. I can't explain right now, I have to think about it for a while to format it in a way you can understand.

Reply #21

Thoth — Mon, 13 Nov 2006 06:41:54 GMT

Thanks for taking your time to help on this one JadeLottery! There is brilliance in your logic and the way you apply those formulas. However, the odds of 1 in 24,939,240,465 for 21 hits in 979 draws seem extraordinarily inflated to me.

I was expecting the odds/probability to be more around one in several hundred thousand or possibly as high as 1 in 2,000,000 or so. From a statistacal perspective, the frequency of boxed combos that show 10 and 11 hits within 979 consecutive trials would s... [ More ]

Reply #20

JADELottery — Mon, 13 Nov 2006 06:01:48 GMT

I did a quick check on my home state WI for the entire run of numbers.

I did not find any 014 anomaly in either the natural or computer picks.

The contributor must be coming from some other state or combination of states.

Reply #19

JADELottery — Mon, 13 Nov 2006 05:14:21 GMT

Also, on the opposite end of the scale. Look for deficient numbers that have a low hit rate.

Reply #18

JADELottery — Mon, 13 Nov 2006 05:04:32 GMT

This is really unusual, if the real count for the total number of draws is 979 for that sample, then there is really something seriously wrong. I would have a better chance of winning a straight pick 3 using a single number of any kind three times in a row. Something is wrong. The first thing I would suggest is separate the computer picked numbers from the natural picked numbers and see if there is some relative difference. Something has to be causing this, or a combination of things could be ca... [ More ]

Reply #17

Bryan — Mon, 13 Nov 2006 04:38:47 GMT

I sure am glad you all are trying to figure out the probability cause I'm as curious as can be. And I am definitely letting you all do the math... LOL. But I did want to tell you it hit again last night as 041, right on cue...It goes no more than 3 days without hitting. Cool!

Thanks Guys,

Bryan

Reply #16

Thoth — Mon, 13 Nov 2006 04:19:14 GMT

Not every state in the list has drawings every day of the week ...and some don't even have midday drawings. 979 should be the right amount of draws (I think).

Reply #15

JADELottery — Sun, 12 Nov 2006 21:11:23 GMT

yet another finger check... I was very tired when I typed it in.

Normal Distribution is actully - y = (1 / s 2 p) e -((x - m)² / 2 s²) missed the 2 in the part s² --> 2 s

Reply #14

JADELottery — Sun, 12 Nov 2006 13:16:40 GMT

i need sleep... i hope this answers the question... i think there was a question, somewhere... just too... sleep... answer..question... zzzzz.

Reply #13

JADELottery — Sun, 12 Nov 2006 13:02:05 GMT

Draws =

1569

Number of hits that might happen for a single box-6

Percent Probability that a single box-6 will produce the given hits

0

0.12

1

0.30

2

0.70

3

1.46

4

2.74

5

4.62

6

7.00

7

9.54

8

11.69

9

12.88

10

12.77

11

11.38

12

9.12

13

6.57

14

4.26

15

2.48

16

1.30

17

0.61

18

0.26

19

0.10

Reply #12

JADELottery — Sun, 12 Nov 2006 12:56:55 GMT

If that is the correct draws, then it's 0.00288490863079851 %, or 2.88490863079851 m% (milli-percent), or 1 in 34,663.

Reply #11

JADELottery — Sun, 12 Nov 2006 12:52:02 GMT

finger check on that last one, it's actually 1 in 24,939,240,465 for 979 draws.

Reply #10

ayenowitall — Sun, 12 Nov 2006 12:38:10 GMT

I certainly don't consider myself to be a mathematician, but I'm counting 1569 draws from October 11th through November 8th.

Reply #9

JADELottery — Sun, 12 Nov 2006 10:37:33 GMT

If this is happening on a regular basis, I'd say something funny is going on and needs further investiagation.

Reply #8

JADELottery — Sun, 12 Nov 2006 10:34:35 GMT

Based on what you were looking for and what we now know.

There is a 0.00000000400974521014437 % probability of 22 boxed-6 draws being the same during those 979 draws.

Put another way, it's 4.00974521014437 n% (nano-percent) probability of 22 boxed-6 draws being the same during those 979 draws.

Or, a 1 in 249,392,404 chance of this happening during those 979 draws.

Reply #7

JADELottery — Sun, 12 Nov 2006 10:18:44 GMT

Based on that equation here's a table for 979 draws.

Number of hits that might happen for a single box-6

Percent Probability that a single box-6 will produce the given hits for 979 draws

0

0.87

1

2.18

2

4.59

3

8.15

4

12.21

5

15.42

6

16.44

7

14.78

8

11.20

9

7.16

10

3.86

11

1.76

12

0.67

13

0.22

14

0.06

15

0.01

16

0.00... [ More ]

Reply #6

JADELottery — Sun, 12 Nov 2006 10:10:05 GMT

The equation I posted was derived this way.

Normal Distribution - y = (1 / s 2 p) e -((x - m)² / s²)

m = (6 / 1000)s = 0.006 s

s = m = 0.006 s

Plugging in the values gives this equation.

y = (1 / 0.012 p s) e -((x - 0.006 s)²/(0.012 s

Reply #5

JADELottery — Sun, 12 Nov 2006 09:49:24 GMT

Thoth,

After thinking about this a while and looking over some of the other posts, I think I have a solution. In order to solve this problem, I had to approach it from a different view. First, the fact that the numbers are being drawn by different states at different times on different days does not matter. Only the draw count of how many selections matter. A sample is a sample, regardless of it's source. If we look at the inverse of what you are doing by state, we can also ask the question,... [ More ]

Reply #4

JADELottery — Sun, 12 Nov 2006 05:37:09 GMT

Also, for clarity, you are referring the permutations related to the basic form A B C, not including the basic forms A A B and A B B

Reply #3

JADELottery — Sun, 12 Nov 2006 04:56:11 GMT

Nevermind, found it, Check this out...

Reply #2

JADELottery — Sun, 12 Nov 2006 04:44:02 GMT

Also, think I've seen something like this in another post, but I can remember where. Can you post a link if you know

Reply #1

JADELottery — Sun, 12 Nov 2006 04:30:03 GMT

Thoth,

Just replying to say I'm working on the problem you stated, Do you know a formula for finding the probability of the same boxed number being drawn in several states over about a one month period of time?

To understand this a little better, I think you are looking for the probability of successfully matching any single boxed permutation in several different pick 3 drawings per day for about 30 drawings.

Does this sound about right

Help JADELottery

Thoth — Sat, 11 Nov 2006 22:51:54 GMT

Do you know a formula for finding the probability of the same boxed number being drawn in several states over about a one month period of time? Bryan made the following post and I can't seem to come up with a true probability for it:

For 30 days, this combo has not gone more than 3 days without being drawn. That is incredible...As matter of fact, it has almost hit solid for the last 3 months...wow!

Draw Date State Game Results

Wed, Nov 8, 2006DelawareMidday 31-0-4

Wed, N... [ More ]