Positional analysis

[johnph77]

Just in case.....

Sum Frequency Chart - 5/55

``15`````1|``66``4032|`116`31616|`166`30412|`216``3507
``16`````1|``67``4317|`117`32201|`167`29796|`217``3266
``17`````2|``68``4612|`118`32760|`168`29159|`218``3034
``18`````3|``69``4925|`119`33312|`169`28522|`219``2818
``19`````5|``70``5248|`120`33836|`170`27867|`220``2611
``20`````7|``71``5590|`121`34350|`171`27214|`221``2418
``21````10|``72``5942|`122`34834|`172`26546|`222``2233
``22````13|``73``6313|`123`35307|`173`25882|`223``2062
``23````18|``74``6694|`124`35748|`174`25205|`224``1898
``24````23|``75``7095|`125`36176|`175`24535|`225``1747
``25````30|``76``7505|`126`36570|`176`23854|`226``1602
``26````37|``77``7935|`127`36950|`177`23181|`227``1469
``27````47|``78``8374|`128`37295|`178`22500|`228``1342
``28````57|``79``8833|`129`37624|`179`21829|`229``1226
``29````70|``80``9301|`130`37917|`180`21152|`230``1115
``30````84|``81``9788|`131`38193|`181`20486|`231``1014
``31```101|``82`10283|`132`38432|`182`19816|`232```918
``32```119|``83`10798|`133`38653|`183`19159|`233```831
``33```141|``84`11320|`134`38836|`184`18500|`234```748
``34```164|``85`11861|`135`39001|`185`17855|`235```674
``35```192|``86`12408|`136`39127|`186`17209|`236```603
``36```221|``87`12974|`137`39234|`187`16579|`237```540
``37```255|``88`13545|`138`39302|`188`15950|`238```480
``38```291|``89`14134|`139`39351|`189`15337|`239```427
``39```333|``90`14727|`140`39361|`190`14727|`240```377
``40```377|``91`15337|`141`39351|`191`14134|`241```333
``41```427|``92`15950|`142`39302|`192`13545|`242```291
``42```480|``93`16579|`143`39234|`193`12974|`243```255
``43```540|``94`17209|`144`39127|`194`12408|`244```221
``44```603|``95`17855|`145`39001|`195`11861|`245```192
``45```674|``96`18500|`146`38836|`196`11320|`246```164
``46```748|``97`19159|`147`38653|`197`10798|`247```141
``47```831|``98`19816|`148`38432|`198`10283|`248```119
``48```918|``99`20486|`149`38193|`199``9788|`249```101
``49``1014|`100`21152|`150`37917|`200``9301|`250````84
``50``1115|`101`21829|`151`37624|`201``8833|`251````70
``51``1226|`102`22500|`152`37295|`202``8374|`252````57
``52``1342|`103`23181|`153`36950|`203``7935|`253````47
``53``1469|`104`23854|`154`36570|`204``7505|`254````37
``54``1602|`105`24535|`155`36176|`205``7095|`255````30
``55``1747|`106`25205|`156`35748|`206``6694|`256````23
``56``1898|`107`25882|`157`35307|`207``6313|`257````18
``57``2062|`108`26546|`158`34834|`208``5942|`258````13
``58``2233|`109`27214|`159`34350|`209``5590|`259````10
``59``2418|`110`27867|`160`33836|`210``5248|`260`````7
``60``2611|`111`28522|`161`33312|`211``4925|`261`````5
``61``2818|`112`29159|`162`32760|`212``4612|`262`````3
``62``3034|`113`29796|`163`32201|`213``4317|`263`````2
``63``3266|`114`30412|`164`31616|`214``4032|`264`````1
``64``3507|`115`31026|`165`31026|`215``3765|`265`````1
``65``3765|

 

[Bertil]

Sum frequency chart for 5/55

Just in case.....

Sum Frequency Chart - 5/55

``15`````1|``66``4032|`116`31616|`166`30412|`216``3507
``16`````1|``67``4317|`117`32201|`167`29796|`217``3266
``17`````2|``68``4612|`118`32760|`168`29159|`218``3034
``18`````3|``69``4925|`119`33312|`169`28522|`219``2818
``19`````5|``70``5248|`120`33836|`170`27867|`220``2611
``20`````7|``71``5590|`121`34350|`171`27214|`221``2418
``21````10|``72``5942|`122`34834|`172`26546|`222``2233
``22````13|``73``6313|`123`35307|`173`25882|`223``2062
``23````18|``74``6694|`124`35748|`174`25205|`224``1898
``24````23|``75``7095|`125`36176|`175`24535|`225``1747
``25````30|``76``7505|`126`36570|`176`23854|`226``1602
``26````37|``77``7935|`127`36950|`177`23181|`227``1469
``27````47|``78``8374|`128`37295|`178`22500|`228``1342
``28````57|``79``8833|`129`37624|`179`21829|`229``1226
``29````70|``80``9301|`130`37917|`180`21152|`230``1115
``30````84|``81``9788|`131`38193|`181`20486|`231``1014
``31```101|``82`10283|`132`38432|`182`19816|`232```918
``32```119|``83`10798|`133`38653|`183`19159|`233```831
``33```141|``84`11320|`134`38836|`184`18500|`234```748
``34```164|``85`11861|`135`39001|`185`17855|`235```674
``35```192|``86`12408|`136`39127|`186`17209|`236```603
``36```221|``87`12974|`137`39234|`187`16579|`237```540
``37```255|``88`13545|`138`39302|`188`15950|`238```480
``38```291|``89`14134|`139`39351|`189`15337|`239```427
``39```333|``90`14727|`140`39361|`190`14727|`240```377
``40```377|``91`15337|`141`39351|`191`14134|`241```333
``41```427|``92`15950|`142`39302|`192`13545|`242```291
``42```480|``93`16579|`143`39234|`193`12974|`243```255
``43```540|``94`17209|`144`39127|`194`12408|`244```221
``44```603|``95`17855|`145`39001|`195`11861|`245```192
``45```674|``96`18500|`146`38836|`196`11320|`246```164
``46```748|``97`19159|`147`38653|`197`10798|`247```141
``47```831|``98`19816|`148`38432|`198`10283|`248```119
``48```918|``99`20486|`149`38193|`199``9788|`249```101
``49``1014|`100`21152|`150`37917|`200``9301|`250````84
``50``1115|`101`21829|`151`37624|`201``8833|`251````70
``51``1226|`102`22500|`152`37295|`202``8374|`252````57
``52``1342|`103`23181|`153`36950|`203``7935|`253````47
``53``1469|`104`23854|`154`36570|`204``7505|`254````37
``54``1602|`105`24535|`155`36176|`205``7095|`255````30
``55``1747|`106`25205|`156`35748|`206``6694|`256````23
``56``1898|`107`25882|`157`35307|`207``6313|`257````18
``57``2062|`108`26546|`158`34834|`208``5942|`258````13
``58``2233|`109`27214|`159`34350|`209``5590|`259````10
``59``2418|`110`27867|`160`33836|`210``5248|`260`````7
``60``2611|`111`28522|`161`33312|`211``4925|`261`````5
``61``2818|`112`29159|`162`32760|`212``4612|`262`````3
``62``3034|`113`29796|`163`32201|`213``4317|`263`````2
``63``3266|`114`30412|`164`31616|`214``4032|`264`````1
``64``3507|`115`31026|`165`31026|`215``3765|`265`````1
``65``3765|

It is generlly assumed that a graph of the frequency data will resemble
a normal curve with a variance of (N+1)N-n)n5/12 for any n/N lotto.
For 5/55 the std.dev. would then be very close to 34.16 and thus
three s.d would range from 37.5 to 242.5 with a mean of 140. How
does this prediction match the actual data?

Stig Holmquist

 

[johnph77]

It is generlly assumed that a graph of the frequency data will resemble
a normal curve with a variance of (N+1)N-n)n5/12 for any n/N lotto.
For 5/55 the std.dev. would then be very close to 34.16 and thus
three s.d would range from 37.5 to 242.5 with a mean of 140. How
does this prediction match the actual data?

Stig Holmquist

The mean, the average and the most common occurance is 140, with a range of 15 to 265.

 

[Bertil]

Frequency of sums for 5/55

The mean, the average and the most common occurance is 140, with a range of 15 to 265.

Perhaps you misunderstood my question. I was hoping you would tell me the
actual variance ( std.dev.) for the sums because I have no ability to do it.
I'm curious about how well the actual matches the formula. In fact, a graph
of the actual data superimposed on a normal curve with mean 140 and
std.dev. 34.16 would be interesting .

In my previous post I suggested that the number of eleven five number
lines might be 1.7x10^50 but this may include duplications and thus should
be divided by 11!(40million).

Stig

 

[johnph77]

I'm coming up with some strange numbers trying for the standard deviation of the sums - something in the area of 14100+. I'm doing something wrong, I think.

On the 11 lines of 5 I came up with the same figure you did - 1.70878335353098E+050. And no division would be necessary - that should be the solution. I took the possibilities in a 5/55 matrix, multiplied by the possibilities in a 5/50 matrix, multiplied by the possibilities in a 5/45 matrix, and so on.

 

[Bertil]

Frequencies for sums

I'm coming up with some strange numbers trying for the standard deviation of the sums - something in the area of 14100+. I'm doing something wrong, I think.

On the 11 lines of 5 I came up with the same figure you did - 1.70878335353098E+050. And no division would be necessary - that should be the solution. I took the possibilities in a 5/55 matrix, multiplied by the possibilities in a 5/50 matrix, multiplied by the possibilities in a 5/45 matrix, and so on.

John,
In case you have all 140 of the last PB draws you could get the sums for
each draw and then the mean and std.dev. Better yet would be to test the last 20,40,60,80,100,120 and140 draws and show how the mean and std.
dev. start to approach a limiting value. Alternatively try the previous 5/53
matrix, which had 300 draws and test every 25 draw and I would expect the
s.d. to approach 34.3

Happy New Year,
Stig

 

[Bertil]

Frequency of sums

Hi John,
My son,who has a degree in cmputer science, came home and helped me
to study the sum and std.dev. of the past 140 PB draws. The results are
in the attached table with a graph superimposed for the last 125 draws.
Please note that the cum. mean of the sums were calculated from the top down first and then the cum. std.dev. was calculated from the bottom up.

The s.d did not vary around 34.16 as I predicted and that is why I'm looking
for the actual s.d for the total set of sums that you submitted.

I plan to invert the draw data to be able to add new draws.

Stig Holmquist

Not allowed to attach file.

 

[johnph77]

I'm coming up with some strange numbers trying for the standard deviation of the sums - something in the area of 14100+. I'm doing something wrong, I think.

My son,who has a degree in cmputer science, came home and helped me to study the sum and std.dev. of the past 140 PB draws. The results are in the attached table with a graph superimposed for the last 125 draws. Please note that the cum. mean of the sums were calculated from the top down first and then the cum. std.dev. was calculated from the bottom up.

The s.d did not vary around 34.16 as I predicted and that is why I'm looking
for the actual s.d for the total set of sums that you submitted.

At this I am not surprised. I was basing my s.d. figures on my spreadsheet numbers listing the sums for all possibilities of the draw rather than a smaller sample of actual drawings In a range of the smaller sample the deviations will tend to be closer to the median, varying less from the norm than will a list of all possibilities. Therefore the s.d. should and would be much smaller. The more I read on s.d (yes, I had to break out the old math books) the more I think I'll stick my neck out and say that this is the s.d. for listing of all possibilities.

Standard deviation of the sums for all possibilities in a 5/55 lottery matrix: 14143.1209652346.

In my previous post I suggested that the number of eleven five number lines might be 1.7x10^50 but this may include duplications and thus should be divided by 11!(40million).

On the 11 lines of 5 I came up with the same figure you did - 1.70878335353098E+050. And no division would be necessary - that should be the solution. I took the possibilities in a 5/55 matrix, multiplied by the possibilities in a 5/50 matrix, multiplied by the possibilities in a 5/45 matrix, and so on

I was in error on the above. It would be necessary to divide 1.70878335353098E+050 by (11! or 39916800) to eliminate duplications. Sorry for any inconvenience.

 

[Bertil]

Sum frequency chart

Hi John,
In case you have the book "Mathematical Statistics" by J. Freund(1962)
you can find the formula I used to predict the s.d. on p.184.

Your value 14143.12 looks like the s.d for the frequencies weighted by their
sums rather than the s.d for the sums weighted by their frequencies.It you
reverse the two columns I think you will get a value close to 34.

We were unable to copy your data to a spreadsheet and thus could not
do the suggested calculation.

If you send me an e-mail I would be able to send you my data for 140 draws
when my son comes home again next week.

Stig Holmquist

 

[johnph77]

I don't have access to your Email address from this forum and you probably can't read mine. I took a stab at other forums and came up with an address - if it works, good - but it wasn't easy.

 

[Paul_b]

Hi johnph77 & stig holmquist,

I purchased this book about a week ago. I have just started to try out prof jones suggestions on the UK lottery. The book seems to explain the positional analysis in two chapters. The first part on page 22 doesnt tell you sort the balls in order! which is explained on page 43. This is why my results appeared to be random! so I was so happy to see this thread, re-read the book and realised that you have to sort the numbers in order as you have in your results.

Also in his book on page 24 he suggests that 'after several months you begin that get a frequency chart' has he suggests. Working on the maximum value of 10 for a frequency of any ball as in figure 5.3 and 6.5 I calculated the number of draws to reach this value. In the UK lottery is was on average around 180 draws. Do you know why he chose the value of 10? Maybe coincidence?
Do you have any suggestions for this value?

 

[Bertil]

There is only one logical explanation for Jones data, they are fake and he
is a con man and as phony as a 3$ bill. If yo need any proof just look at
Fig.6.5 and ask yourself how the total number of hits can vary from one position to the next. Specifically, the totals are: 49, 57,55,54,64 and 70 in
1st, 2nd, 3rd,4th, 5th and 6th resp. They must all be the same if the data were honest, but Jones is a fraud, who could not find any real data to
illustrate his bogus idea.

Your best bet would be to ask for a refund for the book. No national book
seller where I live sells this book after I informed them that Jones is a fraud.

Stig Holmquist

 

[Paul_b]

You sound pretty annoyed! I guess you have spent valuable time reading his book and proving his theories?

It maybe worth putting a review on amazon(Where I bought my book from) to inform people, if you feel that strong. Does he have any contact details in his book? to question his theories. If he was a honest man he would probably get back to you. Not the experience I had when I tried contacting Gail Howard after reading her book!

 

[Bertil]

I'm more than annoyed and at one time I wrote to Jones and asked where
he got his numbers but he refused to tell me. He is a super scam artist and
a charlatan when it comes to statistical analysis.
Also, years ago I wrote to the publisher Cardoza and pointed out the many
errors and false data in the book, but the book was republished with all
errors and false data. I also wrote to Amazon and got no reply. All these
outfit opperate by the slogan : Caveat emptor= the sucker be ware.

Stig Holmquist

 

[Paul_b]

I do sympathise with you. And it sounds like you have had no luck trying to contact the appropriate people. There should be some law against these people.

Myself I feel obliged to try his theories since I have spent the money on the book. Hopefully it wont take long! If I discover anything useful I will post on the forum.

I am new to this forum. Have you found any useful strategies posted on this forum?