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Author Topic: This maths is correct, how can it be wrong?  (Read 10835 times)

Offline Thebox

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This maths is correct, how can it be wrong?
« on: 21/08/2015 18:00:08 »
I am posting this in main because I strongly believe from self taught maths that this is the correct maths for this scenario.  I now believe I am starting to speak your language.

(dx)=52
P(n)/(dx)=(1/52)/t
(dy)=f(^x)
P(n)/(dy)=σ2/t2
P(n)/(dx)≠P(n)/(dy)
[x1∝x2]≠[y1≠y2]

(dy)≠(dx)
t


P(n)/(dy)=σ2/t2=The chance of receiving ~(n) by random choice of set is dependent to the variance of population values by the shuffle of (dX), (the rows), aligning values to p1 , (the output), in a Y-axis (column) and by adding choice, changing the continuous t1 of the dx axis to a ''quantum leap'' of t2 and a (dy) choice bringing the variant in the ^dx position forward in time from of the (dy) axis□

Model:

..(dy)/t2..
nnnnnnnn(dx)/t1
nnnnnnnn(dx)/t1
nnnnnnnn(dx)/t1
nnnnnnnn(dx)/t1
nnnnnnnn(dx)/t1
nnnnnnnn(dx)/t1

P(B | A)=1



« Last Edit: 21/08/2015 18:41:11 by Thebox »


 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #1 on: 21/08/2015 18:49:20 »
Misunderstanding of "random".
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #2 on: 21/08/2015 18:51:12 »
Misunderstanding of "random".

argue the maths please Alan, my maths , your same maths tells me I am correct whether you believe it or not.

added -start with any premise for argument against this P(B | A)=1

probability of event B given event A occurred

Event A=Choice
Event B=Velocity change
« Last Edit: 21/08/2015 19:03:22 by Thebox »
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #3 on: 21/08/2015 19:11:32 »
This is certainly not written in my language--I still don't quite know what you are saying, but if you're using it to prove that the probabilities of drawing specific cards from randomly shuffled decks is time-dependent, then it can't be right.

Anyway, I thought you were leaving...
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #4 on: 21/08/2015 19:16:14 »
This is certainly not written in my language--I still don't quite know what you are saying, but if you're using it to prove that the probabilities of drawing specific cards from randomly shuffled decks is time-dependent, then it can't be right.

Anyway, I thought you were leaving...

I did leave, but I can not rest when I know I am correct 100%.  The maths says so, maths I have self taught , but maths I know is correct.  I have the two pages open now as we speak

http://www.rapidtables.com/math/symbols/Statistical_Symbols.htm

https://en.wikipedia.org/wiki/List_of_mathematical_symbols

It is your maths and you can not understand it, really?

I understand it perfectly now, it seems quite simple.


Which part do you not understand ? I will guide you through it.   

p.s each equation is its own equation, it is not a whole but makes a whole.

« Last Edit: 21/08/2015 19:21:47 by Thebox »
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #5 on: 21/08/2015 19:56:48 »

(dx)=52 what does the d stand for? This shouldn't be calculus or differential equations...
P(n)/(dx)=(1/52)/t Why are we dividing a probability by time?
(dy)=f(^x) what does the d stand for? This shouldn't be calculus or differential equations...
P(n)/(dy)=σ2/t2 what is σ? is it a standard deviation of something?
P(n)/(dx)≠P(n)/(dy) what does the d stand for? This shouldn't be calculus or differential equations...
[x1∝x2]≠[y1≠y2] I don't understand this one at all. how can a proportionality statement and non-equality statement be related like this? what are x1, x2, y1 and y2?

(dy)≠(dx) what does the d stand for? This shouldn't be calculus or differential equations...

 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #6 on: 21/08/2015 20:02:22 »


(dy)≠(dx) what does the d stand for? This shouldn't be calculus or differential equations...
[/quote]

d stands for distance x is a vector and y is a vector, the 1/52 is travelling not you.
« Last Edit: 21/08/2015 20:08:44 by Thebox »
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #7 on: 21/08/2015 20:13:56 »

(dx)=52 what does the d stand for? This shouldn't be calculus or differential equations...
P(n)/(dx)=(1/52)/t Why are we dividing a probability by time?distribution over continuous time time
(dy)=f(^x) what does the d stand for? This shouldn't be calculus or differential equations...it isnt,its  a linear expanded to the power offa function of the power of x
P(n)/(dy)=σ2/t2 what is σ? is it a standard deviation of something?variance of population values
P(n)/(dx)≠P(n)/(dy) what does the d stand for? This shouldn't be calculus or differential equations...
[x1∝x2]≠[y1≠y2] I don't understand this one at all. how can a proportionality statement and non-equality statement be related like this? what are x1, x2, y1 and y2?x1 and x2 are rows, y is colums made by the shuffling of the rows, alignment of a y axis.

(dy)≠(dx) what does the d stand for? This shouldn't be calculus or differential equations...



xxxxxx^0


yyyyy
xxxxx
xxxxx
xxxxx
xxxxx  ^4


Y=function of the power of x, no power of, no y, the ^x is the ingredients of y
« Last Edit: 21/08/2015 20:22:07 by Thebox »
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #8 on: 21/08/2015 20:27:40 »
I will start from the top , I did not explain that very well to you.

(dx)=52


distance of the x axis equals 1-52


x=nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

ok so far?

P(A)/(dx)=(1/52)/t

x=←nnnnnnnnnnnnnnnnnnnAnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
.............................time............................................................

The chance of (A) from distance x is 1 out of 52 over time.

ok so far?


(dy)=f(^x)

Distance Y axis is a function of the  power of x


x=←nnnnnnnnnnnnnnnnnnnAnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn^3
x=←nnnnnnnnnnnnnnnnnnnAnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn^2
x=←nnnnnnnnnnnnnnnnnnnAnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
.............................time............................................................

P(A)/(dy)=σ/t2

The chance of receiving an A from (dy) is dependent to the variance of population values in (dy) over time 2, random choice bringing values forward in time.


P(B | A)=1


The chance, that of event B occurring given that event A has happened is 100%.

event A=^x
event B=A

∑(dX)≠∑(dy)


By adding choice we create a temporal loop of probabilities, it is no  longer science fiction.
« Last Edit: 21/08/2015 21:20:55 by Thebox »
 

Online PmbPhy

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Re: This maths is correct, how can it be wrong?
« Reply #9 on: 22/08/2015 00:02:32 »
Quote from: Thebox
I did leave, but I can not rest when I know I am correct 100%.
Doesn't this seem to imply that he'll never leave?
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #10 on: 22/08/2015 07:03:20 »
Do the maths Pete I am obviously correct like it or not.

x1 is equal to x2 but not equal to y1 which is not equal to y2 etc etc.
« Last Edit: 22/08/2015 07:09:12 by Thebox »
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #11 on: 22/08/2015 08:45:35 »
It is impossible to do this 'maths' because it isn't maths, it is symbol gibberish.
If you truly believe this proves you are right and everyone else is wrong then you will be able to walk away and leave it at that.

You have learnt a lot since being on this forum and it's been to good to see you arguing with some of the new theories.Wishing you all the best for the future, there have been times when it has been good knowing you.
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #12 on: 22/08/2015 09:18:35 »

argue the maths please Alan, my maths , your same maths tells me I am correct whether you believe it or not.


Exactly. "Random" is a mathematical term. You must understand its implications before deploying it.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #13 on: 22/08/2015 10:02:01 »

argue the maths please Alan, my maths , your same maths tells me I am correct whether you believe it or not.


Exactly. "Random" is a mathematical term. You must understand its implications before deploying it.

I understand it very well, 

there is nothing random about X=1/52      were Y is random.  We know there is one in 52 we do not know how many are in y.   I suggest it is you guys who do not truly understand what random is.
My maths is not just symbols,  it reads like words.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #14 on: 22/08/2015 10:04:14 »
It is impossible to do this 'maths' because it isn't maths, it is symbol gibberish.
If you truly believe this proves you are right and everyone else is wrong then you will be able to walk away and leave it at that.

You have learnt a lot since being on this forum and it's been to good to see you arguing with some of the new theories.Wishing you all the best for the future, there have been times when it has been good knowing you.

Not true Colin, it is possible to do the maths for the X axis because it is not random but random at the same time, where as the y axis is absolute random and like you said impossible to calculate.  The position of (A) along X is random, (A) itself is not random.`

I see no errors in my maths where do you see an error?

P(A)/(dy)=σ/t2

scenario - take 100 lottery draw machines , each machine releases one ball of 59 balls, you pick 6 of these balls that have been drawn

do you think the lottery would still work?

We can write the maths this way if you like -

P(A)/(^x)=σ/t2

or this way

P(A)/(^x)=f: X → Y=σ/t2=0_1

maybe it is presentation that you do not understand


P(A)
dy/t2
=
f: X → Y
 =

σ

t2
=0_1

« Last Edit: 22/08/2015 11:20:53 by Thebox »
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #15 on: 22/08/2015 14:35:05 »
........where as the y axis is absolute random and like you said impossible to calculate. 
I never said it was impossible to calculate. Look back at http://www.thenakedscientists.com/forum/index.php?topic=57749.0
Where we have explained a number of times how to calculate it.

I see no errors in my maths where do you see an error?
P(A)/(dy)=σ/t2 is in error.

scenario - take 100 lottery draw machines , each machine releases one ball of 59 balls, you pick 6 of these balls that have been drawn

do you think the lottery would still work?
This is not the same scenario. To be comparable to your scenario you would have to select one of the machines at random for each draw and discard the ones in between.

We can write the maths this way if you like -
All these are incorrect. 
Do you want us to lie to you and say you are right? Would that be honourable of us?
As I said, if you truly believe you are right you will be able to walk away from this forum secure in that knowledge.
Best of luck for the future.
Goodbye
Colin

 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #16 on: 22/08/2015 20:31:59 »

All these are incorrect. 
Do you want us to lie to you and say you are right? Would that be honourable of us?
As I said, if you truly believe you are right you will be able to walk away from this forum secure in that knowledge.
Best of luck for the future.
Goodbye
Colin

Walking away knowing I am right is walking away leaving you with the wrong answer.

You say my calculation is in error, ok what error?

There is no error.  Are you really trying to suggest that the sequences in the Y columns is the same as a sequences in the x columns?

''P(A)/(dy)=σ/t2 is in error.''


n
n
n
n
n


there is no error, y depends on the shuffle of x.
 

''To be comparable to your scenario you would have to select one of the machines at random for each draw and discard the ones in between.''

You are by selecting 1 of the 100 balls,


machine 1/2/3/4/5/6
ball.......1/1/1/1/1/1

Added- your version of the maths might be something to do with xy correlation which I am presently trying to get my head around.

''A correlation coefficient is a coefficient that illustrates a quantitative measure of some type of correlation and dependence, meaning statistical relationships between two or more random variables or observed data values.''

Px,y=cov(x,y)
..........σyσx

'' \operatorname{cov}  is the covariance
 \sigma_X  is the standard deviation of  X ''

A parallel linear bivariate


x=nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/
ct1=,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Y=nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/nnnnn/
rt2=,,,,,,,,,,,,,,,,,,,,,,,,,,

cov = 0_1


y
y
y
y
y
y
y
xxxxxxxxxxxx

Pa
x
=
1/52

pa
y
=cov=σ/(x,y)
=1_0

P(B↓A)=1

Very simply we know x, but we do not know y and could never possibly know because it is absolute random.

x ⊥ y


p.s somebody mentioned mixing calculus with probability, which part is calculus?  I though maths was maths.


« Last Edit: 23/08/2015 08:55:27 by Thebox »
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #17 on: 23/08/2015 14:46:47 »
Time for an experimental verification. If you honestly believe what you are saying, put your money where your mouth is, and make a fortune playing games of pure chance - like the national lottery. Don't come back with tales of great winnings at poker because that involves the unquantifiable skill of the other players, but show us that you can consistently beat the odds in roulette or a one-arm bandit. Or just throwing dice. 
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #18 on: 23/08/2015 21:10:48 »
Time for an experimental verification. If you honestly believe what you are saying, put your money where your mouth is, and make a fortune playing games of pure chance - like the national lottery. Don't come back with tales of great winnings at poker because that involves the unquantifiable skill of the other players, but show us that you can consistently beat the odds in roulette or a one-arm bandit. Or just throwing dice.

I never mention anything about beating the odds or anything of fairy tales. I need not run any experiment when the maths is correct.   I do not understand why you have put fact in new theories, there is nothing theoretical about fact.

I asked you all to contest my maths, none of you h ave been able to do this so far. The ball is not in my court it is in yours to prove my correct maths wrong, which I know You can not or anyone else can not.

I would like to see the scientists put their money where their mouth is and prove me wrong.


xx
xx

yy
yy


x is not equal to y can you prove otherwise?

p.s why are people blatantly lying about this simple process and vector change giving different probabilities?

« Last Edit: 23/08/2015 21:47:03 by Thebox »
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #19 on: 23/08/2015 22:12:42 »
Since you have not explained what your symbols mean, nobody can possibly contest your mathematics.

P(r) = z.dq

Can you challenge that?
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #20 on: 24/08/2015 06:01:52 »
Since you have not explained what your symbols mean, nobody can possibly contest your mathematics.

P(r) = z.dq

Can you challenge that?

I gave two links with the symbols I am using , your symbols from your maths, not made up symbols, and you are asking about the probability of correlation coefficient

P(r)


I presume z is time?

dq?



or is this just something you random made up ?

my maths is real maths. And in my true version P(r)=1


a 100% chance that x and y become a correlation by ^x and adding choice.

« Last Edit: 24/08/2015 06:06:37 by Thebox »
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #21 on: 24/08/2015 11:11:39 »


added -start with any premise for argument against this P(B | A)=1

probability of event B given event A occurred

In plain language, you are saying that B will inevitably happen if A has happened. I cannot think of anything you can do with shuffled cards for which this is true, other than "if A you take away the ace of spades, then B there is no ace of spades left in the pack". This sort of blindingly obvious but utterly useless statement might amuse philosophers but doesn't have much bearing on the game of poker.
[/quote]
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #22 on: 24/08/2015 11:22:36 »
We know there is one in 52 we do not know how many are in y.   I suggest it is you guys who do not truly understand what random is.


We know exactly: it is 1/52 because each sample in the y direction is independent of all the others. The difference is that if you look for AS in any one shuffle, you must find it once and only once among the 52 cards. If you look for AS along any infinite line parallel to the y axis you will eventually find it 1/52 times but in any finite sample it may turn up more or less often on that line. The only "known" is that the sum of all vertical lines must be n where n is the number of shuffles that you have sampled.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #23 on: 24/08/2015 19:53:50 »


added -start with any premise for argument against this P(B | A)=1

probability of event B given event A occurred

In plain language, you are saying that B will inevitably happen if A has happened. I cannot think of anything you can do with shuffled cards for which this is true, other than "if A you take away the ace of spades, then B there is no ace of spades left in the pack". This sort of blindingly obvious but utterly useless statement might amuse philosophers but doesn't have much bearing on the game of poker.
[/quote]

You still are not understanding the idea, you are not reading it right or something.


''n plain language, you are saying that B will inevitably happen if A has happened''


yes exactly, b is a y axis and A is ^x  and b is not 1 a maximum of 52, and b as more 1s than 52



« Last Edit: 24/08/2015 19:57:20 by Thebox »
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #24 on: 24/08/2015 20:01:38 »
Start here

x ⊥ y means x has no factor greater than 1 in common with y.

1/52 over time 1 is not the same as 1/52^?  over time 2

look its easy. bare in mind I know what you are saying and how you are looking at it, I know this way, and that way is incorrect believe me, i know.

if x=1 and 2 and x=1 and 2   what does Y equal?

x ⊥ y

see?

I might be reading this wrong, what I am saying is x has only 1 of each variant , a maximum entropy of 1/52

12
12

xx=1/2
xx=1/2


yy=?/2
yy=?/2


You have to be able to separate the parallel and flip it in your head.

table-xx
table-xx


yy
yy
t.t


how can you not see this?

added- ok I will make it even simpler for you rather than going into a z vector, I will downgrade to a single line





shuffle1-xxxxx/shuffle2-xxxxx/shuffle3-xxxxx/shuffle4-xxxxx
t1................................................................................
t2=random


P(a/x)/t1=1/5

P(a/x)/t2=random

added- sorry I am tired it may seem gibberish.

an independent shuffle of x is 1/52 and so is any other future individual shuffle that follows in concession, however if you add choice and are assigned always the first value, you are no longer playing 1/52 you are playing ?/?
Because the future of x is not written , where by adding choice and creating ^x, your future P is all at once in the present at that specific time of choice.

surely you get this ......


1/52 not written


1/52^y= written.

Do you actually understand what random is?

1/52 is not what random is, 1 is known , all 52 are known, random is all about a specific point in time, an occupying of space at a specific time, a random time.




« Last Edit: 24/08/2015 21:15:02 by Thebox »
 

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Re: This maths is correct, how can it be wrong?
« Reply #24 on: 24/08/2015 20:01:38 »

 

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