This maths is correct, how can it be wrong?

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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 »

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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".
helping to stem the tide of ignorance

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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 »

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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...

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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 »

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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...


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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 »

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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 »

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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 »

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Offline 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?

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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 »

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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.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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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.
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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.

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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 »

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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

and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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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 »

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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. 
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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 »

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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?
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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 »

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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]
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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.
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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 »

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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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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #25 on: 24/08/2015 22:09:46 »
Quote from: Thebox
Do you actually understand what random is?
Why do you keep asking this when its apparent to us that its only you who don't appear to have a firm grasp of it? Alan is a well educated man and knows darn well what random means as do everyone else. You've had a lot of time to learn math and you've never chosen to pick up a good book on math and start learning it from scratch. It's a terrible decision to attempt learning math by first learning about probability and statistics. Your use of the symbols here is so distorted that its next to impossible to follow.

Your choice of not choosing to read those texts on math and physics that I suggested that you read is what made me choose not to converse with you anymore. I speak now only to remind you that you're not going to learn physics correctly the way that you're going.

Pete it is the present symbols science  uses, it should not be that difficult to follow when I have provided the source links to the apparent meaningless symbols.   It reads like a book to me.

I am correct, I may present my maths slightly different to you would but it is easy to read when I h ave given you the very maths dictionary they come from.   It is not me being arrogant, it is me fast tracking.

added- i will translate the maths

P=chance

x=52

y=^x


A=specific variant

σ=variance of population values i.e A

/=in


P(A)/X=1/52



P(A)/Y=σ2

event B=σ2

event A=Y

P(B | A)=1

Yes or no? very simple maths with instruction.



« Last Edit: 24/08/2015 22:23:09 by Thebox »

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Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #26 on: 24/08/2015 23:00:12 »
what does "A is ^x" mean ?
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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #27 on: 25/08/2015 06:39:20 »
what does "A is ^x" mean ?

Where does it say that?

But in answer any specific variant to the power of x,

x
x^1
x^2
x^3

every one of them could be a number 1 in the y column created by the x shuffle.

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #28 on: 25/08/2015 06:42:46 »
scenario - Imagine a 3 lane motorway, you are in the most left lane, above each lane every 5m is a number 1,2, or 3 in no specific order.

123
231
213
231
P

P=you

drive down the motorway in the same lane. Got it now, imagine this scenario,

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Offline PmbPhy

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Re: This maths is correct, how can it be wrong?
« Reply #29 on: 25/08/2015 11:20:55 »
Quote from: Thebox
Pete it is the present symbols science  uses, it should not be that difficult to follow when I have provided the source links to the apparent meaningless symbols.   It reads like a book to me.
Please read what I post more carefully. I said that Your use of the symbols here is so distorted that its next to impossible to follow. That means that its not the symbols that's the problem but your use of them that's the problem. By the way, your constant statement that you're "correct" is also the problem because it shows that you're not willing to consider the notion that you might have made a mistake somewhere and when people can't conceive of making a mistake they rarely listen to the people who are telling them what the mistake is, i.e. they're not paying close enough attention and not keeping an open mind. It's been pointed out to you many times that your posts are confusing. The best example I can give to illustrate my point is the first post in this thread. All you did was to write down symbols without defining how they're being used and what they mean. We know what the symbols mean but its the context that's missing. You don't even use the symbols correctly. E.g. you wrote in part

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

And you didn't state what "(dx)" means. All you did was to assign it a value. Then you have the expression P(n) without defining what "n" is. We know from what the meaning of the symbol is but you're not using it in that context. All you have below is "~(n)" but you didn't specify it's value.

The proper use of the symbol P(A) is that A is an event and therefore P(A) is the probability of that event. An example of an event is "The dealer deals an Ace out of a full deck of cards which are randomly shuffled." Then P(A) = 1/52.

I could go on and on about your poor use and grasp of it but from what I've seen in this thread you won't understand it because you think you're right and therefore aren't willing to consider being wrong and learning what your mistake is.

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Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #30 on: 25/08/2015 15:14:17 »
scenario - Imagine a 3 lane motorway, you are in the most left lane, above each lane every 5m is a number 1,2, or 3 in no specific order.

123
231
213
231
P

P=you

drive down the motorway in the same lane. Got it now, imagine this scenario,
I'm loathed to get back into this as it has been thoroughly covered in the other thread
http://www.thenakedscientists.com/forum/index.php?topic=57749.0

However, I will pick an example to show that your use of maths is confusing.
x ⊥ y means x has no factor greater than 1 in common with y.
This means that y cannot be x^1, x^2, x^3, etc
However, you then say

x=52

y=^x


Quote from: alancalverd on 24/08/2015 23:00:12
what does "A is ^x" mean ?

Where does it say that?

But in answer any specific variant to the power of x,

x
x^1
x^2
x^3
So according to this y=x^1, x^2, x^3, etc

So which is it to be, and how did you decide that y is limited to these values anyway?

This is why I describe your maths as gibberish, you can't just string together symbols you don't understand the meaning of.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #31 on: 25/08/2015 18:49:40 »

Please read what I post more carefully. I said that Your use of the symbols here is so distorted that its next to impossible to follow. That means that its not the symbols that's the problem but your use of them that's the problem. By the way, your constant statement that you're "correct" is also the problem because it shows that you're not willing to consider the notion that you might have made a mistake somewhere and when people can't conceive of making a mistake they rarely listen to the people who are telling them what the mistake is, i.e. they're not paying close enough attention and not keeping an open mind. It's been pointed out to you many times that your posts are confusing. The best example I can give to illustrate my point is the first post in this thread. All you did was to write down symbols without defining how they're being used and what they mean. We know what the symbols mean but its the context that's missing. You don't even use the symbols correctly. E.g. you wrote in part

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

And you didn't state what "(dx)" means. All you did was to assign it a value. Then you have the expression P(n) without defining what "n" is. We know from what the meaning of the symbol is but you're not using it in that context. All you have below is "~(n)" but you didn't specify it's value.

The proper use of the symbol P(A) is that A is an event and therefore P(A) is the probability of that event. An example of an event is "The dealer deals an Ace out of a full deck of cards which are randomly shuffled." Then P(A) = 1/52.

I could go on and on about your poor use and grasp of it but from what I've seen in this thread you won't understand it because you think you're right and therefore aren't willing to consider being wrong and learning what your mistake is.

Ok Pete, I understand why my maths may be confusing you.   You are correct in thinking that I do not know enough maths to tackle this, however in considering I may be wrong, is something I do each and every day, for about 6 years now I have had this idea that I can not prove wrong to myself, and the worse of it I got learnt basic probability by forums and it just confirmed my fears.


(dx) is distance vector x Pete, I thought people would know this, I think you are clever people, so thought you would just get it.

added- Pete I am trying to explain that a multitude of decks creates a y axis that is not equal to any of the x axis's, the y axis has multiples of a specific variant instead of the single of x,


top card
ace diamonds
queen clubs
ace diamonds


y axis choice.

this symbol for y =σ variance of population values compared to x.

Help me please ,  [:(]

« Last Edit: 25/08/2015 19:01:43 by Thebox »

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #32 on: 25/08/2015 18:51:52 »

So according to this y=x^1, x^2, x^3, etc

So which is it to be, and how did you decide that y is limited to these values anyway?

This is why I describe your maths as gibberish, you can't just string together symbols you don't understand the meaning of.

y is infinite Colin not limited, the above was just an example.  It is actually x^∞=y

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #33 on: 25/08/2015 19:13:17 »
scenario - Imagine a 3 lane motorway, you are in the most left lane, above each lane every 5m is a number 1,2, or 3 in no specific order.

123
231
213
231
P

P=you

drive down the motorway in the same lane. Got it now, imagine this scenario,

If you can picture this, you understand it,

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Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #34 on: 25/08/2015 22:45:37 »
OK, let's take it a bit at a time



P=chance

x=52

y=^x





What does y = ^x mean?



Quote
A=specific variant

σ=variance of population values i.e A

/=in

Please explain the meaning of your symbols.

Is A a number? If so, what does it denote? If not, what does A2 mean?

Population of what?

Variance is a description of the distribution of a population along an axis - what axis?

What does "/=in" mean to you?
« Last Edit: 25/08/2015 22:58:14 by alancalverd »
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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #35 on: 26/08/2015 06:18:41 »
OK, let's take it a bit at a time



P=chance

x=52

y=^x





What does y = ^x mean?



Quote
A=specific variant

σ=variance of population values i.e A

/=in

Please explain the meaning of your symbols.

Is A a number? If so, what does it denote? If not, what does A2 mean?

Population of what?

Variance is a description of the distribution of a population along an axis - what axis?

What does "/=in" mean to you?

y=^x means

^x
^x
^x

Y axis.

(A) was representing a specific variant in this instance and squared
/ in, means like a box and all that it contains.


Population is each individual and all values in the box, 


To switch it to your terms after Pete had shown my error in presentation


P(A)=1/52 using a X axis


P(A)=σ  using a Y axis

x=12
x=12
....yy

« Last Edit: 26/08/2015 06:20:36 by Thebox »

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Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #36 on: 26/08/2015 09:10:08 »
P(A)=σ  using a Y axis
This is why I have given up on your theory.
A probability cannot equal a measure of variance or deviation.
Bad maths

y is infinite Colin not limited, the above was just an example.  It is actually x^∞=y
this x^∞=y is also gibberish.

Edit: There is no our maths and your maths, only correct maths and incorrect maths. What we are dealing with here is very basic and covered in secondary schools. In a few year's time your son will be telling you how wrong you are.
« Last Edit: 26/08/2015 10:36:08 by Colin2B »
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Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #37 on: 26/08/2015 18:11:28 »
what does ^x mean? You have used ^ to indicate "to the power of", which is fairly conventional, but

^x
^x
^x
 presumably means something else to you, as do words like "variant", "event", and indeed almost everything you have written.

Mathematics is a universal, formal language. If you were writing in French or German, I'm sure you would use words according to their meaning in France and Germany, so why not use mathematical definitions and symbols the same way as everyone else on the planet?
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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #38 on: 26/08/2015 18:33:28 »
what does ^x mean? You have used ^ to indicate "to the power of", which is fairly conventional, but

^x
^x
^x
 presumably means something else to you, as do words like "variant", "event", and indeed almost everything you have written.

Mathematics is a universal, formal language. If you were writing in French or German, I'm sure you would use words according to their meaning in France and Germany, so why not use mathematical definitions and symbols the same way as everyone else on the planet?


if i rolled a dice twice , the second roll would be ^2 which is 36-1 of a repeat variant from the first roll.

So if we shuffle a deck of cards, played a round then the cards were gathered and reshuffled, the chance of you receiving the same variant as the first shuffle is ^2.


So if we have 2 decks of cards, and used each one, ^2


so x^   

xxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxx

added-think of it like this,


10=100


x^10=10

0123456789
0123456789
0123456789
0123456789
0123456789
0123456789
0123456789
0123456789
0123456789
0123456789

The only possible way x=y was if all values were zero

0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
« Last Edit: 26/08/2015 19:32:26 by Thebox »

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #39 on: 26/08/2015 19:01:28 »

A probability cannot equal a measure of variance or deviation.


Yes Colin exactly , so what is the P of an event from the Y axis Colin?

There is no probabilities which is my whole point, if you add choice and have a multitude of sets creating a Y axis with an alignment of variant to the player there is no P and that is not 1/52 of x  and the symbol represents variance of the values of x in the alignment of y.


P(A)/X=(1/52)/t1


P(A)/Y=(σX+σ)/t2


The only possible answer to Y = a deviation to x ?

0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
↓↓↓↓↓↓↓↓↓↓↓


Both axis's

t1←0000000000/000000000
t1←0000000000/000000000t2
t1←0000000000/000000000t2
t1←0000000000/000000000
t1←0000000000/000000000
t1←0000000000/000000000t2
t1←0000000000/000000000
t1←0000000000/000000000
t1←0000000000/000000000
t1←0000000000/000000000t2
t1←0000000000/000000000
t1...↓↓↓↓↓↓↓↓↓↓↓


P(σX)=1

P(σY≠X)=1

P(B | A)=1

« Last Edit: 26/08/2015 20:22:39 by Thebox »

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Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #40 on: 26/08/2015 22:28:13 »
Yes Colin exactly , so what is the P of an event from the Y axis Colin?
I don't intend to repeat answers that we have already given that you choose not to read
http://www.thenakedscientists.com/forum/index.php?topic=57749.0
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #41 on: 27/08/2015 01:19:29 »
I repeat my questions of reply #37. Please define your terms and symbols.
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Offline PmbPhy

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Re: This maths is correct, how can it be wrong?
« Reply #42 on: 27/08/2015 01:43:52 »
I repeat my questions of reply #37. Please define your terms and symbols.
I'm telling you folks. He's just never going to get it.

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #43 on: 27/08/2015 06:17:52 »
I repeat my questions of reply #37. Please define your terms and symbols.
I'm telling you folks. He's just never going to get it.

I would argue that it is you Pete who is never going to get it, I understand very well, just because my maths may be a bit wayward that does not mean my idea is wayward. I am sure if you got it Pete you would easily do the maths to the idea.



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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #44 on: 27/08/2015 06:19:44 »
Yes Colin exactly , so what is the P of an event from the Y axis Colin?
I don't intend to repeat answers that we have already given that you choose not to read
http://www.thenakedscientists.com/forum/index.php?topic=57749.0

Colin Y has no probability , by adding time 2 it makes a deviation to x. You have not answered it I am afraid.


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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #45 on: 27/08/2015 06:28:44 »
I repeat my questions of reply #37. Please define your terms and symbols.

^x is shuffles or sets, the other symbols mean the same thing as the links provided says they mean. Have you ever played ''duck shoot'' at a fair ground?

where the duck travels left to right on a x axis and you shoot a y axis intercepting the duck.

←0123456789

consider this sequence, you first have 0, if you then had 6 you would be bringing 6 forward in time,





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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #46 on: 27/08/2015 18:05:29 »
To understand is to consider that using a single set, only 1 specific variant occupies the ''top spot'' at a specific time.  Where if you use multiple sets, many variants occupy the ''top spot'' at the same specific time.

1/52 is not a choice, it is an event over a period of time, where as choice is not 1/52 over time, it is lucky interception over time.

I made you a short video , i will make a more detailed one at the week end with motion,

https://www.youtube.com/watch?v=A8FrKH1_mF8
« Last Edit: 27/08/2015 19:09:26 by Thebox »

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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #47 on: 28/08/2015 23:55:41 »
This must be correct now?

If I take a coin and tossed it , you  know the chance of H or T is 1/2, you know this is also the chance for any other coin.
 
 
If I tossed 10 individual coins one after each other and recorded the results of each coins toss, then asked you to pick any of the tosses 1 to 10, you know your chance remains 1/2.
 
This is wrong and a trick your brain is playing on you,
 
Because the event has already happened, you have ten unknown variants aligned to your choice,
 
 
o
o
o
o
o
o
o
o
o
o
 
P(H)=0_1/10
 
P(T)=0_1/10
 
1/2 becomes obsolete and by adding choice, makes a multivariate, and we take a random leap rather than a random walk, bringing values forward in time.

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Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #48 on: 29/08/2015 12:01:24 »
You clearly have no idea what you are talking about.

Toss one coin. What is P(H)?

Put 100 coins in a box, shake the box, and pick one coin with your eyes closed. What is P(H)?

Toss 100 coins, one at a time, put them in a box, then pick one coin with your eyes closed. What is P(H)?

If they are not all the same, explain what physical process is responsible for the change. You are looking for a process whereby the toss of one coin affects the outcome of the toss of another.

Your underlying misconception is the idea that multiplying one random number (say the position of AS in a shuffle) by  another (which shuffle to pick) results in a nonrandom,or at least more predictable, number.
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Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #49 on: 29/08/2015 17:46:53 »




Your underlying misconception is the idea that multiplying one random number (say the position of AS in a shuffle) by  another (which shuffle to pick) results in a nonrandom,or at least more predictable, number.

You still have no idea what my actual idea is sorry Alan,


You are not considering before and after and recorded results,


before a coin toss the result is 1/2 ,

if you recorded the results of the coin toss and kept the results hidden, then asked someone to pick a result, the chance is not 1/2 any more.

the chance is ? out of ?


o
o
o
o
o
o
o


How many heads or tails are in the above number of throws? 

You cant answer it can you?

You can answer left to right which is 1/2. Where as Y could all be Heads or tails or a combination.


Y is a multiple of variants of x, by choice you add standard deviation which is defined by the shuffle of x, the result is set but unknown.

I thank you Alan for maintaining the conversation, I will show you the difference


a coin

HT
HT
HT
HT

pick a coin, you have a chance of 1/2 of any coin because the result is not written



result

?
?
?
?

we have no idea of the result or if it is H or T


« Last Edit: 29/08/2015 17:51:26 by Thebox »