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Author Topic: This maths is correct, how can it be wrong?  (Read 10824 times)

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #75 on: 06/09/2015 09:14:32 »
Talking about the complexity of this, X^100 is always in the future when playing a live game, where as Y a future already written transcends values forward in time to a specific point in space-time, a component of dimensional change of choice. Playing the Y axis your future depends on the values that land on you, values that quantum leap through time to a specific destination and point in time relative to the player, a continuum of countless variants and countless odds, where X is a block of randomness and not really that random, y remains absolute random  for all players.
Anything of block randomness is predictable to a degree, I can accurately predict that if you continued to spin a roulette wheel and spin the ball, turn after turn for 24 hrs, I will predict that zero will always make an appearance within 24hrs.

Random is not what you think it is.

P(0)/24hrs=1

86400/36=2400≈40 mins

86400/52=1661.53846154≈27.6923076923mins

of cause there is boundary of low numbers where the maths breaks down, but that is because the low numbers are definite

and strangely the boundary number is 36, before that 35 downwards, the maths fail and time increases where it should go down.

86400/35=2468.57142857≈41.1428571429mins


sorry i should of done the other way around then it works,


52/86400=0.00060185185
51/86400=0.00059027777
50/86400=0.0005787037
1/86400=0.00001157407


I am at the moment confused, I am not sure what I am counting down or up, I think I am on to something here.



« Last Edit: 06/09/2015 10:41:57 by Thebox »
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #76 on: 06/09/2015 10:34:56 »
I will put it this way, you are dealt a single hand using a single deck, or you pick a hand from 1,000,000 already made hands.

Do you think this sounds like the same game?

Let me ask some questions so I can understand.

1) In a live game the dealer uses a single deck which he shuffles after each hand.

2) If instead he preshuffles 100 decks which he then lays out on the table and asks the first player to choose a deck at random. Is this the same probability as 1)?

3) if the dealer does as 1), but instead of dealing off the top of the deck he deals from the bottom has that affected the game?

Again words only please, no maths.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #77 on: 06/09/2015 10:38:28 »
I will put it this way, you are dealt a single hand using a single deck, or you pick a hand from 1,000,000 already made hands.

Do you think this sounds like the same game?

Let me ask some questions so I can understand.

1) In a live game the dealer uses a single deck which he shuffles after each hand.

2) If instead he preshuffles 100 decks which he then lays out on the table and asks the first player to choose a deck at random. Is this the same probability as 1)?  no

3) if the dealer does as 1), but instead of dealing off the top of the deck he deals from the bottom has that affected the game?yes

Again words only please, no maths.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #78 on: 06/09/2015 10:48:13 »
In the first example you are playing one hand at a time, hands that have not yet been ''written'' , a future


in the second example you have just offered me a choice of 100 already ''written'' hands.   100 hands into my future, approximately 2 hours of play at an instant.   
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #79 on: 06/09/2015 11:14:43 »
Ok, I now understand what you are saying.
From a probability point of view all the games are equal. I would view the 100 decks as following probability rules when they were dealt and equal to the individual shuffled deck. Over a large number of games they are equal. Probability deals with what is most likely to happen under various circumstances and degree of knowledge.
Your theory involves predetermination, which is not part of probability. You are dealing with individual events.
If you want to develop a maths for your theory you have to stop using probability terms as it will only confuse people.
I don't know how you can handle this with maths. Will have a think, but not hopeful.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #80 on: 06/09/2015 11:21:10 »
Ok, I now understand what you are saying.
From a probability point of view all the games are equal. I would view the 100 decks as following probability rules when they were dealt and equal to the individual shuffled deck. Over a large number of games they are equal. Probability deals with what is most likely to happen under various circumstances and degree of knowledge.
Your theory involves predetermination, which is not part of probability. You are dealing with individual events.
If you want to develop a maths for your theory you have to stop using probability terms as it will only confuse people.
I don't know how you can handle this with maths. Will have a think, but not hopeful.

Thank you Colin, if you understand this now you know it is is complex and that is why I was trying to produce new maths, I do not think the maths actually exists.



Was I doing anything known here

86400/86400=1
52/86400=0.00060185185
51/86400=0.00059027777
50/86400=0.0005787037
1/86400=0.00001157407

this seems connected somehow?

added -

52*2/86400=0.0012037037

52*100/86400=0.06018518518

52*1661.53846154/86400=1
« Last Edit: 06/09/2015 12:36:24 by Thebox »
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #81 on: 07/09/2015 12:51:29 »
Colin

I think I am trying to describe this -

SL(n, Z) → SL(n, Z/NZ)

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

 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #82 on: 07/09/2015 14:40:10 »
Colin

I think I am trying to describe this -

SL(n, Z) → SL(n, Z/NZ)

https://en.wikipedia.org/wiki/Congruence_subgroup
Can you explain why you think that and how you link the variables to your problem?


86400/86400=1
52/86400=0.00060185185
51/86400=0.00059027777
50/86400=0.0005787037
1/86400=0.00001157407

this seems connected somehow?

added -

52*2/86400=0.0012037037

52*100/86400=0.06018518518

52*1661.53846154/86400=1
I don't know why you are dividing 52, 51, 50 etc. we agreed that you are not looking at probability but at what is 'written' so you need actual values as what is written is no longer random but fixed. Also your problem has more to do with orientation of the decks that individual cards.

I have to repeat however that for me what is written is irrelevant, it is how it was written that is important and that is what defines the distribution of the cards over a large number of games. Thus until we know the outcome, the rules of probability apply and P(X)=P(Y).
And yes, in your view of the game P(X) cannot =P(Y) because the card values are a single, specific outcome.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #83 on: 07/09/2015 19:17:56 »

Can you explain why you think that and how you link the variables to your problem?

because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant


I don't know why you are dividing 52, 51, 50 etc. we agreed that you are not looking at probability but at what is 'written' so you need actual values as what is written is no longer random but fixed. Also your problem has more to do with orientation of the decks that individual cards.

I have to repeat however that for me what is written is irrelevant, it is how it was written that is important and that is what defines the distribution of the cards over a large number of games. Thus until we know the outcome, the rules of probability apply and P(X)=P(Y).
And yes, in your view of the game P(X) cannot =P(Y) because the card values are a single, specific outcome.

Yes P(X) can not equal P(Y)


presently looking at this.

https://en.wikipedia.org/wiki/Almost_periodic_function
« Last Edit: 07/09/2015 19:34:48 by Thebox »
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #84 on: 07/09/2015 19:57:05 »
presently looking at this.

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

You are in no way prepared to understand that wikipedia page. I have encountered almost periodic functions as applied to models of quasicrystals and there is nothing easy about it. Trust me, spending a few weeks on more fundamental math will help you understand your questions much more fully than if you spent the same amount of time trying to digest this.
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #85 on: 07/09/2015 22:38:37 »
because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant
.
.
.
presently looking at this.

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

These are moving you further from understanding your problem.
You need to concentrate on understanding why P(X)=P(Y). Then you will have your answer.
Let us know when you have done that.
 

Offline alancalverd

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Re: This maths is correct, how can it be wrong?
« Reply #86 on: 07/09/2015 22:40:29 »
Anything of block randomness is predictable to a degree, I can accurately predict that if you continued to spin a roulette wheel and spin the ball, turn after turn for 24 hrs, I will predict that zero will always make an appearance within 24hrs.

And you would be wrong. It is indeed most unlikely that zero would not appear, but there's a difference between unlikely and impossible - as witnessed by the fact that life evolved in a generally hostile universe. 
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #87 on: 08/09/2015 16:32:11 »
because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant
.
.
.
presently looking at this.

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

These are moving you further from understanding your problem.
You need to concentrate on understanding why P(X)=P(Y). Then you will have your answer.
Let us know when you have done that.


X does equal Y if x is a single linearity and y is also a single linearity containing the exact same ingredients.


x
x
x

xxx

is the same without doubt, x can be y from a different orientation, as long as all the values are the same in each set, this was never my argument, I know this, I am not stupid, the problem is I know this and have found an interdependency ,

In algebra would this be correct  A+B=AB?
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #88 on: 08/09/2015 17:55:40 »

In algebra would this be correct  A+B=AB?

This equation is satisfied in the case of A =2 and B = 2. Also A = 0 and B = 0. In General: for any value A, there is a value B = A/(A1) which will satisfy your equation.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #89 on: 08/09/2015 18:43:14 »


This equation is satisfied in the case of A =2 and B = 2. Also A = 0 and B = 0. In General: for any value A, there is a value B = A/(A1) which will satisfy your equation.

I am not sure what you have just said exactly, are you saying that A/B=1/52   , that sort of representation?


PA/(B)=1/52  like this?


what brackets do I need on A to represent a single variant?  is it this [A]?
« Last Edit: 08/09/2015 18:46:20 by Thebox »
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #90 on: 08/09/2015 20:09:47 »
you asked if A+B=AB makes sense algebraically. I understand that statement to mean: "there are two numbers, A and B, for which A plus B equals A times B."

This is a perfectly valid algebraic statement, and can be rearranged to show that every number A has exactly one value of B such that A plus B equals A times B. We can even solve for what B must be for any A, which was I meant to show as B = A/(A1), which you should understand to mean: "B must equal the ratio of A and A minus one."
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #91 on: 08/09/2015 21:45:02 »
you asked if A+B=AB makes sense algebraically. I understand that statement to mean: "there are two numbers, A and B, for which A plus B equals A times B."

This is a perfectly valid algebraic statement, and can be rearranged to show that every number A has exactly one value of B such that A plus B equals A times B. We can even solve for what B must be for any A, which was I meant to show as B = A/(A1), which you should understand to mean: "B must equal the ratio of A and A minus one."


so if I put

x+y=xy


that would mean the same as a+b?


I have been playing around with something, and wish for an opinion,


X^100=XY=10cm

XY^100=XYZ=10cm

E=XYZ-^XYZ=0cm


would this make any sense to you?

« Last Edit: 08/09/2015 22:16:59 by Thebox »
 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #92 on: 08/09/2015 22:48:28 »
so if I put

x+y=xy

that would mean the same as a+b?

x+y=xy is the same as A+B=AB, which is the same as g+q=gq (unless you have defined the letters or added other relationships to them--for instance, if your equation is also accompanied with x+y=6, then it is assumed that the x in one equation is the same as the x in the other, and then numerical values can be found for both x and y)

I have been playing around with something, and wish for an opinion,


X^100=XY=10cm

XY^100=XYZ=10cm

E=XYZ-^XYZ=0cm


would this make any sense to you?

I don't think it makes much sense. I don't know how you're using the ^ symbol. It is also odd to put units (cm) on one side of an equation, but not the other...
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #93 on: 09/09/2015 07:13:48 »
so if I put

x+y=xy

that would mean the same as a+b?

x+y=xy is the same as A+B=AB, which is the same as g+q=gq (unless you have defined the letters or added other relationships to them--for instance, if your equation is also accompanied with x+y=6, then it is assumed that the x in one equation is the same as the x in the other, and then numerical values can be found for both x and y)

I have been playing around with something, and wish for an opinion,


X^100=XY=10cm

XY^100=XYZ=10cm

E=XYZ-^XYZ=0cm


would this make any sense to you?

I don't think it makes much sense. I don't know how you're using the ^ symbol. It is also odd to put units (cm) on one side of an equation, but not the other...

^ to the power of and if you notice I was subtracted the power of contracting XYZ.


Time begun  XYZ-^XYZ=0





« Last Edit: 09/09/2015 07:52:24 by Thebox »
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #94 on: 09/09/2015 09:59:13 »
^ to the power of and if you notice I was subtracted the power of contracting XYZ.


Time begun  XYZ-^XYZ=0
You can't raise a - sign to a power
Also, how can you say 'time begun' = 0cm3

When you write things like this it looks as though you are just pulling our legs (polite version) or trolling.
If that is not the case you really do need to concentrate on learning basic maths, it would make your life, and ours, less frustrating.
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #95 on: 09/09/2015 16:57:56 »
^ to the power of and if you notice I was subtracted the power of contracting XYZ.


Time begun  XYZ-^XYZ=0
You can't raise a - sign to a power
Also, how can you say 'time begun' = 0cm3

When you write things like this it looks as though you are just pulling our legs (polite version) or trolling.
If that is not the case you really do need to concentrate on learning basic maths, it would make your life, and ours, less frustrating.

I am not raising , I am tasking the power of away.   


I am contracting space, or expanding it, try it , it works for me.

 

Offline jeffreyH

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Re: This maths is correct, how can it be wrong?
« Reply #96 on: 09/09/2015 22:01:51 »
You can correctly state it as XYZ^-XYZ or 35dfee07a60159663bf05a846dd783f0.gif. So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000
« Last Edit: 09/09/2015 22:07:09 by jeffreyH »
 

Offline Colin2B

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Re: This maths is correct, how can it be wrong?
« Reply #97 on: 10/09/2015 09:18:42 »
I am not raising , I am tasking the power of away.   

I am contracting space, or expanding it, try it , it works for me.

Is JefferyH's answer what you are trying to do?

You can correctly state it as XYZ^-XYZ or 35dfee07a60159663bf05a846dd783f0.gif. So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000

Also, you still havent explained how can you say 'time begun' = 0cm3

« Last Edit: 10/09/2015 09:20:21 by Colin2B »
 

Offline Thebox

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Re: This maths is correct, how can it be wrong?
« Reply #98 on: 10/09/2015 16:36:17 »



Is JefferyH's answer what you are trying to do?


Also, you still havent explained how can you say 'time begun' = 0cm3

Yes Jeffery's reply reads the same to me as mine did.

Talking about the beginning of time is off topic, but you know me and know I often go off topic , while thinking about one topic often I start to associate other things.

XYZ is the universe has far as we can see, XYZ^-XYZ  shrinks space to a singular point of nothing, a zero point space I have mentioned before.

So arbitrary time = XYZ^-XYZ

contracting space to nothing, or the opposite ^XYZ expanding space from nothing. Because time is a measurement based on another measurement, i.e 24 hrs = 0.0288m/s

 

Offline chiralSPO

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Re: This maths is correct, how can it be wrong?
« Reply #99 on: 10/09/2015 19:50:42 »
Because time is a measurement based on another measurement, i.e 24 hrs = 0.0288m/s

you can't equate any number of hours with any value of m/s. they have different units. didn't we go over this when you first joined the forum?? time ≠ speed!
 

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Re: This maths is correct, how can it be wrong?
« Reply #99 on: 10/09/2015 19:50:42 »

 

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