Post by: guest39538 on 21/08/2015 18:00:08
(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
Post by: alancalverd on 21/08/2015 18:49:20
Post by: guest39538 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
Post by: chiralSPO on 21/08/2015 19:11:32
Anyway, I thought you were leaving...
Post by: guest39538 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.
Post by: chiralSPO 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...
Post by: guest39538 on 21/08/2015 20:02:22
[/quote]
(dy)≠(dx) what does the d stand for? This shouldn't be calculus or differential equations...
d stands for distance x is a vector and y is a vector, the 1/52 is travelling not you.
Post by: guest39538 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
Post by: guest39538 on 21/08/2015 20:27:40
(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)=σ²/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=A²
∑(dX)≠∑(dy)
By adding choice we create a temporal loop of probabilities, it is no longer science fiction.
Post by: PmbPhy 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?
Post by: guest39538 on 22/08/2015 07:03:20
x1 is equal to x2 but not equal to y1 which is not equal to y2 etc etc.
Post by: Colin2B on 22/08/2015 08:45:35
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.
Post by: alancalverd 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.
Post by: guest39538 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.
Post by: guest39538 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)=σ²/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)=σ²/t2
or this way
P(A)/(^x)=f: X → Y=σ²/t2=0_1
maybe it is presentation that you do not understand
P(A)
dy/t2
=
f: X → Y
=
σ²
t2
=0_1
Post by: Colin2B 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)=σ²/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 drawnThis 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.
do you think the lottery would still work?
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
Post by: guest39538 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)=σ²/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)
..........σ²yσx
'' \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=σ²/(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.
Post by: alancalverd on 23/08/2015 14:46:47
Post by: guest39538 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?
Post by: alancalverd on 23/08/2015 22:12:42
P(r) = z.dq
Can you challenge that?
Post by: guest39538 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.
Post by: alancalverd 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]
Post by: alancalverd 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.
Post by: guest39538 on 24/08/2015 19:53:50
[/quote]
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.
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
Post by: guest39538 on 24/08/2015 20:01:38
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.
Post by: guest39538 on 24/08/2015 22:09:46
Quote from: TheboxDo 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
σ²=variance of population values i.e A²
/=in
P(A)/X=1/52
P(A)/Y=σ2²
event B=σ2²
event A=Y
P(B | A)=1
Yes or no? very simple maths with instruction.
Post by: alancalverd on 24/08/2015 23:00:12
Post by: guest39538 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.
Post by: guest39538 on 25/08/2015 06:42:46
123
231
213
231
P
P=you
drive down the motorway in the same lane. Got it now, imagine this scenario,
Post by: PmbPhy 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.
Post by: Colin2B 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.I'm loathed to get back into this as it has been thoroughly covered in the other thread
123
231
213
231
P
P=you
drive down the motorway in the same lane. Got it now, imagine this scenario,
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=52So according to this y=x^1, x^2, x^3, etc
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 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.
Post by: guest39538 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 =σ² variance of population values compared to x.
Help me please , [:(]
Post by: guest39538 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
Post by: guest39538 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,
Post by: alancalverd on 25/08/2015 22:45:37
P=chance
x=52
y=^x
What does y = ^x mean?
Quote
A=specific variant
σ²=variance of population values i.e A²
/=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?
Post by: guest39538 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?QuoteA=specific variant
σ²=variance of population values i.e A²
/=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)=σ² using a Y axis
x=12
x=12
....yy
Post by: Colin2B on 26/08/2015 09:10:08
P(A)=σ² using a Y axisThis 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^∞=ythis 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.
Post by: alancalverd on 26/08/2015 18:11:28
^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?
Post by: guest39538 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,
10²=100
x^10=10²
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
Post by: guest39538 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+σ²)/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(σ²Y≠X)=1
P(B | A)=1
Post by: Colin2B 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
Post by: alancalverd on 27/08/2015 01:19:29
Post by: PmbPhy 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.
Post by: guest39538 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.
Post by: guest39538 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.
Post by: guest39538 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,
Post by: guest39538 on 27/08/2015 18:05:29
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,
Post by: guest39538 on 28/08/2015 23:55:41
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.
Post by: alancalverd on 29/08/2015 12:01:24
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.
Post by: guest39538 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
Post by: alancalverd on 30/08/2015 00:24:19
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 are beginning to see the light. Please keep your eyes open.
Nobody will accept a bet after the race is over. But I can tell you that if the coin is fair and you made a very large number of throws, about half of them would be heads, and the more throws you made, the closer you would get to 0.5. So my best guess at the number of heads in a finite number n throws would be n/2, and the probability of guessing the next throw would be 0.5 too, because the trials are fair and independent.
The essence of Bayesian statistics is that if all trials are independent, the particular trial you are considering is not special. It actually goes further than that:
You have captured one of the enemy's guns. It bears the serial number 1846. How many of these guns does the enemy have?
Post by: guest39538 on 30/08/2015 09:07:19
You are beginning to see the light. Please keep your eyes open.
Nobody will accept a bet after the race is over. But I can tell you that if the coin is fair and you made a very large number of throws, about half of them would be heads, and the more throws you made, the closer you would get to 0.5. So my best guess at the number of heads in a finite number n throws would be n/2, and the probability of guessing the next throw would be 0.5 too, because the trials are fair and independent.
The essence of Bayesian statistics is that if all trials are independent, the particular trial you are considering is not special. It actually goes further than that:
You have captured one of the enemy's guns. It bears the serial number 1846. How many of these guns does the enemy have?
What the hell? you are completely diverting from the topic and point Alan, like the entire internet is doing, you all are avoiding my simple maths. Science told me if i want to prove anything learn maths, I did learn maths and now you alll say its wrong, my head is seriously confused, what is the world trying to cover up?
Pa/x=1/52
pa/y=0_1
I know I am correct so why is everyone lying to me?
Post by: guest39538 on 30/08/2015 09:15:00
Take 2 sets of 2 sweets, in each set there is a blue sweet and a red sweet,
So we have two people with two sweets each a red and a blue sweet, ok so far?
Each person swaps shuffles the sweets between their hands behind their back and then both persons hold out their left closed hand and ask you to pick a one of the peoples left hands
the odds of a blue sweet are?........
Post by: alancalverd on 30/08/2015 17:09:48
Post by: Colin2B on 30/08/2015 18:14:10
Don't tell him he is wrong because if you can understand the answer you will take a big step forward in understanding probability.
... what is the world trying to cover up?No one is trying to cover anything up.
...........why is everyone lying to me?
Why would we lie to you, far easier to lie and say you are right?
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?
Probability is not about predicting the exact result of a particular group of throws, only what the likelyhood is of a particular mix of H & T.
Think about a shuffled deck of cards. You agree that the probability of the top card being ace of clubs is 1/52 yet, like the coins you have recorded above, the order of the deck is already determined after it has been shuffled. What probability tells you is the likelyhood that when shuffled the first card might be ace of clubs (or any other card).
With the coins tossed and recorded the probability that a particular toss came down H or T is 1/2. We have explained elsewhere how you can then calculate the probability of combinations and permutations of the 7 coins. But, it is only ever a probability, never a precise prediction.
Post by: alancalverd on 30/08/2015 21:52:22
Post by: guest39538 on 31/08/2015 07:49:51
What probability tells you is the likelyhood that when shuffled the first card might be ace of clubs
Probability tells us the likelihood of an event happening, Yes the probability of an ace of clubs being a top card after a random shuffle is 1/52. The problem is if we took 1000 decks, they are all equally as likely to have the ace of clubs as the top card. So if you were to pick one of the 1000 decks, you are equally likely to pick a deck with the same card as the top card you previously had.
I understand this, it is all of you that is not understanding me.
top card
top card
top card
top card
do you still believe your chance is 1/52?
pick one. In this scenario you can not define what 1 is neither can you define 52, it is something out of 4.
You guys are truly not seeing the point,
1
2
3
1
2
3
1
2
3
above is 3 sets that are randomly shuffled, you get the first value.
x
x
x
pick one and tell me your odds of getting a number 3?
None of you are accounting for the deviation from x to y and when considering the maths divert back to the x axis and independence rather than the collective of sets and repeat values of the sets occupying the same position of order in the sequence.
try it with 1 and 2
12
21
12
21
12
12
21
12
are you really suggesting this would not happen?
In a normal distribution we would receive 12121121121, in a random leap distribution we could receive 1111111111.
added- I would like to call the theory, The random leap distribution.
and here is the model
0....................................................................................100 normal distribution
t>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
0....................................................................................100 random leap distribution
Post by: guest39538 on 31/08/2015 08:39:52
cdefgabc does not sound the same as cfac
Post by: alancalverd on 01/09/2015 23:11:08
Quote
In a normal distribution we would receive 12121121121, in a random leap distribution we could receive 1111111111.
There being 11 members in the first sequence, the probability of generating it is 1 in 2^11. The second sequence contains 10 members so the probability is 1 in 2^10, so you are twice as likely to receive the second sequence as the first.
Post by: guest39538 on 02/09/2015 18:22:15
If the trials are independent, the odds do not change, by definition.QuoteIn a normal distribution we would receive 12121121121, in a random leap distribution we could receive 1111111111.
There being 11 members in the first sequence, the probability of generating it is 1 in 2^11. The second sequence contains 10 members so the probability is 1 in 2^10, so you are twice as likely to receive the second sequence as the first.
Yes if the trials are independent the chance would remain 1/52 , but when we have a superset and all the results are dependent of each other by offering choice the odds change,
You say that by random leaping you are twice as likely to receive the second sequence, So if I have 100 decks of cards and we choose a deck, and we will get the top card, are you saying we are twice as likely to receive the same value?
Post by: Colin2B on 02/09/2015 22:32:36
Post by: guest39538 on 03/09/2015 06:03:31
I think you have misunderstood the point Alan was making.
I am not sure then Colin, everyone misses my point,
I am tired now of trying and may just give up now and leave science to it although they are completely wrong about several things. Another forum just banned me again for discussing what was before the big bang, it is hilarious that they ban me if I do not comply to discipline and accept their version.
It has become a religion, how can people argue a theory to be absolute true and fact? it is funny that people do this , even funnier that the mods play along with it because they do not dare to upset long standing members.
I thing stuff them now Colin , wasted breath on deaf ears is just stealing my time.
Post by: alancalverd on 03/09/2015 18:39:00
You say that by random leaping you are twice as likely to receive the second sequence, So if I have 100 decks of cards and we choose a deck, and we will get the top card, are you saying we are twice as likely to receive the same value?
Nothing to do with random leaping. You are twice as likely to get the second set because it contained one less binary element!
If you had said 1212112112, which contains 10 elements, it would have had exactly the same probability as 1111111111, which also contains 10 elements, i.e. 1 in 2^10. That said, even if you made 1024 trials of 10 coin tosses, there is no certainty that either sequence would turn up once or once only, you'd just be a bit surprised if it didn't. And you have no prior knowledge of where in that 1024 trials it would turn up, so looking at random or sequentially would have exactly the same probability of finding it.
Here's a real-world example of the "significant number delusion". Many years ago I rebuilt a laboratory rig that originally included an analog voltmeter. For about 10 years, people had recorded its readings to 3 figures, e.g. 38.6V. I replaced it with a digital voltmeter and printer that displayed 5 figures, up to 99.999 volts. My boss complained that the rig wasn't working properly because it often reported "repeated digits" like 16.885, 17.793, 22.541 and so forth. It was working of course perfectly: if you take 5 random digits, the probability that the second matches the first is obviously 1/10, and that the third matches the second, 1/10.... i.e. there is a 40% probability that any 5 digit random number (such as a digital voltmeter reading) will include a repeated digit.
There is a similar problem with "significant sequences" and "significant cards". It is noticeable that throughout these discussions you have concentrated on the probability of finding aces whereas form the point iof view of the cards themselves, and indeed of the dealer (who can't see the faces) an ace is no more significant than a 6. So whilst 123 or AAA may mean something to you, it means nothing to the shuffled pack. Furthermore, you are not signifcant to the dealer who is also supplying cards to Alf, Bill and Charlie
Post by: guest39538 on 03/09/2015 20:01:39
Nothing to do with random leaping. You are twice as likely to get the second set because it contained one less binary element!
If you had said 1212112112, which contains 10 elements, it would have had exactly the same probability as 1111111111, which also contains 10 elements, i.e. 1 in 2^10. That said, even if you made 1024 trials of 10 coin tosses, there is no certainty that either sequence would turn up once or once only, you'd just be a bit surprised if it didn't. And you have no prior knowledge of where in that 1024 trials it would turn up, so looking at random or sequentially would have exactly the same probability of finding it.
Here's a real-world example of the "significant number delusion". Many years ago I rebuilt a laboratory rig that originally included an analog voltmeter. For about 10 years, people had recorded its readings to 3 figures, e.g. 38.6V. I replaced it with a digital voltmeter and printer that displayed 5 figures, up to 99.999 volts. My boss complained that the rig wasn't working properly because it often reported "repeated digits" like 16.885, 17.793, 22.541 and so forth. It was working of course perfectly: if you take 5 random digits, the probability that the second matches the first is obviously 1/10, and that the third matches the second, 1/10.... i.e. there is a 40% probability that any 5 digit random number (such as a digital voltmeter reading) will include a repeated digit.
There is a similar problem with "significant sequences" and "significant cards". It is noticeable that throughout these discussions you have concentrated on the probability of finding aces whereas form the point iof view of the cards themselves, and indeed of the dealer (who can't see the faces) an ace is no more significant than a 6. So whilst 123 or AAA may mean something to you, it means nothing to the shuffled pack. Furthermore, you are not signifcant to the dealer who is also supplying cards to Alf, Bill and Charlie
Thank you for the reply, I still do not think you even understand my idea , I will try for one last time to clearly explain the idea. I will explain it in experiment form.
We take a single deck of 52 cards, we will define this deck (A) , the deck has 52 individual variants, each individual variant we will define as (a).
We randomly shuffle (A) leaving (a) unseen .
We then stop shuffling, having a 52*(a) unknown set sequence, the order of the sequence can not change unless by interference.
We then in order top (a) first, lay the variants horizontally left to right still unseen , we will define this as vector X.
X=52*(a)=aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
P(a)/X=1/52
Do you understand this far and is the chance maths correct?
I will continue because I know you understand this.
We have a second deck that is equal to deck (A) in every way, we will call this (B)
We repeat the same process as we did with (A), with (B) ,and multiple sets making a second vector we will define Y
X(A)=52*(a)=aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
X(B)=52*(a)=aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
P(a)/X(A)=1/52
P(a)/X(B)=1/52
Is this correct so far showing you I understand the present information?
So my question , you have a choice of X(A) or X(B) and will receive the first (a) of the sequence which is the value on the left
X(A)=1*1=a
X(B)=1*1=a
P(a)/[X,Y]=?
Post by: alancalverd on 03/09/2015 23:32:02
It is clear that the probability of N being the first card in the second deal is also 1/52.
So the probability of N being the first card in both deals is (1/52)2
But if you don't name the card in advance, the first card in the first deal defines a value n, and the probability of n being the first card in the second deal is 1/52.
Thus if I want the ace of spades to be the first card, the probability of getting AS in any one deal is 1/52 and the probability of getting it in two consecutive deals is (1/52)2. However suppose I get 10C onhe first deal, and manage to win the hand with it. 10C has now become a "special" card and the chance of getting "my special card" on the next deal is 1/52
Now I see your delusion. You think that by choosing "nonconsecutive" deals you can increase the probability because 1/52 + 1/52 = 2/52. There's the error. The probability of getting one AS in two deals is indeed 2/52 because the desired result is "a OR b". But the probability of getting two AS in two deals is 1/522 because the desired result is "a AND b". George Boole is credited with laying the mathematical foundations for machine logic, which starts with the realisation that "+" describes "or" relations and "x" describes "and" relations.
Post by: guest39538 on 04/09/2015 07:46:33
What I think you are trying to say by "P(a)/X(A)" is that the probability of a named card N being first in the first deal is 1/52.
It is clear that the probability of N being the first card in the second deal is also 1/52.
So the probability of N being the first card in both deals is (1/52)2
But if you don't name the card in advance, the first card in the first deal defines a value n, and the probability of n being the first card in the second deal is 1/52. yes
Thus if I want the ace of spades to be the first card, the probability of getting AS in any one deal is 1/52 and the probability of getting it in two consecutive deals is (1/52)2. However suppose I get 10C onhe first deal, and manage to win the hand with it. 10C has now become a "special" card and the chance of getting "my special card" on the next deal is 1/52 yes
Now I see your delusion. You think that by choosing "nonconsecutive" deals you can increase the probability because 1/52 + 1/52 = 2/52. There's the error. The probability of getting one AS in two deals is indeed 2/52 because the desired result is "a OR b". But the probability of getting two AS in two deals is 1/522 because the desired result is "a AND b". George Boole is credited with laying the mathematical foundations for machine logic, which starts with the realisation that "+" describes "or" relations and "x" describes "and" relations.no
I think that that removing all the other variants except the first column, and changing alignment from the x axis to the Y axis changes the probabilities, nothing to do with improving chance or lessening chance, more a multivariate. Because obviously and factually if we in the end only have two unknown variants,
a
a
the chance is now certainly not 52 any more because it is 2. and we certainly do not know the identity of the 2 variants, so the P would be ?/2 and not 1/52
The answer I asked for should of been 2/2 , I asked what it the chance of receiving a (a) from this scenario. We started with 52 in each but we no longer have 52 we have 2. two that we do not the values of at all, we know that 1 of 52 could be an ace diamonds or a ten of clubs, we only know this because we know what the 52's ingredients are, to define a chance we have to know the ingredients, if we do not know the ingredients we can not have a clue of the chance. Both top cards could be the ace of diamonds or the ten of clubs, they have an equal chance to be so. Consider 100 decks and take the 100 top cards , a subset made by x^100, a column that is multivariate and unknown in values, so if I ask you to choose 1 of these 100 cards, what are you choosing from? you know longer are playing the set you are playing the subset with a variation of values made by the shuffle of the x's
added I have just thought of this
P(n)=[X,Y}^100/t=δ
a
aa
x=y in this scenario
aa
aa
x ≠ Y in this scenario considering that any (a) is a specific variant and the rows are randomised making a column subset Y to choose from.
PPaa
PPaa
...PP
Now we have added aligned outputs to X and Y, we can clearly see that Px is 1/2 and the rows and x always remain at 1/2.
But we can also see that by adding choice, and diverting from x to y, we are no longer playing 1/2 . we are playing ?/2 and player 1 and player 2 in the Y alignment both have different chances compared to the chances of x. X is dependent of 52 known variants where Y is dependent to choice and alignment. Each player every turn has a dependent chance that is different to all players playing the x axis having the same chance.
12
12
by showing values we can clearly see that the columns of the y axis can have repeat values, we can shuffle 1 and 2 as many times as we like, and we can never align x with y unless the values were all the same.
11
11
We can clearly observe this
12
21
22
11
21
12
21
21
12
12
x is not equal to Y , we could never ever have 22 aligned to x or 11 where y we can.
Post by: alancalverd on 04/09/2015 17:07:04
Beyond that, I have no idea what you are talking about.
Post by: guest39538 on 04/09/2015 17:43:40
I'm sorry you disagree with Boolean algebra. It's the basis for all computer hardware, and seems to work for everyone else.
Beyond that, I have no idea what you are talking about.
I have no idea where computer hardware comes into this, I have looked at Boolean Algebra but cannot read it because I do not know what the symbols represent, I will start to learn about this.
Maybe you will understand this, at any specific point in time whilst playing live texas holdem poker there is only ever 1 of 52 variants aligned to your seat.
On the internet playing internet texas holdem poker, there is 1,000,000 unknown variants aligned to your seat at any specific time.
So if you have a choice of 1,000,000 unknown variants, because these are aligned to you, that would be impossible to be a 1/52 chance.
Post by: Colin2B on 05/09/2015 09:35:53
Maybe you will understand this, at any specific point in time whilst playing live texas holdem poker there is only ever 1 of 52 variants aligned to your seat.OK, I understand where 52 variants comes from in live game = 52 different card faces.
On the internet playing internet texas holdem poker, there is 1,000,000 unknown variants aligned to your seat at any specific time.
I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.
Post by: guest39538 on 05/09/2015 10:59:31
OK, I understand where 52 variants comes from in live game = 52 different card faces.
I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.
Yes you do understand, because in bold on the internet that is exactly what you do by the alignment of seat to card order,
I get we have 80,000* chances of receiving an ace from the proportion of 1,000,000 top cards
I have been stating the question wrong, that is why you do not understand.
I will restate the question
If we have 100 decks of cards and randomly shuffle each deck individually, what proportion of the 100 top cards of each deck will be an ace, the answer is approximately 8, so what is the chance of intercepting one of these aces out of the 100, the answer is 8/100.
So what is the chance of incepting an ace from an aligned column of 1,000,000 top cards?
see new model here
page 652
http://badscience.net/forum/viewtopic.php?f=3&t=36878&start=16275
Consider that by adding choice of deck, we change axis to an alignment of unknown face values,
Post by: guest39538 on 05/09/2015 11:40:20
12
12
12
12
I offer you a choice of set and you will get the first value, your choice is from a created subset of x,
∄∀=P(Y)∀⇒≦≠∀
∃f (x)≠f(⊂ Y)∴∄=P
Post by: Colin2B on 06/09/2015 00:13:45
OK, I understand where 52 variants comes from in live game = 52 different card faces.
I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.
Yes you do understand, because in bold on the internet that is exactly what you do by the alignment of seat to card order,
No I don't understand.
What do you mean by "the alignment of seat to card order"
How does this differ from live game.
You are surely not suggesting that you play with a deck of 1,000,000 cards?
And please, describe this in words don't try to use maths.
Post by: guest39538 on 06/09/2015 08:16:36
OK, I understand where 52 variants comes from in live game = 52 different card faces.
I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.
Yes you do understand, because in bold on the internet that is exactly what you do by the alignment of seat to card order,
No I don't understand.
What do you mean by "the alignment of seat to card order"
How does this differ from live game.
You are surely not suggesting that you play with a deck of 1,000,000 cards?
And please, describe this in words don't try to use maths.
In texas holdem poker we have what is called a small blind and a big blind, an opening bet/ante by two players, the blinds rotate every turn clockwise, the small blind always receives the first card, the big blind always receives the second card, and so on in order clockwise around the table, this is the same as a live game.
Well actually if there is 1,000,000 top cards there is 52,000,000 cards in total, If I asked you to pick any card from any position of any deck, I have asked you to choose a card of 52,000,000 cards.
If I tell you to choose a card from the top cards, you are choosing from 1,000,000 cards effectively playing a 1,000,000 card dec
Think about this diagram Colin
12
12
12
12
12
If we play an x axis, we always play 1/2, if I ask you to pick a card from the x axis of any set it is 1/2 , you are playing a 2 card deck, if I asked you to pick from the vertical Y axis you are playing a 5 card deck. If you can consider the axis change, this is the main part to understanding and I am sure you will understand, concentrate on the axis change, look and observe the difference.
2 meter is not the same as 5 meters, the vertical is a greater distance that contains more space and more variants in that space compared to x, the deviation of velocity x to velocity y, changes x to y a greater distance and a greater amount of values, x can only have 1 of each variant over a set distance of x, where y can have multiple of the same variant or none of a variant over the greater distance .
first odds is to pick a deck
12
12
12
12
12
p
second odds once the deck choice is made
p12
However the second odds become irrelevant because the first odds defines your card.
then consider two players, the small blind and big bang
12
12
12
12
12
py
p is not equal to y
columns Colin, not a linearity like x
Py12
1
2
p
y
that would be equal
py1-52
1-
52
p
y
is not the same as
py1-52,1-52,1,52
1-
52
1-
52
1-
52
p
y
1 hand at a time, is not the same as picking from 3 hands at the same time.
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?
Post by: alancalverd on 06/09/2015 08:47:39
Quote
Well actually if there is 1,000,000 top cards there is 52,000,000 cards in total, If I asked you to pick any card from any position of any deck, I have asked you to choose a card of 52,000,000 cards.
But there are only 52 face values, so the probability of getting any given card is 1/52
Quote
If I tell you to choose a card from the top cards, you are choosing from 1,000,000 cards effectively playing a 1,000,000 card dec
But there are only 52 face values, so the probability of getting any given card is 1/52
Post by: guest39538 on 06/09/2015 09:03:20
But there are only 52 face values, so the probability of getting any given card is 1/52
There are only 52 face values in the x-axis Alan, consider the proportion of face values in the created subset Y-axis. This is not equal to the X-axis Alan.
There is a range from 0.01 to 1 per example 100 face values of the Y-axis.
Consider 100 decks of cards is not just decks of cards, they are hands, it would be 100 hands Alan, a choice of 100 Hands per column .
Then consider continuous time 1 a linearity, a path that has not yet been written because the hands have not been created or distributed, it is played 1 hand at a time.
Then consider random time 2 and interception, playing 100 hands at a specific time that are already written, lets say 1 hours of play all in space at 1 second of play, its complex, but you should be able to understand the basic things easy enough.
1 at a time is not 100 at a time.
Post by: guest39538 on 06/09/2015 09:14:32
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.
Post by: Colin2B 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.
Post by: guest39538 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.
Post by: guest39538 on 06/09/2015 10:48:13
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.
Post by: Colin2B on 06/09/2015 11:14:43
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.
Post by: guest39538 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
Post by: guest39538 on 07/09/2015 12:51:29
I think I am trying to describe this -
SL(n, Z) → SL(n, Z/N·Z)
https://en.wikipedia.org/wiki/Congruence_subgroup
Post by: Colin2B on 07/09/2015 14:40:10
ColinCan you explain why you think that and how you link the variables to your problem?
I think I am trying to describe this -
SL(n, Z) → SL(n, Z/N·Z)
https://en.wikipedia.org/wiki/Congruence_subgroup
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.
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 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.
Post by: guest39538 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
Post by: chiralSPO 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.
Post by: Colin2B 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.
Post by: alancalverd 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.
Post by: guest39538 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?
Post by: chiralSPO 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/(A–1) which will satisfy your equation.
Post by: guest39538 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/(A–1) 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]?
Post by: chiralSPO on 08/09/2015 20:09:47
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/(A–1), which you should understand to mean: "B must equal the ratio of A and A minus one."
Post by: guest39538 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/(A–1), 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=10cm²
XY^100=XYZ=10cm³
E=XYZ-^XYZ=0cm³
would this make any sense to you?
Post by: chiralSPO 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=10cm²
XY^100=XYZ=10cm³
E=XYZ-^XYZ=0cm³
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...
Post by: guest39538 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=10cm²
XY^100=XYZ=10cm³
E=XYZ-^XYZ=0cm³
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
Post by: Colin2B on 09/09/2015 09:59:13
^ to the power of and if you notice I was subtracted the power of contracting XYZ.You can't raise a - sign to a power
Time begun XYZ-^XYZ=0
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.
Post by: guest39538 on 09/09/2015 16:57:56
^ to the power of and if you notice I was subtracted the power of contracting XYZ.You can't raise a - sign to a power
Time begun XYZ-^XYZ=0
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.
Post by: jeffreyH on 09/09/2015 22:01:51
. So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000
Post by: Colin2B 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
Also, you still havent explained how can you say 'time begun' = 0cm3
Post by: guest39538 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
Post by: chiralSPO 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!
Post by: guest39538 on 10/09/2015 20:07:05
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!
yes we did go over this a while back, but regardless of how science defines it, arbitrary time is based on motion of the sun relative to the earth's spin, and also the dating of the big bang is based on expansion which is a distance travelled of mass from an observation point.
Time what we know is arbitrary and that is why a time dilation does not occur to anything that really changes time, it is nothing more than an arbitrary clock failure.
Post by: Kenyonm on 19/12/2015 20:57:05
Post by: guest39538 on 20/12/2015 14:05:27
Is probability a product of the human brain trying to order what is of course chaos?
Probability is the amount of something in something, i.e 1/6 explains there is 6 individual variants.
Quote
Is anything actually random or is it actually the outcome of a set of historic actions?
Yes there is a proper random, but something like 1/6 is only semi-random dependent to time.
Quote
Is everything going to be predictable in the future where computer processing speeds are as fast as light in the time to see a path with infinitely dense vision with an accuracy to be able to see the outcome. For example watch the lottery balls spinning around chaotically and then predict which balls will come out before they actually do.
No, you can't predict what exactly will come out but you can narrow it down to certain possibilities.