# The Naked Scientists Forum

### Author Topic: Is this maths correct?  (Read 12608 times)

#### Thebox

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##### Is this maths correct?
« on: 18/06/2015 19:12:40 »
x={1/52}/t
y={1/∞}/t
X=player seat
t=time
P=probabilities
∅=random

P(x)=P(X∩y)/∅t
« Last Edit: 18/06/2015 19:20:45 by Thebox »

#### chiralSPO

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##### Re: Is this maths correct?
« Reply #1 on: 18/06/2015 19:23:18 »
I have no idea what you're trying to derive here...

I assume it has something to do with cards (1/52) and probability.
1/∞ is not defined itself, and so isn't a good part of a deffinition.
what does "player seat" mean?
what does "∅ = random" mean? is it a random number (generated how/chosen from what set with what probability function) is ∅ a randomizing function?

#### Thebox

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##### Re: Is this maths correct?
« Reply #2 on: 18/06/2015 19:40:24 »
I have no idea what you're trying to derive here...

I assume it has something to do with cards (1/52) and probability.
1/∞ is not defined itself, and so isn't a good part of a deffinition.
what does "player seat" mean?
what does "∅ = random" mean? is it a random number (generated how/chosen from what set with what probability function) is ∅ a randomizing function?

I have just realised that ''y'' is wrong, i think the set is {∞/∞}  , in this situation y axis has an infinite number of a single x, x being any one of 52 variants, a deck of cards.

52 in a x axis
then a sequence on top of that of 52 and so on for infinite times y axis,

123456789
123456789
123456789
123456789
Like that......1-52 being x axis and any x value being infinite y axis.

x is finite but y is infinite ,

Player seat is the receiver of x from x axis, but by random timing of events, we skip rows and receive a random row of x from y.

so player X, intercepts Y and between there i space time, but the gap is random ,

∅ just means random symbol,  that the distribution of y is by random timing of x

P(x/t)≠P(y/∅t)

P(x/t)=221

P(y,x/∅t)=var(X)

x~=1/52

corr(X,Y)={1/∞}/var(X)~t

{x=y}≠{y/∅t}

« Last Edit: 18/06/2015 20:47:22 by Thebox »

#### PmbPhy

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##### Re: Is this maths correct?
« Reply #3 on: 19/06/2015 06:30:00 »
x={1/52}/t
y={1/∞}/t
X=player seat
t=time
P=probabilities
∅=random

P(x)=P(X∩y)/∅t
It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.

#### Colin2B

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##### Re: Is this maths correct?
« Reply #4 on: 19/06/2015 07:36:49 »
It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.
To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!

#### PmbPhy

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##### Re: Is this maths correct?
« Reply #5 on: 19/06/2015 10:18:25 »
It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.
To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!
Quite right my friend.  And as ChiralSPO mentioned, 1/∞ is total nonsense. Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it. Something like this rooted in complete ignorance.
« Last Edit: 20/06/2015 12:11:56 by PmbPhy »

#### Thebox

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##### Re: Is this maths correct?
« Reply #6 on: 19/06/2015 17:40:39 »
It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.
To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!
Quite right my friend.  And as ChiralSPO mentioned, 1/∞ is total nonsense. Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it. Something like is rooted in complete ignorance.

I know very well that ∞ represents infinite, I have a Y axis that is infinite and contains an infinite number of variant x's, where my x axis only contains 52 variants of x.

I am trying to associate maths, and represent 52 variants of x, and 52 variants stacked for infinite time.

52
52
52
52

like that, I need the for Y, to conclude my formula.

#### alancalverd

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##### Re: Is this maths correct?
« Reply #7 on: 19/06/2015 17:49:48 »
1/x → 0 as x → ∞, so why not just write 0?

#### Thebox

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##### Re: Is this maths correct?
« Reply #8 on: 19/06/2015 18:31:43 »
1/x → 0 as x → ∞, so why not just write 0?

y=0 ?  that would not make any sense?

Probability of X from x is 1/52
probability of X from y is infinite

p580

« Last Edit: 19/06/2015 19:55:15 by Thebox »

#### PmbPhy

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##### Re: Is this maths correct?
« Reply #9 on: 19/06/2015 23:53:38 »
Quote from: Thebox
y=0 ?  that would not make any sense?
It's an abuse of infinity because you're treating it as a number and not using it in the proper sense, i.e. in terms of limits. Infinity is not a number and division is only defined for numbers.

#### Colin2B

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##### Re: Is this maths correct?
« Reply #10 on: 20/06/2015 00:19:42 »
y=0 ?  that would not make any sense?
Correct, it does not make any sense, but that is what you wrote!
Nothing else makes sense either.

probability of X from y is infinite
Probability can only take values from 0 to 1. Infinite probability is meaningless.

Why are you using time as a value. To paraphrase Tina Turner "What's time got to do with it?"??

#### Thebox

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##### Re: Is this maths correct?
« Reply #11 on: 20/06/2015 09:31:26 »
y=0 ?  that would not make any sense?
Correct, it does not make any sense, but that is what you wrote!
Nothing else makes sense either.

probability of X from y is infinite
Probability can only take values from 0 to 1. Infinite probability is meaningless.

Why are you using time as a value. To paraphrase Tina Turner "What's time got to do with it?"??

I am using time, because time decides what point of Y you receive. Intersection of x,y by random timing of a tables hand.

But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.

x ⊇ ∞y

every element of x is an infinite element of y.

[x : y]=∞

« Last Edit: 20/06/2015 10:24:34 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #12 on: 20/06/2015 10:51:10 »
I am using time, because time decides what point of Y you receive. Intersection of x,y by random timing of a tables hand.
It doesn't change the probability. Elements are not events.

every element of x is an infinite element of y.
"Infinite element" is meaningless unless you explain what you mean by it.

But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.
This is a brick wall of your own building. You seem committed to building it as thick and high as you can.

Please stop to think before you post.
And do try to learn basic maths. Probability is not a good place to start as your misunderstandings show. You are also misusing set notation. Don't pick symbols and think you understand their meaning without studying their maths theory first.
People are willing to help you, but you do need to put some effort in, it's not easy but I'm convinced you can do it.

« Last Edit: 20/06/2015 11:17:18 by Colin2B »

#### PmbPhy

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##### Re: Is this maths correct?
« Reply #13 on: 20/06/2015 12:24:28 »
Quote from: Thebox
I know very well that ∞ represents infinite, ..
You misunderstood what I said. I wrote
Quote
Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it.
Did you see me say or imply that you don't know what ∞ means? No. You didn't.  Your problem with it has always been that you're treating infinity as if it was a number, and it's not.

Quote from: Thebox
I have a Y axis that is infinite ...
This statement is meaningful because the y-axis in a Cartesian coordinate system is unbounded and never ends and that's the meaning of infinite. So you got that part correct. Then you went downhill from there and said
Quote from: Thebox
and contains an infinite number of variant x's, where my x axis only contains 52 variants of x.
when you never defined what "variant x's" means. In the context that you used it, it's not a mathematical term so you must have defined it yourself. Perhaps its just poor English on your part.

Quote from: Thebox
I am trying to associate maths, and represent 52 variants of x, and 52 variants stacked for infinite time.
But what you've ended up doing is posting things which are completely meaningless. Nobody besides yourself knows what the hell you're talking about.

Quote from: Thebox
But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.
That's up to you but you never gave science a shot. All you've done is made erroneous claims as well as claim that physics is wrong when you couldn't understand it. You've never taken the time to learn it. I gave you the opportunity to do so. But instead of reading a physics text you used what little time that you claim to have posting all this nonsense. Any reason that you give for not reading is bogus. Everyone who's serious about learning physics can most certainly take 15 minutes out of their day to read a few pages in a text. You claimed you couldn't but I believe that you've been lying to me. That's probably why you refuse to send me a message every day telling me what you could accomplish that day, even if it was nothing. What's your excuse for not doing that? I've asked you that almost every single day and you ignored my question every single day. That's why you were suspended from my forum. I'll give you two weeks to start doing that after which you'll be permanently banned.

The worst part about your attitude in this particular thread is that you asked us if the math was correct and when we explain that it isn't you claim that we're wrong.
« Last Edit: 20/06/2015 12:55:53 by PmbPhy »

#### Thebox

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##### Re: Is this maths correct?
« Reply #14 on: 21/06/2015 06:27:36 »

The worst part about your attitude in this particular thread is that you asked us if the math was correct and when we explain that it isn't you claim that we're wrong.

I  assume you know what it means without explanation.  X is any one of the 52 variants of x axis. X in the y axis is any one of the 52 variants of x*∞.  There is an infinite amount of rows of 52, y axis.
So lets say row 10 , column 1, there is an X with the value of being an ace.
By random timing this could be intercepted of the distribution.  Do you agree?

12345
23145
53241
12345

In the above I have a 2/4 chance of receiving a 1 if my go is first of the distribution.

So if we increased the rows, lets say to 500 rows, and lets say in column one there was 150 ones spread out over y, I have 150/500 chance of receiving a 1.

Now do you see?

shuffle a deck of cards, there is a 4/52 chance that the top card is an ace.

shuffle another deck of cards, there is also a 4/52 chance the top card is an ace.

have a choice of either decks,

that is two chances that the top card is an ace, 4/52^2 or 4/52*2 or something else.

I have a 1/52 chance of any card being in my distribution place, so consider that I also have like my example,

also I have a 150/500 chance from Y being inline,

I have an easy explanation now, it came to me.

shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.

shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52

This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

You then have a choice of any deck, your odds are 15/100 of getting an ace and not 4/52.

Now this is as simple explanation as it gets, surely you all understand this?

« Last Edit: 21/06/2015 07:00:53 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #15 on: 21/06/2015 10:54:59 »

Quote
shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.

shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52

This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

and there's your problem. If you know that 15/100 are aces, then you have good reason to believe that the cards were not randomly shuffled. If they were properly shuffled the most probable number of aces would be between 7 and 8. You can use a Gauss curve to determine the probability of 15 aces in 100 random shuffles: I don't have the time to do right now it but my guess is way less than 5%.

#### PmbPhy

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##### Re: Is this maths correct?
« Reply #16 on: 21/06/2015 13:15:58 »
Quote from: Thebox
I  assume you know what it means without explanation.
Therein lies your problem. You can't make assumptions of what people know in either math or physics. It's just bad juju.

Quote from: Thebox
X is any one of the 52 variants of x axis.
Here's a good example of what you're doing wrong. The term variant means either
Quote
1:  manifesting variety, deviation, or disagreement

2: varying usually slightly from the standard form
But you haven't said what it is a variant of. The number "52" reminds us of poker but you didn't mention poker so you're forcing us to make assumptions and that's a very poor thing for someone to do in either physics or math. And if its 52 different things what are those different things? Are the letters? Numbers? If they're numbers are they real or complex? Are they vectors, tensors, 1-forms, etc?

Quote from: Thebox
X in the y axis is any one of the 52 variants of x*∞.
A mathematician wouldn't even talk to you if you didn't stop writing things like this nonsense "x*∞."  And the statement " X in the y axis" is meaningless.

Quote from: Thebox
There is an infinite amount of rows of 52, y axis.
Also meaningless.

Quote from: Thebox
So lets say row 10 , column 1, there is an X with the value of being an ace.
By random timing this could be intercepted of the distribution.  Do you agree?
Nobody can agree with something that's so poorly stated as to have little meaning. I.e. the phrase this could be intercepted of the distribution is meaningless, just plain nonsense.

What you've posted here is so poorly phrased that you won't find anybody who will be able to understand it. Your problem is that you want to talk about a more difficult area of math, i.e. probability, when you have no understanding of algebra yet,

#### Thebox

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##### Re: Is this maths correct?
« Reply #17 on: 23/06/2015 07:09:05 »
xxx
xxx
xxx
xxx
xxx

each row has 3 units 1-3 that are randomly shuffled, the odds of any row having a 1 in position one (the left) is 1/3.

What are the odds of choosing a row from the five rows, that number one fell into that position one?

#### alancalverd

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##### Re: Is this maths correct?
« Reply #18 on: 23/06/2015 07:48:03 »
1/3, obviously. Since the rows are independent, it doesn't matter which you choose.

#### Thebox

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##### Re: Is this maths correct?
« Reply #19 on: 23/06/2015 18:56:53 »
1/3, obviously. Since the rows are independent, it doesn't matter which you choose.

consider the random sequence is set, you are choosing from the Y axis, how can it be 1/3 when there is clearly 5 choices?

........................y
........................y
x axis  123456789
........................y
........................y
........................y

x ≠ y

P(x)=1/9

p(y)={1/9}+{1/5}=?

x∩y
« Last Edit: 23/06/2015 19:10:51 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #20 on: 23/06/2015 19:07:15 »
You only make one choice. None of the groups knows anything about any of the other groups.

Consider a simpler example. I have tossed a coin 1,000,000 000 times. What are the odds that it will come down heads next time?

Now I roll a die 1,000,000,000 times. What are the odds of getting a six on the next roll?

#### Thebox

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##### Re: Is this maths correct?
« Reply #21 on: 23/06/2015 19:17:34 »
You only make one choice. None of the groups knows anything about any of the other groups.

Consider a simpler example. I have tossed a coin 1,000,000 000 times. What are the odds that it will come down heads next time?

Now I roll a die 1,000,000,000 times. What are the odds of getting a six on the next roll?

I understand what you are saying , but you are not considering after the role or toss.

I toss a coin 1,000,000 times, and record the results, Heads is a win and tails is a loss.

The sequence of the first ten throws are

hhththttth

This is all good, based on only you playing, now lets consider that there is 2 players, I and you, except by randomness, the already results, are distributed to us both.

you get the already tossed 3rd toss, and the 5th toss, and the ninth toss.

You receive 3 tails in a  row.

can you understand that?

the rows are different to the columns.  The Y deck contains repeat variants.

see model here, surely you understand this one page 581

« Last Edit: 23/06/2015 19:40:19 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #22 on: 23/06/2015 23:03:21 »
Why don't I get the 1st and 7th?

#### Thebox

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##### Re: Is this maths correct?
« Reply #23 on: 24/06/2015 06:20:49 »
Why don't I get the 1st and 7th?

You may well get the 1st and 7th because it is random, but that does not matter, what matters is that x is not equal to Y.

you do not get 1/52 from the Y axis, you would get ?/52 if 52² and 52 is x axis, and 52 high Y axis.

123
231
123
132

all x axis would be 1/3 where y axis you can clearly observe is different.

randomly mix the x axis as much as you like it will remain 1/3 always. where Y is subject to constant change .
Y axis is not 52 variants, it is ? variants by alignment.

there could be 50 variants, having 6 aces in the Y axis.

« Last Edit: 24/06/2015 06:48:10 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #24 on: 24/06/2015 07:58:24 »
Again, think about the coin toss.

What is the probability that you get 4 heads and 1 tail?*

What is the probability that you get 40000 heads and 10000 tails?

You can't discern or prove randomness from a small number of trials. 4:1 is unlikely but would not suggest a biassed coin, whilst 40000:10000  looks like a bent penny.

If you study any table of random numbers you will find a few short sequences, but the mere presence of sequences is not proof of predictability.  Indeed the absence of sequences is evidence that the numbers are not random.

*these numbers are not chosen at random! Test cricket fans keep a record of captains winning the toss, and anyone who wins more than 3 in a 5-match series is honoured in the record books even if the team performs disastrously.
« Last Edit: 24/06/2015 08:00:44 by alancalverd »

#### The Naked Scientists Forum

##### Re: Is this maths correct?
« Reply #24 on: 24/06/2015 07:58:24 »