# I am a bit confused about something. Watch

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(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

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

(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

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I did post the question in the first post. I suppose you mean a more detailed question,

In a game of chance that has 52 variants such as a game like texas holdem poker, does offering a choice of pre-shuffled decks change the probability function of using a single deck?

My maths says yes is does.

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#6

Those are random messes of symbols. That's not a question. Nor is it a proof of anything. It's just some symbols. Start by defining them.

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(Original post by

Those are random messes of symbols. That's not a question. Nor is it a proof of anything. It's just some symbols. Start by defining them.

**BlueSam3**)Those are random messes of symbols. That's not a question. Nor is it a proof of anything. It's just some symbols. Start by defining them.

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

d = distance or dimension

P=probability

x=vector

y=vector

f=function

^x=number of sets/random shuffles

t1=continuous time

t2=random choice of set bringing variants forward in time

∝=equally proportionate

≠=not equal to

σ²=variance of population values

n=any specific variant

P(n)/(dy)=σ²/t2=0_1

....yyyyy

x=nn

**n**nn

x=n

**n**nnn

x=nnn

**n**n

x=nn

**n**nn

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,

oooooooooo 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?

*P*(B | A)=1

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#8

(Original post by

(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]

t

d = distance or dimension

P=probability

x=vector

y=vector

f=function

^x=number of sets/random shuffles

t1=continuous time

t2=random choice of set bringing variants forward in time

∝=equally proportionate

≠=not equal to

σ²=variance of population values

n=any specific variant

P(n)/(dy)=σ²/t2=0_1

....yyyyy

x=nn

x=n

x=nnn

x=nn

**AlbertXY**)(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

d = distance or dimension

P=probability

x=vector

y=vector

f=function

^x=number of sets/random shuffles

t1=continuous time

t2=random choice of set bringing variants forward in time

∝=equally proportionate

≠=not equal to

σ²=variance of population values

n=any specific variant

P(n)/(dy)=σ²/t2=0_1

....yyyyy

x=nn

**n**nnx=n

**n**nnnx=nnn

**n**nx=nn

**n**nn
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,

oooooooooo 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?

*P*(B | A)=1
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(Original post by

This is gibberish. Explain your point.

This is wrong. You are making a decision based on limited information. The information that is available to you absolutely does have baring on the probability of things, and in this case, the probability absolutely is 1/2.

This, again, is gibberish.

**BlueSam3**)This is gibberish. Explain your point.

This is wrong. You are making a decision based on limited information. The information that is available to you absolutely does have baring on the probability of things, and in this case, the probability absolutely is 1/2.

This, again, is gibberish.

we know the values of x but we do not know the values of y.

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

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?........

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#10

(Original post by

It is not gibberish the world is just not understanding it, limited information exactly.

we know the values of x but we do not know the values of y.

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

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?........

**AlbertXY**)It is not gibberish the world is just not understanding it, limited information exactly.

we know the values of x but we do not know the values of y.

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

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?........

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You know Im correct, the world knows im correct , I have produced a piece of brilliance, yet the world wants to ignore me, Einstein would be turning in his grave at the thought of it, simple maths and it is avoided, this is hilarious from all the forums, why?

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#12

(Original post by

It is not gibberish the world is just not understanding it, limited information exactly.

**AlbertXY**)It is not gibberish the world is just not understanding it, limited information exactly.

we know the values of x but we do not know the values of y.

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

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?........

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

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?........

(Original post by

You know Im correct, the world knows im correct , I have produced a piece of brilliance, yet the world wants to ignore me, Einstein would be turning in his grave at the thought of it, simple maths and it is avoided, this is hilarious from all the forums, why?

**AlbertXY**)You know Im correct, the world knows im correct , I have produced a piece of brilliance, yet the world wants to ignore me, Einstein would be turning in his grave at the thought of it, simple maths and it is avoided, this is hilarious from all the forums, why?

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(Original post by

No, it's poorly written gibberish. You are explaining it. If people don't understand it, that is your fault and nobody else's.

One half. Obviously. Probability is relative to information availability.

Cut the garbage. Explain your point. Properly. In English that people can actually understand. You're nowhere near as intelligent as you seem to think that you are. To quote Einstein: "If you can't explain it to a six year old, you don't understand it yourself". Given that you've not only failed to explain it to a six year old, but also to actual mathematicians. What does that tell you about your (complete lack of) understanding?

**BlueSam3**)No, it's poorly written gibberish. You are explaining it. If people don't understand it, that is your fault and nobody else's.

One half. Obviously. Probability is relative to information availability.

Cut the garbage. Explain your point. Properly. In English that people can actually understand. You're nowhere near as intelligent as you seem to think that you are. To quote Einstein: "If you can't explain it to a six year old, you don't understand it yourself". Given that you've not only failed to explain it to a six year old, but also to actual mathematicians. What does that tell you about your (complete lack of) understanding?

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#14

(Original post by

X=(52)

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

Y=f(^X)

P(A)/Y=σ

X≠Y

**AlbertXY**)X=(52)

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

Y=f(^X)

P(A)/Y=σ

^{2/t2}X≠Y

^{How can this be incorrect?}
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#15

(Original post by

The point is that using many sets instead of one set gives a different probability to using one set of variants.

**AlbertXY**)The point is that using many sets instead of one set gives a different probability to using one set of variants.

(Original post by

Is this A level maths?

**Mason_**)Is this A level maths?

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#16

(Original post by

No. It's poorly explained drivel.

**BlueSam3**)No. It's poorly explained drivel.

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#17

(Original post by

Serious answer please because if it is I may have to reconsider the subjects I'm taking.

**Mason_**)Serious answer please because if it is I may have to reconsider the subjects I'm taking.

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#18

**Mason_**)

Serious answer please because if it is I may have to reconsider the subjects I'm taking.

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#19

(Original post by

That is a serious explanation. Note the first word. This is not, in any way, A-level maths. It's not even maths, at this point. It's an incoherent mess of symbols.

**BlueSam3**)That is a serious explanation. Note the first word. This is not, in any way, A-level maths. It's not even maths, at this point. It's an incoherent mess of symbols.

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(Original post by

Now state it clearly, explain it, and prove it. Recall that proofs are written in English.

**BlueSam3**)Now state it clearly, explain it, and prove it. Recall that proofs are written in English.

The simplified maths what you will know is

P(A)/x=1/52

P(A)/y=σ

_{X=1_0}

_{model}

_{....yy}

x=00

x=00

*P*(B | A)=1

∑P(A)/y=(0)&(.5)&(1)=0_1

*A*⊥

*B*

*x*⊢

*y*

*μx=1*

*μy=*σ²

x

__<<__y

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