# This maths is correct, how can it be wrong?

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#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #50 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?
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #51 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?

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #52 on: 30/08/2015 09:15:00 »
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?........

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #53 on: 30/08/2015 17:09:48 »
1/2
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#### Colin2B

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##### Re: This maths is correct, how can it be wrong?
« Reply #54 on: 30/08/2015 18:14:10 »
Alan beat me to it.
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?

...........why is everyone lying to me?
No one is trying to cover anything up.
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.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #55 on: 30/08/2015 21:52:22 »
PS the answer to my question in reply #50 is "probably 3692, so knowing people's predilection for round numbers, assume 3500 - 4000 until we capture another one". Military intelligence is not a selfcontradiction, it is the life-or-death application of good statistics.
« Last Edit: 30/08/2015 21:57:35 by alancalverd »
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #56 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

« Last Edit: 31/08/2015 08:36:36 by Thebox »

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #57 on: 31/08/2015 08:39:52 »
anyone play a piano?

cdefgabc  does not sound the same as cfac

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #58 on: 01/09/2015 23:11:08 »
If the trials are independent, the odds do not change, by definition.

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.
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #59 on: 02/09/2015 18:22:15 »
If the trials are independent, the odds do not change, by definition.

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.

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?

#### Colin2B

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##### Re: This maths is correct, how can it be wrong?
« Reply #60 on: 02/09/2015 22:32:36 »
I think you have misunderstood the point Alan was making.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #61 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.

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #62 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
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #63 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]=?

« Last Edit: 03/09/2015 20:34:15 by Thebox »

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #64 on: 03/09/2015 23:32:02 »
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.

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.
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #65 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.

« Last Edit: 04/09/2015 08:27:13 by Thebox »

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #66 on: 04/09/2015 17:07:04 »
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.
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#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #67 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.

#### Colin2B

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##### Re: This maths is correct, how can it be wrong?
« Reply #68 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.
On the internet playing internet texas holdem poker, there is 1,000,000 unknown variants aligned to your seat at any specific time.
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.
« Last Edit: 05/09/2015 09:37:32 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #69 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

Consider that by adding choice of deck, we change axis to an alignment of unknown face values,

« Last Edit: 05/09/2015 11:01:03 by Thebox »

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #70 on: 05/09/2015 11:40:20 »
I think people are close to understanding

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
« Last Edit: 05/09/2015 14:03:37 by Thebox »

#### Colin2B

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##### Re: This maths is correct, how can it be wrong?
« Reply #71 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.

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

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #72 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

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

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?

« Last Edit: 06/09/2015 08:45:41 by Thebox »

#### alancalverd

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##### Re: This maths is correct, how can it be wrong?
« Reply #73 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
helping to stem the tide of ignorance

#### Thebox

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##### Re: This maths is correct, how can it be wrong?
« Reply #74 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.
« Last Edit: 06/09/2015 09:07:44 by Thebox »

#### Thebox

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

Random is not what you think it is.

P(0)/24hrs=1

86400/36=2400≈40 mins

86400/52=1661.53846154≈27.6923076923mins

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

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

86400/35=2468.57142857≈41.1428571429mins

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

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

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

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

#### Colin2B

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

Do you think this sounds like the same game?

Let me ask some questions so I can understand.

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

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

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

Again words only please, no maths.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### Thebox

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

Do you think this sounds like the same game?

Let me ask some questions so I can understand.

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

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

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

Again words only please, no maths.

#### Thebox

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

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

#### Colin2B

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

#### Thebox

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

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

Was I doing anything known here

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

this seems connected somehow?

52*2/86400=0.0012037037

52*100/86400=0.06018518518

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

#### Thebox

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

I think I am trying to describe this -

SL(n, Z) → SL(n, Z/N·Z)

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

#### Colin2B

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

I think I am trying to describe this -

SL(n, Z) → SL(n, Z/N·Z)

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

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

this seems connected somehow?

52*2/86400=0.0012037037

52*100/86400=0.06018518518

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

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

#### Thebox

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

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

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

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

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

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

presently looking at this.

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

#### chiralSPO

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

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

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

#### Colin2B

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

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

These are moving you further from understanding your problem.
You need to concentrate on understanding why P(X)=P(Y). Then you will have your answer.
Let us know when you have done that.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### alancalverd

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

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

#### Thebox

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

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

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

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

x
x
x

xxx

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

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

#### chiralSPO

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

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

This equation is satisfied in the case of A =2 and B = 2. Also A = 0 and B = 0. In General: for any value A, there is a value B = A/(A–1) which will satisfy your equation.

#### Thebox

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

This equation is satisfied in the case of A =2 and B = 2. Also A = 0 and B = 0. In General: for any value A, there is a value B = A/(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]?
« Last Edit: 08/09/2015 18:46:20 by Thebox »

#### chiralSPO

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

This is a perfectly valid algebraic statement, and can be rearranged to show that every number A has exactly one value of B such that A plus B equals A times B. We can even solve for what B must be for any A, which was I meant to show as B = A/(A–1), which you should understand to mean: "B must equal the ratio of A and A minus one."

#### Thebox

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

This is a perfectly valid algebraic statement, and can be rearranged to show that every number A has exactly one value of B such that A plus B equals A times B. We can even solve for what B must be for any A, which was I meant to show as B = A/(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?

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

#### chiralSPO

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

x+y=xy

that would mean the same as a+b?

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

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

X^100=XY=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...

#### Thebox

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

x+y=xy

that would mean the same as a+b?

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

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

X^100=XY=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

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

#### Colin2B

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

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

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

#### Thebox

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

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

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

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

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

#### jeffreyH

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##### Re: This maths is correct, how can it be wrong?
« Reply #96 on: 09/09/2015 22:01:51 »
You can correctly state it as XYZ^-XYZ or $$\frac{1}{XYZ^{XYZ}}$$. So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000
« Last Edit: 09/09/2015 22:07:09 by jeffreyH »
Fixation on the Einstein papers is a good definition of OCD.

#### Colin2B

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

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

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

You can correctly state it as XYZ^-XYZ or $$\frac{1}{XYZ^{XYZ}}$$. So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000

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

« Last Edit: 10/09/2015 09:20:21 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

#### Thebox

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

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

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

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

#### chiralSPO

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

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