# The Naked Scientists Forum

### Author Topic: Is this maths correct?  (Read 13374 times)

#### jccc

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##### Re: Is this maths correct?
« Reply #100 on: 05/07/2015 19:18:54 »
I will give up , although I know very well I am correct,
You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/53 is correct for every deck.

100 cards you get 100/13  aces.

i see your fishing trip is very successful, good job bro!

#### Thebox

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##### Re: Is this maths correct?
« Reply #101 on: 05/07/2015 19:23:58 »
I will give up , although I know very well I am correct,
You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/53 is correct for every deck.

100 cards you get 100/13  aces.

i see your fishing trip is very successful, good job bro!

7.69230769231≠0.07692307692

#### chiralSPO

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##### Re: Is this maths correct?
« Reply #102 on: 06/07/2015 02:16:01 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces.

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.

7.69230769231≠0.07692307692

No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...

#### jccc

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##### Re: Is this maths correct?
« Reply #103 on: 06/07/2015 02:29:19 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces.

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.

please save that line for my gravity/light theory.

#### Thebox

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##### Re: Is this maths correct?
« Reply #104 on: 06/07/2015 06:15:31 »

oh but I understand.

You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?

4/52 is correct for every deck.

100 cards you get 100/13  aces.

i see your fishing trip is very successful, good job bro!

jccc is absolutely correct here.

7.69230769231≠0.07692307692

No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...

No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.

x is not equal to  y. I believe I am correct in my idea?

#### Colin2B

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##### Re: Is this maths correct?
« Reply #105 on: 06/07/2015 09:18:31 »
jccc is right
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss  ∞ times  still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.

Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw.  Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.

Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.

Happy fishing by the way.

PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc

EDIT:

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
These are not the same question. On the 1st jccc is right.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.

I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

« Last Edit: 06/07/2015 11:21:04 by Colin2B »

#### Thebox

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##### Re: Is this maths correct?
« Reply #106 on: 06/07/2015 15:52:28 »
jccc is right
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss  ∞ times  still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.

Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw.  Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.

Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.

Happy fishing by the way.

PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc

EDIT:

I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
These are not the same question. On the 1st jccc is right.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.

I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

I am not making no mistakes, my logic is accurate, but from this post I now know why you are looking always from the wrong angle.
You are not considering that the coin is already tossed 100 times.

You are not considering the already set unknown sequence of a deck of cards.

The top card is not random, it is a unknown value.  Once the top card is set, it is set and nothing can change this.

You are making a mistake and still looking a the wrong angle and perspective.

Take two decks of cards for simplicity, try it at home, shuffle both decks individually,

both decks have a probability that an ace is the top card of 4/52.

I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.

I have a quick look, and can confirm that one of the cards is an ace.

Please state the probability of the two cards now?

100*4/52 = 7.69, which is roughly 8.

a
x
x
x
x
a
x
x
x
x
a
a
a
x
x
x

quantum leap , skip some space time, get lucky and land on another a.

it is very simple

pick a card anywhere from in a deck. 1/52

pick a card from another deck, 1/52

and continue this 100 times.

x
x
x
x
x
x
x
x
x
and so on to 100.

now what is the probability of drawing an ace from the new made y axis?

You are player 1, you will get card 1, pick a deck

1-52
1-52
1-52
1-52
and so 100 times in total.

You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.

« Last Edit: 06/07/2015 20:06:32 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #107 on: 07/07/2015 00:07:46 »

You are not considering that the coin is already tossed 100 times.
It doesn't matter. Previous tosses do not affect future tosses. This is a common error, the coin has no memory. Probability is 1/2 every time

You are not considering the already set unknown sequence of a deck of cards.
Probability isn't about a specific set sequence, it is the likelihood of an event occurring or having occurred and is based on lots of trials not just the one or two you have carefully selected to make your point.

The top card is not random, it is a unknown value.
A card can be both random and unknown, in fact it usually is both until looked at. So I don't understand what you mean by this.

Once the top card is set, it is set and nothing can change this.
No one is saying you can change it, but probability only deals with what is likely or unlikely not with what is actually there.

You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.
This is wrong as already explained.

I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.
No the probability is not 4/52, ChiralSPO and I explained this in posts #54 & 55.

I have a quick look, and can confirm that one of the cards is an ace.

Please state the probability of the two cards now
Again, this was explained fully in those 2 posts. This is condition probability which I have also explained in other posts. It is a mistake to assume P(AB)=P(B|A) as I previously explained.

If you are not going to bother reading and understanding our posts there really isn't any point continuing. I have persisted this far because I am aware that in a few years your son will be starting secondary school. His first exposure to probability in maths will be through these very basic first year examples using coins and cards. If you teach him your pseudomaths he will fail his coursework and be considered a fool by his friends and teachers. This will be unfair. You owe it to him to learn real maths, not myth.

Go back, read what Alan, ChiralSPO and I have written. If you have specific questions on understanding what we have written I will try to answer, but I am unable to deal with your delusion regarding the 'timing' of decks and it's effect on outcomes.

Whatever it is you think you are doing it isn't probability.

Like Alan, I'm out.

#### Thebox

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##### Re: Is this maths correct?
« Reply #108 on: 07/07/2015 15:50:19 »
We are in just chat , chatting, out of what exactly?

And like every other forum on the internet, again your versions are being forced onto me. The things you defend are not even your things, they were some ones from history things.

I ask where is your own thought and consideration for the talking point?  without resorting back to present information.
Does nobody on this planet have the ability to think about what is actually said?
I wish to discuss x is not equal to y, I have shown axiom logic and models and maths. All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.

#### Colin2B

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##### Re: Is this maths correct?
« Reply #109 on: 07/07/2015 17:53:28 »
I ask where is your own thought and consideration for the talking point?
These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.

I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.
And we have shown by logic, models and maths that we disagree with you

All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.

I have to be true to myself, I cannot agree with false 'facts',  false logic or false maths.

You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.
No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.

#### Thebox

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##### Re: Is this maths correct?
« Reply #110 on: 07/07/2015 19:01:05 »
I ask where is your own thought and consideration for the talking point?
These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.

I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.
And we have shown by logic, models and maths that we disagree with you

All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.

I have to be true to myself, I cannot agree with false 'facts',  false logic or false maths.

You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.
No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.

I am not telling you that there is no fish in the sea, I am telling you that there is too may fish in the sea.

100 decks of randomly shuffled cards will contain approx 8 aces as the top card of the 100 decks.

The small blind is in alignment with card 1 of each deck.

''There are 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 arrangements of a set of 52 cards. 7.69% of them have an ace at the top.''

So if I have 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000  decks of cards pre-shuffled, column 1 to the small blind contains 7.69% worth of aces.  compared to 4/52

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
« Last Edit: 07/07/2015 19:02:51 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #111 on: 08/07/2015 04:39:03 »
the small blind contains 7.69% worth of aces.  compared to 4/52
My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x

Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.

There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.

#### Thebox

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##### Re: Is this maths correct?
« Reply #112 on: 08/07/2015 16:31:24 »
the small blind contains 7.69% worth of aces.  compared to 4/52
My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x

Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.

it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.

There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.

Thank you Colin, interesting that probabilities may not be the answer I am looking for, although alternative scenarios must apply?

I do not why you think I am agreeing with the maths and that Y=X.  Not for one second do I think Y=X.

Forget probabilities for a while and please answer me this, in the diagram below , does Y=X?

.
.
.
.  .  .  .

#### alancalverd

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##### Re: Is this maths correct?
« Reply #113 on: 08/07/2015 17:25:51 »
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!

#### Thebox

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##### Re: Is this maths correct?
« Reply #114 on: 08/07/2015 17:31:13 »
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.

Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!

^5  reverse bet not fallacy, I would bet you could not throw another heads, I am not betting 1/2 of heads or tails

and I forgot science likes simple

x
x
x
a
b
c

x=1 or 2

x~abc random over time.

what is the chance that (a) will receive a 1.
« Last Edit: 08/07/2015 17:33:02 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #115 on: 09/07/2015 00:04:51 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.

#### Thebox

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##### Re: Is this maths correct?
« Reply #116 on: 11/07/2015 09:33:52 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.

You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.  And this has nothing to do with my poker ideas.
I understand randomness, I understand sequence, and I certainly understand x and y. It is not my misconception, it is the entire human races perspective view of just about everything,  from atoms to probability.

The human race looks at things in a 1 dimensional perspective, where everything in my space is see through.

#### Thebox

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##### Re: Is this maths correct?
« Reply #117 on: 11/07/2015 10:30:18 »
I refuse to back down from this when I know I am correct and in this post I will concentrate on what I am writing and try to get through an understanding of mine that science seems to be missing.

It has come to my attention by playing online poker, and by my observations and knowledge of the distribution process that there is a flaw in the process. The flaw is a probability function, that at the start of any hand, all players have  uneven probabilities, a flaw caused by a multitude of pre-shuffled decks, that are randomly issued to game tables each and every hand.
A time based distribution, based on when a table finishes it's hand, it gets a new deck from the system that is pre-waiting in storage.
This makes an anomaly, where  standard random sequences over time from using a single deck, is altered by periodic ''jumping'' of spacial dividers.

We can assert that of a multitude of individual pre-shuffled decks, that  the  top cards of each deck will all have a 4/52 chance of it being an ace, we can with a certainty be assured that a single deck only contains 4 aces. However we can also also assert that the top card of each deck could all be an ace, or all could be a different card to an ace.

If we consider a diagram form and a variation of top cards,

ace
queen
four
ace

the four cards representing 4 individual decks and the top card of each deck, if a table received deck 1, being at the top , and deck 4 being the at the bottom, then that table card 1 is both times an ace.  This is the spacial dividers over time ''jumping'' I refer to.

ace,queen,four,ace

ace........time........ace

it is not hard to consider an oblong format of random sequences of rows,

Obviously x axis does not equal y axis,

123
231
123

X makes y, and Y is different to x, x remains 1/3 whilst Y remains unknown ,

I am not saying ''that the probability of getting an ace decreases as the number of decks increases''.

I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

If we call 1, (n), P(n)/x=1/3 but P(n)/y=?, because we could never possibly know what values had fell in place of the Y axis columns, example

123
123
123

In this situation column one on the left, the P(1)/y=3/3 where as x always remains 1/3.
And if we hide the values, the result could still be the same.

nnn
nnn
nnn

we know left to right of each row is still 1/3, but we do not know the columns values.

« Last Edit: 11/07/2015 10:35:23 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #118 on: 11/07/2015 19:45:03 »

Quote
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.

However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.

#### Thebox

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##### Re: Is this maths correct?
« Reply #119 on: 11/07/2015 22:18:50 »

Quote
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.

This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.

However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.

I do not see how advancing forward is deplorable in a science forum?

I have no idea where you have 1/169 from?

or 1/13?

I have one deck of cards 4/52 chance of an ace following my x axis, you you can make the x a Y if you like and the result is the same.

I have the y axis, at  ?/? .

12
21

x=y

21
21

x≠y

only 2 values are needed to show this.

12
21

xy≠xy
« Last Edit: 11/07/2015 22:21:59 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #120 on: 12/07/2015 00:46:52 »
4/52 = 1/13

If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.

So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169

A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.

For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.

Belief in anything else is the first step on the road to gambling bankruptcy.

#### Thebox

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##### Re: Is this maths correct?
« Reply #121 on: 12/07/2015 10:41:10 »
4/52 = 1/13

If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.

So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169

A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.

For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.

Belief in anything else is the first step on the road to gambling bankruptcy.

I know Alan you still do not understand my argument and what I am arguing about.  My argument is not about the probabilities of independent decks.

I know each deck has a 4/52 chance of the top card being , but I  also know this only applies using an x-axis scenario.

To say Y is equal to X when talking probability is absurd when it is evidentially not, no matter how many digits is in use.

xxxxx
xxxxx
xxxxx
xxxxx
xxxxx

5², each row as the values of 1 to 5, these values have been displaced of any order by a random shuffle, when the shuffle stops, in alignment with a Y axis and of 5 columns, each column also has 5 values, can you tell me what values are in the columns?

I can tell you that in every row there is 1 to 5 in no order.

#### Colin2B

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##### Re: Is this maths correct?
« Reply #122 on: 12/07/2015 17:38:01 »
I do not why you think I am agreeing with the maths
Your interpretation of the probabilities in the x&y directions has always been a sticking point in discussing you theory. In particular you think that the probability of an ace when picking from the top cards of 100 decks is different to that of picking from a single deck of 52. Well, in one post you said:
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?

i bet you say 4/52 which is incorrect.
However, later when talking about 100 decks you say:
..........column 1 to the small blind contains 7.69% worth of aces.  compared to 4/52
So, because 7.69%=4/52=8/104=0.769 you are agreeing that probability in columns = probability in rows

I forgot science likes simple
No, science is comfortable with complex. What it does like is clarity. If you ask vague, woolly, unclear or confusing questions you can expect to be asked "what do you mean by...", "define ....", "what are you really asking". You have to be clear, something you are usually not.
However, because you have been much clearer in post #117, I will make one last attempt to show why the probability in x & y are the same, but the sequences are not.

nnn
nnn
nnn

we know left to right of each row is still 1/3, but we do not know the columns values.
We have previously pointed out that the probability of an ace in the y direction is the same as in the x see posts.#54, 55 and 105. What differs, as we have explained, are the sequences.
However, these are not unknown because probability helps us to understand the unknown.

OK, let's take your example as it is fairly simple.

nnn
nnn
nnn

You have 3 decks y1, y2, y3. Rather than use numbers for the cards I am going to use letters because otherwise we get confused when talking about 1st cards, 2nd card etc. So the 3 cards are A, B, C = Ace, Bravo, Charlie
So in the x direction P(x1)=P(x2)=P(x3)=1/3 and we agree on that.
So if you shuffle a deck of these cards and pick the first there is a 1/3 chance it is A, but if you had picked the bottom card that would also be probability of 1/3, same with the middle card.
So in the x direction we are agreed 1/3 1/3 1/3.
If we shuffle your 3 decks and take the top 3 cards y1 y2 y3 then there are 3 possibilities for each y and so the number of possible ways these cards can fall = 3*3*3=27 possibilities.
Let's look at the probability that in these 3 cards the 1st card only is A.
We can show this with maths using an X to show “not an A” = a 2 or a 3
So, AXX
here the probability of A is 1/3, but there are 2 possible cards in y2, a B or a C so the probability is 2/3, same with y3. So total probability = 1/3*2/3*2/3= 4/27.
We can  also do this by writing out the possibilities for this deal:
A in y1
C   B   B   C
B   C   B   C
A   A   A   A
a total of 4 ways out of 27 =4/27

We can also look at the probability that one of the cards out of the 3 is an A regardless of position. To do this we need to look at  Ain y2 and A in y3

A in y2
We can also do this for y2 = A
C    B    B    C
A    A    A    A
B    C    B    C
Another 4 ways

A in y3
and for y3 = A
A    A    A    A
C    B    B    C
B    C    B    C
yet another 4 ways
Total 12

So if we want to know the probability of a single A anywhere in the 3 cards it is = 12/27=4/9

We would also want to look at the other ways for completeness:
A    B    C
A    B    C
A    B    C
3 more ways
and the ones we missed:
A   C   B   C
A   C   C   B
B   B   B   B

A   A   B   C
B   C   A   A
A   A   A   A

B   A   B   C
B   A   C   B
C   C   C   C
Total 27

In order to do this many ways we have had to lay out 3x27 decks = 81
so in all those top cards how many were As?
Count them A = 27
So probability of A occurring in y = 27/81 = 1/3
the same as in x.
You will also note that AAA occurs once so probability = 1/27
but so does CBC, and BCC and BBB. All have equal probability.
Only humans decide that A is special. That an ace wins over a 6 or a 5. In reality their probabilities are all the same!

Now, the chances are that you would not get this exact match in exactly 81 decks, but as Alan said if you do enough games (trials) say >1000, then the actuality will get close to these probabilities.

What this also tells us is that for any 3 decks all arrangements are equally likely, so it doesn't matter if you skip decks or do them in sequence, the probability is the same. Out of a 1000 decks you could choose 3 from anywhere and the probability would be the same for each. If you think otherwise, you do not understand what probability is telling us.

You can obviously do the calculation shown above for any number of decks and cards, but the numbers become large, so I'll leave that to you.

Thank you Colin, interesting that probabilities may not be the answer I am looking for,
I would agree. Probability tells us that there is no difference between shuffled decks, so skipping some of the decks has no effect on the outcome of games from a probability point of view.
Yes, there are other scenarios. Your's sounds closer to predestination, or "that one was intended for me" type of luck or fortune. I know something about probability, permutation and sequences, but nothing about predestination, fortune or luck, so I can't help you with that.
« Last Edit: 12/07/2015 17:43:11 by Colin2B »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #123 on: 12/07/2015 18:41:23 »
, I would bet you could not throw another heads,

Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.

You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.

But the evidence to date suggests that it doesn't! The probability of scoring 5H in a row in inependent trials of a fair coin is (1/2)^5, around  3%, so the evidence suggests a 97% probability that the coin is not fair. On that basis it would be foolish to bet against it.

#### Thebox

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##### Re: Is this maths correct?
« Reply #124 on: 13/07/2015 14:52:42 »
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.

Colin calls it pre destined, I call it probabilities over a time period.

This is my whole point and the only part you are not understanding, the part about interception.

y1,y2,y3,y4,y5 where y is deck
t1,t2,t3,t4,t5  where t is table

randomly distributing y to each table is not the same.

You say you know sequences Colin, then you understand spacing between values, spacing being equivalent to time.

ace.........end.......ace..............end.................ace..end.  end being end of sequence.

Above I show 3 shuffles and 3 decks, Player 1 table one receives on average 1/221 aces/t

divided by time being a key element when considering probabilities and certainly poker.

#### The Naked Scientists Forum

##### Re: Is this maths correct?
« Reply #124 on: 13/07/2015 14:52:42 »