# The Naked Scientists Forum

### Author Topic: Is this maths correct?  (Read 12620 times)

#### Thebox

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##### Re: Is this maths correct?
« Reply #75 on: 02/07/2015 18:02:23 »
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?4/51 if you discard the card you have separated or 4/52 if you include it.

No, the probability is 4/52 in both situations, as long as you don't know what the first card is.
It's important you understand this before thinking about timing because it is fundamental to your understanding of the y axis.
Remember I said that for all the points in your xy matrix -the sample space - there is an equal probability of it being any card from the pack, 1/52.
You have a concern that the 1st card in each deck (y diRection) can have repeat values eg 2 ace of diamonds, which would not occur in the x direction as you are limited to one deck.
Because of this you see the y direction as being different from the x. But, remember what I said, the probability of any card in any space x or y of your matrix - the sample space - is the same, 1/52. This is no different to a 52 sided dice! You would be very happy to accept that a Dice rolled many times might have multiple 6s occuring in succession, but after a large number of games it would be even. Cards in the y direction are the same because each card has a probability of 1/52, so there is no problem if you skip decks, the probability for any deck given to you is the same as any other.
There is no point talking timing, because there is nothing timing can influence.

Timing is what defines what deck you get out of the Y axis.

The Y axis columns are not 1/52, only x has 52 individual values of a set.

123
123
123
123

x axis 1/3

y axis 1/3 in 4

''You have a concern that the 1st card in each deck (y diRection) can have repeat values eg 2 ace of diamonds, which would not occur in the x direction as you are limited to one deck.
Because of this you see the y direction as being different from the x.''

You understand

And Y axis is obviously different to x, take any 3 variants of a set, add more sets of 123,

123
123
123
123
123
123

randomly shuffle all the sets,

x≠y

This is evidential , we can all see this to be true, an axiom.  X≠Y is the correct formula, x does not equal y, we can all observe this.

player 1y≠player 2y≠player 3y, all the probabilities of Y are differential.  Where the probability of x is constant to all observers.

in reality player 1,2,3 is multiplied  in the y axis

123
123
123
123
123
ppp
ppp
ppp
ppp
ppp

p=player

i did say it was complex and all this is divided by random time.

« Last Edit: 02/07/2015 18:45:19 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #76 on: 03/07/2015 13:29:39 »
Quote
The Y axis columns are not 1/52, only x has 52 individual values of a set.

WRONG! Since each pack is randomly shuffled, each point on the y axis is random, therefore

A. in an infinite number of hands you will find each card value appears 1/13 in each position.

B. in a finite number of hands it would not be surprising (1/6.5) if a given card value appeared in the same position twice in succession

C. but 3 times in succession is pretty unlikely

D. and if you nominate the position (say the first card) then 3 in succession would suggest a fix.

#### Thebox

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##### Re: Is this maths correct?
« Reply #77 on: 03/07/2015 15:29:13 »
Quote
The Y axis columns are not 1/52, only x has 52 individual values of a set.

WRONG! Since each pack is randomly shuffled, each point on the y axis is random, therefore

A. in an infinite number of hands you will find each card value appears 1/13 in each position.

B. in a finite number of hands it would not be surprising (1/6.5) if a given card value appeared in the same position twice in succession

C. but 3 times in succession is pretty unlikely

D. and if you nominate the position (say the first card) then 3 in succession would suggest a fix.

Hi Alan, I know from your post that you are still not understanding and still playing the x axis.

In simple explanation I will say this.

In the x axis of infinite rows there can be a possibility of only 1 ace of diamonds per row.

In the Y axis of an infinite column, the x axis shuffle makes the Y axis, and the Y axis contains ? ace of diamonds.

The top card of each individual deck as a 1/52 chance of being an ace of diamonds considering rows.

so lets us put this into a reality situation

1/52
1/52
1/52
1/52
1/52

In the above there is a 1/52 chance of the x axis's top card being an ace of diamonds.

The Y axis obviously reads (1/52)*5

#### alancalverd

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##### Re: Is this maths correct?
« Reply #78 on: 03/07/2015 18:09:36 »
That's why I said card value - the assumption is that all aces, queens or whatever are of equal value. It just makes the maths a lot easier as we are now dealing with 13 cards instead of 52.

As I interpret your model, you lay a single deck along the x axis, then shuffle and lay another deck above it, thus building up the y axis.

Then there is a 1/13 chance of the card in any given position being the value you state, in each row.

#### Colin2B

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##### Re: Is this maths correct?
« Reply #79 on: 03/07/2015 18:18:30 »
The Y axis columns are not 1/52, only x has 52 individual values of a set.
The 52 values in x are what define the probability in y.
You either have not read my post properly or have not understood. Similarly you are not understanding what Alan is saying, because he is saying the same as me.
I will explain again with a simpler example below with 123.

123
123
123
123

x axis 1/3

y axis 1/3 in 4
This does not reflect what happens in reality.

x≠y

This is evidential , we can all see this to be true, an axiom.  X≠Y is the correct formula, x does not equal y, we can all observe this.
No we can't all observe this. The probability in y is the same as in x
Let me explain again using 123.

If P(x) is the probability in the x axis then for x1, x2, x3 P(x1)=P(x2)=P(x3)=1/3
This is as you describe above.
However, x1 for any deck is the same as yn in that line - this you understand.
So P(x1)=P(y1)=P(y2)=P(y3)=P(y4)............P(y)

1/3 1/3 1/3
1/3 1/3 1/3
1/3 1/3 1/3
1/3 1/3 1/3
etc

So the probability for any position is 1/3 and so P(x)-P(y)=1/3
You do not compute the rows as (1/3)*3
Why do you compute the columns as (1/3)*4?

Because this is what you are doing in your response to Alan

1/52
1/52
1/52
1/52
1/52

In the above there is a 1/52 chance of the x axis's top card being an ace of diamonds.

The Y axis obviously reads (1/52)*5
If the probability in y is (1/52)*5 then probability in x is (1/52)*52 which is nonsense.

i did say it was complex
No, it's not complex it is very simple. This is very basic probability and if you cannot understand this you will never understand probability.
There is no difference in the probability in the x or y directions and until you read carefully what we have written and fully understand, there is little value in continuing this.

EDIT: I notice Alan has replied while I was typing. Again we are both saying the same thing. Do try to understand.
« Last Edit: 03/07/2015 18:20:35 by Colin2B »

#### Thebox

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##### Re: Is this maths correct?
« Reply #80 on: 03/07/2015 19:13:35 »

If the probability in y is (1/52)*5 then probability in x is (1/52)*52 which is nonsense.

Where are you getting (1/52)*52 from?

In my example I put 5 rows of 1/52

5*52=260

5/260 in the correct position

(1/52)*5

1/52
1/52
1/52
1/52
1/52

there is 5 chances of 1/52 that the top card is an ace of diamonds

in fact there is a 0-∞/t chance an ace of diamonds is the top card.

52x²=(1/52)*52=1

52x²=2704/13/4=52

Y=52a

x=52b

« Last Edit: 03/07/2015 20:00:54 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #81 on: 04/07/2015 00:03:51 »
Where are you getting (1/52)*52 from?

I'm getting it from the same place you get (1/52)*5, just to show how you are wrong in what you are doing. If you do that to the column you have to do it to the rows as well.

there is 5 chances of 1/52 that the top card is an ace of diamonds
These 5 events are not mutually exclusive so you can not say (1/52)*5, but must use (1/52)5, so the probability of 5 decks in a row all yielding a specific card, eg ace of diamonds, as first card is 1/380204032, in other words unlikely. It is worth noting that this probability is the same for any selected group of 5 decks, whether consecutive or spaced eg decks 2 3 4 5 6 or  2 4 5 8 15 both these sequences have the same probability.
However, this is not what you are doing in the games you play, you are only ever taking one deck at a time and therefore the probability  is always 1/52. This again is independent of the order of the decks so y=x.

in fact there is a 0-∞/t chance an ace of diamonds is the top card.

52x²=(1/52)*52=1

52x²=2704/13/4=52

Y=52a

x=52b
All of this is mathematical gibberish, and wrong.

Please look back at the posts by Alan, ChiralSPO and myself and try to understand where you are going wrong.
When you do understand, let me know and we can talk some more.

Edit: if you understand what we have written, you will then be able to understand why timing ie sequence is irrelevant.
« Last Edit: 04/07/2015 05:20:56 by Colin2B »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #82 on: 04/07/2015 06:43:05 »
It is certainly true that if you shuffle 5 decks of cards, the probability that the first card in at least one deck is the ace of diamonds is 5/52 but as you can't predict which deck, you still lose money because you have to bet 5 times as much in order to win, so the return on winning remains 1/52.

It is also obvious that if you shuffled a very large number N of decks, the number of times you would draw AD as the top card should approach N/52 as N increases, but you still have no means of predicting which shuffle will achieve this.

#### Thebox

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##### Re: Is this maths correct?
« Reply #83 on: 04/07/2015 09:02:31 »
Hi, I am confused,I thought Colin was close to understanding and agreeing with me, and now Colin is seemingly back off track.
Now Alan seem's close to understanding.
I suggest the problem is with communication, I am translating your posts and you are translating my posts differently to there meanings.
The messages intentions and meanings are being lost in translation, can I please suggest we start over and try to keep it simple, I understand in my head exactly what I mean, people I speak to I know understand me.

So in starting new, let us define some values and stick to the values.

One deck of cards, containing 52 individual variants we will call this {N} and put on set brackets to define a set.

A single deck of cards spread out in a row, we will call this the (x) axis.Brackets defining it is a scalar direction

more rows of cards, we will call this (y) axis.Brackets defining it is a scalar direction
t=time
R=random
P=probabilities
p=player

Each deck has  subsets of 4 values, i.e 4 aces, we will call this {_N}

For a single variant, we will call this {_n}

Can we agree that P{_n}=1/52 and the P{_N}=4/52 using a single deck?

If p1 receives the top card and it is an ace, the probability of the top card being an ace  after the next shuffle  is P{_N}=4/52^2?

P{_N} from any singular (x) from (y) is 4/52?

P{_n} from any singular (x) from (y) is 1/52?

We also need to add {_sn}   , which represents a specific variant of the 52.

~=distribution of
#=not equal to
•=repeat
(1)(2)(3) ect= deck numbers from the (y) axis.

ok so far?

Ok first scenario,

we take the first 3 cards of each deck and leave them face down unseen.

(p1)nnn
(p2)nnn
(p3)nnn
(p4)nnn
(p5)nnn

You are player 3, can you confirm the probabilities of the first {_sn} you will receive?

P{_sn}=?

secondly ....we will take the first 5 {_n} of each {N} from (y)

nnnnn
nnnnn
nnnnn
nnnnn
nnnnn
ppppp

Can you please now confirm the P{_sn} from the (y)  , you are still p3.

P{_sn} from (y)=

« Last Edit: 04/07/2015 09:37:45 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #84 on: 04/07/2015 09:32:12 »
"scalar direction" is meaningless.

Quote
If p1 receives the top card and it is an ace, the probability of the top card being an ace  after the next shuffle  is P{_N}=4/52^2?

No. The shuffles are independent so P{_N} = 4/52 every time.

But the probability of drawing two aces in successive shuffles is obviously (1/13)^2

Hence "beginner's luck". You remember your first win, whether it was your first game or your 13th, but the probability of winning n games in a short sequence thereafter is the square, cube.... nth power of 1/13  so it looks as though your luck runs out even though it hasn't changed at all.

And of course it's more complicated in a real poker game because "first to draw an ace" is not the winning hand, and a good bluffer can win with an empty hand.
« Last Edit: 04/07/2015 09:34:37 by alancalverd »

#### Thebox

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##### Re: Is this maths correct?
« Reply #85 on: 04/07/2015 09:39:19 »
"scalar direction" is meaningless.

Quote
If p1 receives the top card and it is an ace, the probability of the top card being an ace  after the next shuffle  is P{_N}=4/52^2?

No. The shuffles are independent so P{_N} = 4/52 every time.

But the probability of drawing two aces in successive shuffles is obviously (1/13)^2

Hence "beginner's luck". You remember your first win, whether it was your first game or your 13th, but the probability of winning n games in a short sequence thereafter is the square, cube.... nth power of 1/13  so it looks as though your luck runs out even though it hasn't changed at all.

And of course it's more complicated in a real poker game because "first to draw an ace" is not the winning hand, and a good bluffer can win with an empty hand.

we take the first 3 cards of each deck and leave them face down unseen.

(p1)nnn
(p2)nnn
(p3)nnn
(p4)nnn
(p5)nnn

You are player 3, can you confirm the probabilities of the first {_sn} you will receive?

P{_sn}=?

secondly ....we will take the first 5 {_n} of each {N} from (y)

nnnnn
nnnnn
nnnnn
nnnnn
nnnnn
ppppp

Can you please now confirm the P{_sn} from the (y)  , you are still p3.

P{_sn} from (y)=

added - ''Chaos theory is the field of study in mathematics that studies the behavior of dynamical systems that are highly sensitive to initial conditions—a response popularly referred to as the butterfly effect. Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for such dynamical systems, rendering long-term prediction impossible in general.''

https://en.wikipedia.org/wiki/Chaos_theory#/media/File:Chaos_Sensitive_Dependence.svg
« Last Edit: 04/07/2015 09:57:37 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #86 on: 04/07/2015 11:39:04 »
Hi, I am confused,I thought Colin was close to understanding and agreeing with me,
No, there is a big difference between understanding and agreeing. I do understand the mistaken assumptions you are making, but I have never agreed with you about your random timing and your interpretation of probabilities in x&y.  You frequently misunderstand what I am writing.

Now Alan seem's close to understanding.
Alan also understands the mistakes you are making.
You are right about misunderstandings as many of your posts are very confusing, and it is very easy to misunderstand your questions.

If p1 receives the top card and it is an ace, the probability of the top card being an ace  after the next shuffle  is P{_N}=4/52^2?
No, this has been explained before. It is 4/52.
It is only 4/52^2 if you play both decks together and turn over the top 2 cards.
Once you have played a deck and know the outcome the probability for the next shuffle is 4/52. This is called conditional probability

Those posts also give you the method for determining the probability that at least one of the top cards of 5 packs is an ace.
As explained there, the probability for 2 packs that one and only one of the top cards is an ace is 2*(4/52)*(48/52)
For 5 decks the probability that one and only one of the top cards is an ace is 5*(4/52)*(48/52)4
Also as I have explained above, the order in which you select the decks - what you call random timing - does not affect the probabilities.

'Edit: Don't try to understand chaos theory until you thoroughly understand probability. Chaos theory is not appropriate here because we are not dealing with approximations and this example is a simple probability problem.'

If you are not reading and understanding my posts I see no point continuing.
So I'll leave it to Alan to see if his explanations can help you understand.

Hence "beginner's luck". ...
Alan, even though you are armed with much more than beginner's luck, you are going to need more luck than I have had.
I wish you ..... luck?

« Last Edit: 04/07/2015 13:55:00 by Colin2B »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #87 on: 04/07/2015 14:55:10 »
1. I have absolutely no idea what {_sn} means

2. Poker has nothing to do with chaos theory. The fall of the cards is pure linear statistics.

#### Thebox

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##### Re: Is this maths correct?
« Reply #88 on: 05/07/2015 07:34:16 »
1. I have absolutely no idea what {_sn} means

2. Poker has nothing to do with chaos theory. The fall of the cards is pure linear statistics.

Huh?  I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.

P{_sn} from x is always 1/52

P{_sn} from Y is ?

It is unknown and can never be known, it is chaos.

I really do understand what I am on about.

take 123 and mix it randomly, take another 123 and mix it randomly, the mixing of 123 and 123 generates Y axis.

231
213

In this example x remains 1/3 where is Y is differential.    Just consider rows and columns.

P(x)#P(y)

post#14608

page 585

P{_sn}/(x)
t
=1

P{_sn)/(y)
t
=0-∞

x=123

y=
1
1
1

or

0
0
0
« Last Edit: 05/07/2015 08:38:19 by Thebox »

#### Colin2B

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##### Re: Is this maths correct?
« Reply #89 on: 05/07/2015 09:41:42 »
Huh?  I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.
No you didn't explain at all. Your use of set notation is confusing, better not to use it unless you are intending to manipulate sets.
The way you wrote your subset n as being sets of 4 cards means P(n)=1/4 - which is not what you intended.

P{_sn} from x is always 1/52
Agreed

P{_sn} from Y is ?

It is unknown and can never be known, it is chaos.
Rubbish. It is also 1/52

I really do understand what I am on about.
No you don't.

However, I am beginning to see where you are confusing yourself. If I get time I will put together an explanation later today.

I'm not going to comment on the rest of your post as it is so confusing, both to you and others.

#### Thebox

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##### Re: Is this maths correct?
« Reply #90 on: 05/07/2015 09:49:06 »
Huh?  I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.
No you didn't explain at all. Your use of set notation is confusing, better not to use it unless you are intending to manipulate sets.
The way you wrote your subset n as being sets of 4 cards means P(n)=1/4 - which is not what you intended.

P{_sn} from x is always 1/52
Agreed

P{_sn} from Y is ?

It is unknown and can never be known, it is chaos.
Rubbish. It is also 1/52

I really do understand what I am on about.
No you don't.

However, I am beginning to see where you are confusing yourself. If I get time I will put together an explanation later today.

I'm not going to comment on the rest of your post as it is so confusing, both to you and others.

if Y has 5o rows it can not be 52,    it would be /50 .

and it just came to me, players by random timing of table hands are quantum leaping through space-time and intercepting values by timing luck.

p.s you keep turning the visual around in your head, so you keep getting 1/52

left to right is x,

bottom to top is y

keep the perspective the same.

123
231
123

the rows are not the same as the columns.

P(1)/x=1/3

P(1)/y=?/3

x is a constant of 123 where as Y is not a constant of 123, x creates unknown y.

« Last Edit: 05/07/2015 10:02:04 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #91 on: 05/07/2015 10:01:08 »
Let's get to the nub of this.

If you play any game of pure chance against n -1  other players, the probability of your winning is 1/n.

Poker cards are randomly shuflled for each hand so if all players are of zero skill and stake the same amount, your longterm return will be 1/n of everyone's stake - i.e, your stake (less the house commission, of course). And everyone else will receive exactly the same.

But there is a considerable element of skill in poker, so in real life your return will be x/n (where x represents your fraction of the total skill around the table) as long as everyone plays. But people drop out and the final head-to-head is effectively a winner-takes-all contest of skill, but because it involves a large element of chance it takes longer than a darts or boxing match to resolve.

No difficult maths involved. As with any game from chess to cricket, if you play a lot, against better players, you will learn a bit and lose a lot. At least in chess and cricket there are leagues and ratings tables so you can choose an opponent of your own level and have fun.

I prefer backgammon.

#### Thebox

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##### Re: Is this maths correct?
« Reply #92 on: 05/07/2015 10:09:33 »
Let's get to the nub of this.

If you play any game of pure chance against n -1  other players, the probability of your winning is 1/n.

Poker cards are randomly shuflled for each hand so if all players are of zero skill and stake the same amount, your longterm return will be 1/n of everyone's stake - i.e, your stake (less the house commission, of course). And everyone else will receive exactly the same.

But there is a considerable element of skill in poker, so in real life your return will be x/n (where x represents your fraction of the total skill around the table) as long as everyone plays. But people drop out and the final head-to-head is effectively a winner-takes-all contest of skill, but because it involves a large element of chance it takes longer than a darts or boxing match to resolve.

No difficult maths involved. As with any game from chess to cricket, if you play a lot, against better players, you will learn a bit and lose a lot. At least in chess and cricket there are leagues and ratings tables so you can choose an opponent of your own level and have fun.

I prefer backgammon.

Yes Poker is a game of skill, and in the long term it is +ev for good players.

But all the skill in the world can not prevent you being dominated by ''quantum leaping'',

Timing is everything in poker, if you make a final table you hope you get some good hands, you do not want to be bad ''quantum leaping'' and landing bad hands.
I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.

#### alancalverd

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##### Re: Is this maths correct?
« Reply #93 on: 05/07/2015 10:19:42 »

123
231
123

the rows are not the same as the columns.

P(1)/x=1/3

P(1)/y=?/3

Wrong.

Since the rows are independent, the probability of anything happening anywhere in the y axis is exactly the same as in the x axis. If it were not so, you would be asserting that choosing an element in row 1 affects the distribution of row 2, i.e. the rows are not independent.

In the case of online poker, I think the promoters have gone to great lengths to ensure that each shuffle is indeed independent.

Your problem of understanding is this:

By shuffling from a limited set, you obviously can't get two identical cards in sequence on the x axis in any one shuffle. Each card has only one degree of freedom in a shuffle.

But it is entirely possible that he same card will appear in the same position in two rows in the y axis PRECISELY because the shuffles are random.

The a priori probability of finding a given card in a given position is obviously 1/52 for each row, but less demonstrably 1/52 for each column because each card has one degree of freedom in EVERY shuffle so you need an infinite number of shuffles to prove it.

1. If you find an ace at position p in one shuffle, what is the probability Pp of finding an ace at p in the previous or subsequent shuffle?

2. If you nominate position q, what is the probability Pq of finding an ace at q in two subsequent shuffles?

#### alancalverd

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##### Re: Is this maths correct?
« Reply #94 on: 05/07/2015 10:23:41 »

I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.

The butterfly effect is the consequence of a trigger in an unstable system. In poker the system is stable but contains a random element. The art is to ride the wave you are given, but unlike surfing, you can't see it coming.

#### Thebox

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##### Re: Is this maths correct?
« Reply #95 on: 05/07/2015 10:44:10 »

I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.

The butterfly effect is the consequence of a trigger in an unstable system. In poker the system is stable but contains a random element. The art is to ride the wave you are given, but unlike surfing, you can't see it coming.

Unfortunately  the waves of y are different to x, you are not considering random choice of Y and the set sequence.

are you really persisting your argument saying that x is equal to y?

123
321
123

X is obviously not equal to Y,

All observers looking left to right observe 1/3.

All observers looking up observe ?/3

column 1, 131 =player 1 the small blind
column 2, 222 =player 2 the big blind
column 3, 313 =player 3 under the gun

so how is x =y?

it obviously is not.
« Last Edit: 05/07/2015 10:50:48 by Thebox »

#### alancalverd

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##### Re: Is this maths correct?
« Reply #96 on: 05/07/2015 12:07:20 »
There is no set sequence in poker. You cannot usefully compare three carefully chosen groups wth an infinity of random ones. It's another part of the Gambler's Delusion.

It's your money and your life. Gamblers Anonymous have more experience and patience than me in dealing with your problem.

Alfa Charlie out.

#### Thebox

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##### Re: Is this maths correct?
« Reply #97 on: 05/07/2015 18:16:51 »
There is no set sequence in poker. You cannot usefully compare three carefully chosen groups wth an infinity of random ones. It's another part of the Gambler's Delusion.

It's your money and your life. Gamblers Anonymous have more experience and patience than me in dealing with your problem.

Alfa Charlie out.

It is called gamblers fallacy and not gamblers delusion,

and if you shuffle any deck, and stop, the sequence is set although unknown.

I will give up , although I know very well I am correct,

#### Colin2B

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##### Re: Is this maths correct?
« Reply #98 on: 05/07/2015 18:36:28 »
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.

#### Thebox

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##### Re: Is this maths correct?
« Reply #99 on: 05/07/2015 18:48:13 »
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?

#### The Naked Scientists Forum

##### Re: Is this maths correct?
« Reply #99 on: 05/07/2015 18:48:13 »