Advanced statistics.

added - lets draw a picture of how set card sequences look from a pre-shuffled deck, and sequencing stacking.

Based using cards and the top five card sequences of each deck to further define, this we will call a virtual number variants position grid (vnvpg), the second part of on-line process after the RNG and before the deal.

X/(xt)
.deck 1 - ah jd 6c 7d ks
.deck 2 - 4s 2d kh jc 8d
.deck 3 - ts qd 8s 3h ah
.deck 4 - kc 9s ac 6d js
..................................y/(t)

X being the vertical columns and y being the horizontal rows. (t) being time and x being random

In example X1,y1=ah

The second part we call it the players variant position (pvp)

.X/(xt)
T1 - p1 p2 p3 p4 p5
T2 - p1 p2 p3 p4 p5
T3 - p1 p2 p3 p4 p5
T4 - p1 p2 p3 p4 p5
..................................y/(t)

Do any of you even care that it does not work online?
analogy - you have been playing for 7 hours and have made the final table, the dealer now tells you that there is two decks of cards, both decks will deal you aces, one deck is set to lose and one deck is a winning aces.

He asks you to choose a deck. ...............................would you think this would be a fair conclusion to game of poker....?

4524/1,000,000 pocket aces

what percentage are winning aces and losing aces.

standard 82%-18%

what insurance is in place to make sure by timing of the decks that I do not get all 18% and miss all the 82%

P(A ∩ B) this is why online poker fails by a multi-deck distribution.

Originally posted by shatteredaces
P(A ∩ B) this is why online poker fails by a multi-deck distribution.

I think we need more details to understand.

Hi, shatteredaces,
Let's assume that you are correct and that online poker rooms have a multi-deck system.

I don't see how that makes even the slightest difference to game play.

Originally posted by shatteredaces
P(A ∩ B) this is why online poker fails by a multi-deck distribution.

Can you explain P(A ∩ B) and explain how it applies to this situation?

There is a vast number of possible decks -- 8.07 x 10^67
In a full ring poker game, only 23 (or 25) cards of a deck are used, so of that number, several will yield the exact same distribution of cards.

I also suppose that the method a room uses to select the deck is relevant too.

Do you have any information on how this is done?

Furthermore, regardless of HOW the deal is arrived at the critical thing is that ALL PLAYERS EXPERIENCE THE SAME RISK.

As long as that is true, the game is fair.

Here is what I mean by that:
 Assume that the winning hand is known for each deck.
 Assume that the winning player is chosen in a truly random fashion
 A deck that gives that player the winning hand is used.

In on-line poker this is impossible of course, because the players have a choice. How many times have you folded 83o in MP only to see an 883 flop? I chose this because it happened to me yesterday.

As long as no player is favoured, and as long as the RNG meets the accepted standards for RNGs the game is as fair as it can get.

Cheers,
VS

Originally posted by VorpalF2F
Hi, shatteredaces,
Let's assume that you are correct and that online poker rooms have a multi-deck system.

I don't see how that makes even the slightest difference to game play.

Originally posted by shatteredaces
P(A ∩ B) this is why online poker fails by a multi-deck distribution.

Can you explain P(A ∩ B) and explain how it applies to this situation?

There is a vast number of possible decks -- 8.07 x 10^67
In a full ring poker game, only 23 (or 25) cards of a deck are used, so of that number, several will yield the exact same distribution of cards.

I also suppose that the method a room uses to select the deck is relevant too.

Do you have any information on how this is done?

Furthermore, regardless of HOW the deal is arrived at the critical thing is that ALL PLAYERS EXPERIENCE THE SAME RISK.

As long as that is true, the game is fair.

Here is what I mean by that:
 Assume that the winning hand is known for each deck.
 Assume that the winning player is chosen in a truly random fashion
 A deck that gives that player the winning hand is used.

In on-line poker this is impossible of course, because the players have a choice. How many times have you folded 83o in MP only to see an 883 flop? I chose this because it happened to me yesterday.

As long as no player is favoured, and as long as the RNG meets the accepted standards for RNGs the game is as fair as it can get.

Cheers,
VS

Hi Vorpal. it is well known that for each hand played on any table online , a new deck is brought to the table from a queue of pre-shuffled decks by timing of which ever table needs a new hand first.
Deck (a) has P(x)=1/52
Deck (b) also has P(x)=1/52

(a)+(b)=P(x)=2/104

(a)*1000,000=P(x)=1000,000/52,000,000

axis x=1/52 (horizontal)

axis y=1000,000/52,000,000 (vertical)

.......................x..............
.........x............................
..............x.......................
......................x...............
...x.................................

P(x ∩ y)

see model here -

http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&start=14450

page 579 - post towards the bottom

Hi, shatteredaces,
I *think* I get where you're going with this.

First off, though, let's look at the basic assumptions.
First,

Originally posted by shatteredaces
Hi Vorpal. it is well known that for each hand played on any table online , a new deck is brought to the table from a queue of pre-shuffled decks by timing of which ever table needs a new hand first

It may be well-known, but it was not know to me at all. Do you have a reference for this?

What is a "pre-shuffled deck"? To a computer a deck of cards is a sequence of numbers. It is really irrelevant how or even when the sequencing is done as long as all possible sequences are possible.

At the moment you are dealt into a hand, the cards for that hand are fixed -- nothing can change them.

It does not matter in the slightest which deck is used -- your skill determines what you do with the cards you get. It is HOW you play, not the cards you are dealt that determine your long term success at poker.

As for intersection of probabilities, I can't see any intersection at all, since the choice of deck A instantly precludes all other possible decks from that hand. There is therefore nothing to intersect.

In the thread you linked to please link to the exact post.
I *think* you mean this one:
http://www.badscience.net/forum/viewtopic.php?p=1380138&sid=2ca3ce4df23983ca9a1c1cc77945bad8#p1380138

Which as far as I can tell doesn't shed much light on the matter.

Cheers,
VS

Originally posted by VorpalF2F
Hi, shatteredaces,
I *think* I get where you're going with this.

It may be well-known, but it was not know to me at all. Do you have a reference for this?

I have a confirmation email of the entire process from the biggest online company,

What is a "pre-shuffled deck"? To a computer a deck of cards is a sequence of numbers. It is really irrelevant how or even when the sequencing is done as long as all possible sequences are possible.

In short there is a shuffle server that all the ''decks'' go into where the rng does its work, totally random.
The ''decks'' are then put into a queuing system, and when a table need a deck, they get one, which is every hand played.
Shuffle a deck of cards, once the shuffle stops, the top 18 cards are an unknown set sequence, the order does not change of the sequence.

so imagine this

1...............x...........................
2...x.....................................
3........................x................
4..............x..........................

We can clearly see that by timing I can intersect x, if I receive distribution 1 and 4.

Using geometrics P(x ∩ y)

In simple terms it is not a uniform distribution or even distribution, timing of decks being the discrepancy factor.

I think this may apply x²(y)

https://en.wikipedia.org/wiki/Chi-squared_distribution

and if you now consider the model link, an intersection probability applies.

Hi, shatteredaces,

I have a confirmation email of the entire process from the biggest online company,

Which company?
Would you be willing to post the relevant section of the email?

In this diagram copied from your post:

1...............x.........................
2...x.....................................
3........................x................
4..............x..........................
No two x's line up. I switched to monospaced font courier new to be sure.
If you want to use a monospaced font use to start it, and close it with tags.
I don't know if all fonts will work.

Are you actually trying to represent this situation:

1..............x..........................
2...x.....................................
3........................x................
4..............x..........................
where the x's in lines 1 and 4 line up, representing (I presume) the same card in the same position?

If so, then I fail to see how this is relevant to poker.
It may accurately represent the actual situation, however it is not possible to know the timing, nor is it possible to know which deck comes next, therefore it is not possible to use it to make decisions.

My last thought is that any two decks are mutually exclusive. Therefore "intersecting" an occurrence from one deck with an occurrence in another deck has no actual meaning in real life.

Let's image a real world casino there is a paranoid millionaire at a high-roller table is a paranoid millionaire that insists on a new deck each hand. The casino would like to keep his business, so they have a team people who pre-shuffle decks for use at this table.

It is clear that just because deck #2 and deck #5 have the A♠: in position 7 nothing changes in how the game is played. It is no less fair, and represents no failing of online poker.

Or am I missing the point entirely?
VS

Originally posted by VorpalF2F
Hi, shatteredaces,

I have a confirmation email of the entire process from the biggest online company,

Which company?
Would you be willing to post the relevant section of the email?

In this diagram copied from your post:

1...............x.........................
2...x.....................................
3........................x................
4..............x..........................
No two x's line up. I switched to monospaced font courier new to be sure.
If you want to use a monospaced font use to start it, and close it with tags.
I don't know if all fonts will work.

Are you actually trying to represent this situation:

1..............x..........................
2...x.....................................
3........................x................
4..............x..........................
where the x's in lines 1 and 4 line up, representing (I presume) the same card in the same position?

If so, then I fail to see how this is relevant to poker.
It may accurately represent the actual situation, however it is not possible to know the timing, nor is it possible to know which deck comes next, therefore it is not possible to use it to make decisions.

My last thought is that any two decks are mutually exclusive. Therefore "intersecting" an occurrence from one deck with an occurrence in another deck has no actual meaning in real life.

Let's image a real world casino there is a paranoid millionaire at a high-roller table is a paranoid millionaire that insists on a new deck each hand. The casino would like to keep his business, so they have a team people who pre-shuffle decks for use at this table.

It is clear that just because deck #2 and deck #5 have the A♠: in position 7 nothing changes in how the game is played. It is no less fair, and represents no failing of online poker.

Or am I missing the point entirely?
VS

Imagine in a live game there is multiple tables, and a stack of decks already shuffled, and by random timing of the table hands, each table receives a new deck every hand from the deck stack/queue.
deck 1 comes to your table, in position 1 and 10 in the pre-shuffled unknown sequence is an ace. so you are dealt pocket aces.
The next hand you have a new deck, and by luck of timing you receive deck 11, in deck 11 in the correct sync, is again pocket aces for you.
The next hand......deck 52, again you get pocket aces.
etc, etc,
shuffle 1
shuffle 2
shuffle 3
standard.

shuffle 1,
shuffle 11,
shuffle 51,
internet by random timing.

''where the x's in lines 1 and 4 line up, representing (I presume) the same card in the same position?''

exactly that ,

consider that what you said. intersecting cards by random timing making probabilities void and meaningless.

Because of the button moving, the actual context is diagonal, xyz, but it is simpler to explain in a linear form.

also it was pokerstars, and also these emails are already on the net, I will dig them out of my emails again.

just consider this , 1 is a winner.

00010000000000100000010000
10001000001000010000000100
00010000010000101011110000
00000010000100000000000100
00001000100100001000000010
10000001000001000001000000

you get L2 and L6

The actual version is diagonal because of the button move. Your seat is static.
P(x ∩ y)/∅t

I can't believe you're still stuck on this problem. As I pointed out several weeks ago, all decks are the same. They are all shuffled randomly. It doesn't matter which deck you get, 'by timing' or by any other technical means. 'Timing' decides which random deck you get your cards from, but since the decks are all random, it doesn't matter which deck is selected. 'Timing' doesn't change anything.

You couldn't increase your EV even if you had the option of selecting which deck you get your cards from, because all that is known about the decks is that they are shuffled randomly. If it were known, that deck A would give you a winning hand, and deck B would give you a losing hand, sure, you could gain an advantage by choosing deck A, but this is not the case.

Well put, redskwerl,
I have had pocket Aces thrice in succession once (that I noticed) all the time I've been playing.
So what? I have probably had other hands three times in succession also, but never noticed.
Once long ago I did a query on my database to determine the number of times I had received each possible set of hole cards. Most were at the correct frequency, but 92o was considerably over-represented. I was getting all paranoid until I went several days w/o seeing 92o more than a few times and it evened out.

@shatteredaces: Although I understand the concept that you are illustrating, I contend that it is meaningless because it is comparing events in mutually exclusive populations.

The positions of each card in each deck is random, and the deck used in any given hand is random, hence randomness is preserved.

Cheers,
VS

confusing but i will again try to figure out.

Guys you are missing the picture, I am continuing to learn science and maths, I will get you to understand in the end.

consider vertical and horizontal,

x=horizontal

y=vertical

an L shape.

x=finite variables of 52

y=infinite variables

each deck has a set sequence of unknown variants after a shuffle.

This sequence does not alter of the deck.

The SB will get the top card of the sequence always.

shuffle 1 sb gets a first card of the sequence of a King of diamonds.

shuffle 2 of the same deck the sb gets dealt an ace of hearts from the top card being issued.

OK so far?

The BB player receives the second card of the sequence always.

To each player the cards they receive come from a diagonal path.

You are on the SB, in order of circuit

card 1
card 2
card 3
card 4
card 5
card 6
card 7
card 8
card 9

for nine shuffles of the singular deck, that is the order you receive your first card from the set sequence.

Ok so far?

grid form

123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789

ok so far you can understand the diagonal path?

look what happens if we randomise..

123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789

Random timing rather than the shuffle defining what card we get, I.e by timing luck we can intercept above or below the probabilities in a game, or bad timing losing hands, i.e row one is a losing hand for you, you receive by bad timing another 20 losing hands in a row.

If you are dealt pocket aces in a game, the sequence is already set, and if every player went all in, the aces are already predestined to win or lose by the set sequence.Nothing can change this.

Ask yourself this, how can I ever hope that by random timing of decks, I will ever receive an MTT winning sequence?

How many times must I receive a losing big hand at the wrong time?

1w3456789
1w3456789
1l3456789
1w3456789
1l3456789
1l3456789
1w3456789
1l3456789

2 is not the same as 2

w=win

l=loss

Originally posted by shatteredaces
Guys you are missing the picture, I am continuing to learn science and maths, I will get you to understand in the end.

Hi, shatteredaces,
Sorry, but I believe I understand now.
What you are saying is 100% correct, but 100% irrelevant.
HOW the randomness is achieved does not matter as long as randomness IS achieved.

Start with this thought:
 The position of cards in one deck have no influence on the position of cards in any other deck.

Continue with:
 All decks are equally random.

finish with:
 The timing between hands, and thus the deck used is random.

Originally posted by shatteredaces
Ask yourself this, how can I ever hope that by random timing of decks, I will ever receive an MTT winning sequence?

In an MTT of 50,000 players the odds of 1 player winning are 100%. In other words, there is always a winner.

(note , is a grouping separator, not a decimal separator)
If poker were purely a game of chance, your odds of winning are thus 1 in 50,000, or 49,999:1

If poker were purely a game of chance, you could expect to win once out of every 50,000 tournaments you play. Poker is NOT purely a game of chance. What you do with your cards is important. Odds are long term averages. You may go 150,000 MTTs without winning one, then win 3 in a row. You may start out winning 3 in a row and never win another one in your life.

You can improve your chances in a number of ways:
 Fold when the odds are against you.
 Raise when the odds are in your favour (even if you don't have currently the best hand)
 Pretend that you have the best hand, and bet accordingly (bluff).
 Recognize when another player is bluffing and don't fold.

Since poker is not purely a game of chance, the odds of getting any particular hand at any particular time really don't matter. With AA, (for example) you can expect to win 81% of the time, however not all pots are the same. It frequently happens that in the course of a night I'll get AA a bunch of times and just pick up the blinds. What difference does it make that I have AA or 72o if I raise and just get the blinds?

In poker it's not what you're dealt, but what you do with it that counts.

Peace,
VS

Originally posted by VorpalF2F

Originally posted by shatteredaces
Guys you are missing the picture, I am continuing to learn science and maths, I will get you to understand in the end.

Hi, shatteredaces,
Sorry, but I believe I understand now.
What you are saying is 100% correct, but 100% irrelevant.
HOW the randomness is achieved does not matter as long as randomness IS achieved.

Start with this thought:
 The position of cards in one deck have no influence on the position of cards in any other deck.

Continue with:
 All decks are equally random.

finish with:
 The timing between hands, and thus the deck used is random.

Originally posted by shatteredaces
Ask yourself this, how can I ever hope that by random timing of decks, I will ever receive an MTT winning sequence?

In an MTT of 50,000 players the odds of 1 player winning are 100%. In other words, there is always a winner.

(note , is a grouping separator, not a decimal separator)
If poker were purely a game of chance, your odds of winning are thus 1 in 50,000, or 49,999:1

If poker were purely a game of chance, you could expect to win once out of every 50,000 tournaments you play. Poker is NOT purely a game of chance. What you do with your cards is important. Odds are long term averages. You may go 150,000 MTTs without winning one, then win 3 in a row. You may start out winning 3 in a row and never win another one in your life.

You can improve your chances in a number of ways:
 Fold when the odds are against you.
 Raise when the odds are in your favour (even if you don't have currently the best hand)
 Pretend that you have the best hand, and bet accordingly (bluff).
 Recognize when another player is bluffing and don't fold.

Since poker is not purely a game of chance, the odds of getting any particular hand at any particular time really don't matter. With AA, (for example) you can expect to win 81% of the time, however not all pots are the same. It frequently happens that in the course of a night I'll get AA a bunch of times and just pick up the blinds. What difference does it make that I have AA or 72o if I raise and just get the blinds?

In poker it's not what you're dealt, but what you do with it that counts.

Peace,
VS

Correct ,poker is a game of skill and also a game of probabilities, the average of you getting pocket aces are 1/221 .
This is based on a sole decks distribution over a period of time. This is not based on random decks chosen by random timing.
Although yes indeed any card from any deck you will receive will be a random unknown card, the card you receive by a random deck is not the card you would of received from a single deck , a single deck that from the first hand you play, your probabilities come into effect, we know that roughly some time in a game or games, we will receive on average 1/221 pocket aces.
It is important to understand the fundamentals of time, timing being a key factor in poker that decides losers or winners. Random is co-efficient with time, i.e on a roulette wheel over a period of random time , out of the block/group of finite variants, it is guaranteed 100% that all the numbers will make an appearance. Over infinite time all the numbers will be equally distributed the same amount of times.
The big point and big picture you are not considering is this -
A single deck distributes your own dependent probabilities over time.
Random decks chosen by random timing distributes your own dependent probabilities randomly over time.

Understand this and it will be like an eureka moment for you and you will understand this very advanced and complex situation.
If you understand science at all?
''deck skipping'' produces space-time gaps of probabilities.
I do understand that sounds complex, but it is simple once you understand.

added
x={1/52}/t
y={∞/∞}/t
X=player seat
t=time
∅=random
∩=intersect

P(x)=P(X∩y)/∅t

P(x/t)≠P(y/∅t)

P(x/t)=221

P(y/∅t~)=var(X)

x~=1/52

corr(X,Y)={1/∞}/~t

Hi, shatteredaces,

If you understand science at all?
''deck skipping'' produces space-time gaps of probabilities.

I have a degree in it, if that means anything, but I'm guessing it doesn't.

Comparing pre-shuffled decks one to another is like trying to correlate events in parallel universes -- there is no connection between them.

All of your charts and diagrams are completely correct, but have no application to poker at all, since the cards in one deck do not affect the cards in the next.

I've tried my best to explain myself, but it isn't working.

If you can figure out HOW the timing of the decks influences the outcome of a hand, let me know.

if you do though, please don't quote my entire post -- just the part you are commenting on.

All the best,
VS

This is madness. I can't even decipher your maths, it's all over the place. It reads like something generated by mathgen. Also I still don't understand what you mean by time. You sit down at a poker table and they deal you a card from a random deck. The card has a 1/13 chance of being an ace. The chance of you getting an ace is the same, regardless of when they start dealing the cards. It's going to be the same tomorrow and a million years from now. Also it's the same if there are a million randomly shuffled decks in the universe, and the same if there's only one deck in existence. Nothing is being changed over time. The fact that there are other decks you are not getting cards from is also completely irrelevant.

Originally posted by redskwerl
This is madness. I can't even decipher your maths, it's all over the place. It reads like something generated by mathgen. Also I still don't understand what you mean by time. You sit down at a poker table and they deal you a card from a random deck. The card has a 1/13 chance of being an ace. The chance of you getting an ace is the same, regardless of when they start dealing the cards. It's going to be the same tomorrow and a million years from now. Also it's the same if there are a million randomly shuffled decks in the universe, and the same if there's only one deck in existence. Nothing is being changed over time. The fact that there are other decks you are not getting cards from is also completely irrelevant.

I know by your posts you are still missing the point entirely. It is complex , but once you get it, you will understand it very clearly and the simplicity of it.

My maths is a bit all over the show, to explain something new is not easy when the maths does not exist in the first place.

I will try it this way, imagine 52 cards face up in a line horizontally, axis x containing a start point of card 1 on the left and an end point of the last card 52, and containing 52 variants of x. We can say that x axis is finite, 1-52 being a ''distance'' , a linear dimension or vector.

x axis =card 1..................................................................card 52.

ok so far?

Now we want to increase the x axis representing multiple decks pre-shuffled making a Y axis.
......................Y
x axis =card 1..................................................................card 52
x axis =card 1..................................................................card 52.
x axis =card 1..................................................................card 52.
x axis =card 1..................................................................card 52.
x axis =card 1..................................................................card 52.

We can clearly see that x axis is finite and Y axis is infinite. I.e there is an infinite amount of the ace of diamonds in Y axis, there is only one in the x axis.

so far?

I will continue when I know you have understood this part.