Is this maths correct?

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Offline Thebox

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Re: Is this maths correct?
« Reply #50 on: 28/06/2015 15:34:22 »



At times it is very difficult to understand what he is trying to solve, as some examples appear to be related to selecting from rows and columns.
However, I believe the crux of the problem is as follows:
If you and I are playing a face to face card game and we need to change the deck, we gather the cards together and shuffle the deck. Probability of an ace 1st card is 4/52,
In online poker at change of deck the player apparently selects, at random, from a group of previously shuffled decks. This appears to be where Box derives the infinite number of decks.
However, his concern is with the unselected decks. Let's say you select decks 2, 4 and 7 for your next 3 games. I would say that for each deck the probability of an ace 1st card is 4/52, but box does not. His view is that decks 1, 3, 5 and 6 might have given better hands, hence the probability should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.
This is more confused by some of the other scenarios he gives, and his poor understanding of maths.
See also his lack of understanding in this reply to you


No. You are not understanding , we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks.
So does every other table.
and yes

''should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.''


table one table 2 table 3 table 4 table 5

deck one deck 2 deck 3  deck 4 deck 5
........................deck 6......................


table 3 finishes their hand first so get the next deck.
« Last Edit: 28/06/2015 15:37:19 by Thebox »

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Offline PmbPhy

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Re: Is this maths correct?
« Reply #51 on: 28/06/2015 17:07:13 »
TB - I haven't followed this thread but have been informed by others about it. From what I gather you need to stop thinking about this and start developing a strong mathematical background and that can't be done by starting with attempting to study probability. You first have to study the basics and study them hard. Eventually you can move onto things like combinatorics since combinatorics arises in the study of probability theory. See: https://en.wikipedia.org/wiki/Combinatorics

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Offline Colin2B

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Re: Is this maths correct?
« Reply #52 on: 28/06/2015 23:02:57 »
....we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks.
So does every other table.
It doesn't matter who selects the new deck or in what order, whether it is the player, the dealer, an onlooker or a computer.

and yes

''should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.''
and no

should not take account of this 'infinite vertical y axis' and so the probability for a particular deck is not a complex combination of infinity, time and the x, y axes, but the probability of an ace 1st card is 4/52 (ie odds in favour 1:12) for any deck however selected.

table one table 2 table 3 table 4 table 5

deck one deck 2 deck 3  deck 4 deck 5
........................deck 6......................


table 3 finishes their hand first so get the next deck.
Doesn't matter.

Please give up this pseudomaths and follow PmbPhy's advice.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #53 on: 29/06/2015 18:18:46 »




should not take account of this 'infinite vertical y axis' and so the probability for a particular deck is not a complex combination of infinity, time and the x, y axes, but the probability of an ace 1st card is 4/52 (ie odds in favour 1:12) for any deck however selected.


Hi Colin , I still can see you are not understanding what I am saying exactly.  You and I are talking about two entirely different things.
For some reason you keep resorting back to thinking x axis and disregarding the Y axis.
Do you not  understand that I understand basic probabilities, and I am saying to you there is more with a 100% certainty that I know of and am trying to explain.

One deck of cards, we will call this (a), and in (a) there is 4 aces, which we will call an ace, (n).

A second deck of cards, any we will call this (b) , and in (b) is also 4 (n)'s.



Both (a) and (b) have a 4/52 chance of (n) being the top card after a shuffle.

If we were to (a)+(b)=104 then (n) would be 8/104 chance of (n) being the top card.

So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.

This is to show you that I understand.   

added to clarify, imagine looking at 2 decks of card, there is 8 aces in total in the two decks, and a total of 104 cards, there is an 8/104 chance that one of , or both of the decks, has an ace as the top card.


added -

machine (a) randomly displaces variants on a horizontal conveyor belt for 1 minute.

machine (b) randomly displaces variants on a horizontal conveyor belt for 1 minute.

machine (c) randomly displaces variants on a horizontal conveyor belt for 1 minute.

they stop, creating y axis.


machine (a) randomly variants on a displaces  horizontal  belt for 1 minute conveyor .

machine (b) displaces randomly  variants on a horizontal conveyor belt for 1 minute.

machine (c) randomly horizontal displaces variants on a 1 minute conveyor belt for .



« Last Edit: 29/06/2015 20:23:30 by Thebox »

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Offline chiralSPO

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Re: Is this maths correct?
« Reply #54 on: 29/06/2015 20:38:56 »
There is a 4/52 chance (≈7/7%) chance that the top card of a deck is an ace.

If you shuffle two decks together, there is still the same chance that the top card is an ace (4/52 = 8/104 ≈ 7.7%) Same if you shuffled 1000 decks together (4000/52000 ≈ 7.7%)

However, if you have two sheffled decks side-by-side, each has a 4/52 chance of having an ace on top. The probability of BOTH decks having an ace on top is (4/52)*(4/52) ≈ 0.59%. The probability that only one of them has an ace on top (and the other has some card other than an ace on top) is 2*(4/52)*(48/52) ≈ 14.2% the probability that neither of them has an ace on top is (48/52)*(48/52) ≈ 85.2%.  (quick reality check: the only options in this scenario are two aces on top, one ace on top, and no aces on top, so the probability of one of those three options occurring must be 100%. 0.59 + 14.2 + 85.2 = 99.9, which is within my rounding error of 100%)

If you have these two decks side by side, and you look at the top card of one deck, and it is an ace, you still don't know anything about the other deck, so it still has 4/52 chance of having an ace on top. But if you have the two decks shuffled together, there is an 8/104 chance of the top card being an ace, and if that first card is an ace, then you know the next card has a 7/103 chance of being an ace.

Does that make sense?

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Offline Colin2B

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Re: Is this maths correct?
« Reply #55 on: 29/06/2015 21:04:52 »
To add to what ChiralSPO said, putting a different viewpoint on it but saying the same thing.

So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.

This is to show you that I understand.   
Unfortunately you don't understand as much as you think you do.
This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52


added to clarify, imagine looking at 2 decks of card, there is 8 aces in total in the two decks, and a total of 104 cards, there is an 8/104 chance that one of , or both of the decks, has an ace as the top card.
Again you don't understand probability.

Two decks of cards A and B
Let's call probability (not odds, that's different) that the top card of deck A is an ace [tex]P(A)[/tex], and probability that it's B [tex]P(B)[/tex]. If it's not an ace we show it as [tex]\bar A[/tex] or [tex]\bar B[/tex].
If we put the 2 decks side by side turn over the 2 top cards there are 4 possible events:
[tex]A B[/tex]
[tex]A \bar B[/tex]
[tex]\bar A B[/tex]
[tex]\bar A \bar B[/tex]

Because these are mutually exclusive events there are 2 ways of handling this.
We can either use[tex]P(A \cap B)+P(A\cap\bar B)+P(\bar A \cap B)[/tex] which gives the probability of either 1 or 2 aces
or more easily [tex]1- P(\bar A \cap\bar B)[/tex]
either way the probability is around 14% as ChiralSPO said not 8% as you seem to think, you can follow the arithmetic in his post.

However, ChiralSPO also asked you if that made sense, and given your poor understanding of probability, I suspect not.

Edit: please stop writing probabilities and calling them odds, they are not the same.
The probability of tossing a coin is 1/2 or 50%, or 0.5
The odds are 1:1 or 50:50
Don't confuse the 2

« Last Edit: 29/06/2015 22:08:23 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Colin2B

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Re: Is this maths correct?
« Reply #56 on: 30/06/2015 07:33:33 »

So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.

This is to show you that I understand.   
Unfortunately you don't understand as much as you think you do.
This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52
Sorry, in my post above I forgot to answer this part of your question fully.
If you have 2decks and are going to choose between them, then before you choose the probability is 1/2*4/52=4/108 not 8/108 as you think.
However, as I pointed out, you have already decided to choose one deck and never make a different decision, so this selection follows the rules of conditional probability. As you seem to want to use symbolic notation for these situations, it is shown as P(A|B)=4/52.

Hint, learn maths.

EDIT: Added

Steve
I'm addding this because I'm going to try and explain one last time.
Forget your roads, conveyer belts etc and concentrate on cards.

I do understand that you are trying to arrange your decks as 52 cards in the x axis and decks 1 through to infinity in the y axis. Let's forget infinity for the moment.
When we write that the probablity of an ace is 4/52 we should really treat this as a decimal fraction in order to compare different probabilities. In this case ≈0.077.
Probabilities range from 0 to 1 with 0 being never happens to 1 being a certainty. So if we toss a coin 1/2=0.5 mid way on the probability scale (odds 50:50). If we toss a coin the probability of either a head or a tail =1 certainty because you have to get one or the other. If you toss a coin the probability of 6=0 = will never happen.

So if you have 2 decks of cards probability of ace first card = 8/104 = 0.077, same as 4/52 in other words not very likely to happen.
For 10 decks 40/520=0.077
100 decks 400/5200=0.077
All the same all the way up to infinity.
No difference when you include the y axis, which is why I said it was irrelevant.

You however are playing a little game of retrospective probability = what if. What if the packs you didn't choose had a better hand.
A similar situation occurs every time there is an aircrash. Someone misses the plane and then asks "what if I had caught that plane"? Well, probability says it doesn't matter which one you caught, that one, the one before, the one after, the probability for that aircraft, that company that route is the same. We are not concerned about the single event, it's probability is the same as all other events.

But let's look at your missed decks, those better hands. If we consider an ace first card, probability tells us that for every favourable deck there 12 unfavourable, so unlikely you find a better one. Also probability as I explained above is not concerned with the outcome of specific games, but with what happens when you play a large number of games. In that case the probabilities for all decks are as described no matter how the decks are chosen. It really doesn't matter which deck you choose, 12 will be unfavourable for every 1 in your favour.

If you don't agree I suggest you write up your cross odds theory and post it in New Theories. I'll stay away and see what others think. OK

« Last Edit: 30/06/2015 15:16:55 by Colin2B »
and the misguided shall lead the gullible,
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Offline Thebox

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Re: Is this maths correct?
« Reply #57 on: 30/06/2015 16:42:17 »
Thank you all for some great answers and maths. This as given me a greater knowledge, Thank you Colin for provided extra effort in explaining, I will be writing up a new theory , cross odds.  I understand your end of the discussion completely, however it still is not what I am precisely on about, although you Colin do understand where  I am coming from, and understand x and y that I mention.

Time plays a key role, it important to the conclusion, I just need to find the correct wording and steal some of the provided maths.

added - it just came to me how I can explain, (n) is a winner, I need two players,


player 1 . please pick from the x axis,

player 2, please pick from the y axis,


a
c
b
n
e
f
g
n
h
n
i
k
l
n
abcdefghijklmnopqrstuvwxyz
m
n
o
p
q
r
n
s
t
u
z


of cause you can not see the values in reality,

player 1 or player 2 has the better chance of picking an (n)?

« Last Edit: 30/06/2015 16:58:35 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #58 on: 30/06/2015 17:07:28 »
of cause you can not see the values in reality,

player 1 or player 2 has the better chance of picking an (n)?
Yes, the way you have 'rigged' it, player 2 has a better chance.
But you have given y 6 times what would occur from a random distribution and as you play more and more games the sequence tends towards a random distribution.
Also, the way the decks are shuffled you will tend to get 4/52 in the y direction as well as x, so in the long run ( which is what probability is all about) you will get the same result.

PS, I think you know, but I forgot to say that probability is often shown as % ie 1=100%=certainty etc

Have fun with your new theory
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #59 on: 30/06/2015 17:59:26 »
of cause you can not see the values in reality,

player 1 or player 2 has the better chance of picking an (n)?
Yes, the way you have 'rigged' it, player 2 has a better chance.
But you have given y 6 times what would occur from a random distribution and as you play more and more games the sequence tends towards a random distribution.
Also, the way the decks are shuffled you will tend to get 4/52 in the y direction as well as x, so in the long run ( which is what probability is all about) you will get the same result.

PS, I think you know, but I forgot to say that probability is often shown as % ie 1=100%=certainty etc

Have fun with your new theory

x axis 1-52

y axis ?

52²

1-52 is randomly shuffled along the x axis.   When the shuffle ends, the unknown sequence is set of each variant in position within the square,
the y axis contains  1-52 variants in  rows where the x axis contains columns of unknown values


stupid me , I had it back to front, sort of.

It does not matter if I un-rig it becuase x is not equal to y


x
x
x
x
x
x
xxxxxxxxxxxxxxxx
x
x
x
x

we know the vertical or Y axis as repeat values over a large quantity.

lets do this properly 52²

....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<52 players


What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?


« Last Edit: 30/06/2015 18:14:35 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #60 on: 30/06/2015 23:55:10 »
What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?
So eyewateringly small that it is impossible.
The number of ways of arranging a deck of 52 cards is 52!=1068 so the probability of dealing a particular ordering of 52 cards is too small to consider possible. So don't even consider doing it for 52 packs! You won't live to see it even if you deal 24 hrs a day for your lifetime.
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #61 on: 01/07/2015 05:55:29 »
What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?
So eyewateringly small that it is impossible.
The number of ways of arranging a deck of 52 cards is 52!=1068 so the probability of dealing a particular ordering of 52 cards is too small to consider possible. So don't even consider doing it for 52 packs! You won't live to see it even if you deal 24 hrs a day for your lifetime.

Thank you Colin for your help.

on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows.   What is the chance of there being more or less than 4/52 in the columns aligned to players?

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Offline Colin2B

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Re: Is this maths correct?
« Reply #62 on: 01/07/2015 10:29:35 »
on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows.   What is the chance of there being more or less than 4/52 in the columns aligned to players?

Before I answer this let me check what you are trying to do otherwise the answer will mislead you.

player 1 . please pick from the x axis,

player 2, please pick from the y axis,

For simplicity lets take a example you have given of 4 decks a b c d

a 132
b 123
c 132
d 321

Each player will receive 2 cards

Player 1 is going to use the x direction and is given deck a and draws 13

Player 2 is going to use the y direction chooses 2 decks b&c and hence draws 11

This seems a funny sort of game and I can't see how it relates to your real life game!

Note, these numbers do not relate to reality and player 2 could also have lost with a different distribution

EDIT:
Just to make sure I understand what you mean by using the y axis.
If we have 4 decks of cards:

y4   d   d  x
y3   d   d  x
y2   d   d  x
y1   d   d  x
     x1 x2 x3 x4 x5 x6 etc ............ x52

If you want to use the y axis to deal 4 cards to a player, from x3 as shown, you would have to take decks y1.....y4 discard the top 2 cards from these packs =d and deal the cards x to the player.
« Last Edit: 01/07/2015 15:22:50 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline Thebox

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Re: Is this maths correct?
« Reply #63 on: 01/07/2015 16:19:32 »
on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows.   What is the chance of there being more or less than 4/52 in the columns aligned to players?

Before I answer this let me check what you are trying to do otherwise the answer will mislead you.

player 1 . please pick from the x axis,

player 2, please pick from the y axis,

For simplicity lets take a example you have given of 4 decks a b c d

a 132
b 123
c 132
d 321

Each player will receive 2 cards

Player 1 is going to use the x direction and is given deck a and draws 13

Player 2 is going to use the y direction chooses 2 decks b&c and hence draws 11

This seems a funny sort of game and I can't see how it relates to your real life game!

Note, these numbers do not relate to reality and player 2 could also have lost with a different distribution

EDIT:
Just to make sure I understand what you mean by using the y axis.
If we have 4 decks of cards:

y4   d   d  x
y3   d   d  x
y2   d   d  x
y1   d   d  x
     x1 x2 x3 x4 x5 x6 etc ............ x52

If you want to use the y axis to deal 4 cards to a player, from x3 as shown, you would have to take decks y1.....y4 discard the top 2 cards from these packs =d and deal the cards x to the player.


Ok,

yes Y represents the individual sets of decks,

y1=deck 1
y2=deck 2

and so on.

expanding this


1 to 52 along the  x axis. 

123456789 and so on.

so in Y and x form it looks like this

123
123
123

I am sure you understand this part from your post.


Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axis

y
p3 {1. 2.  3}
p2 {1 .2.  3}
p1 {1. 2. 3}
....p1 p2 p3


p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.

they have a 1/3 chance of guessing where the 2 is using the correct axis of x.


p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.

So now consider poker using one deck,


and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.

My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite.  Every time a deck is gone, a new one appears. It is infinite for as long as you play.

X and Y alignment is not the same as just x.

In saying pick a deck, the deck is randomly given to your table, after every hand, each table gets a new deck, timing of table hands deciding what deck you get out of the system.

y1
y2
y3
y4


would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.

Imagine a tournament

time=...


Now imagine a player having luck by timing


time=aaa..........aaa.........a.a...a....a....a................a...................a..........a..........a.a.a..a


a=ace


now imagine a player having bad timing luck

time=......................................................I stop here because they ran out of blinds and were forced to fold every hand because they was bad.













« Last Edit: 01/07/2015 16:29:44 by Thebox »

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Offline Thebox

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Re: Is this maths correct?
« Reply #64 on: 01/07/2015 16:41:06 »
And here is what my argument is about.

''Where a deck of cards is shuffled has absolutely no bearing as to
whether it is shuffled randomly or not.  As we have explained to
your repeatedly, the cards are shuffled randomly.  They *are*
dealt to your table alone, in order, without being altered in any
way.  As I write this, there are more than 25,000 tables in
action, and each of them requires a new, shuffled deck of cards
foe every hand it deals.  Each table requires a deck more than
once per minute, on average.  It makes no sense to have the cards
shuffled at each table.  That would require tens of thousands of
individual shuffle servers, which is just silly.
 
Instead, there are servers that do nothing but shuffle decks of
cards.  They have no idea how those cards will be used, they just
shuffle decks and line them up.  When your table needs a deck, it
pings the server, and the next deck in line is delivered to that
table.  The cards are then dealt in order.  There is no such
thing as a "cool deck"; every deck is shuffled randomly, so it is
a fair deck.  It cannot *possibly* favor any particular player. 
NOTHING is ever done to alter the decks -- they are simply dealt
in order, fairly.
 
The "user input" we refer to is not done at your table and does
not result in "control" of the hands.  It is a steady stream of
data which is used as one part of the shuffling process.  No
*individual* alters the dealing.  NOBODY can alter the dealing. 
The user input is the entire mass of data coming in from every
single ''edited out name of company'' user in the world, all at once.  Understand
that your ''edited out name of company'' table is a representation.  As far as the
client and our server are concerned, your table is just a grid of
numbers.  When you move your mouse or click a button, it is
transmitting pairs of numbers to our server which represent the
location you are clicking on.  For example, a "Fold" might be
something like 28,604x58,789 -- two *large* numbers which
represent a location.  Right now, as I am writing this, there are
more than 160,000 players on our site, and *every one of them* is
transmitting a string of these pairs of numbers to our server --
that's how it knows what you want to do.
 
That *MASSIVE* amount of information is turned into an even more
*MASSIVE* strings of binary data, which is *one* of the sources
of data which our shuffle servers use to shuffle decks of cards. 
That string of data has no *meaning* whatsoever.  If you move
your mouse left or right, or click a slightly different spot,
that string of data does *change*, but it doesn't influence the
dealing in any particular way.  There is no specific control. 
Even if you *wanted* to somehow influence the dealing, you would
need to know what every single poker player on ''edited out name of company'', in
every part of the world, is doing at that moment, *AND* you would
have to know how every single bit of that data will be used,
which is *impossible* to know.
 
Furthermore, that string of binary data is combined with another
string of complete random binary data which comes from the
Quantis Random Number Generator, as described to you already and
linked to on our website.  Between those two completely
unpredictable sources, a *massive* number of randomly shuffled
decks of cards are generated every single day -- many tens of
millions of them.
 
Yes, it could be argued that you moving your mouse a millimeter
to the left somehow *changed* something.  But there is absolutely
no way to know *HOW* it changed anything.  You don't know the
algorithms used to shuffle, or how the two sources of data are
intertwined, or anything else about that change.  You don't even
know the pairs of numbers *you* are generating.  It's all
completely impenetrable.  That's the most basic quality of
randomness -- it's unpredictable.
 
The truth is, the shuffling on''edited out name of company''  is likely *MORE* random
than the shuffling of a deck of cards at a live poker table. 
That's because the dealers at a live table have a good shuffle
method (scramble, shuffle, cut, shuffle, box, shuffle, cut), but
it's not perfect.  It's *POSSIBLE* to track visible cards through
that shuffle.  It's *POSSIBLE* for a shuffling expert to
manipulate that shufflie.  It's *POSSIBLE* that the dealer will
flash a couple of cards when he's shuffling or cutting.  Or it
will "waffle" while it's being pitched across the table.  It's
*POSSIBLE* that some player will get an advantage via such dealer
error.  It's *POSSIBLE* that the cards are marked.  *Usually*
nobody knows what's coming, or any of the cards, but I don't know
any casino player who has not seen a flashed card, or experienced
a mis-deal.  On PokerStars, that *NEVER* happens.  You *NEVER*
know what's coming, and cards are *NEVER* flashed.  It really is
a completely random deal.''
 
User input is not a "terrible flaw" in our dealing, it's an
*extra* bit of random, unpredictable information.  This is used
*along with* another random source.  This methodology has been
inspected and approved by Cigital, a company whose job is it to
analyze such things.  The *results* of this method of shuffling
and dealing have also been analyzed by countless people for more
than *eleven years* now, and not one person has ever found
anything wrong.  The math in our games *is* correct.  That's not
a matter of opinion, it's a *fact* and it's completely
verifiable.  That's the whole power of hand histories -- you
don't have to have beliefs about the dealing, you can actually
check the results.
 
As to your claim that "the math is not working out" for you, you
talk about checking your stats.  Have you actually done this
math?  How did you gather your raw data?  What stats have you
checked?  How did you do this calculation?  What software did you
use?  How big is your sample?  What method did you use to measure
the significance of your findings?  Please send us the details of
your findings.
 
Or is this all just observations?  Are you just frustrated with
losing, and venting that frustration?
 
Yes, you take bad beats when you play poker.  *Every* person who
plays poker takes bad beats.  But bad beats are *NOT* why you are
losing.  Even *if* it were true that you always bust out on bad
beats (which is *not* true) That wouldn't prove anything.  That's
how poker tournaments work -- you play and play and play until
you either make a mistake or take a bad beat.  It's the same for
everyone.  You just have to lose *ONE* hand and you're busted.
 
To win a big poker tournament you have to survive a dozen or more
all-in hands.  Even if you get your money in ahead every single
time, you're *still* a favorite to lose one of them.  For
example, if you get your money in as an 80% favorite (like pair
over pair) just *four* times, your odds of winning all four are
.8x.8x.8x.8 = .4096 -- you're a 2:3 *underdog* to win all of
them.  Yes, you remember the time you lose, but you *do* win
those contests 80% of the time.  And like I said, you have to win
*many* more than four to win an MTT.  Plus, sometimes you don't
get your money in as an 80% favorite -- sometimes it's a flip,
sometimes it's 60%, sometimes you're even an underdog.  You *DO*
win most of the time with your good hands, and when you do, you
just keep on playing.  *Every* all-in hand is an *opportunity* to
bust out.  Complaining about the ones you lose is just silly.  OF
COURSE you lost those hands -- that's why you busted out, and why
you remember it.
 
The last MTT you played here a few days ago had 4,501 players in
it.  If you're an average player, your odds of winning a
tournament that big are 4,500:1.  Those are *very* long odds. 
You did *NOT* bust out of that tournament on a bad beat -- you
busted out with a good hand, AKs, but you lost to a better hand
-- Aces.
 
In your entire history here, you've only played 643 MTTs with a
cash buy-in.  The *average* player count in those events is
3,996.  Yet you're upset that you haven't won one yet?  Even the
best poker player in the world would not be a favorite to win 1
of 643 tries in a tournament with almost 4,000 players.  You
cashed in 90 of them, which is 14% -- that's a normal amount,
since we usually pay 12-15% of the field.  Your results are
perfectly normal.
 
You don't lose every time you're ahead.  You win *and* lose
hands, just like everyone else does.  You've won a bunch of Sit &
Go's, which of course are easier to win because they have fewer
players in them.  When you win them, it's because you *don't*
take a bad beat on an all-in hand, or because *you* got lucky. 
The last heads-up S&G you won, just a few days ago, you won when
your opponent pushed all-in with King High, and you called with
Jacks and he *didn't* hit a King.  Your good hand *won*.  In the
one after that, you lost with KJ to AJ -- *not* a bad beat.  In
the 180-player S&G you played just before that, you busted out
when you called *two* pre-flop all-ins with 8-7 (Eight High!),
and lost to pocket Aces.  Obviously that's not a bad beat,
either.  The truth is, you only bust out of tournaments on bad
beats *sometimes*.  *Most* of the time, you bust out when you get
your money in behind.  You *don't* "lose to the bad beat always".
 
Attached to this email is a chart of *every* all-in hand you've
been involved in for the last month in tournaments.  Go through
it carefully.  Count the times you were ahead and the times you
were behind.  Count the wins *and* the losses.  What you will
find is that when you have the better hand, you *do* win most of
the time.  When you have the worse hand, you lose most of the
time.  You take some bad beats and you give some bad beats.  It's
all just normal poker.  You only have to glance at the "Was
Allin" column and the "Won" column to see the All-In hands where
you lost.  It's *obvious* that most of the time you did *not*
bust out on a bad beat.  Starting from the top:  AK vs AA, KJ vs
AJ, A9 vs 66, QJ va AT, 78 vs AA, Q9 vs A7, J7 va A5, T7 vs 89
(you started ahead but got the money in behind), AQ vs QT... 
Finally a true bad beat.  Like I said, *most* of the time you do
*not* bust out on a bad beat.  You are *not* losing because of
bad beats or some flaw in our dealing.  You're losing because
you're getting your money in as an underdog too often.
 
That's what real poker analysis is about -- checking the
percentages that hands win in similar situations.  It's not about
the painful beats.  Winning at poker is about recognizing the
mistakes that you make and correcting them.  It's *not* about
blaming the dealer because you took a few beats when you had a
good hand.
 
We are happy for you to make your "findings" public as long as
they are mathematically sound.  Your opinion that there is a flaw
in our dealing methodology is incorrect, and your assertions
about how you lose are inaccurate.  If you would like to do a
complete, unbiased analysis of your hands, we welcome it.  We
have 243,557 of your hands on file, and we will send them all to
you if you ask.  There are many software packages available
online which you can use to do a full analysis.  It is 100%
certain that if you go through *all* of your hands, you will find
that you have been treated completely fairly.  You have won and
lost according to the normal poker math percentages.  Here are
some of the hand analysis tools which are available online:

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Offline alancalverd

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Re: Is this maths correct?
« Reply #65 on: 01/07/2015 18:23:42 »
1. A bad workman blames his tools.

2. In poker, you are playing against a whole lot of other people.

3. In online poker you don't know who they are.

4. The casino is not a charity

5. Each hand is winner-takes-all

6. Very good players can win money in the long term

7. Therefore anyone who is less than very good is likely to lose money.

See? No complicated maths required!
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Offline Thebox

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Re: Is this maths correct?
« Reply #66 on: 01/07/2015 19:51:29 »
1. A bad workman blames his tools.timing is not a tool

2. In poker, you are playing against a whole lot of other people.yes

3. In online poker you don't know who they are.yes

4. The casino is not a charitypoker is not the same as a casino game,

5. Each hand is winner-takes-allnot really ,

6. Very good players can win money in the long termcrap players win money online

7. Therefore anyone who is less than very good is likely to lose money.not true

See? No complicated maths required!untrue

The relevant part.

''As I write this, there are more than 25,000 tables in
action, and each of them requires a new, shuffled deck of cards
foe every hand it deals.  Each table requires a deck more than
once per minute, on average.  It makes no sense to have the cards
shuffled at each table.  That would require tens of thousands of
individual shuffle servers, which is just silly.
 
Instead, there are servers that do nothing but shuffle decks of
cards.  They have no idea how those cards will be used, they just
shuffle decks and line them up.
  When your table needs a deck, it
pings the server, and the next deck in line is delivered to that
table. ''

y≠x

123
231
213

P(n)={4/52}/x
...............t


P(n)=?/y
...........t


« Last Edit: 01/07/2015 20:18:04 by Thebox »

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Offline chiralSPO

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Re: Is this maths correct?
« Reply #67 on: 01/07/2015 20:12:31 »
Box, it sounds like you have played some online poker and lost. Then it sounds like you complained to the administrators, who sent you a very long, very detailed, and very awesome message. I would not recommend that you continue your argument with them (I might also recommend you stop playing online poker, but that is entirely up to you  [:)])

These guys spent a lot of time making sure their algorithms are fair, and you are not going to prove otherwise with your current level of mathematical prowess.

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Offline Thebox

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Re: Is this maths correct?
« Reply #68 on: 01/07/2015 20:32:15 »
Box, it sounds like you have played some online poker and lost. Then it sounds like you complained to the administrators, who sent you a very long, very detailed, and very awesome message. I would not recommend that you continue your argument with them (I might also recommend you stop playing online poker, but that is entirely up to you  [:)])

These guys spent a lot of time making sure their algorithms are fair, and you are not going to prove otherwise with your current level of mathematical prowess.

I will give you an idea of my poker ability,

45 man tournaments, 8-9 final tables in a row, on several occasions, hat tricks of first places on several occasions, 180 man 1st places, huge amount of players, cashes. including 3rd places.

   HENL   Just Lika   $1.10   3659   /4   $201 
  30 Apr 10   No Limit Hold'em   HENL   Zwane69   $1.10   3802   /5   $162


I am a skilled player, want to see?

http://www.boomplayer.com/poker-hands?search=holdemace486


By the way the company never did send my hand audit, and if you type my player name in google you will see some crazy posts, you will also see if they still exist posts where I am experimenting with online poker.
What he says in his post to me , was when I was on a rant and trying to time a win.





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Offline Colin2B

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Re: Is this maths correct?
« Reply #69 on: 01/07/2015 21:57:38 »
I have no question about your poker ability, nor your understanding of how the games are played. I am however having difficulty understanding how your xy matrix relates to real card games. Let me explain



Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axis

y
p3 {1. 2.  3}
p2 {1 .2.  3}
p1 {1. 2. 3}
....p1 p2 p3


p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.

they have a 1/3 chance of guessing where the 2 is using the correct axis of x.


p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.
In order to make this work you would have to lay out all the cards from the y decks into the xy matrix you show. Players would then have to choose a row or column. This is not how card games are played.

Many people are confused when told that the probability of an ace 1st card is 4/52. What they don't realise is that this is the same probability for a 2, a 3, a 9, Q, J etc. In other words all the cross points of your matrix have the same probability of being an ace (or any other card).


So now consider poker using one deck,

and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.
No, your scenario doesn't show this, it shows only a matrix which the players cannot access for a single deck.


My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite.
Again you cannot access the y axis for a single deck

 
would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.
As I have said before, for every favourable deck there are 12 unfavourable so little point trying to change.
If the decks are distributed at random then each deck has the same probability of a winning hand, the order in which you receive them is irrelevant.
You are playing a game here of 'what if', which as I have said before is not what probability is about.

If you genuinely think that the choice of deck order influences the probability of winning I suggest you set up a Monte Carlo simulation for the 2 types of game and see what happens.
« Last Edit: 02/07/2015 00:44:40 by Colin2B »
and the misguided shall lead the gullible,
the feebleminded have inherited the earth.

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Offline alancalverd

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Re: Is this maths correct?
« Reply #70 on: 02/07/2015 00:14:14 »
1. A bad workman blames his tools.timing is not a tool

2. In poker, you are playing against a whole lot of other people.yes

3. In online poker you don't know who they are.yes

4. The casino is not a charitypoker is not the same as a casino game,

5. Each hand is winner-takes-allnot really ,

6. Very good players can win money in the long termcrap players win money online

7. Therefore anyone who is less than very good is likely to lose money.not true

See? No complicated maths required!untrue



This is known as The Gambler's Delusion. It's remarkably common and the reason that very few people can make a living from playing poker.
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Offline Thebox

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Re: Is this maths correct?
« Reply #71 on: 02/07/2015 05:47:53 »
I have no question about your poker ability, nor your understanding of how the games are played. I am however having difficulty understanding how your xy matrix relates to real card games. Let me explain



Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axis

y
p3 {1. 2.  3}
p2 {1 .2.  3}
p1 {1. 2. 3}
....p1 p2 p3


p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.

they have a 1/3 chance of guessing where the 2 is using the correct axis of x.


p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.
In order to make this work you would have to lay out all the cards from the y decks into the xy matrix you show. Players would then have to choose a row or column. This is not how card games are played.

Many people are confused when told that the probability of an ace 1st card is 4/52. What they don't realise is that this is the same probability for a 2, a 3, a 9, Q, J etc. In other words all the cross points of your matrix have the same probability of being an ace (or any other card).


So now consider poker using one deck,

and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.
No, your scenario doesn't show this, it shows only a matrix which the players cannot access for a single deck.


My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite.
Again you cannot access the y axis for a single deck

 
would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.
As I have said before, for every favourable deck there are 12 unfavourable so little point trying to change.
If the decks are distributed at random then each deck has the same probability of a winning hand, the order in which you receive them is irrelevant.
You are playing a game here of 'what if', which as I have said before is not what probability is about.

If you genuinely think that the choice of deck order influences the probability of winning I suggest you set up a Monte Carlo simulation for the 2 types of game and see what happens.

Hi Colin, you do not pick columns, the columns are a set order and inline with each player always, the rows are distributed over random timing of hands, in simple terms there is gap spacing between the decks your table receives.

pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.

................................................. unbroken/t


.      .      .                   .               . broken/t

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Offline Colin2B

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Re: Is this maths correct?
« Reply #72 on: 02/07/2015 10:07:48 »
pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.

................................................. unbroken/t


.      .      .                   .               . broken/t

As I've said before the order of the packs or the gaps in between doesn't make a difference to the probability, if you believed that what about all the packs that pass by while you are sleeping?

To check you understand what I am saying:

Q1 shuffle a deck and turn over the top card, what is probability it is an ace?
Q2 shuffle the deck but do not look at the top card, what is probability of an ace?
Q3 still without looking, waste this pack, shuffle and turn over top card, probability of an ace?
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?


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

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Offline Thebox

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Re: Is this maths correct?
« Reply #73 on: 02/07/2015 17:03:51 »
pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.

................................................. unbroken/t


.      .      .                   .               . broken/t

As I've said before the order of the packs or the gaps in between doesn't make a difference to the probability, if you believed that what about all the packs that pass by while you are sleeping?

To check you understand what I am saying:

Q1 shuffle a deck and turn over the top card, what is probability it is an ace?4/52
Q2 shuffle the deck but do not look at the top card, what is probability of an ace?4/52
Q3 still without looking, waste this pack, shuffle and turn over top card, probability of an ace?4/52
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.

I think Colin you are still unsure of what I am on about in full,


You are wiser than most and close to understanding in full I feel.

Lets us discuss the time issue,


Lets say I and you sit down at the same poker table in a game.


You have £1000 and I have £1000 in starting chips , we are playing a small  tournament.

In a tourney we post blinds and antes, so we can say that £1000 is equal to so many blinds and antes

If we sat out and did not play a hand, we will both get blinded out.  We can say and by using measurement,

that there would be an uncertainty of time it will take before we get blinded out, but we could put an  approx time value to this, a range between ?hrs-?hrs based on the blind increase time, and an estimate means of how long a table takes to play a hand.

Ok so far?



« Last Edit: 02/07/2015 17:17:55 by Thebox »

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Offline Colin2B

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Re: Is this maths correct?
« Reply #74 on: 02/07/2015 17:47:41 »
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.
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Offline 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.

added - add 3 players to the bottom of the columns



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 »

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Offline 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.
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Offline 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





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Offline 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.
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Offline 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 »
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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 »

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Offline 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 »
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Offline 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.
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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=radius
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 »

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Offline 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 »
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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.

please answer

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 »

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Offline 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

Please reread posts #56 and #57 where this is also explained.
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 »
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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.
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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)

advanced model

http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1386649#p1386649

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 »

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Offline 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.
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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 »

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Offline 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. 
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Offline 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.

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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.



Now to add to your confusion!

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?
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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.
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Offline 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 »

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Offline 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.

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Offline 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,

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Offline 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.
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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?