Advanced statistics.

Originally posted by VorpalF2F
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If you can figure out HOW the timing of the decks influences the outcome of a hand, let me know.

All the best,
VS

You have not understood, nothing to do with the outcome of a hand, the random timing of decks influences the probabilities over time.

I am glad you understand science.

In a game of poker the shuffle determines what card you get, in online poker random timing of decks determines what cards you get.

Let us say you receive ten random numbers from a single distribution that shuffled and repeated.

1367197647

consider now a y axis of stacked pre-set sequences of x .

P={y,x}/t

It would be possible to receive an ace every single hand by timing luck or a 3 clubs.

Probability of X from x is 1/52
probability of X from y is infinite

Originally posted by shatteredaces
You have not understood, nothing to do with the outcome of a hand, the random timing of decks influences the probabilities over time.

Hi, shatteredaces,
Absolutely nothing influences the probability. This is the concept that redskwerl was trying to convey to you.
The probability of any outcome is fixed and immutable once the deck is chosen.
It never matters which deck is chosen nor when it is chosen, nor how it is chosen.

Originally posted by shatteredaces
I am glad you understand science.

Me too ♠_biggrin:

Originally posted by shatteredaces
In a game of poker the shuffle determines what card you get, in online poker random timing of decks determines what cards you get.

Let us say you receive ten random numbers from a single distribution that shuffled and repeated.

All your diagrams are just as valid for this scenario as for the randomly pre-shuffled decks -- the only difference is that we are re-shuffling from a presumably different starting point. The outcome is equally unpredictable.

In the case of online poker, there is no deck, there is no starting point, there are no cards -- a "deck" is merely an ordering of the numbers 1 to 52, each of which is the abstraction of a card.

Originally posted by shatteredaces
It would be possible to receive an ace every single hand by timing luck or a 3 clubs.

Yes, and it is just as possible if you shuffle physical cards. Possible does not equal probable.

Let's assume we are playing Hold'em. Fixed Limit, Pot Limit, No Limit does not matter.
There are 1326 2-card combinations in a 52 card deck. (reference: https://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29)
51 of them contain 3♣:, 1275 of them do not, therefore the odds that you receive 3♣: in any one had are 25:1 or approximately 38.5%

In the next hand regardless of whether the deck was shuffled or whether you use a pre-shuffled deck the odds of receiving 3♣: are EXACTLY THE SAME. There is no avoiding this.

The odds of receiving a 3♣: two hands in a row, is the product of the chances of each independent event.
So in this case 0.3846 * 0.3846 or about 0.15%.
For 3 in a row, it is 0.3846 cubed and so on and so on.
The probability of receiving 3♣: (or any card) 8 times in a row are 4.78 * 10^-12 percent -- but not 0 so still possible.

All of this is mathematically correct and is not influenced the the least way by HOW the randomness happened.

Cheers,
VS

Originally posted by VorpalF2F

In the case of online poker, there is no deck, there is no starting point, there are no cards -- a "deck" is merely an ordering of the numbers 1 to 52, each of which is the abstraction of a card.

.

VS

see complete model here

http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1382194#p1382194

p580
#14498

Hi shatteredaces,
Yes, I read that thread on badscience.net when you first posted it.
That is how I determined that you were trying to correlate independent data sets.

Correlating independent data sets is invalid in reality.

You can certainly calculate the overall probability of similar events in different data sets, but there is no correlation at all between the events in one data set and the events in another.

Let's reduce this to the most simple case possible.
In Case #1 we flip a single coin 10 times.
In the other, we pre-flip 10 coins and lay them out on the table.

Can you see that the odds of any one coin being heads is fifty-fify?

If so, you understand.

Peace,
VS

Originally posted by VorpalF2F
Hi shatteredaces,
Yes, I read that thread on badscience.net when you first posted it.
That is how I determined that you were trying to correlate independent data sets.

Correlating independent data sets is invalid in reality.

You can certainly calculate the overall probability of similar events in different data sets, but there is no correlation at all between the events in one data set and the events in another.

Let's reduce this to the most simple case possible.
In Case #1 we flip a single coin 10 times.
In the other, we pre-flip 10 coins and lay them out on the table.

Can you see that the odds of any one coin being heads is fifty-fify?

If so, you understand.

Peace,
VS

Hi Vorpal, you are still not seeing the point, x axis a player as only one chance out of 52 chances of receiving any specific card. Y axis a player as infinite chance of receiving any specific card.

To simplify consider 100 decks of cards that are individually random shuffled. 100 decks decks contains 400 aces Y axis, while x axis only contains 4 aces,
We have 4 chances that there is an ace in our seat position order after the shuffle in the unknown set sequence of a single deck , x axis.
We have 400 chances there is an ace in our seat position if we are playing all 100 decks y axis.

Originally posted by shatteredaces
Hi Vorpal, you are still not seeing the point, x axis a player as only one chance out of 52 chances of receiving any specific card. Y axis a player as infinite chance of receiving any specific card.

Hi shatteredaces,
Yes, I understand what you are trying to say.

What I am trying to say is that there is no y axis.
You can line up those decks for ever, but the events on one row are independent of the events on any other.

Any correlation you make between the rows is meaningless.

It seems that neither of us is going to get their point across.

I suggest we leave it at that.

Cheers,
VS

Originally posted by shatteredaces
http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1382194#p1382194

Oh wow, I should've looked at this link earlier. I say this with the best of intentions, you probably need professional help.

Hey
I've been a long time lurker here at PS, but I'll try to take a stab at this.
Firstly my background, I'm an engineering student currently on a masters degree programme. I've been playing poker for a while, but not until recently really started to study the game.

Hopefully I'll manage to make some sense and not make anything worse. Out of curiosity, and if I may ask, Shattered what is your background, for instance do you have any degree in maths, for evaluating these kinds of problems? from your thread at badscience I see that you do try to understand a quite wide grasp of science. It would also help if you state which statistical distributions and models you are using and why you are using them. Also some of your simplifications might not always work in your favor and actually make it a bit more confusing to understand.

Please have an open mind while reading what I'll be writing.

The scenario in which you're describing with lined up decks which some of them give you a winning hand, and some don't, only makes sense if we're talking about a type of game where human intuition and interaction has no say on the outcome of the game. And the only version of Hold'Em this is possible is an All-inn tournament. This is a situation we all can agree on that the winner is decided purely by luck. The winning hand you're then describing would also be the hand that wins at the river, called the nuts. In this type of tournament, what is the probability of winning a hand?

Let's now consider that we are playing at a 6 seated table, before looking at the hand we are dealt, the probability of winning the hand is 1/6 ~16.66%. This again means that the probability of losing the hand is 5/6. Now this is where I feel you might have a bit of misunderstanding of probability. It seems to me that you treat probability as a kind of guarantee of being dealt a winning hand after a sertain number of hands. But this means that on average over a huge sample size the average will be a win 1/6 times, it might take 100 hands but in some cases it might take as many as 100 million hands.
Probability is cruel. Without considering what cards that are dealt the probability of losing 9 hands in a row: (5/6)^9 = 19,38%. Which is higher than being dealt a winning hand. The winning hand sometimes might be AKs or 83o.

Originally posted by shatteredaces

Consider aces lose 18% of the time, suppose by timing of the decks you were dealt the 18% of aces , 100% of the time.....or 100% at the wrong time

Now still considering the situation of a 6-seat all inn every hand table with pocket aces. The probability for pocket aces heads up is somewhere in the area that you described, that they lose 18% of the time. this is not the case against 5 other players. The probability to win before any action is made with pocket aces against all the possible hands your opponents may have is actually closer to 49,17%. (ref. propokertools) and probably even lower at a 9 seated table. it is in fact the best possible starting hand, but that does not translate into always winning.

Now let's assume the decks are preshuffeled and waiting in line. You are drawing the top card, what would the probability of drawing a spade be? It would be 1/4.
Now looking at all the decks, the probabiliy of drawing a spade for each deck is 1/4. Now even if we remove five decks, because the decks are independent the probability of drawing a spade from the remaining decks does not change. The probability of drawing two spades in a row would still be 1/16. This would still be the case of the probability of being dealt a winning hand in poker from one of these decks. Even if they are being dealt to the first table that needs a new deck, the probability of being dealt a winning hand is still going to be unchanged. In our case of All-inns it would be 1/6. What I'm trying to point out is two things, first there are no guarantees for being dealt a winning hand with a certain interval. Secondly the probability of a being dealt a winning hand does not change even if the number of decks we draw from is being altered.

Another thing with probability is that it states a theoretical value of how often something should happen whith the information available.

Originally posted by shatteredaces

It is very simple, if I have two decks of cards pre-shuffled, and you have to pick one deck , how many opportunities do you have of aces been in your seat order in the deck sequence?

The truth is if you pick one of the two decks, no longer is the shuffle of the deck defining what cards you get, your choice defines what cards you get, the same as timing on the internet defines which cards you get, and not an even probability function of a single decks shuffle.

Yes it is very simple, the proability of getting aces is 1/221 for each deck. Why is it so? well you do not know anything about each deck. Thus you also do not know anything more than the standard probability of hitting aces. Which deck you chose does not do anything to the probability of hitting aces in the chosen deck becuase they also are independent.

You talk a lot about the decks after being shuffeled, they have a winner and loser no matter what. This is true, but this does not change the probability of you winning or losing, no matter which sequence of decks and which are removed or not. Why is it so? Well, you do not know anything about the decks and all the decks are independent and the probability of getting a new deck is 1. I'll use your dice example, The probability of throwing two of a kind is 1/6, why is it 1/6? Well the probability of throwing a number is 1, then the probability of throwing a matching number is 1/6. This is the same for picking a deck out of 100, the probability of picking a deck is 1, that you're dealt aces is 1/221.

I'm trying to come to the fact that mixing time into the probability of being dealt a hand is not correct for these kind of probability distributions, because you are guaranteed to be dealt a hand. Time based probability distributions are used for situations when an event do happen or do not. Like the probability of a passing car each minute in a given location. If we return back to the cards and decks, your probability over time of being dealt a winning hand will not change over time, due to removal of independent decks. you might include the probability of hitting aces withing a time interval if you look at a different rate of hands played per hour.

Now let's consider a normal game of hold'em 6-handed. The probability of being dealt the winning hand is the same. Though now human actions are controlling the game. And often the hand is won pre-flop, sometimes at the flop, etc. It often does not come to showdown. So you would not know if you just folded the winning hand or not. What we can do is look at the probability of the hand we're dealt and play it in that way. We know that 72o has the lowest winrate and therefore do not play it. unless we're the SB and the pot odds makes it such that even going all-inn vs the BB makes it a profitable move (this is not my strongest area, but it should be described in an article somewhere in the strategy pages on this site). All that we can do as poker players are to put or money in profitable situations over time, sure they might lose money short term, but in the long run they pay off. That's why we play different ranges against UTG raises or if we raise as the first one in the pot at the BTN. If you haven't studied poker very much, I suggest you do it is quite interesting how certain situations are profitable.

Originally posted by shatteredaces

Originally posted by Tomaloc

Originally posted by shatteredaces
1/221 is not the same as 4524/1,000,000

looks the same to me

also wow

In comparison how is a 221 number roulette wheel with one winning number the same as a 1,000,000 number roulette wheel with 4524 winning numbers?

These two are mathematical equivalent of each other with the same porbability of winning. If you do not get this it might be wise to question your own understanding of probability.

To conclude this post of reasoning. What I've been trying to say is that the probability of getting a deck that would deal you a winning hand does not change by timing, it is simple independent probability of each deck that matters. Also there is a difference between a winning hand and a playable hand that hits your range. Folding is a part of poker and no one wins a tournament on pure luck, unless they are all inn on every hand.

It seems to me that you are trying to find a pattern in a patternless system, like the decimals in pi, to explain a bad beat losing several hands in a row.

Hope that this makes sense, because your theory doesn't make sense to me unless you have information about all the decks in line and know which one would be dealt, and then you would have an advantage and probably win with insane rate. Considering the probability of being dealt a hand. Remember that probability dictates that if it can happen, it will happen, even if it is being dealt aces 10 times in a row and losing with all of them. The probability is extremely low, but at some point in time it will happen.

Best regards
Oddiee

Originally posted by redskwerl

Originally posted by shatteredaces
http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1382194#p1382194

Oh wow, I should've looked at this link earlier. I say this with the best of intentions, you probably need professional help.

No insult taken, often this has been said, you may want to consider that forum is a moron baiting forum.

I have an easy explanation now, it came to me.

shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.

shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52

This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

You then have a choice of any deck, your odds are 15/100 of getting an ace and not 4/52.

Now this is as simple explanation as it gets, surely you all understand this?

Originally posted by shatteredaces

Originally posted by redskwerl

Originally posted by shatteredaces

........lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

You then have a choice of any deck, your odds are 15/100 of getting an ace and not 4/52.

Now this is as simple explanation as it gets, surely you all understand this?

Cheers Shattered, I have highlighted the incorrect part of this chain of thought in red

Because the decks are defined as random, you cannot say "for example" because the whole point of being random is that we cannot know this.

To elaborate you would have to give the full range of possibilities for how many aces were on the top of each deck, and because it is random if you had a very large number "examples" the answer approaches 1 in 13
:-)

Later Adds
I don't know how I got stuck inside this quote box :-)

In this universe Chaos rules
Randomness cannot be added, subtracted or multiplied by itself.
Just like infinity.
Hyperrandom systems have been conceptualised (where nothing is fixed) but have no practical use in this universe.

BTW : How did the new job go ?
Regards Michael
:-)

Originally posted by GoOnCal1

Originally posted by shatteredaces

Originally posted by redskwerl

Originally posted by shatteredaces

........lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

You then have a choice of any deck, your odds are 15/100 of getting an ace and not 4/52.

Now this is as simple explanation as it gets, surely you all understand this?

Cheers Shattered, I have highlighted the incorrect part of this chain of thought in red

Because the decks are defined as random, you cannot say "for example" because the whole point of being random is that we cannot know this.

To elaborate you would have to give the full range of possibilities for how many aces were on the top of each deck, and because it is random if you had a very large number "examples" the answer approaches 1 in 13
:-)

Later Adds
I don't know how I got stuck inside this quote box :-)

In this universe Chaos rules
Randomness cannot be added, subtracted or multiplied by itself.
Just like infinity.
Hyperrandom systems have been conceptualised (where nothing is fixed) but have no practical use in this universe.

BTW : How did the new job go ?
Regards Michael
:-)

My new job is going well thanks, and the point is P(x)=P(X∩y)/∅t, an interception ruling that I have clearly shown by basic maths and models.

{4/x}/t≠{1/y}/random timing

clearly 4/52 is not equal to 1 out of a possible infinite amount of times.

In simple terms it means luck of the timing dictates the game and play, rather than a consistent distribution .

I understand, but you've made two mistakes.

Firstly, the top cards are not known. All you can assume about the top card is that it's an ace 1/13 of the time. If we have 100 decks, the randomly selected deck's top card will be an ace 1/13 of the time. The number of decks doesn't matter at all, because it's true for all decks that the top card is 1/13 ace. All decks are the same.

Secondly, let's look at your scenario. If, say, it was known that 15 of the top cards happened to be an ace:
1) Your chances of getting an ace would indeed be 15/100, when you're getting dealt a card from a randomly selected deck. This is correct.
2) When you're getting cards from a single deck with a known top card however, your chances of getting an ace would be either 1 or 0, not 1/13. If we take the top card to be known, well, then we know if it's an ace or not. If it is an ace, we'll get an ace 100%. If it isn't, we won't.

Originally posted by redskwerl
I understand, but you've made two mistakes.

Firstly, the top cards are not known. All you can assume about the top card is that it's an ace 1/13 of the time. If we have 100 decks, the randomly selected deck's top card will be an ace 1/13 of the time. The number of decks doesn't matter at all, because it's true for all decks that the top card is 1/13 ace. All decks are the same.

Secondly, let's look at your scenario. If, say, it was known that 15 of the top cards happened to be an ace:
1) Your chances of getting an ace would indeed be 15/100, when you're getting dealt a card from a randomly selected deck. This is correct.
2) When you're getting cards from a single deck with a known top card however, your chances of getting an ace would be either 1 or 0, not 1/13. If we take the top card to be known, well, then we know if it's an ace or not. If it is an ace, we'll get an ace 100%. If it isn't, we won't.

Hi Red, you are thinking in terms of x axis and not in terms of the Y axis, it does not matter that we do not know what the cards are, in an infinite column of y the odds switch direction, and the odds change randomly of Y.
It could be 0/500 or 500/500 , it is certainty not 1/13

Wtf did I just read.

Unless I'm mistaken the problem you have is "what if the actual probability distribution doesn't match the theoretical distribution" which is just as likely live. The 15/100 example is basically just the same as saying what if you ran bad/good in online poker (eg 0 aces in 1k hands, or 20 aces in 1k hands)? What if you ran bad in live? What if live shuffles resulted in 15/100 card decks having an ace on top? They absolutely can.

It's like, would you have a problem live if the dealer burnt two cards instead of one each time? It would make a difference to the results, which might be good or bad for you in the particular hand (though not long term) but the cards would still be random.

It really really seems like you just have a problem with short term variance being different to the theoretical probabilities, this happens in live poker too, and i'm not sure how you're not getting it, I think you've overthought it too much and confused yourself.

Originally posted by shatteredaces
you are thinking in terms of x axis and not in terms of the Y axis, it does not matter that we do not know what the cards are, in an infinite column of y the odds switch direction, and the odds change randomly of Y.
It could be 0/500 or 500/500 , it is certainty not 1/13

These are things you made up, not real things. There isn't a y axis. There aren't an infinite number of decks. The odds don't have a direction and certainly don't change randomly. Just because it's possible to type something, it doesn't mean it has real meaning. All of what you wrote is nonsense and doesn't apply to this situation at all.

Originally posted by redskwerl

Originally posted by shatteredaces
you are thinking in terms of x axis and not in terms of the Y axis, it does not matter that we do not know what the cards are, in an infinite column of y the odds switch direction, and the odds change randomly of Y.
It could be 0/500 or 500/500 , it is certainty not 1/13

These are things you made up, not real things. There isn't a y axis. There aren't an infinite number of decks. The odds don't have a direction and certainly don't change randomly. Just because it's possible to type something, it doesn't mean it has real meaning. All of what you wrote is nonsense and doesn't apply to this situation at all.

It is not made up. I have spent years trying to show this, it is a new concept,I am on science forums trying to get their perspective of this, it is complex.

A set of 52 variants get placed into the shuffler, they are shuffled and then become 1 set among multiple sets in a queue system.

Giving an x and y axis.(grid).

Spacing of variants of the Y axis playing a key role in the failure of online poker.

I get deck 1 , card 1, and get lucky and receive an ace

I then by timing get deck 100 the next hand , and lucky me card 2 was an ace.

the next hand i get deck 500, and card 3 is an ace. so on and so on,

Let's say I have a hundred coins and throw them all up in the air at once. What is the chance of any one individual coin landing heads?

Originally posted by redskwerl
Let's say I have a hundred coins and throw them all up in the air at once. What is the chance of any one individual coin landing heads?

50/50

Okay. Let's say I have a hundred 6-sided dice and throw them all up in the air at once. What is the chance of any one individual dice landing on 1?