Firstly my notion proves it is impossible to rig.


?/3 is easy . I have 3 boxes , how many dead cats are in the boxes?


https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation

Can the dead cats melt steel?

I don't know because they might not be dead.

I'm really dumb so for simplicity's sake let's just use 2 decks for your demonstration, x1 and x2. Please proceed.

About 1.92%, which can also be written as 1/52.

Does that mean you believe live cats can melt steel?

Let's flip this around and take the online situation to the casino.

Table 1 - One deck is used, shuffled after every hand. 10 hands are played this way.
Table 2 - Ten decks are used, each shuffled independently when you first sit down at the table. For the next 10 hands, the dealer uses those decks one at a time for each hand.

pkdk, do you expect different results, in terms of randomness or patterns, on the two tables?

Would your answer changed based on how the decks are chosen on table 2 (for example, using some kind of randomization factor, or letting a player choose)?

Ok, so you are going to have the top card, of one of the decks.


so your choice is

x
x


So what is your chance of an ace of diamonds in this setup?

Try to realise the Truth... There is no spoon decks.






EDIT:

Why is my strikethrough code not working? Anyone?

If you think anyone who's been debating with you in this thread would say 1/3, you're clearly misunderstanding a lot of things.

It doesn't matter how you're picking those cards (as long as it's done randomly) and what you're doing with the remainder of them, the answer is the same. Take out 7, light the rest on fire, and then pick one card at random. Take the deck skydiving, throw out 17 on the way down (without looking at them), then pick one from the remainder when you get to the ground. Remove one card and shred it, then pick one from the remaining 51. The answer is always 1/52.

Ok, good scenario , well devised.

The randomness never changes on either table in your setup. The randomness never changes, let me make that clear that random is random.
In your setup , table two would have a slightly different sequencing/pattern, but because the decks were coming only to that table, it would not be as bad as the decks going to all different tables in the same game/tourney.

In your table two setup , there is no jumps to different decks in the que.


example a deck order 1,2,3,4,5,6,7 distribution is not the same sequence /pattern as a deck order 1,4,6,7


It is literally a quantum leap doing that. Deck 4 for example should of been received later in time ,

The decks are virtual pieces of information, data occupies the space, that data is an energy, the information exists therefore the virtual deck exists.

Are you saying that in online poker the deck in play CHANGES mid hand?

If I give you a choice of x1 or x2 , I have just created a relationship. X1 and x2 are no longer independent .

No, you get a new deck every hand.

Thanks for the response.

Of course, it is absolutely & indisputably incorrect. Probably a waste of time explaining why, as you've shown a complete and utter refusal to listen to anyone, but what the hell.

Each deck is independent of one another, just like each deal from a single deck after it has been shuffled is independent of one another. The order you distribute those shuffled decks is irrelevant, and how you distribute them among different tables is irrelevant.

Also, this is not possible:

"The randomness never changes on either table in your setup. The randomness never changes, let me make that clear that random is random.
In your setup , table two would have a slightly different sequencing/pattern"

Either it's random, or it's not.

I think this incorrect belief is at the core of all your misunderstanding. They are most certainly still independent. If you could get your head around this, I think it would clear everything else up.

1/52 as Lego said.
Let's switch to a simpler example than a deck of cards: dice.

If I have two 6 sided dice, and roll them both, what are my odds of rolling a 6?

I thought you had understood for a minute, understand ostensible, do not be objective without looking in depth at this, withhold judgement .


The randomness of the game is not the problem. Let me try to explain a quantum leap to you. This stuff I am on about is advanced physics, so I will try to explain in simple terms.


Let us look at a simple sequence of 1,2,3,4,5


Now in order left to right you receive 1,2,3,4,5/t where t is time


Now what happens if we decide to stop the ''conveyor'' belt and randomly pick values?


Now we can have any order example 2,1,4,3,5/t



Do you agree with this so far?

Dice do not explain it , see my other post with the sequence and stay objective please.

Now we will be in a factory, we are order picking boxes 1-5 off several stopped conveyor belts. It is a lucky dip prize draw, number 3 is the star prize on each conveyor belt but somebody as forgot to put the numbers on the boxes.

{1,2,3,4,5} {2,3,4,5,1,} {1,3,5,4,2} {5,3,4,1,2}

spike the order picker , picks box 3,box 2, box 2, box 2, some lucky winner had all the star prizes.

I smell a containment thread in your future. Probably need to include the thread you started as pokerready from 2015.

The problem, of course, is where to contain you. There is quite an array to the left to choose from.

Oh and.... Butterfly Effect.

PKDK:

I have a proposal for a bet for you. If you win the bet, I will give you $100,000. If I win the bet, you will give me $50,000. We have to escrow the money before arranging to effectuate the bet.

The bet will be based upon the scenario you laid out earlier in this thread.


We will take 3 Jacks, 3 Queens and 3 Kings and arrange them into a 3 by 3 grid as follows:

JQK
JQK
JQK

Then we will shuffle each row so that the Jack, Queen and King is arranged randomly in each row. We will then have the following grid:

xxx
xxx
xxx

where we do know that each row contains exactly 1 Jack, 1 Queen and 1 King, but the columns could contain more than 1 or less than 1 of each card.

We will then flip over all of the cards in the first column and note whether or not we see at least one Jack. (For the record, earlier in the thread you said that the chance of seeing a Jack is ?/3. I disagreed with that and said the chance is roughly 70.4%. You said 70.4% is wrong and the correct answer is ?/3.)

We will do the above 1,000 times. If we see at least one Jack 550 through, and including, 850 times, then I win. If we see at least one Jack less than 550 times or more than 850 times, then you win.

I am giving you 2 to 1 on the money ($100,000 vs. $50,000) and I am giving you 700 numbers to my 300 numbers.

What do you say? Do you accept?

If you accept, then if you would agree to do so with a simulation rather than actually dealing cards, we could be done relatively quickly.

I have not got 50k lol, what do you mean by this part?


We will do the above 1,000 times. If we see at least one Jack 550 through, and including, 850 times, then I win


Are you saying you have a range of 550 - 850 ?

Because if you are, that is ?/3 being nowhere near exact.

No, x is not equal to y, but it can be explained by using something similar to the butterfly affect.

So you will happily ignore Schrodinger's cat?

I have said it is ostensible, ostensible is hard to show so you have to use different forms of logic etc to explain.