I do not why you think I am agreeing with the maths

Your interpretation of the probabilities in the x&y directions has always been a sticking point in discussing you theory. In particular you think that the probability of an ace when picking from the top cards of 100 decks is different to that of picking from a single deck of 52. Well, in one post you said:

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.

However, later when talking about 100 decks you say:

..........column 1 to the small blind contains 7.69% worth of aces. compared to 4/52

So, because 7.69%=4/52=8/104=0.769 you are agreeing that probability in columns = probability in rows

I forgot science likes simple

No, science is comfortable with complex. What it does like is clarity. If you ask vague, woolly, unclear or confusing questions you can expect to be asked "what do you mean by...", "define ....", "what are you really asking". You have to be clear, something you are usually not.

However, because you have been much clearer in post #117, I will make one last attempt to show why the probability in x & y are the same, but the sequences are not.

nnn

nnn

nnn

we know left to right of each row is still 1/3, but we do not know the columns values.

We have previously pointed out that the probability of an ace in the y direction is the same as in the x see posts.#54, 55 and 105. What differs, as we have explained, are the sequences.

However, these are not unknown because probability helps us to understand the unknown.

OK, let's take your example as it is fairly simple.

nnn

nnn

nnn

You have 3 decks y1, y2, y3. Rather than use numbers for the cards I am going to use letters because otherwise we get confused when talking about 1st cards, 2nd card etc. So the 3 cards are A, B, C = Ace, Bravo, Charlie

So in the x direction P(x1)=P(x2)=P(x3)=1/3 and we agree on that.

So if you shuffle a deck of these cards and pick the first there is a 1/3 chance it is A, but if you had picked the bottom card that would also be probability of 1/3, same with the middle card.

So in the x direction we are agreed 1/3 1/3 1/3.

So what about y.

If we shuffle your 3 decks and take the top 3 cards y1 y2 y3 then there are 3 possibilities for each y and so the number of possible ways these cards can fall = 3*3*3=27 possibilities.

Let's look at the probability that in these 3 cards the 1st card only is A.

We can show this with maths using an X to show “not an A” = a 2 or a 3

So, AXX

here the probability of A is 1/3, but there are 2 possible cards in y2, a B or a C so the probability is 2/3, same with y3. So total probability = 1/3*2/3*2/3= 4/27.

We can also do this by writing out the possibilities for this deal:

A in y1

C B B C

B C B C

A A A A

a total of 4 ways out of 27 =4/27

We can also look at the probability that one of the cards out of the 3 is an A regardless of position. To do this we need to look at Ain y2 and A in y3

A in y2

We can also do this for y2 = A

C B B C

A A A A

B C B C

Another 4 ways

A in y3

and for y3 = A

A A A A

C B B C

B C B C

yet another 4 ways

Total 12

So if we want to know the probability of a single A anywhere in the 3 cards it is = 12/27=4/9

We would also want to look at the other ways for completeness:

A B C

A B C

A B C

3 more ways

and the ones we missed:

A C B C

A C C B

B B B B

A A B C

B C A A

A A A A

B A B C

B A C B

C C C C

Total 27

In order to do this many ways we have had to lay out 3x27 decks = 81

so in all those top cards how many were As?

Count them A = 27

So probability of A occurring in y = 27/81 = 1/3

the same as in x.

You will also note that AAA occurs once so probability = 1/27

but so does CBC, and BCC and BBB. All have equal probability.

Only humans decide that A is special. That an ace wins over a 6 or a 5. In reality their probabilities are all the same!

Now, the chances are that you would not get this exact match in exactly 81 decks, but as Alan said if you do enough games (trials) say >1000, then the actuality will get close to these probabilities.

What this also tells us is that for any 3 decks all arrangements are equally likely, so it doesn't matter if you skip decks or do them in sequence, the probability is the same. Out of a 1000 decks you could choose 3 from anywhere and the probability would be the same for each. If you think otherwise, you do not understand what probability is telling us.

You can obviously do the calculation shown above for any number of decks and cards, but the numbers become large, so I'll leave that to you.

Thank you Colin, interesting that probabilities may not be the answer I am looking for,

I would agree. Probability tells us that there is no difference between shuffled decks, so skipping some of the decks has no effect on the outcome of games from a probability point of view.

Yes, there are other scenarios. Your's sounds closer to predestination, or "that one was intended for me" type of luck or fortune. I know something about probability, permutation and sequences, but nothing about predestination, fortune or luck, so I can't help you with that.