Assuming that (i) each ? represents a random card from a randomly shuffled deck of standard cards and (ii) one selects one ?, then the chance of such selected ? being an A is approximately 7.69% aka 1/13. The chance of such selected card being a particular A, such as the Ad, is approximately 1.92% aka 1/52.

If you don't understand the math that yields the above calculations, you could just do it a million times or so and keep track of the number of times you draw an A and/or an Ad.

I do not agree with you because you are wrong. I'm sorry.

@Lego

How many ad are in x? 1/52

How many ad are in y? ?/10


Do you disagree?


Your earlier answer

What is the chance of drawing ad from x?

What is the chance of drawing ad from y?

p.s Yes lego I am very correct. x ≠ y

Each z represents one card randomly selected from a different standard 52 card playing deck.

z
z
z
z
z
z
z
z
z
z

and
are different questions, hence they have different answers.







Sorry, man. I'm gonna go do something else now.

NO, it is NOT RELATED.

Here's why:
Lets say you have 52 decks of cards, each deck containing the standard 52 cards, all of which are shuffled 93 times by 18 different people to ensure complete randomness.

The 52 decks, in turn, are laid out in a line and rearranged 75 times by 20 different people so the order of the decks is also completely random.

Then you randomly pick a deck. The odds of picking the deck you got are 1/52. You happen to pick up deck #16

Then you take the top card from the deck you picked.

Here's where your connection falls apart - EACH DECK is a self contained entity. It contains 52 cards in a random order. Deck 15 also contains 52 cards in an random order and so does deck 17 but you picked up deck 16 and it doesn't matter.

Because deck 16 consists of 52 cards in random order.

The odds of you drawing A from deck 16 is 1/52. The fact that it was randomly drawn from an absurd collection of other decks doesn't matter - because that set of cards is wholly self contained and NOTHING YOU DO to the other decks changes the order of that deck.

What you're trying to do is link two unrelated items. The ordering of the set of sets - ie: the collection of decks - that has NO impact on the sets within. The internal set you picked doesn't have any connection to any other set - only it's CONTAINER set (the box containing hte cards) does.

No matter what you say or do, no amount of re-arranging the boxes of decks will change the fact that any 1 given deck will have exactly a 1/52 chance of giving you A as the first card.

Yes they have different answers and are different questions, however it shows the ostensibly content i mentioned. At first glance the chance of drawing ad would be 1/52, but it is not , it is ?/10

NO, honestly, you are making the same mistake over and over with all due respect.

Each of the 52 decks are independent decks.
Each top card would be independently 1/52 in being any value

But you are not still not considering y and the dependency to the choice.

Take 52 decks of standard cards, shuffle all the decks independently ,

take a top card of each deck , throw the rest away

52
.
.
.
.
1


How many ad are there in this column?

Yeah, but if you have 10 decks to choose from, the top cards are only Ad ?/10 times, not 1/52 if you have one deck.

Explain that!

I call it cross odds.

OK, peekydeeky. I think I've got it.

You agree that if you pick the top card of one deck the probability of it being Ad is 1/52.

If you have a choice of two decks, although the probs are the same for each, individually, because you're looking only at the top cards available on the the two decks the chance of the one you pick being the Ad is ?/2.

Is that right?

You got it Mike, good maths skills and understanding.

You might have said it slightly wrong but i think you got it.


added- after reading it again, you said it right.

@Mike

Imagine what you just said with a billion decks in a crate.

This is all we need to know.

Randomness of each deck presented yields the same chance (1/52) at receiving an Ad as the first card dealt to the SB.

Poker being a game using one deck. And each deck being shuffled into a random order.

The random choice of decks is NOT required. They can be sequential because it would not impact that the hand (next hand if you prefer) is played with a randomly order set of 52 individual cards.

It does not matter if the SB receives the Ad as the first card dealt for the hand, next hand or the next dozen hands. Over time, the Ad will be dealt as the first card to the SB very very very close to 1 in every 52 deals (if not exactly 1/52).

Not a problem

There is still a normal distribution of cards that will fall to the SB.

With all due respect King, you are still looking at this from the wrong angle and not ''seeing'' the problem.

Maybe in time myself or somebody else will be able to explain it better than I do now. I understand it is hard to understand something that as not been presented well.
My apologies for the poor explanations, it is not a simple problem to explain. One questions sounds so similar to the other question but they are asking two different things.
It is complex and involves time and quantum leaping over time. But that stuff is just a bit too much for a poker thread.

1) I don't care, it doesn't matter.
2) You cannot say how many A there are.
3) You can predict what the odds are that at least 1 A will appear.
4) You may even be able to predict how many A are likely but you cannot say how many there are but ultimately
5) It doesn't matter because the probability is exactly 100% identical whether it's live or online, 1 deck shuffled 1000 times, or 1000 unique decks

Do you know you are also agreeing with me but do not realise you have ?


Consider your answer I have highlighted. You will start to see the problem more clearly in time.

No.

There is no angle to miss.

There is no problem other than you haven't the math skills to "see" what every one else is saying about your theory.

There is still a normal distribution of cards that will fall to the SB.

There is no problem

No , there isn't a normal distribution of cards to the sb, there is a cumulative distribution to the sb.

P [a] /x = 1/52

P [a] /y = ?/?

No, I don't see the problem.

I see what you are saying you believe is a problem.

I am telling you, it isn't.

You tell me - WHY do you need to know how many aces there are when you've taken 1 card at random from 52 sets of 52 cards?

Can you reword your question, sorry I do not understand what you are asking ,

No, it's an absolutely normal distribution.

The distribution is based on 52 independent draws, not sets of the first card.

For one thing - the person who is SB on each hand is different. The first SB gets card 1 on deck 1, then card 10 on deck 2, 9 on 3, 8 on 4, etc.

But none of it matters because the distribution is based on the cumulative probability across multiple unique instances - each deal being 1 unique instance.

Cumulative probability comes into play on things like flops where you get 3 cards from 1 set and you want to know the chances of the A coming out on the flop, so the odds are 1/50+1/49+1/48.

You do not know this....plus you won't believe me when I tell you this....

There is a normal distribution of the cards. Cumulatively is implied.

The Y you keep referring to has NO VALUE in the game of poker.

There is nothing inherently wrong with the game if the SB receives the Ad ten hands in a row.

There is no problem

There is still a normal distribution of cards that will fall to the SB.

I have told you that, given a set of 52 cards taken from the first card in 52 randomly shuffled decks, you cannot predict how many A will be in that stack.

You have said - that is the problem.

I am saying, WHY is that a problem? Why do you think you can know that? Why does it matter? Why do you think that is relevant to a single hand?

I have no problem if you want to continue with your belief that it is a normal distribution. I know the sb gets card 1 and card 10 , card 10 would also be subject to the y array and be ?/?

It is just easier to show using the first card though.


You are missing the point honestly, so we will just have to agree to disagree.

I'm not missing the point pkdk I'm telling you that I think it's completely irrelevant.

I'm ASKING YOU to tell me why you think the point matters.