yes of course you know that bit

well actually you dont know because you think it is 4/52 or a 1/4/52

Sorry for my intrusion, I have a question for you pkdk.

Simplifying a bit the problem, what do you think about considering only red cards or black cards?
I guess the answer is 26/52 and that the new deck is just the 52 step composition of perfectly 50/50 random events.

Then?

Thanks in advance.

as.

Hmmm ...

Well surely:

Chance of card being Ad = Chance of card being Ad


But if we add the Ac, As and Ah to one side of the above equation, then wouldn't such equation no longer be true? After all we have already agreed that there is a greater than 0 chance the card is the Ac; a greater than 0 chance the card is the As; and a greater than 0 chance the card is the Ah. So, since these values are not 0 and the above equation is already true in stating that the terms on both sides of equal sign are equal, then adding these values to only one side of the equation must make the equation no longer true. Correct?

So the below equation must be wrong. Do you agree?

Chance of card being Ad + Chance of card being Ac + Chance of card being Ah + Chance of card being As = Chance of card being Ad

suits would be 1/4 any card, when you draw a card from a deck you have 1/4 chance of suit, so if you place this card on the counter it remains 1/4 to be any suit. But also remains 1/4 to be the same suit.

I am not sure what you are saying, but there is a 1.6% chance there is no aces on your counter at all. There is a 0%, and this is where it differs from what you are asking.


Your questions are based on the present information, you have not gone into the idea.

The card you have your hand on , has a 1/52 chance of being the Ah, move your hand to another card, this card has a 1/52 chance of being the Ah, look at the card, lets say its the 3 clubs, move your hand to the next card, the next card has a 1/52 chance of being the 3 clubs.

.Ac .......Ad.......As.......Ah......Any A...Any A Simplified
1/52 + 1/52 + 1/52 + 1/52 = 4/52 = .....1/13

Hope you enjoyed your day.





EDIT:

Everyone, enjoy.

Ok, but disregarding the suit and considering only black and red cards, our new deck will follow a binomial distribution.
Of course it will be very unlikely to get a new deck containing exactly 26 red cards and 26 black cards as many other deviated situations, still I cannot understand the point of it.

as.

There is no adding up to equal 4/52 , each card is independent to the deck it originally come from, remaining 1/52. there is not 4 aces of different suits on your counter.

The point is of a timely nature , and of repeat values in periods of time, and clusters of cards. Very complex to understand if I am honest.

.Ac .......Ad.......As.......Ah......Any A...Any A Simplified
1/52 + 1/52 + 1/52 + 1/52 = 4/52 = .....1/13


See ya .... long weekend starts .... well after sitting in traffic for a while probably.



Happy Thanksgiving to all who care. Consider getting drunk and then reading this thread.

No you still have it wrong, there is no 4/52, that you have done is a standard deck.

1/52=1/52=1/52=1/52=1/52............=1/52


for 52 times.

X²=X,Y=52*52=2704²


P(A)/{X}=4/52

P(A)/{Y}=1/52 even if Y=100000000000000000000000000000000000000000000000 000000000000000000000000000000000000000

That's a good answer.

as.

Thank you, the simplicity of it is this,

Pick a card from a standard deck, we know there is 4 aces in the deck so we can say the chance is 4/52 known.

Pick a random card from a selection, we do not know how many aces there is or any values, the card is just one card of 52 cards, so we can say the odds are 1/52 unknown, we can't say 4/52.

Please answer these questions in order with your reasoning for each answer. Assume all decks are standard 52 cards and are randomly shuffled.

1. A deck is sitting on a table in front of you. You slide the top card off the deck and place it on the table. What is the chance this card is an ace?

2. There are 52 decks sitting on a table in front of you. You randomly choose one of the decks and slide the top card off the deck and place it on the table. What is the chance this card is an ace?

3. There are 52 decks sitting on a table in front of you. You slide the top card off all of the decks and place them in front of the decks on the table. You randomly choose one of these cards. What is the chance this card is an ace?

4. There are 52 decks sitting on a table in front of you. You slide the top card off all of the decks and place them in front of the decks on the table. You combine these cards into one pile and shuffle it (assume the pile is now random). You slide the top card off the deck and place it on the table. What is the chance this card is an ace?

Thank you for your consideration.

Since this is the only thread that is still "going", here I wish everyone a Happy Thanksgiving!

(Of course, that is the best contribution I can make at this time to this train wreck of a thread.)

You will not understand how the answer to question 1 can change by question 2, that is the paradox....

question 1, 4 in 52 known

question 2, 1 of 52 unknown only the top cards are relevant.

In the first instance you offer me the top card of a single deck, 1 in a known 52 values, in the second question you offer me a choice of 52 random top cards, 1 of 52.


Now if for question 2 you had said , pick any deck and pick any card from within the deck, it would be 4/52


question 2,3,4 ask the same thing, question one is a different game.

To express in scientific terms, there is two events rather than one.

Event (A) pick a deck/top card, your question 2


Event (B) receive the top card, your question 1


In a normal game of live poker, event (B) is event (A) and event (B) does not exist.

OP - you answered 1/52. That means you would have answered 1/52 for a king, and 1/52 for a queen, and ... and 1/52 for a 3 and 1/52 for a 2. Since these events are mutually exclusive, that means the probability the top card is either and ace or a king or a ... or a 3 or a 2 is 1/52 + 1/52 + ... +1/52 = 13/52.

That of course is impossible for the probability you select one of the 13 ranks has to be 1.0.

I can't believe we're still responding to this guy.

A trivial simulation or live test will quickly disprove this idiocy. I think you know this and are just seeing how long people will bite.

Even without a simulation, plain axioms about set properties and probability prove you wrong with no math required.

It's a paradox!

Is it possible that nobody has already explained it thusly:

Number the decks from 1 to 52. Number small boxes from 1 to 52. Select one card from deck 1 and put it in box 1. Same for deck 2 and box 2. And so on.

What are the chances that the card in, say Box 15 is an ace? Obviously its 1/13/Since it is a randomly selected card from deck 15. But this would be true for any of those 52 boxes. So the answer is always going to be 1/13.

Both the answer and the explanation is so obvious it makes me wonder if I misunderstood the question or missed someone else answering it this way.

Thank you to the 2 members who have now understood. Yes it is a Paradox.