Hello all, I will try to make this clear and understandable as I can.


If we have 52 decks of cards

from each deck we randomly pick an unknown card


we then have 52 random cards, (a new deck)


what is the chance? of an ace from this new deck not knowing any of the values in the new deck, i.e there could be more or less than four aces.


?/52

X=52

(A)=?

I get this P(A)/X=0_1

Which means, the chance of an ace from the new deck are minimum 0% and a limit of 100%

It's 1/13 because it's the same as picking a card from a deck of 52^2 cards with 4*52 aces. Since the ratio of aces to other ranks hasn't changed (compared to a single deck), the probability is the same.

Edit: You can say that about any probability question (but I guess your point was that you couldn't rule any answer out).

It can't be the same, 1/13 comes from knowing the value of 4/52, in this scenario we do not know the values in the new deck, so how can you denote a value to something that is absolute random?


In comparison , I have a deck of xmas cards, how many aces is in the deck?

You do not know the values, 1 may not exist, the only value we are certain of is 52, so where do you get your 1/13 from? there is neither 1 known or 13, we have 52 unknown ''xmas cards''.

Because it's analogous to the 52^2 card deck, whose values we do know (namely, 4*52 of each rank).

Look at it this way. By picking a random card from the newly constructed deck, all you're doing is picking a card from a random one of those 52 decks. Beforehand, if you were to pick from one of the 52 decks, then no matter which deck you picked from your chance would be 1/13. Making the new deck simply randomizes which deck you pick from. The fact that you're picking from a random one of those decks doesn't change the probability.

But if you still don't believe that reasoning, you can solve it the long way:

P(draw ace from new deck) = P(new deck has 1 ace)/52 + P(has 2 aces)*2/52 + ... + P(has 52 aces)

P(new deck has x aces) = (1/13)^x * (12/13)^(52-x)

Edit: hm I guess there's a flaw in my analogy, because Wolfram says the sum is this, which is not 1/13.

No , I am sorry the power of 2 does not come into use, you are picking from a new deck of unknown variants, absolute random, 4/52 is not random, we know the values because we looked before hand.

Try it this way,

I have 52 pieces of a4 paper, I am going to randomly colour each piece in a single colour using 4 colours, red,blue, green, orange. how many of the papers are orange?

P(new deck has 0 aces upto 52 aces,


quote a science forum member.

''Does this help:

If you draw a card from 52 decks to make a new deck then ask a friend to look through it and count how many aces there are, the 4/52 odds will change the instant he tells you the new number.

If he says there are five, your odds are now 5/52.

Your odds depend on what you know.''

Fmp, with the bolded it's now correct. I messed up on binomial distribution, how embarrassing!

And now, as expected, it sums to 1/13.

I'll try once more to explain my analogy. You have decks 1 to 52. From each deck you pick a random card to form the new deck. Suppose the card from Deck #1 is the top card in the new deck, and the one from Deck #2 is the 2nd card, and so on. So then:
top card = random card from Deck #1
2nd card = random card from Deck #2
...
No matter which card you choose, you're picking a random card from one deck. Now if we shuffle this new deck, maybe the card from Deck #43 is on the top. So what, it's still a random card from one full deck. Now suppose we shuffle it and don't know which deck's card is on top. So what, all we're doing is randomly deciding which single deck to randomly pick from. This is no different than lining up all 52 decks at the beginning, then randomly picking which deck to draw from.

Ok, bare in mind I do already know your views on this, I know how it is presently thought. However the odds of picking 52 random cards from 52 decks, and the new deck being the same in exact values as an original deck, the odds are near impossible , almost negligible, so the odds of this tells us the odds of the new deck change, without looking.

i.e it is possible the new deck is 100% the 2 of clubs.

True, but it all averages out. Similar to how, if you have a flush draw on the flop in 9max, and it's a heads-up pot, you don't worry about what the 7 folded players have. You count yourself as having 9 outs despite the fact that most likely some of those players have hearts. That's because if you calculate it the long way, it averages out to 9/47.

Yes, but that possibility is less likely than the possibility of the new deck having, say, 4 aces. All the possibilities average out.

I already showed the long formula, so you don't have to believe my analogy, just believe my math which is very simple. The chance the deck has x # of aces is just a binomial distribution. Then if the deck has x # of aces, the chance of picking an ace from it is x/52. So I summed up all of the possibilities from x=1 to x=52 (except on the wolframalpha page I used k instead of x).

I understand you think the maths works, but it doesn't , it fails in many aspects. Strict definition flaws the maths ,

try it this way and look at it in this context, imagine 100 random cards,


we have no idea how many aces are in those 100 cards, we have a value of 100 , so we can certainly say ?/100.

We could not say 4/52, by strict definition the values do not match to ?/100.

Showing this falseness of the maths in itself.


We know the odds of the new deck being the same in values as an original deck is almost impossible and negligible ,


Ok I feel I may have to go over to a roulette wheel explanation,

Imagine a 52 number roulette wheel , each number a different variant,

can we both agree we have a 1/52 chance of any number?

This belongs in the Science, Math and Probability forum. I'm too lazy to do all the math, but you have a 1.6% chance that the new deck will have no aces in it.

I assure you it works. This is a basic AP Stats level problem. (But what's fun about this problem is that it can be figured out without math.)

You keep reiterating that we don't know the composition of the new deck. But that's precisely why it's 4/52. If we knew that the composition were, say, 5 aces, then the probability would be 5/52. When we don't know the composition, we have to take the average one, and the average is a deck with 4 aces.
E(aces) = sum of k * C(52,k) * (1/13)^k * (12/13)^(52-k) from k=1 to 52
= 4

Earlier I showed a sum that took it one step further and gave the weighted average of all the probabilities that could occur (after being told the makeup of the deck). The probability without knowing the makeup is equal to the average known-makeup probability.

Again, see my flush draw example. If you knew how many of your suit was already taken by other players, you'd know you don't have 9 outs. But since you don't know, you have to take the average, which happens to be 9 outs. (Of course, with that example too, there is an easy math-free way of knowing that it's 9/47.)

Yes , but it is also poker related and belongs here too, who knows about poker than poker players?


And thank you for the 1.6% which I did not know.

can you explain where 1.6% comes from please?

I assure you , you only think it works, by strict definition it fails.


a 52 card deck as 0% chance there is no aces in it.

x=52

σ²(X)=y

the possibility Y has no aces is 1.6%

x=constant=4/52

Y=variate of x

4/52

52(x)Δ52(y)=Δ4=σ²(x,y)

(12/13)^52 = 1.557% or about 1.6%

Why ask the question if you think you already know better than the people answering? And if you know better, then why don't you solve it yourself? What formula would you use if not the one I showed?

You don't believe me then go ahead and write a simulation in the programming language of your choice. Or if you don't know a language then do the experiment with 52 decks. See what happens. I'd be willing to bet everything I own that about 1/13 of the time you'll draw an ace (and that it will get closer to 1/13 the more times you run the experiment).

Why ask the question? because it is broken!

I would use this formula,

P(A)/ X= 1/52

P(A)/Y=σ²(x)=0_1

You probably don't need to waste your time trying to further explain.

http://forumserver.twoplustwo.com/15...proof-1525137/

http://forumserver.twoplustwo.com/32...proof-1525138/

http://forumserver.twoplustwo.com/15...ation-1556539/

http://forumserver.twoplustwo.com/15...fails-1571909/





I wonder if this is the guy who went on and on about how there was extra "variance" or whatever in online poker because they didn't have one "deck" that they they dealt from and instead every table would randomly get a new "deck" every hand.




EDIT:

But I'm going to go ahead and move it to the probability forum.

Take 52 random cards, from 52 random decks, place them on a roulette wheel and spin the wheel, what is the chance of intercepting an ace?

each card as a 4/52 chance of being an ace, but what is the chance of intercepting a 4/52 chance that is an ace?


We know that the 52 cards on the wheel could have no aces, a 1.6% chance. we know all the cards could be aces as they have equal chance.

surely there is new odds.

Assuming that all 52 random decks are standard playing card decks, we are taking 1 random card from each random deck, we haven't looked at any of the cards that were selected, the roulette wheel has 52 spaces and each is taken up by one of the cards, the cards are affixed to a spot on the wheel so they won't fly off or move while it spins and the spin of the roulette wheel is random, 1/13.

So you can explain how you are turning 52 into 13 and a range of zero to 52 being changed into 1. That is incorrect and not according to strict definition.


52 cannot become 13, and 0-52 can not become 1 when the values are ?/52


1/13 is base don knowing 4/52, you have no prior knowledge to what cards are on the wheel.

If there is no aces, a 1.6% chance, then where do you get 1 from?

If there is 52 aces, where do you get 1 from?

Whoops, yes now I remember OP, silly me for engaging/feeding.

Theories improve with time, I can not see how it is possible to deny the maths I have provided, I feel it is on you to disprove my functional maths of strict definition.

The average of all possibilities.

There are no A's 1.6% of the time (assuming that math was correct; I did not check it). Also, a certain percentage of the time the deck is all A's. A certain percentage of the time the deck has every number of A's possible, which is 0 through 52, inclusive.



The above includes the correction from the original quoted post that was made in the following post:

http://forumserver.twoplustwo.com/sh...69&postcount=6




I'm pretty sure that this isn't an actual term and that you made it up and further that you did not provide a definition for this term you made up.

5000 years from now, the answer will still be 1/13, and 1+1 will still be 2, and the quadratic formula will still be true. This isn't Physics, where theories are never quite proven and someone like Einstein can come along and prove F=ma wrong. Once something in math is proven (such as high school probability concepts), it can never be disproven. That's the point of a proof.

Yes exactly so you understand that the P(A)/x is different to the P(A)/y


where x is a standard 52 deck and Y is a 52 variate of x and σ² is a variance in the population values.

σ²(X)=y

denoted by this simple formula.

I beg to differ, I think you will find in time if this thread remains open that I am correct. you only just do not understand it yet.