1/13.

You keep replying 1/13, this is not the answer, you are incorrect, there is neither 1 or 13, there is 52 unknown, the answer by strict definition is something out of 52.


1.6% chance there is no aces may I remind you, you obviously can not say 1 when there may not be one to start off with.

4/52

{x}Δ{σ²X}={Y}=Δ4=σ²P[A]/{Y}

1/13 is the average of all possibilities and, therefore, is the answer to your original question regarding the chance of drawing an A in the situation you described.

Nothing more to say. Just read this post again after your reply to it.

You can keep saying that and you will be always incorrect. You have not once explained how you are denoting 1 when Y may have no aces, so you obviously are not very good at maths, I will wait until someone with a better understanding of poker physics and dynamics comes along. Thanks for the input but you keep giving the wrong answer.

...

Thanks for the input but you keep giving the wrong answer.

I'm probably going to hate myself for getting into this, but I think there is a potential for confusion here.

Let's say we construct a deck such as in the OP, and draw a card. Then we construct it again, and draw another card. Every time, we reset to the original decks, construct a new deck, and draw a card. Hopefully everyone can agree that after doing this a bunch of times, you will have drawn an A close to 1/13 times.

However, if you construct a deck this way, and draw from it many many times, never reconstructing the deck, the percentage of times you draw an A will not (necessarily) converge to 1/13. If we had to make a bet on the outcome, before seeing any draws, then our best bet would probably still be to guess that it would be 1/13. And indeed if, say, a million people performed this experiment, I would expect the average distribution of aces among the decks to be 1/13.

In any case, this is not a case where the constructed deck represents an unknown distribution, because it's contents are randomly selected from decks of a known distribution, and therefore we can make quantitative statements about the distribution of the new deck, in the form of probability distribution functions.

It would be different if someone constructed a deck for us by hand, looking at all the cards in each deck and picking some and not picking other. This decks probability distribution can not be modelled without more information, because the mechanism for selecting cards is not known.

But the mechanism of random selection IS known and it's quite simple to arrive at mathematical truths regarding it. Spouting a bunch of symbols is not "doing math" or proving anything. I still can't tell if OP is an elaborate troll, a crazy person, or someone who knows just enough math to not understand what he doesn't know (I have known many of the latter)

But there is new information, we know that 52 random variants, (unknown values) have a 1.6% chance there is no ace at all, so to presume 1/13 is no different to presuming 0/13 or 13/13, a guess is not a known function.

f(x) is not equal to f(y), f(Y) is a multifunction/multi-value of f(x).

f=function

triangles=change (physics maths symbols)

σ²=variance in population

4/52

{x}Δ{σ²X}={Y}=Δ4=σ²P[A]/{Y}

OK well, I am done. The thing you did there is not math. It's meaningless garbage.

I take back what I said to you the other day - do not attempt to post this in Poker Theory, under any circumstances. It is not welcome.

P(Y |σ² X)=1


Fact is not theory

http://www.rapidtables.com/math/symb...al_Symbols.htm

4th one down on the list

f (x)=4/52

f (y)=0_52/52

I get it now, you're Dr. Steve Brule!

I do not see how a silly video is going to rule out the maths I have provided.

I am awaiting confirmation from a maths forum of this ,

σ²{X}=Y


which is pretty much an axiom in my opinion.

Look, you had your little run, but you need a new gimmick troll account because this one's used up. Next time see if you can get to 100 posts.

This is not religion it is a discussion, you seem very biased can I ask you why sir?

What concern is this to you personally? You seemingly are taking this personally when I am here to discuss. I have heard often claims of you are a troll, when it is your own ability to fail to continue in the discussion.

It is no shame if the maths is beyond your level of ability of understanding it.

μ{X}=1/52

μ{Y}=?/52

this is very basic maths .

Heehaw is one of the most accomplished probabilists on this site so your comments about him are ludicrous. For this particular problem we don’t actually need his expertise for reasonably simple logic will yield the answer of 1/13.

Let me suggest you do the following. Consider a 4 card deck consisting of 2 black aces and 2 black kings. Assume 4 such decks. Now randomly select a card from each.. You will find the probability of picking an ace from the newly constructed deck is ½. Try it.

Even easier would be a 2 card deck of one ace and one king. You can then easily show what the new 4 card deck distribution will be.

AA ¼
AK ½
KK ¼

The probability of getting an ace from this new deck is 1/4*1 + 1/2*1/2 +1/4*0 = ½. Gee, maybe heehaw and the others are right!

Recycled troll confirmed. Whether he believes his nonsense is debatable.

I will put it really simply for you all



Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.


Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

Pick a card from a known 52 , even thought the card is unknown it has a 1/52 chance to be any card.

do this for 52 times


you now have 52 unknown cards, you knew that each deck had a 1/52 chance of any card,

what is the standard on the new deck ? /52

your new deck <1 or >1 or probability unlikely =1

μ{X}=1/52

μ{Y}={<1.>1.=1}

What are the odds to get a flush on the river assuming you have four to your suit on the turn? Some of your outs may be in the muck!!!

Trolling a troll!

Imagine in a live game, the dealer shuffles the deck, then before he deals he says ''right lads, pick any deck you like from any other table that as just been shuffled''.

Completely changing timely events,

I am no troll.

You have 2704 cards (52x52).
You are choosing one card (it doesn't matter that you take an intermediate step of picking a card from each of the 52 cards, as long as you do it randomly.
There are 208 aces (4x52).
Therefore the chance of the card being an ace is 208/2704... which is the same as 1/13.

SOMETIMES
... you will make a deck with 40 aces (very unlikely) and choose an ace from that deck (very likely)
....you will make a deck with all 2c (very very unlikely) and have no chance of pulling an ace
... etc

But the bottom line is that all of these probabilities add up to 208/2704, because we are blindly and randomly choosing a single item from 2704 items and there are 208 items that meet our criteria.

The chance remains the same if there is new information known, except there is new information known.

{Y}=σ²{X} is known information like you have yourself just stated


''SOMETIMES
... you will make a deck with 40 aces (very unlikely) and choose an ace from that deck (very likely)
....you will make a deck with all 2c (very very unlikely) and have no chance of pulling an ace
... etc''


{X} contains [A]=1/52

μ[A]/{X}=1

{^σ²X}={Y}=σ²[A]=μ[A]{<1.>1.=1}

P[A]/{X}=1/52

P[A]/{Y}}=0_52/52

I like this explanation. Perhaps it is simpler to grasp than the average of all possibilities in the new deck being that there are 4 A's out of the 52 cards in it.

We are effectively just choosing 1 card out of 2704 cards which include 208 A's. We just move 52 of the cards first before choosing the one.

OP's argument is essentially the same as those who believe that moving unknowns from the deck stub to the muck changes the chance to draw a particular card off the deck. They too are often fervent in their mistaken belief.

Referring to burn cards, having no burn cards is still equal to all, although yes it creates more suck outs. But we all have equal chance of the suckouts.