You've bamboozled us with something, but certainly not maths. Whatever it is you're doing isn't math.

pkdk wrote:
Well! the point is there is no accurate way to work out how many aces are in the cards that are in the box.
It is true that we can't know the exact number but we can determine the probability. Isn't that what we are discussing?

The values are unknown and impossible to know,
See above.

In a equivalent question, I post 100 Christmas cards into a box, how many of them are addressed to you?
This is not an equivalent question. We have no information about the population you are drawing these Christmas cards from or the relation of my address to that population.

I can see now that this is the place where we will get stuck for agreement.

In basic principle , we know that there could be no aces, 1 ace, 2, aces, 3 aces, 4aces..........................100 aces in the box.

do you agree with this?
Yes.

P(A)/X=4/52

P(A)/Y=0_100/100
You're going to need to educate me on what these formulas mean. I am not following.

To save time how about I just agree with your maths and say on average there will be 7-8 aces among 100 random cards in the box?
Now we're getting somewhere!

Even this will not solve the problem.
Isn't that what the problem is getting at? I am confused.

Even if you knew the values in the box, it would still not solve the problem.
This makes no sense to me. If you know the population in the box of course you can work out the probabilities.

Added - I do not want to try to bamboozle you with maths , but to express in simplicity is this -

/=in

\=of

∆=change

1/52∆1\52
...
1/52∆1\52

This is the entire process we discussed and have done by placing the 100 top cards in a box and shuffling them,
None of this makes any sense to me, is this your new math? Maybe you could walk me through it?

By shuffling the top cards we un-align the card dependency to its own deck.
This is a very curious statement. I'm not sure how shuffling the cards in the box can change their physical properties. Perhaps you could expand on it. I'm sure it has something to do with quantum mechanics and the changing nature of observed items - kind of a Shrodinger's Cards situation. I would like to know more.
[/QUOTE]

[/QUOTE]


Ok, yes it does go has deep as the world of Quantum, but I think for now we are probably best off avoiding this in the discussion.

Lets us start with explanation of this

In basic principle , we know that there could be no aces, 1 ace, 2, aces, 3 aces, 4aces..........................100 aces in the box.

do you agree with this?
Yes.

P(A)/X=4/52

P(A)/Y=0_100/100
You're going to need to educate me on what these formulas mean. I am not following.


You agreed with the maths with the yes.


P is probability.


(A) is an ace

X is total values/distance of 52

Y is a distance of Y with unknown values


The probability of the top card in any X being any ace is 4/52

The probability of the top card of any Y being any ace is none to all out of 100


Do you understand this first part maths ,thanks?

Sigh...

pkdk - would a Perl script that does the following settle this matter for you?

1. Create a standard deck of 52 playing cards.
2. Draw one card at random and save it in a virtual pile.
3. Repeat the above 2 steps 100 times until we have a pile of 100 cards drawn independently from 100 different decks of cards.
4. Draw a card randomly from the new set of 100 cards (or if you prefer, we can always just draw the top card).
5. Let this run several million times and record how often the drawn card is an Ace, and how often it is a specific Ace.

Everyone but you is 100% certain that we will draw an ace very close to 4/52. And that we will draw the Ace of Spades very close to 1/52.

An interesting idea, but you are not considering the importance of timely intervention. Also I have not finished explaining, the Paradox is a strange one.

Didace is doing very well so far in understanding, and once we have past this maths step, I am sure Didace will understand. Maybe then he may be able to explain better than me,

Just consider 1 in 52 and 1 of 52, that is the Paradox.

So you no longer disagree with the probability then? And if you do, then state what you expect the outcome of the test to be and we'll run it.

Bare in mind I know what we all think will happen etc, we all know what we think is the probability etc, your test will not show the problem.

1 in 52 and 1 of 52, surely you can understand this?



There is no point posting the present information, I know this, we all know this, I am trying to explain possible new information, a Paradox, which means with further investigation it may turn out to be true.

The thread discussion is investigating. The only way you will ever understand the Paradox, is if you listen and discuss it. like the other member is doing, then when he understands what I perceive, If I am wrong he can point out why I am wrong , Until you understand me, how do you know I am wrong if you don't understand what I perceive. ?

No, we don't think this. It established fact how probability works, and we know what the result will be.

You are being patronized for amusement, nothing more. Probability rules are not something complex that need to be "perceived". The math is simple, the axioms are simple, the result is predictable, and your handwaving about paradox and complexity is irrelevant to that.

You understand the Paradox then? You can explain this in this thread what I am saying the Paradox is?

You are saying something is wrong when you do not even know what the something is?


Please explain the Paradox back to me and explain where the Paradox fails.

What is the difference between "in" and "of" in this context? I think that if you could give a detailed explanation of that it would go a long way towards understanding.

Also, isn't "I just agree with your maths and say on average there will be 7-8 aces among 100 random cards in the box" the very definition of probability? It looks like it's case closed - no paradox, no disagreement. I'm glad you came around.

:
Originally Posted by Didace You’ve been pointed out wrong a number of times. You conveniently ignore those postings or respond with a bunch of nonsense that noone can understand. The above is an example where you answered that if you select a card from a randomly selected deck out of 52 decks the probability the card is an ace is 1/52. This is obviously wrong as explained above. BTW - as was pointed out very early, this question is exactly equivalent to constructing your new deck from 52 decks and then picking a card.

I have a bucket , in this bucket are 52 balls each numbered 1-52


each different value ball is 1 ball in the bucket , 1 in 52. 1/52

I take one ball from the bucket, the chance of any value is 1/52 dependent to the balls in the bucket. The ball remains one in 52 because we keep the dependency related to the bucket.


If we do not reveal the value, the ball is an unknown value, 1 of 52. removing the dependency of the original bucket.


So we place this ball in a separate bucket. 1 of 52


We take a second bucket that also has 52 balls of 52 numbers in it, we repeat this, placing 1 of 52 in the new bucket,


We do this for 52 times in total , until we have 52 balls that are all 1 of 52 in the new bucket.

draw a ball out of the new bucket, the chance is 1 of 52 rather than 1 in 52.

Say you draw number 4, the chance the next ball is also a 4 is 1 of 52.


Which can be expressed has this simple formula

/=in

\=of

∆=change

bucket 1 ........... 1/52 ∆ 1\52...new bucket
bucket 2 ........... 1/52 ∆ 1\52...new bucket
bucket 3 ........... 1/52 ∆ 1\52...new bucket

.

Expressing the situation of something that is or appears to be enclosed or surrounded by something else changing by motion of the variant to another place expressing the relationship between a scale or measure and a value.

In and of.


1 of 52 cards is not the same as 1 in 52 different cards.

1 of because the value is not known

1 in when the value is known.

In this case, a distinction without difference.

We may not know the exact values, but we know the probabilities.

Earlier on you agreed there could be no aces, there could be 100 aces, there is no constant set amount, so to say 7-8 would be an utter outright lie.


The only information we have is the card is 1 of 52 cards,

I draw a number out , lets say 6, you draw a number out , you have a 1 of 52 chance that is also a 6.

I draw a third ball out, that also has a 1 of 52 chance to be a 6.


The first card has two odds running simultaneous, 1 in 52 and 1 of 52, that is the paradox.


Before you try to rule this out , consider the earlier agreement , any card in the 100 cards has an equal chance of 1/52 to be the same card.


Consider this grid

XXXXX
XXXXX
XXXXX
XXXXX
XXXXX


The red X is your choice alignment, we will call this the Y-axis.
If you was on the left of the x's playing a single X row . you have a 1 in 5 choice, if you stand under the X's and play the Y-axis column choices, you have a 1 of 5 choice.


to reveal the content

12345
12345
12345
12345
12345

I sincerely doubt whether this will help anyone, but let's return to the original problem of a single card being pulled randomly from 52 standard shuffled decks to form a new (non-standard) 52-card deck.

It is straightforward "math" to show that the number of Aces in the new deck will take on the following probabilities:

Prob(X aces in new deck) = C(52,X) * ((4/52)^X) * ((48/52)^(52-X) for X=0,1,2,...,52, where C(Y,Z) is the number of ways to choose Z items out of a total of Y items (combinations), which is easily seen to be a standard binomial distribution with N=52 and p=4/52.

Prob(0 aces in new deck) = 0.01557
Prob(1 aces in new deck) = 0.06749
Prob(2 aces in new deck) = 0.14340
Prob(3 aces in new deck) = 0.19917
Prob(4 aces in new deck) = 0.20332
Prob(5 aces in new deck) = 0.16265

Prob(6 aces in new deck) = 0.10618
Prob(7 aces in new deck) = 0.05814
Prob(8 aces in new deck) = 0.02726
Prob(9 aces in new deck) = 0.01110
Prob(10 aces in new deck) = 0.00398

Prob(11 aces in new deck) = 1.266E-3
Prob(12 aces in new deck) = 3.605E-4
Prob(13 aces in new deck) = 9.243E-5
Prob(14 aces in new deck) = 2.146E-5
Prob(15 aces in new deck) = 4.530E-6

Prob(16 aces in new deck) = 8.729E-7
Prob(17 aces in new deck) = 1.540E-7
Prob(18 aces in new deck) = 2.496E-8
Prob(19 aces in new deck) = 3.722E-9
Prob(20 aces in new deck) = 5.118E-10

Prob(21 aces in new deck) = 6.499E-11
Prob(22 aces in new deck) = 7.631E-12
Prob(23 aces in new deck) = 8.295E-13
Prob(24 aces in new deck) = 8.353E-14
Prob(25 aces in new deck) = 7.796E-15

Prob(26 aces in new deck) = 6.746E-16
Prob(27 aces in new deck) = 5.414E-17
Prob(28 aces in new deck) = 4.028E-18
Prob(29 aces in new deck) = 2.778E-19
Prob(30 aces in new deck) = 1.775E-20

Prob(31 aces in new deck) = 1.050E-21
Prob(32 aces in new deck) = 5.740E-23
Prob(33 aces in new deck) = 2.899E-24
Prob(34 aces in new deck) = 1.350E-25
Prob(35 aces in new deck) = 5.786E-27

Prob(36 aces in new deck) = 2.277E-28
Prob(37 aces in new deck) = 8.205E-30
Prob(38 aces in new deck) = 2.699E-31
Prob(39 aces in new deck) = 8.074E-33
Prob(40 aces in new deck) = 2.187E-34

Prob(41 aces in new deck) = 5.333E-36
Prob(42 aces in new deck) = 1.164E-37
Prob(43 aces in new deck) = 2.256E-39
Prob(44 aces in new deck) = 3.845E-41
Prob(45 aces in new deck) = 5.697E-43

Prob(46 aces in new deck) = 7.224E-45
Prob(47 aces in new deck) = 7.685E-47
Prob(48 aces in new deck) = 6.671E-49
Prob(49 aces in new deck) = 4.538E-51
Prob(50 aces in new deck) = 2.269E-53

Prob(51 aces in new deck) = 7.415E-56
Prob(52 aces in new deck) = 1.188E-58

It is straightforward to show that the expected value of any binomial distribution is equal to N*p. Thus, in our case above, the "expected" number of Aces in the newly-created 52-card deck is 52*(4/52)=4.

Even more simply, the prob of any of the cards in the newly-created 52-card deck being an Ace is simply p=4/52.

pkdk - You're not too bright, are you. Sorry for wasting your time.

The probability of an ace from the new deck is 0_52/52 all of the below

Prob(0 aces in new deck) = 0.01557
Prob(1 aces in new deck) = 0.06749
Prob(2 aces in new deck) = 0.14340
Prob(3 aces in new deck) = 0.19917
Prob(4 aces in new deck) = 0.20332
Prob(5 aces in new deck) = 0.16265

Prob(6 aces in new deck) = 0.10618
Prob(7 aces in new deck) = 0.05814
Prob(8 aces in new deck) = 0.02726
Prob(9 aces in new deck) = 0.01110
Prob(10 aces in new deck) = 0.00398

Prob(11 aces in new deck) = 1.266E-3
Prob(12 aces in new deck) = 3.605E-4
Prob(13 aces in new deck) = 9.243E-5
Prob(14 aces in new deck) = 2.146E-5
Prob(15 aces in new deck) = 4.530E-6

Prob(16 aces in new deck) = 8.729E-7
Prob(17 aces in new deck) = 1.540E-7
Prob(18 aces in new deck) = 2.496E-8
Prob(19 aces in new deck) = 3.722E-9
Prob(20 aces in new deck) = 5.118E-10

Prob(21 aces in new deck) = 6.499E-11
Prob(22 aces in new deck) = 7.631E-12
Prob(23 aces in new deck) = 8.295E-13
Prob(24 aces in new deck) = 8.353E-14
Prob(25 aces in new deck) = 7.796E-15

Prob(26 aces in new deck) = 6.746E-16
Prob(27 aces in new deck) = 5.414E-17
Prob(28 aces in new deck) = 4.028E-18
Prob(29 aces in new deck) = 2.778E-19
Prob(30 aces in new deck) = 1.775E-20

Prob(31 aces in new deck) = 1.050E-21
Prob(32 aces in new deck) = 5.740E-23
Prob(33 aces in new deck) = 2.899E-24
Prob(34 aces in new deck) = 1.350E-25
Prob(35 aces in new deck) = 5.786E-27

Prob(36 aces in new deck) = 2.277E-28
Prob(37 aces in new deck) = 8.205E-30
Prob(38 aces in new deck) = 2.699E-31
Prob(39 aces in new deck) = 8.074E-33
Prob(40 aces in new deck) = 2.187E-34

Prob(41 aces in new deck) = 5.333E-36
Prob(42 aces in new deck) = 1.164E-37
Prob(43 aces in new deck) = 2.256E-39
Prob(44 aces in new deck) = 3.845E-41
Prob(45 aces in new deck) = 5.697E-43

Prob(46 aces in new deck) = 7.224E-45
Prob(47 aces in new deck) = 7.685E-47
Prob(48 aces in new deck) = 6.671E-49
Prob(49 aces in new deck) = 4.538E-51
Prob(50 aces in new deck) = 2.269E-53

Prob(51 aces in new deck) = 7.415E-56
Prob(52 aces in new deck) = 1.188E-58

A perfect example of what I mean:

So you have pulled out of the discussion now the diagrams and maths is coming out to show I am correct?

it doesn't make it none sense just because you can not understand it. If it is beyond you in discussion, don;t discuss.

Consider this grid

XXXXX
XXXXX
XXXXX
XXXXX
XXXXX


The red X is your choice alignment, we will call this the Y-axis.
If you was on the left of the x's playing a single X row . you have a 1 in 5 choice, if you stand under the X's and play the Y-axis column choices, you have a 1 of 5 choice.

X≠Y

Y=σ²{X}


to reveal the content

p12345
p12345
p12345
p12345
p12345
..ppppp
I do not feel it takes any maths to spot the problem with this situation,p is player

My maths does not lie either.

This is one of the clues. Nothing is showing you are correct. In fact, everything is showing you are wrong.

Do you not understand maths?

The maths formula and diagram in the last post shows I am correct. Can you argue the simple formula which is accurate and true?

The maths does not lie.


f{Y}=σ²f{X}