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12-02-2015 , 12:48 PM
Quote:
Originally Posted by pkdk
You continue to post and re-explain what we already know
You don't get to include yourself in "we". You don't know basic probability, which is why you need to watch that Khan series. If Khan can't make you see the light then there's no hope for you and math just isn't for you.
A strange question of values.
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A strange question of values.
12-02-2015 , 01:55 PM
Quote:
Originally Posted by heehaww
You don't get to include yourself in "we". You don't know basic probability, which is why you need to watch that Khan series. If Khan can't make you see the light then there's no hope for you and math just isn't for you.
Say that after I have proved it, I will prove it, I am on the correct path, I am on a maths forum and science forums, I will show this

''In statistics, the autocorrelation of a random process describes the correlation between values of the process at different times, as a function of the two times or of the time lag. Let X be some repeatable process, and i be some point in time after the start of that process. (i may be an integer for a discrete-time process or a real number for a continuous-time process.) Then Xi is the value (or realization) produced by a given run of the process at time i. Suppose that the process is further known to have defined values for mean μi and variance σi2 for all times i. Then the definition of the autocorrelation between times s and t is''
12-02-2015 , 02:53 PM
Moderator (I suppose we have one). This thread gets more nonsensical by the hour. If OP is serious then the nonsense he posts can only warp unsuspecting newbies’ thinking about basic probability. Actually,though, I think OP is toying with us for it is clear that his posts are so out of the realm of logic and reality, it’s hard to believe he is serious.

In any event, I hope you can see it is in everybody’s best interest to close this thread before it further denigrates the usefulness of the Probability Forum.
12-02-2015 , 03:00 PM
Quote:
Originally Posted by Lego05
Why do you always capitalize the letter "p" in the word "paradox"?
ANSWER THE QUESTION!!! This might explain it all!
12-02-2015 , 03:07 PM
Quote:
Originally Posted by statmanhal
Moderator (I suppose we have one). This thread gets more nonsensical by the hour. If OP is serious then the nonsense he posts can only warp unsuspecting newbies’ thinking about basic probability. Actually,though, I think OP is toying with us for it is clear that his posts are so out of the realm of logic and reality, it’s hard to believe he is serious.

In any event, I hope you can see it is in everybody’s best interest to close this thread before it further denigrates the usefulness of the Probability Forum.
What are you so concerned about? I thought I was so wrong it was not a worry!



Y't2=σXt1λ




Co-variance and cross correlation of time.

Newbies will learn a whole lot about probabilities in this one thread.

Last edited by pkdk; 12-02-2015 at 03:27 PM.
12-03-2015 , 06:57 AM
Hello all, I now believe I have a completed the formula to show the problem and a differential formula that highlights the paradox. A paradox that I am now believing with further investigation to be a new set theory rather than a paradox.

λ(X)=φt1=ab0294b94eb253246e368db2032b6e2e.gif

λ(Y)=σab0294b94eb253246e368db2032b6e2e.gif=t2

sorry this site can not do the frac maths symbols.
12-03-2015 , 12:19 PM
Why are you no longer capitalizing the letter "p" in the word "paradox"?
12-03-2015 , 01:01 PM
Quote:
Originally Posted by statmanhal
Moderator (I suppose we have one). This thread gets more nonsensical by the hour. If OP is serious then the nonsense he posts can only warp unsuspecting newbies’ thinking about basic probability. Actually,though, I think OP is toying with us for it is clear that his posts are so out of the realm of logic and reality, it’s hard to believe he is serious.

In any event, I hope you can see it is in everybody’s best interest to close this thread before it further denigrates the usefulness of the Probability Forum.
Don't bother with closing the thread, just ban OP and be done with him.
12-03-2015 , 02:29 PM
Quote:
Originally Posted by R Gibert
Don't bother with closing the thread, just ban OP and be done with him.
You want to ban me for discussing probability and a possible paradox?

I guess you are one of these internet bingo type internet players. ..


The software of this sight will not allow me to post the relevant maths.

You can find it here -

https://en.wikipedia.org/wiki/Cross-correlation

Under time series analysis.
12-03-2015 , 02:33 PM
Why do you not address the issue of the capitalization or non-capitalization of the letter "p" in the word "paradox"?
12-03-2015 , 02:36 PM
Quote:
Originally Posted by Lego05
Why do you not address the issue of the capitalization or non-capitalization of the letter "p" in the word "paradox"?
P or p is hardly a problem, I changed it back to p to give you nothing to flame at.
12-03-2015 , 02:39 PM
I ask you to calculate the chance of only 4 aces being in 52 random cards from 52 shuffled decks..
12-03-2015 , 02:41 PM
Quote:
Originally Posted by pkdk
P or p is hardly a problem, I changed it back to p to give you nothing to flame at.
You cannot have changed it back to the lower case "p" because consistently and from the very beginning you were spelling it with the upper case "P". I don't know why you always spelled "paradox" with a capital "P" and I don't know why you have all of a sudden switched to spelling it with a lower case "p". It seems to be a mystery that you mostly will not expand upon with any sort of explanation.
12-03-2015 , 02:43 PM
Quote:
Originally Posted by Lego05
You cannot have changed it back to the lower case "p" because consistently and from the very beginning you were spelling it with the upper case "P". I don't know why you always spelled "paradox" with a capital "P" and I don't know why you have all of a sudden switched to spelling it with a lower case "p". It seems to be a mystery that you mostly will not expand upon with any sort of explanation.
When you mentioned it, I realised I was making a paradox a noun by putting a capital letter.

Good enough?

Or I considered it was no longer a paradox after further investigation, so took away the emphasise. Meaning it is now theory and not a paradox at all.

Last edited by pkdk; 12-03-2015 at 02:49 PM.
12-03-2015 , 02:59 PM
Quote:
Originally Posted by pkdk
When you mentioned it, I realised I was making a paradox a noun by putting a capital letter.

Good enough?

Or I considered it was no longer a paradox after further investigation, so took away the emphasise. Meaning it is now theory and not a paradox at all.

The word "paradox" is a noun. Capitalization of the letter "p" has nothing to do with that.

Maybe this issue of capitalization is the paradox and has been all along.
12-03-2015 , 03:23 PM
Quote:
Originally Posted by Lego05
The word "paradox" is a noun. Capitalization of the letter "p" has nothing to do with that.

Maybe this issue of capitalization is the paradox and has been all along.
Well it would take someone smarter than Einstein to have capitalized the ''p'' in Paradox, also there would be no real purpose in doing so.
12-03-2015 , 03:27 PM
Quote:
Originally Posted by pkdk
Well it would take someone smarter than Einstein to have capitalized the ''p'' in Paradox, also there would be no real purpose in doing so.
Hmmm ...... no real Purpose you say ....


Interesting.
12-03-2015 , 03:31 PM
Quote:
Originally Posted by Lego05
Hmmm ...... no real Purpose you say ....


Interesting.
Presuming we are talking about the same thing, you cant random time an oil rig by random waves at sea. The oil companies get their money either way, and the roulette lucky timing seamen get the rewards and filter into the main live feed disrupting the entire process of champions. Give a donkey the means for support, and your fighting the donkey in a live fight.
12-03-2015 , 06:55 PM
Quote:
Originally Posted by pkdk
I ask you to calculate the chance of only 4 aces being in 52 random cards from 52 shuffled decks..
Having 4 in there is but one conditional probability of many, that sum up to the unconditional probability of drawing an ace from that new deck. Which is still 4/52 in every universe but yours.


Why is it so hard for us to stop? This is like pulling your eyes away from a train wreck.
12-03-2015 , 07:38 PM
Quote:
Originally Posted by NewOldGuy
Why is it so hard for us to stop? This is like pulling your eyes away from a train wreck.
It's a Paradox!
12-03-2015 , 08:07 PM
Quote:
Originally Posted by NewOldGuy
Having 4 in there is but one conditional probability of many, that sum up to the unconditional probability of drawing an ace from that new deck. Which is still 4/52 in every universe but yours.


Why is it so hard for us to stop? This is like pulling your eyes away from a train wreck.
It's a Paradox, I am awaiting for confirmation of this maths, but the chance of 52 top cards (y), from 52 shuffled decks having 4 aces in them is only 20.3%

added confirmation of maths from science forum moderator

''These equations look good to me''

Last edited by pkdk; 12-03-2015 at 08:35 PM.
12-03-2015 , 08:38 PM
20.3% seems way too high to me if you mean exactly 4.
12-03-2015 , 08:52 PM
Quote:
Originally Posted by Lego05
20.3% seems way too high to me if you mean exactly 4.
The exact number to the maths is this P(4)=0.203317


The maths was beyond me and I had help. Then he later replied 20.3%.


Because 100% is 1 in probability, and I do not think this means 4 different suits either, just any 4 aces.

Then I had a second forum confirm the maths.
12-03-2015 , 09:00 PM
Quote:
Originally Posted by pkdk
The exact number to the maths is this P(4)=0.203317


The maths was beyond me and I had help. Then he later replied 20.3%.


Because 100% is 1 in probability, and I do not think this means 4 different suits either, just any 4 aces.
Yes, we already had this.

Quote:
Originally Posted by whosnext
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.
12-03-2015 , 09:00 PM
Guess I was wrong; it wasn't too high.
A strange question of values.
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A strange question of values.

      
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