Anybody good at maths?

    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Is there anybody on here who knows how to do this equation?

      Time series analysis[edit]
      In time series analysis, as applied in statistics and signal processing, the cross-correlation between two time series describes the normalized cross-covariance function.[clarification needed]

      Let (X_t,Y_t) represent a pair of stochastic processes that are jointly wide-sense stationary. Then the cross-covariance and the cross-correlation are given by

      cross-covariance \gamma_{XY}(\tau) = \operatorname{E}[(X_t - \mu_X)(Y_{t+\tau} - \mu_Y)],
      cross-correlation \rho_{XY}(\tau) = \operatorname{E}[ (X_t-\mu_X)\,(Y_{t+\tau}-\mu_Y)]/(\sigma_{X} \sigma_{Y}),


      https://en.wikipedia.org/wiki/Cross-correlation
  • 64 replies
    • metalmonkey80
      metalmonkey80
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      Hello shatteredaces,

      I had a quick look on the net and found a good reference that may be some help on dell computers text book section. If you require a link pm me.
      Sadly I am not good at math but I hope this may be of some help.

      Regards

      Matt
    • shatteredaces
      shatteredaces
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      Joined: 15.04.2015 Posts: 141
      Originally posted by metalmonkey80
      Hello shatteredaces,

      I had a quick look on the net and found a good reference that may be some help on dell computers text book section. If you require a link pm me.
      Sadly I am not good at math but I hope this may be of some help.

      Regards

      Matt
      Thank you for your reply, I am struggling to find private messages, could you possibly post the link in this thread?
    • shatteredaces
      shatteredaces
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      Joined: 15.04.2015 Posts: 141
      P.s this maths shows online poker is broken....
    • DarkNeo1
      DarkNeo1
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      Joined: 17.02.2008 Posts: 164
      Originally posted by shatteredaces
      P.s this maths shows online poker is broken....
      Explain?
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by DarkNeo1
      Originally posted by shatteredaces
      P.s this maths shows online poker is broken....
      Explain?
      They will close the thread if I explain, 2+2 have just shut a 4k view in a few days thread, when I have provided the maths and have had people agree.

      Mods- Can I run the thread and explain the proof?
    • la55i
      la55i
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      Pokerstrategy isn't like North-Korea. Please explain how online poker is broken. I have seen these funny calculations before but they were wrong :f_pleased: :f_pleased:
    • Harrier88
      Harrier88
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      Joined: 01.05.2012 Posts: 1,971
      Hi shatteredaces,

      Earlier this year, I believe he had a very similar discussion in this forum. As you may remember, we did unfortunately not really manage to find any common ground.

      You may also recall this post by one of my fellow mods, specifically this part:
      Originally posted by VorpalF2F
      Please -- no more analogies, diagrams or nonsensical examples.
      Please stick with established mathematics. If you can find a reference that supports your theory please provide a link that we can all check.

      You claim that odds differ between pre-shuffled decks and a single deck shuffled multiple times.
      Please provide a link to a recognized authoritative article that supports this claim.
      Please do not copy/paste the article itself without the source reference.

      Until you post that link, please post nothing else.
      Unless you have anything new to add this time, we're not really interested in re-starting this discussion all over again.

      I hope you understand.
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by Harrier88
      Hi shatteredaces,

      Earlier this year, I believe he had a very similar discussion in this forum. As you may remember, we did unfortunately not really manage to find any common ground.

      You may also recall this post by one of my fellow mods, specifically this part:
      Originally posted by VorpalF2F
      Please -- no more analogies, diagrams or nonsensical examples.
      Please stick with established mathematics. If you can find a reference that supports your theory please provide a link that we can all check.

      You claim that odds differ between pre-shuffled decks and a single deck shuffled multiple times.
      Please provide a link to a recognized authoritative article that supports this claim.
      Please do not copy/paste the article itself without the source reference.

      Until you post that link, please post nothing else.
      Unless you have anything new to add this time, we're not really interested in re-starting this discussion all over again.

      I hope you understand.
      I do have things to add, I have had agreement from other people, I have already provided the link with the maths of time correlation.


      I have confirmed maths that 4/52 changes.

      Please pick 52 top cards from 52 shuffled decks, leaving the values unseen. Shuffle the 52 cards, you think your chance of an ace is 4/52. This is incorrect, there is only a P=0.203317 that your 52 random's contain 4 aces, a P=0.8 approx there is not 4 aces.
    • la55i
      la55i
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      I remember that thread.. It was hilarious :f_biggrin: I agree with the mods.. Please no more. If you haven't learned to calculate.
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by la55i
      I remember that thread.. It was hilarious :f_biggrin: I agree with the mods.. Please no more. If you haven't learned to calculate.
      I have just give you the results, please feel free to do the maths yourself.

      [tex]P(4) = \left(\frac{4}{52}\right)^4 \times \left(1-\frac{4}{52}\right)^{52-4} \times \frac{52!}{4!(52-4)!} = 0.203317[/tex]
    • Harrier88
      Harrier88
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      Joined: 01.05.2012 Posts: 1,971
      I had a brief look at your thread over at 2+2. I did not really spot anything that struck me as new, nor did I see a lot of approval for your theories.

      If you have nothing further to add apart from what you already mentioned in this thread, I don't really see any reason to keep this one going, to be honest.
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by Harrier88
      I had a brief look at your thread over at 2+2. I did not really spot anything that struck me as new, nor did I see a lot of approval for your theories.

      If you have nothing further to add apart from what you already mentioned in this thread, I don't really see any reason to keep this one going, to be honest.
      I had about 4 people in that thread discussing it, two of them understood. I have people on science forums who have done me the maths to calculate the chance that Y contains 4 aces. I was told If I could provide a link to show the problem, then to return. I have done this.
      λ(X)=φ[sub]t[/sub]

      λ(Y)=σ[tex]\frac{n}{y}[/tex]/[tex]\frac{d}{t}[/tex]


      λ(X)=φ[sub]t1[/sub]=[tex]\frac{t1}{Z}[/tex]

      λ(Y)=σ[tex]\frac{t1}{Z}[/tex]=t[sub]2[sub]


      Copy and paste this in a forum that supports it, it explains the problem and the link supports this. Science will soon support this, some scientists already understand.

      alternative see here -

      http://www.thenakedscientists.com/forum/index.php?topic=62950.0

      and here

      https://theoristexplains.wordpress.com/2015/12/01/apparently-this-is-wrong/


      This effects science, never mind just poker.
    • Zhusy
      Zhusy
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      Joined: 17.01.2010 Posts: 382
      I saw this problem bugging people on MIT, a friend of mine came across it. It was a coding issue which has too many details so that I would go into it.

      It's explained as a bug in logic. What you are trying to do is redefine a well defined logic with maths, such logic can't be explained with maths as you are trying to do.

      You should send this to some more experienced math students/doctors, they like helping with such stuffs.

      Actually a probability and statistics theorist is obviously something you need.

      I wish you the best of luck
    • la55i
      la55i
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      And eventually none of this matters. Those calculations are irrelevant. We play with one deck and the randomizer works. My Holdem Manager shows after hundreds of thousands of hands that I get AA as often as I should.
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by Zhusy
      I saw this problem bugging people on MIT, a friend of mine came across it. It was a coding issue which has too many details so that I would go into it.

      It's explained as a bug in logic. What you are trying to do is redefine a well defined logic with maths, such logic can't be explained with maths as you are trying to do.

      You should send this to some more experienced math students/doctors, they like helping with such stuffs.

      Actually a probability and statistics theorist is obviously something you need.

      I wish you the best of luck
      Thank you, and yes it is complex
    • shatteredaces
      shatteredaces
      Basic
      Joined: 15.04.2015 Posts: 141
      Originally posted by la55i
      And eventually none of this matters. Those calculations are irrelevant. We play with one deck and the randomizer works. My Holdem Manager shows after hundreds of thousands of hands that I get AA as often as I should.
      Hello la55, your holdem manager will show that in time you get the expected amount of aces, the Hypothesis is neither anything to do with the randomness of the cards themselves.
      The Hypothesis is a complex matter involving ''time dilation'' and probably only about 10% of the population will even comprehend the hypothesis because it's complex nature is ''Einstein'' complex. The calculations are formulas to express the problem.
      If you are interesting in wanting to or trying to understanding, I will be happy to try to explain to you and go through some simple thought examples to try and show you the problem if you are unable to grasp the formula descriptions expressing the problem.
    • YohanN7
      YohanN7
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      Joined: 15.06.2009 Posts: 4,086
      Lol.

      Your calculation,
      P(4 aces in top cards picked) = (4/52)^4 * (1 - 4/52)^(52-4) * 52!/(4!(52 - 4)!) = 0.203317
      is entirely correct. That is, the left hand side is correct. (I haven't evaluated to confirm the r h s. I'll just take it at face value, doesn't matter.)

      The flaws are that you require that there are 4 aces among the 52 top cards and assign that as the probability of picking an A (two errors, the requirement and the assignment). Look again!
      Please pick 52 top cards from 52 shuffled decks, leaving the values unseen. Shuffle the 52 cards, you think your chance of an ace is 4/52. This is incorrect, there is only a P=0.203317 that your 52 random's contain 4 aces, a P=0.8 approx there is not 4 aces.
      There can be anything between 0 and 52 aces in the top cards you picked, each with a probability p(i) (with p(0) + p(1) + p(2) + ... + p(52) = 1). But this is irrelevant. What is relevant is that you take one card from a shuffled 52-deck shoe. The intermediate shuffling (pick top cards, blah, blah,) is just more shuffling. The probability that it is an ace is (4*52)/(52*52) = number of A/total number of cards = 1/13.
    • la55i
      la55i
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      I think I'm gonna pass.. This sounds like a scientific problem and doesn't concern me at all as a poker player. I understand what you are calculating here and the result doesn't seem logical but it doesn't mean that online poker is broken/rigged. Lets assume that we are playing live poker with one deck of cards. The probability of me getting dealt aces is 0.045. When I play online poker the probability is the same in the long run. So I don't really care. That is why I think this is the wrong forum for this discussion.
    • YohanN7
      YohanN7
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      And oh, if you want to do it the hard way, you'd need to sum over conditional probabilities picking an ace given there are N aces with N going from 0 to 52. Unsurprisingly, you'll get 1/13 for the total probability.