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Thread: Robert V. Lux: Math formula?

  1. #14
    Zenfighter
    Guest

    Zenfighter: Re: Normal Approach

    > That's what I did with my abacus

    >
    > (8/104)*(7/103)*(6/102)*(96/101)*(95/100)*(94/99)*
    >
    > (93/98)*(92/97)*(91/96)*(90/95)*(89/94)*(88/93)*
    > (87/92)*(86/91)*(85/90)*455=

    > =0.07332209549650789440

    > Sincerely,
    > Cacarulo

    That's right, Sir. I've got also:

    0.073321885

    So probably Don is right and the final probability
    is 100/7.33 = 13.64 or one in 13.64 odds.

    This figure makes a lot of sense, btw it doesn't
    mean that your approach using the RC tricks or
    mine, with the binomial distribution for an ace
    count typically with sampling without replacement,
    are both wrong or producing erroneous data. It's
    just that we are reading the results under the
    Normal curve wrong.

    My z =1.92 is yours (1.96 + 1.88)/2 don't forget.

    For z=1.9245 you get Qz = 0.0271 but that's to the
    right in other words, that's the probability for
    more than 3 aces, so is not the correct answer and I was correct not to 'gamble' on it. Too much
    speed for a not so easy problem.

    Have a look at your figure 0.0051. That will mean
    you'll have 3 aces in the first 15 cards only once every 196 deals. Doesn't make sense, does it?

    Are we both wrong Cac? I guess probably not, maybe just blind under the Gaussian curve.

    I need more time to think about this. Let's
    improve the normal approach.

    Regards
    Z

    Regards

  2. #15
    Don Schlesinger
    Guest

    Don Schlesinger: Re: Normal Approach

    > That's what I did with my abacus

    (8/104)*(7/103)*(6/102)*(96/101)*(95/100)*(94/99)*
    (93/98)*(92/97)*(91/96)*(90/95)*(89/94)*(88/93)*
    (87/92)*(86/91)*(85/90)*455=

    > =0.07332209549650789440
    I redid the multiplication and do get your answer.

    So, it happens once it 13.64 chances. Sorry for any inconvenience.

    Don

  3. #16
    Don Schlesinger
    Guest

    Don Schlesinger: Re: Normal Approach

    > That's right, Sir. I've got also:

    > 0.073321885

    > So probably Don is right and the final
    > probability
    > is 100/7.33 = 13.64 or one in 13.64 odds.

    > This figure makes a lot of sense, btw it
    > doesn't
    > mean that your approach using the RC tricks
    > or
    > mine, with the binomial distribution for an
    > ace
    > count typically with sampling without
    > replacement,
    > are both wrong or producing erroneous data.
    > It's
    > just that we are reading the results under
    > the
    > Normal curve wrong.

    > My z =1.92 is yours (1.96 + 1.88)/2 don't
    > forget.

    > For z=1.9245 you get Qz = 0.0271 but that's
    > to the
    > right in other words, that's the probability
    > for
    > more than 3 aces, so is not the correct
    > answer and I was correct not to 'gamble' on
    > it. Too much
    > speed for a not so easy problem.

    > Are we both wrong Cac? I guess probably not,
    > maybe just blind under the Gaussian curve.

    > I need more time to think about this. Let's
    > improve the normal approach.

    It's all quite simple. There are three relevant numbers. The first is for 3 or fewer aces. The second is for exactly 3 aces, and the third is for strictly more than 3 aces.

    The .0271 number is for strictly more than 3 aces. The .073 number is for exactly 3 aces, and the other number, which you don't have, namely, about 0.972, is for 3 or fewer aces.

    Another number, about 0.10, would represent 3 or more aces, and is the sum of 0.073 and .0271.

    Clear to everyone??

    Don

  4. #17
    Zenfighter
    Guest

    Zenfighter: Re: Epilog

    > It's all quite simple. There are three
    > relevant numbers. The first is for 3 or
    > fewer aces. The second is for exactly 3
    > aces, and the third is for strictly more
    > than 3 aces.

    > The .0271 number is for strictly more than 3
    > aces. The .073 number is for exactly 3 aces,
    > and the other number, which you don't have,
    > namely, about 0.972, is for 3 or fewer aces.

    > Another number, about 0.10, would represent
    > 3 or more aces, and is the sum of 0.073 and
    > .0271.

    > Clear to everyone??

    > Don

    Despite how lengthy or tedious the calcs might
    be, combinatorics seems to me a sine qua non to
    solve these type of problems, as has been demonstrated here, looking under the curve can be fun and a shortcut somehow, but I'm afraid we only get at the end with this method, an approximately figure, of the type between this and this.

    I would like to see, as a Christmas wish, to watch
    all the BJ talented guys out there, post more
    frecuently on this and on the other S&S page.
    One of the benefits of being wrong, is having the
    opportunity to get corrected. A 'luxus', I'm not
    quite sure, many youngers do fully understand.

    Sincerely
    Z


  5. #18
    Cacarulo
    Guest

    Cacarulo: Re: Normal Approach

    > It's all quite simple. There are three
    > relevant numbers. The first is for 3 or
    > fewer aces. The second is for exactly 3
    > aces, and the third is for strictly more
    > than 3 aces.

    > The .0271 number is for strictly more than 3
    > aces. The .073 number is for exactly 3 aces,
    > and the other number, which you don't have,
    > namely, about 0.972, is for 3 or fewer aces.

    > Another number, about 0.10, would represent
    > 3 or more aces, and is the sum of 0.073 and
    > .0271.

    > Clear to everyone??

    The question is: What is it wrong with Griffin's approach? 0.0051 should be the approximated answer for exactly 3 aces, no? I can't believe it is so far from 0.0733. What am I missing?

    Sincerely,
    Cacarulo

  6. #19
    Don Schlesinger
    Guest

    Don Schlesinger: Re: Normal Approach

    > The question is: What is it wrong with
    > Griffin's approach? 0.0051 should be the
    > approximated answer for exactly 3 aces, no?
    > I can't believe it is so far from 0.0733.
    > What am I missing?

    You don't seem to be missing anything. I've gone over the math several times, but something just doesn't add up. I'll try to find it. Doing it Griffin's way doesn't produce the correct answer, so we need to understand why either that approach is incorrect for this problem, or, somehow, the math isn't making sense.

    Don

  7. #20
    Don Schlesinger
    Guest

    Don Schlesinger: I deleted my post

    For the few of you who read an earlier post I had up here, forget about it! I'm obviously losing my mind. What I put up was silly, and the original calculation I have up above is correct, as is.

    Sorry for any inconvenience.

    Don

  8. #21
    Don Schlesinger
    Guest

    Don Schlesinger: Two sites you can enjoy

    Everyone should be aware of the two sites, below. When attempting to work out precise binomial distributions, they can be very helpful:

    http://faculty.vassar.edu/lowry/binom_stats.html
    http://faculty.vassar.edu/lowry/zbinom.html

    Now, in our problem, we have sampling WITHOUT replacement, so the calcuators do not apply. Nonetheless, they would give answers that are "ballpark" correct, assuming that we had sampling WITH replacement, so that the chances of getting an ace wouldn't diminish as actual aces appeared.

    In our example, we begin with 8/104, which is 1/13. But, the next fraction is 7/103, which is closer to 1/15, while the third fraction, 6/102 is precisely 1/17.

    If we use the first calculator, with n = 15, k = 3, and p = 1/13 = .076923, we find that P of exactly 3 out of 15 is .07928, which, expectedly, is just a bit more than the answer I posted above.

    But, if we use p = 1/15, as a kind of average of the actual situation, then the calculator gives .05888, which is too low. The problem, of course, is that p+q = 1, for the binomial situation, whereas, here, without replacement, that relationship simply doesn't hold.

    In any event, we get an answer that is of the correct order of magnitude and, unfortunately, is very different from Griffin's approach, which I'm still trying to reconcile.

    Don

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