What is the Probability that N Randomly chosen Individuals will have at least TWO with a matching bi

Discussion in 'Scientific Statistics Math' started by Reef Fish, Apr 6, 2006.

  1. Reef Fish

    Reef Fish Guest

    This of course is an easy analogue of the matching birthDATE problem,
    of at least two people (out of N randomly chosen ones) having birthdays
    on the same MONTH and same DAY.

    So, I gave the OP of that birthdate question the solution that it is 1
    minus
    the probability that they ALL have different birthdates,

    1 - (365/365)(364/365)...((365-N+1)/365). (1)

    and went on to give the analogue of at least two born on the same
    MONTH,

    1 - (12/12)(11/12) ... ((12-N+1)/12) , for N < 13. (2)


    Of course both are approximations where 365 assumes the probability
    of a person born on any given day is 1/365, and ignoring Feb 29.
    <Statistics showed that the probability of a person born on any
    particular
    day of the year is slightly different from 1/365> but nobody quibbles

    about that.

    Afonso first tried to argue that (1) was wrong, by giving the "all
    different" part as

    Afonso> _ p(n , day) = (1-1/365)(1- 2/365)...(1- n/365) for n<=
    365


    It took Jack Tomsky less than 2 hours from Afonso's post to point out
    that Afonso's formula is wrong, for n = 1, and

    JT> Bob's formula

    JT> P(n) = (365/365)(364/365)...((365-N+1)/365)

    JT> is the correct one.

    JT> Jack

    Actually Afonso's formula is wrong, not only for n = 1, but
    wrong for ALL n, because of the absence of the (365/365) factor.

    I had left that dead dog of Afonso's lie, after Jack's correction.
    This
    is the first time since Afonso's blunder on March 31 that I am pointing
    out that he formula was wrong for ALL n.

    ------------------------------------------------------------

    The second part of Afonso's BLUNDER was his accusation that
    my formula (2) was wrong.

    (2) was so OBVIOUSLY correct (as an approximation and an analogue
    of the birthDATE problem) that I didn't bother to point out Afonso's
    error in his arguement!

    Having been ignored by me, Afonso was getting the false idea
    that he had found a flaw in my solution, and he went out of his
    way to have made that point in AT LEAST THREE different
    threads, besides the original one, including ones he created,

    Subject: "Bob and the birth-months"

    Subject: "Waiting till Bob admits his blunder about the birth-months
    formula."

    now spilt his noise about the birth MONTH formula into the thread

    Subject: "The <exact> C.I. for the Pearson´s Product Moment
    Coefficient"

    in which Jack Tomsky point out one error and one silliness in Afonso's
    post.


    Well, Afonso's gag has been left long enough without me stuffing it
    down his throat.

    He said my approximate formula (2) was wrong because I didn't
    count the numbers of days in each month.

    The 1/12 approximation of the 2nd person born on the same MONTH
    as the randomly chosen 1st person is no different from 1/365 used as
    the approx. probability of the 2nd person born on the same DAY.

    Now let's examine Afonso's argument which he didn't realize was an
    ERROR of his that cannot be used (correctly) for n =2, or any other
    n < 13.

    Just as his matching birthDAY formula was wrong for ALL n.


    Mastro, .... let's have some MUSIC please, for the event Afonso
    has been begging for SIX DAYS, in FOUR different threads!

    This was Afonso's BLUNDER and false balloon that was left
    untouched and unpuntured until NOW:


    LA> What we immediately conclude is that Bob did not understand
    LA> HOW the formula is derived: to change 365 by 12 is
    LA> unmistakably the proof.

    LA> ____p(n, from January till month u inclusive) =
    LA> ________product (1 - v*k(v)*/ 365)

    LA> is the probability to have not a month, from January , v=1,
    LA> to the month u with at least two persons with the same
    LA> birth-month.

    LA> This is a corrected formula:
    LA> _____k(1) = 31 ... January
    LA> _____k(2) = 28 + ¼... February (leap-years included)
    LA> _____k(3) = 31...Mars, etc.

    Take two randomly selected persons. My approx. solution that they
    have
    the same birth MONTH is

    1 - (12/12)(11/12).

    Afonso thinks that the probability for each month must be based on
    the number of days, such as 31/365.25, 28.25/365.25, 31/365.25, etc.

    If he had tried to carry that argue one tiny step further, he should
    have seen the silliness of his argument and the INABILITY to use
    his probability numbers for any 1 < n.

    If TWO people are picked at random, how does Afonso, or anyone
    else know, the birth MONTH of those two people to use a
    probability other than 1/12 for "a different month" for the 2nd
    person?

    In short, Luis A. Afonso not only does NOT have a workable
    formula (even if it's a wrong formula -- to know what to plug
    into the formula for n = 2, 3, 4, ...), he is blithering all this
    time, creating NUISANCE thread titles, and making his
    incessant noise over his own blunder that it only goes to
    show WHY I have developed the tendency to just IGNORE
    him, as I've done in NOT pointing his TWO errors since
    he posted them on March 31, until now April 5, and only so
    because he was making too much noise making up new
    derogative subjects and making accusatory remarks in
    OTHER threads unrelated to the birthdate, birth-month one.

    Luis A. Afonso, I hope you are happy now, that you no longer
    have to wonder "HOW LONG" you have to wait for me to
    show, in explicit details what a blithering IDIOT you have been.

    -- Bob.
     
    Reef Fish, Apr 6, 2006
    #1
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