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[MATH for FUN] Re: What would be % chances to win?

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  • video_ranger
    Yes the probability of a large random number being divisible by N would be expected to be about 1/N. The last 114 could be left off since it s divisible by
    Message 1 of 13 , May 1, 2010
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      Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.

      The last 114 could be left off since it's divisible by 19. That is

      96 1 v2 v3...................v113 v
      96 1 v2 v3...................v113 v 114

      have the same remainder when divided by 19. I did do a computer check just to make sure using random permutations and there's no measurable difference of the probabiliy (of numbers of the above form being divisible by 19) from 1/19.


      --- In mathforfun@yahoogroups.com, ynineteen <no_reply@...> wrote:
      >
      > 1/19?
      > You mean every 19th trail hits the target?
      > here we have fixed values start with 96 1 and variable "v" and at the end we have fixed 114.
      > 96 1 v2 v3...................v113 v114
      > Valuve for variable cards is ame 1 to 114 but 96 is talen out before shuffling.
      > 19 cycle here is locked with fixed number 114
      > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
      > >
      > > There are too many permutations to calculate the exact probability by enumeration on a computer but I'm 99% sure that the probability of that 468 digit number being divisible by 19 is very close to 1/19.
      > >
      > > In other words it's virtually a random number mod 19. This could be verified using a Monte Carlo program.
      > >
      > >
      > > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
      > > >
      > > > He's talking about string the digits of the cards together - count the number of digits between 1 and 114, and double it.  I get 468, not 458.
      > > >
      > > > Ed
      > > >
      > > >
      > > >
      > > >
      > > > ________________________________
      > > > From: clooneman2000 <clooneman2000@>
      > > > To: mathforfun@yahoogroups.com
      > > > Sent: Wed, April 28, 2010 1:09:09 PM
      > > > Subject: [MATH for FUN] Re: What would be % chances to win?
      > > >
      > > >  
      > > > So the second set of cards contains numbers from 1 to 114? Now forgive me if I'm misunderstanding this, but don't the numbers from 1 to 114 total to 6555, and not 458?
      > > >
      > > > --- In mathforfun@yahoogro ups.com, ynineteen <no_reply@ .> wrote:
      > > > >
      > > > >
      > > > >
      > > > >
      > > > > Either set are printed 1 to 114.
      > > > > one set in row from one to 114 and other set shuffled
      > > > > I forgot to indicate that we pick number 96 before shuffelling.
      > > > > 96 can be front or after fixed card number 1
      > > > > 96 1 or
      > > > > 1 96
      > > > > Thank you for your concern
      > > > >
      > > > > --- In mathforfun@yahoogro ups.com, "clooneman2000" <clooneman2000@ > wrote:
      > > > > >
      > > > > > What are the values contained on the other set of 114 cards?
      > > > > >
      > > > > > --- In mathforfun@yahoogro ups.com, ynineteen <no_reply@> wrote:
      > > > > > >
      > > > > > >
      > > > > > > There is 114 cards on the table in one row set from 1 to 114 in order,
      > > > > > > All the card are displaced for inserting another card . Now we have
      > > > > > > another set of 114 cards which we shuffle it and insert one at the times
      > > > > > > starting before card or after . We insert all cards and now we have
      > > > > > > large figure 458 digits . If this figure by multiple of 19 (114=19x6)
      > > > > > > we are winner.
      > > > > > >
      > > > > > > 1-Would you show what would be % chances to win with one trial.
      > > > > > >
      > > > > > > 2- It it possible to use computer program to find the winner figure?
      > > > > > >
      > > > > > > Thank you
      > > > > > >
      > > > > > >
      > > > > > >
      > > > > > >
      > > > > > >
      > > > > > > [Non-text portions of this message have been removed]
      > > > > > >
      > > > > >
      > > > >
      > > >
      > > >
      > > >
      > > >
      > > >
      > > >
      > > >
      > > > [Non-text portions of this message have been removed]
      > > >
      > >
      >
    • video_ranger
      Correction: I shouldn t have said 96 1 v2 v3...................v113 v 96 1 v2 v3...................v113 v 114 have the same remainder when divided by 19
      Message 2 of 13 , May 1, 2010
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        Correction: I shouldn't have said

        96 1 v2 v3...................v113 v
        96 1 v2 v3...................v113 v 114

        have the same remainder when divided by 19 (unless it's zero).
      • bogaduck
        ... It probably doesn t alter your general reasoning but those two above generally do not have the same remainder. eg if 96 1 v2 v3...................v113 v
        Message 3 of 13 , May 1, 2010
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          --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...> wrote:
          >
          > Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.
          >
          > The last 114 could be left off since it's divisible by 19. That is
          >
          > 96 1 v2 v3...................v113 v
          > 96 1 v2 v3...................v113 v 114
          >
          > have the same remainder when divided by 19.
          >

          It probably doesn't alter your general reasoning but those two above generally do not have the same remainder.

          eg if 96 1 v2 v3...................v113 v
          has a remainder of 8
          then 96 1 v2 v3...................v113 v 114
          has a remainder of 1.

          The fact they both have a remainder of some sort is the most important thing however.

          Not sure whether the option of placing the second pack of cards AFTER the first; to give

          1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option

          makes any difference.

          Peter
        • bogaduck
          ... OK ignore the first part of my previous reply then. Peter
          Message 4 of 13 , May 1, 2010
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            --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...> wrote:
            >
            > Correction: I shouldn't have said
            >
            > 96 1 v2 v3...................v113 v
            > 96 1 v2 v3...................v113 v 114
            >
            > have the same remainder when divided by 19 (unless it's zero).
            >

            OK ignore the first part of my previous reply then.

            Peter
          • ynineteen
            1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option 96 1 ? 2 ? 3 ? 4..............................and ? 114 96 is fixed and 1...114 fixed . None of these
            Message 5 of 13 , May 1, 2010
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              "1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option "
              96 1 ? 2 ? 3 ? 4..............................and ? 114
              96 is fixed and 1...114 fixed . None of these 114 numbere can be relocated or changed as it is structure of case.
              We are allowed to insert random numbers between fixed with other ste which is shuffled.
              My first question is what % chances is there to have the pttern mutiple of 19 (no reminder) by "one time" inserting shuffled random cards?


              --- In mathforfun@yahoogroups.com, "bogaduck" <pmaxotzen@...> wrote:
              >
              >
              >
              > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
              > >
              > > Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.
              > >
              > > The last 114 could be left off since it's divisible by 19. That is
              > >
              > > 96 1 v2 v3...................v113 v
              > > 96 1 v2 v3...................v113 v 114
              > >
              > > have the same remainder when divided by 19.
              > >
              >
              > It probably doesn't alter your general reasoning but those two above generally do not have the same remainder.
              >
              > eg if 96 1 v2 v3...................v113 v
              > has a remainder of 8
              > then 96 1 v2 v3...................v113 v 114
              > has a remainder of 1.
              >
              > The fact they both have a remainder of some sort is the most important thing however.
              >
              > Not sure whether the option of placing the second pack of cards AFTER the first; to give
              >
              > 1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option
              >
              > makes any difference.
              >
              > Peter
              >
            • video_ranger
              ... I tried it that way first before ynineteen specified the other way:(96 1 v2 v3...................v113 v 114) and it made no difference (using random
              Message 6 of 13 , May 2, 2010
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                --- In mathforfun@yahoogroups.com, "bogaduck" <pmaxotzen@...> wrote:
                >
                >
                >
                > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
                > >
                > > Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.
                > >
                > > The last 114 could be left off since it's divisible by 19. That is
                > >
                > > 96 1 v2 v3...................v113 v
                > > 96 1 v2 v3...................v113 v 114
                > >
                > > have the same remainder when divided by 19.
                > >
                >
                > It probably doesn't alter your general reasoning but those two above generally do not have the same remainder.
                >
                > eg if 96 1 v2 v3...................v113 v
                > has a remainder of 8
                > then 96 1 v2 v3...................v113 v 114
                > has a remainder of 1.
                >
                > The fact they both have a remainder of some sort is the most important thing however.
                >
                > Not sure whether the option of placing the second pack of cards AFTER the first; to give
                >
                > 1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option
                >
                > makes any difference.
                >
                > Peter
                >

                I tried it that way first before ynineteen specified the other way:(96 1 v2 v3...................v113 v 114) and it made no difference (using random permutations). The fraction divisible by 19 is always about 1/19
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