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Re: [PrimeNumbers] Re: Cute dice problem

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  • Jack Brennen
    Show the values that you can have after rolling the first 3 dice, and then for each of those values, show how many primes can be reached with the last die:
    Message 1 of 16 , Apr 15, 2013
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      Show the values that you can have after rolling the first 3 dice, and
      then for each of those values, show how many primes can be reached with
      the last die:

      First three == 3: 2 primes (5 and 7)
      First three == 4: 2 primes (5 and 7)
      First three == 5: 2 primes (7 and 11)
      First three == 6: 2 primes (7 and 11)
      First three == 7: 2 primes (11 and 13)
      First three == 8: 2 primes (11 and 13)
      First three == 9: 2 primes (11 and 13)
      First three == 10: 2 primes (11 and 13)
      First three == 11: 2 primes (13 and 17)
      First three == 12: 2 primes (13 and 17)
      First three == 13: 2 primes (17 and 19)
      First three == 14: 2 primes (17 and 19)
      First three == 15: 2 primes (17 and 19)
      First three == 16: 2 primes (17 and 19)
      First three == 17: 2 primes (19 and 23)
      First three == 18: 2 primes (19 and 23)

      The rest is left as an exercise for the reader in use of the
      distributive property.



      On 4/15/2013 9:30 AM, John wrote:
      >
      >
      > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >>
      >>
      >>
      >> --- In primenumbers@yahoogroups.com,
      >> Chris Caldwell <caldwell@> wrote:
      >>
      >>> if you roll four standard 6-sided dice and add up the resulting values,
      >>> what is the probability of rolling prime number?
      >>
      >> Help! I did this a dumb way, but got a simple answer:
      >> (4+20+104+140+104+56+4)/6^4 = 1/3
      >> Is there a smarter way, please?
      >>
      >> David
      >>
      >
      > Not all that dumb in my book. In fact, it is the most logical. Just work out the number of permutations for n=5,7,11,13,17,19,23 and add them up, and divide by 1296.
      >
      >
      >
      > ------------------------------------
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      > The Prime Pages : http://primes.utm.edu/
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      >
    • woodhodgson@xtra.co.nz
      ... values, ... work out the number of permutations for n=5,7,11,13,17,19,23 and add them up, and divide by 1296. ... It seems to me there is a very short
      Message 2 of 16 , May 16, 2013
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        --- In primenumbers@yahoogroups.com, "John" wrote:
        >
        >
        >
        > --- In primenumbers@yahoogroups.com, "djbroadhurst" d.broadhurst@
        wrote:
        > >
        > >
        > >
        > > --- In primenumbers@yahoogroups.com,
        > > Chris Caldwell wrote:
        > >
        > > > if you roll four standard 6-sided dice and add up the resulting
        values,
        > > > what is the probability of rolling prime number?
        > >
        > > Help! I did this a dumb way, but got a simple answer:
        > > (4+20+104+140+104+56+4)/6^4 = 1/3
        > > Is there a smarter way, please?
        > >
        > > David
        > >
        >
        > Not all that dumb in my book. In fact, it is the most logical. Just
        work out the number of permutations for n=5,7,11,13,17,19,23 and add
        them up, and divide by 1296.
        >

        It seems to me there is a very short proof available. The prime numbers
        in the range (4,24) are precisely those equivalent to 1 or 5, modulo 6.

        Now it can very easily be shown by mathematical induction (**) that the
        6^N values (modulo 6) obtained by adding N numbers selected "one from
        each" of N copies of the 6-moduli {0,1,2,3,4,5} are represented the same
        number of times {i.e. 6^(N-1) times}.

        It then follows trivially that the required probability is 2/6 = 1/3.

        I have only run this "through my head", but it seems correct to me.


        (**) I suspect it is a trivial consequence of some algebraic or
        number-theoretic theorem.



        [Non-text portions of this message have been removed]
      • woodhodgson@xtra.co.nz
        ... I have worded part of that clumsily and possibly misleadingly: it should read ... the set of 6^N results (modulo 6) obtained by adding N numbers
        Message 3 of 16 , May 16, 2013
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          --- In primenumbers@yahoogroups.com, "woodhodgson@..." <rupert.weather@...> wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com, "John" wrote:
          > >
          > >
          > >
          > > --- In primenumbers@yahoogroups.com, "djbroadhurst" d.broadhurst@
          > wrote:
          > > >
          > > >
          > > >
          > > > --- In primenumbers@yahoogroups.com,
          > > > Chris Caldwell wrote:
          > > >
          > > > > if you roll four standard 6-sided dice and add up the resulting
          > values,
          > > > > what is the probability of rolling prime number?
          > > >
          > > > Help! I did this a dumb way, but got a simple answer:
          > > > (4+20+104+140+104+56+4)/6^4 = 1/3
          > > > Is there a smarter way, please?
          > > >
          > > > David
          > > >
          > >
          > > Not all that dumb in my book. In fact, it is the most logical. Just
          > work out the number of permutations for n=5,7,11,13,17,19,23 and add
          > them up, and divide by 1296.
          > >
          >
          > It seems to me there is a very short proof available. The prime numbers
          > in the range (4,24) are precisely those equivalent to 1 or 5, modulo 6.
          >
          > Now it can very easily be shown by mathematical induction (**) that the
          > 6^N values (modulo 6) obtained by adding N numbers selected "one from
          > each" of N copies of the 6-moduli {0,1,2,3,4,5} are represented the same
          > number of times {i.e. 6^(N-1) times}.
          >
          > It then follows trivially that the required probability is 2/6 = 1/3.
          >
          > I have only run this "through my head", but it seems correct to me.
          >
          >
          > (**) I suspect it is a trivial consequence of some algebraic or
          > number-theoretic theorem.
          >
          >
          >
          > [Non-text portions of this message have been removed]
          >

          I have worded part of that clumsily and possibly misleadingly: it should read " ... the set of 6^N results (modulo 6) obtained by adding N numbers selected "one from each" of N copies of the 6-moduli {0,1,2,3,4,5} contains each of the values 0,1,2,3,4,5 the same number of times {i.e. 6^(N-1) times}."
          >
        • djbroadhurst
          ... Indeed: http://tech.groups.yahoo.com/group/primenumbers/message/25015 David
          Message 4 of 16 , May 16, 2013
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            --- In primenumbers@yahoogroups.com,
            "woodhodgson@..." <rupert.weather@...> wrote:

            > It seems to me there is a very short proof available ....
            > by mathematical induction ....

            Indeed:
            http://tech.groups.yahoo.com/group/primenumbers/message/25015

            David
          • John
            ... In this little interesting problem, David s dumb answer was the dumbness of ordinary arithmetic compared with the elegance of modular arithmetic and
            Message 5 of 16 , May 19, 2013
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              --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
              >
              > --- In primenumbers@yahoogroups.com,
              > "woodhodgson@" <rupert.weather@> wrote:
              >
              > > It seems to me there is a very short proof available ....
              > > by mathematical induction ....
              >
              > Indeed:
              > http://tech.groups.yahoo.com/group/primenumbers/message/25015
              >
              > David
              >

              In this little interesting problem, David's "dumb" answer was the dumbness of ordinary arithmetic compared with the elegance of modular arithmetic and induction.

              But, though both give the right answer, I have to "dip the lid" to elegance, even if it requires a bit of effort to bring it about.
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