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Re: Cute dice problem

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  • John
    ... 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
    Message 1 of 16 , Apr 15, 2013
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      --- 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.
    • 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 2 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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        > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
        > 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 3 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 4 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 5 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 6 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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