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Re: [PrimeNumbers] Re: Cute dice problem

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  • Jack Brennen
    The math required to understand the easy solution to the dice problem is easily within the realm of 8th grade mathematics. The insight required to come up
    Message 1 of 16 , Apr 5, 2013
      The math required to understand the "easy" solution to the dice problem
      is easily within the realm of 8th grade mathematics.

      The insight required to come up with the easy solution, while under time
      pressure... That's another issue entirely. To get to the insight
      requires recognizing that for 4 dice (and for 4 dice only!), you can
      replace "prime" with "coprime with 6". Transforming a problem into
      an equivalent easier problem -- that's something that is really hard
      to master. (And almost certainly something that most 8th graders have
      not been taught in school.)



      On 4/5/2013 12:52 PM, Chris Caldwell wrote:
      > I also enjoyed that test (the ACM-12) as a child, but of course that
      > test is for students 4 years older than this test for Tennessee youth
      > (only) is. Also this test is supposed to be limited to the state
      > standards, so should not be the level of the ACM-8 either . But you
      > are right in that they are designed to not allow the dreaded perfect
      > scores (or even tied high values).
      >
    • 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 2 of 16 , Apr 15, 2013
        --- 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 3 of 16 , Apr 15, 2013
          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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          >
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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 4 of 16 , May 16, 2013
            --- 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 5 of 16 , May 16, 2013
              --- 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 6 of 16 , May 16, 2013
                --- 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 7 of 16 , May 19, 2013
                  --- 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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