## Re: Cute dice problem

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• ... 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 1 of 16 , May 16, 2013
>
>
>
wrote:
> >
> >
> >
> > 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
Message 2 of 16 , May 16, 2013
--- In primenumbers@yahoogroups.com, "woodhodgson@..." <rupert.weather@...> wrote:
>
>
>
> --- In primenumbers@yahoogroups.com, "John" wrote:
> >
> >
> >
> 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}."
>
Message 3 of 16 , May 16, 2013
"woodhodgson@..." <rupert.weather@...> wrote:

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

Indeed:

David
• ... In this little interesting problem, David s dumb answer was the dumbness of ordinary arithmetic compared with the elegance of modular arithmetic and
Message 4 of 16 , May 19, 2013
>
> "woodhodgson@" <rupert.weather@> wrote:
>
> > It seems to me there is a very short proof available ....
> > by mathematical induction ....
>
> Indeed: