- 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.

>

>

>

> ------------------------------------

>

> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

> The Prime Pages : http://primes.utm.edu/

>

> Yahoo! Groups Links

>

>

>

>

> - --- In primenumbers@yahoogroups.com, "John" wrote:
>

wrote:

>

>

> --- In primenumbers@yahoogroups.com, "djbroadhurst" d.broadhurst@

> >

values,

> >

> >

> > --- In primenumbers@yahoogroups.com,

> > Chris Caldwell wrote:

> >

> > > if you roll four standard 6-sided dice and add up the resulting

> > > what is the probability of rolling prime number?

work out the number of permutations for n=5,7,11,13,17,19,23 and add

> >

> > 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

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] - --- In primenumbers@yahoogroups.com, "woodhodgson@..." <rupert.weather@...> wrote:
>

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}."

>

>

> --- 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]

>

>

- --- In primenumbers@yahoogroups.com,

"woodhodgson@..." <rupert.weather@...> wrote:

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

Indeed:

> by mathematical induction ....

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

David - --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

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

> --- 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

>

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.