## RE: [PrimeNumbers] Re: Cute dice problem

Expand Messages
• From: Jack Brennen ... Yes! So four dice is a trivial problem! Sadly five does not divide six, so five dice is much harder (it is 137/432 for five dice and
Message 1 of 16 , Apr 5, 2013
From: Jack Brennen
> Throw the first three dice and figure your sum.
> How many sides of the fourth die will give you a sum congruent to 1 or 5 modulo 6?

Yes! So four dice is a trivial problem! Sadly five does not divide six, so five dice is much harder (it is 137/432 for five dice and 235/864 for six). The possibility of rolling 2 or 3 makes 2 or 3 dice a little harder. But the 4 dice case can be done instantly.

But why would a middle school child recognize that? Surely they did not want a long calculation. (One of my colleagues said generating functions give you the answer very quickly--definitely not a middle school solution!)

CC

-----Original Message-----
Sent: Friday, April 05, 2013 12:22 PM
Subject: Re: [PrimeNumbers] Re: Cute dice problem

Throw the first three dice and figure your sum.

How many sides of the fourth die will give you a sum congruent to 1 or 5 modulo 6?

On 4/5/2013 10:11 AM, djbroadhurst wrote:
>
>
> Jack Brennen <jfb@...> wrote:
>
>> Since the range of sums is 4 to 24, the sum is prime if and only if
>> it is coprime to 6
>
> So the sides can be taken as {-2,-1,0,1,2,3} and we want a total of 0,
> from 4 throws.
>
>> Once you have that insight, it's pretty easy.
>
> I still had to some work to get the very simple answer.
>
> David
>
>
>
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• ... Jack s first three seems to be a red herring. Solution: First n give equipartition mod 6, by induction. So the probability of (1 or 5) mod 6 is /always/
Message 2 of 16 , Apr 5, 2013
Jack Brennen <jfb@...> wrote:

> Throw the first three dice and figure your sum.

Jack's "first three" seems to be a red herring.

Solution: First n give equipartition mod 6, by induction.
So the probability of (1 or 5) mod 6 is /always/ 1/3,
no matter how may dice you throw.

Doh!

David (not taught induction until high school)
• ... Sort of a red herring. You could replace throw the first three dice and figure your sum with pick any integer . :) ... Most wouldn t recognize it. I d
Message 3 of 16 , Apr 5, 2013
On 4/5/2013 11:20 AM, djbroadhurst wrote:
>
> Jack Brennen <jfb@...> wrote:
>
>> Throw the first three dice and figure your sum.
>
> Jack's "first three" seems to be a red herring.
>

Sort of a red herring.

You could replace "throw the first three dice and figure your sum"
with "pick any integer". :)

On 4/5/2013 10:28 AM, Chris Caldwell wrote:
>
> But why would a middle school child recognize that? Surely they did
> not want a long calculation. (One of my colleagues said generating
> functions give you the answer very quickly--definitely not a middle
> school solution!)

Most wouldn't recognize it. I'd like to think I would have recognized
it at that age, but I can't be sure. In seventh grade, I scored a 105
out of 150 on the contest that is now called the AMC-12, which was a
pretty decent score. If you believe what you read online, Noam Elkies,
who is three months younger than me, scored a 112 on that test, and then
he proceeded to beat my math competition scores routinely over the next
three years.

All of which is to say... a decent math competition should have some
problems which are well beyond the capabilities of the average
competitor. In ninth grade, I think I was a national co-champion of
the Continental Math League competition, because I didn't get a single
question wrong all year. It wasn't a particularly satisfying result.
That league was fairly new at the time, and I hope they learned that
you need to put some "stretch" questions in with the routine ones.
• 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
Message 4 of 16 , Apr 5, 2013
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).

-----Original Message-----
From: Jack Brennen [mailto:jfb@...]
Sent: Friday, April 05, 2013 2:30 PM
Subject: Re: [PrimeNumbers] Re: Cute dice problem [by induction]

On 4/5/2013 11:20 AM, djbroadhurst wrote:
>
> Jack Brennen <jfb@...> wrote:
>
>> Throw the first three dice and figure your sum.
>
> Jack's "first three" seems to be a red herring.
>

Sort of a red herring.

You could replace "throw the first three dice and figure your sum"
with "pick any integer". :)

On 4/5/2013 10:28 AM, Chris Caldwell wrote:
>
> But why would a middle school child recognize that? Surely they did > not want a long calculation. (One of my colleagues said generating > functions give you the answer very quickly--definitely not a middle > school solution!)

Most wouldn't recognize it. I'd like to think I would have recognized it at that age, but I can't be sure. In seventh grade, I scored a 105 out of 150 on the contest that is now called the AMC-12, which was a pretty decent score. If you believe what you read online, Noam Elkies, who is three months younger than me, scored a 112 on that test, and then he proceeded to beat my math competition scores routinely over the next three years.

All of which is to say... a decent math competition should have some problems which are well beyond the capabilities of the average competitor. In ninth grade, I think I was a national co-champion of the Continental Math League competition, because I didn't get a single question wrong all year. It wasn't a particularly satisfying result.
That league was fairly new at the time, and I hope they learned that you need to put some "stretch" questions in with the routine ones.
• 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 5 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).
>
• ... 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 6 of 16 , Apr 15, 2013
>
>
>
> 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.
• 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 7 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:
>
>
>>
>>
>>
>> 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/
>
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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 8 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 9 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 10 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 11 of 16 , May 19, 2013
>
> "woodhodgson@" <rupert.weather@> wrote:
>
> > It seems to me there is a very short proof available ....
> > by mathematical induction ....
>
> Indeed: