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Cute dice problem
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I ran across a version of the following (with fewer dice) on a middle school math contest (so we want a quick easy answer):
if you roll four standard 6sided dice and add up the resulting values, what is the probability of rolling prime number?
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 In primenumbers@yahoogroups.com,
Chris Caldwell <caldwell@...> wrote:
> if you roll four standard 6sided dice and add up the resulting values,
Help! I did this a dumb way, but got a simple answer:
> what is the probability of rolling prime number?
(4+20+104+140+104+56+4)/6^4 = 1/3
Is there a smarter way, please?
David 0 Attachment
Since the range of sums is 4 to 24, the sum is prime if and
only if it is coprime to 6  we very conveniently dodge
the largest prime which is not coprime to 6 (3) and the
smallest composite which is coprime to 6 (25).
Once you have that insight, it's pretty easy.
On 4/5/2013 9:04 AM, Chris Caldwell wrote:
> I ran across a version of the following (with fewer dice) on a middle school math contest (so we want a quick easy answer):
>
> if you roll four standard 6sided dice and add up the resulting values, what is the probability of rolling prime number?
>
>
> [Nontext portions of this message have been removed]
>
>
>
> 
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
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 In primenumbers@yahoogroups.com,
Jack Brennen <jfb@...> wrote:
> Since the range of sums is 4 to 24, the sum is prime if and
So the sides can be taken as {2,1,0,1,2,3}
> only if it is coprime to 6
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 0 Attachment
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:
>
>
>  In primenumbers@yahoogroups.com,
> 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
>
>
>
> 
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
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From: Jack Brennen> Throw the first three dice and figure your sum.
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.
> How many sides of the fourth die will give you a sum congruent to 1 or 5 modulo 6?
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 quicklydefinitely not a middle school solution!)
CC
Original Message
From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of Jack Brennen
Sent: Friday, April 05, 2013 12:22 PM
To: djbroadhurst
Cc: primenumbers@yahoogroups.com
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:
>
>
>  In primenumbers@yahoogroups.com,
> 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
>
>
>
> 
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
> Yahoo! Groups Links
>
>
>
>
>

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 In primenumbers@yahoogroups.com,
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) 0 Attachment
On 4/5/2013 11:20 AM, djbroadhurst wrote:>
Sort of a red herring.
>  In primenumbers@yahoogroups.com,
> Jack Brennen <jfb@...> wrote:
>
>> Throw the first three dice and figure your sum.
>
> Jack's "first three" seems to be 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 quicklydefinitely 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 AMC12, 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 cochampion 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. 0 Attachment
I also enjoyed that test (the ACM12) 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 ACM8 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
To: primenumbers@yahoogroups.com
Cc: djbroadhurst; Chris Caldwell
Subject: Re: [PrimeNumbers] Re: Cute dice problem [by induction]
On 4/5/2013 11:20 AM, djbroadhurst wrote:
>
>  In primenumbers@yahoogroups.com,
> 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 quicklydefinitely 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 AMC12, 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 cochampion 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. 0 Attachment
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 ACM12) 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 ACM8 either . But you
> are right in that they are designed to not allow the dreaded perfect
> scores (or even tied high values).
> 0 Attachment
 In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:>
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.
>
>
>  In primenumbers@yahoogroups.com,
> Chris Caldwell <caldwell@> wrote:
>
> > if you roll four standard 6sided 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
>
 0 Attachment
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 6sided 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: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
> Yahoo! Groups Links
>
>
>
>
> 0 Attachment
 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 6sided 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 6moduli {0,1,2,3,4,5} are represented the same
number of times {i.e. 6^(N1) 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
numbertheoretic theorem.
[Nontext portions of this message have been removed] 0 Attachment
 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 6moduli {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^(N1) 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 6sided 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 6moduli {0,1,2,3,4,5} are represented the same
> number of times {i.e. 6^(N1) 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
> numbertheoretic theorem.
>
>
>
> [Nontext portions of this message have been removed]
>
>
 0 Attachment
 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 0 Attachment
 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.
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