Problem

[Yog]

So I was working on this little exercise today and I'm pretty sure I missed one permutation. So if anyone can figure out the one I missed I'd love to hear it.

The rules are as follows:

Each entry must consist of six numbers
Each number must be an integer between 3 and 18
The sum of all six numbers must be 31

The ones I have verified are as follows:

16 3 3 3 3 3
6 5 5 5 5 5
15 4 3 3 3 3
14 5 3 3 3 3
14 4 4 3 3 3
13 6 3 3 3 3
13 5 4 3 3 3
13 4 4 4 3 3
12 7 3 3 3 3
12 6 4 3 3 3
12 5 5 3 3 3
12 5 4 4 3 3
12 4 4 4 4 3
11 8 3 3 3 3
11 7 4 3 3 3
11 6 5 3 3 3
11 6 4 4 3 3
11 5 4 4 4 3
11 5 5 4 3 3
11 4 4 4 4 4
10 9 3 3 3 3
10 8 4 3 3 3
10 7 5 3 3 3
10 7 4 4 3 3
10 6 5 4 3 3
10 6 4 4 4 3
10 5 4 4 4 4
10 6 6 3 3 3
10 5 5 4 4 3
10 5 5 5 3 3
9 8 5 3 3 3
9 8 4 4 3 3
9 7 6 3 3 3
9 7 5 4 3 3
9 7 4 4 4 3
9 6 5 4 4 3
9 6 5 5 3 3
9 6 4 4 4 4
9 6 6 4 3 3
9 5 5 4 4 4
9 5 5 5 4 3
9 9 4 3 3 3
8 7 6 4 3 3
8 7 5 5 3 3
8 7 4 4 4 4
8 6 5 4 4 4
8 6 5 5 4 3
8 5 5 5 4 4
8 5 5 5 5 3
8 8 6 3 3 3
8 8 5 4 3 3
8 8 4 4 4 3
7 6 5 5 4 4
7 6 6 5 4 3
7 6 6 4 4 4
7 6 6 6 3 3
7 7 6 4 4 3
7 7 6 5 3 3
7 7 5 4 4 4
7 7 5 5 4 3
7 7 7 4 3 3
6 6 5 5 5 4
6 6 6 5 4 4
6 6 6 5 5 3
6 6 6 6 4 3
8 6 6 4 4 3
8 6 6 5 3 3
8 7 7 3 3 3
7 6 5 5 5 3
7 5 5 5 5 4

[smorzando]

Nope... I dont think you missed any. Not that I can see or think of anyway. Its impossible for the number to be between 3 and 18 anyway. The biggest possible integer is - like you have there - 16. Let me know if you figure it out though.

smorz.

[Yog]

Thanks for taking a look. It's part of fairly complex probability problem. For instance, with the above set of numbers that add up to 31, there are a total of 7668 possible combinations that those numbers can be arranged into. That number seems slightly low, but I've checked and can't seem to find an error. I'll probably set up a histogram in a couple of days and take a look at it visually. At any rate, thanks again for looking it over. *scratches head*

[smorzando]

Yick... I hated probability when we did it in math, even though it was only as a two week time filler. There are some parts of math I find fascinating, but probability isn't one of them.

smorz.

[madkill]

So...all of these can't have the exact same numbers in them but in a different order?

for example:

16 3 3 3 3 3
couldn't be rearranged as like...
3 3 3 3 3 16

because in all it'd work it's way backwards which makes it incredibly longer, unless it can and to save time all you do is add up how many of these there are and times it by 6.

but in all Da-amn!..talk about determination! *applauds*

[Yog]

Yep Madkill I've already taken those into account.

For instance:

16 3 3 3 3 3
3 16 3 3 3 3
3 3 16 3 3 3
3 3 3 16 3 3
3 3 3 3 16 3
3 3 3 3 3 16

I count as 6 permutations of 16 3 3 3 3 3. There are 30 permutations of 15 4 3 3 3 3, and there are 270 permutations of 8 7 6 4 3 3. So all I'm looking for is a combination of 6 integers between 3 and 18 that add up 31 and not in my list already.

[madkill]

but what I'm curious about is this.

Once you have all of them in the way you have it now without the rearrangement of numbers etc. Are you planning on simply (gonna put it into simple algibra)

X = Y x Z = W

Where X sums them all up and thus equals Y which is the total and that gets multiplied by Z which gives the grand total of.

Hmm...I'm curious how one counts them all in the easiest and fastest possible way.
I would probably have it as 3 to 16 laid out as a normal tally chart and cross them off and mark it down on the tally etc.
then total how many there of that number specifically, then add up the totals to one overall total then count as to how many entries there are and take that number to divide with the overall total, then multiply it again by 6.
That's the easiest way, not sure if it'd be the fastest, but you'd get the right answer as too the probability of how many entries there are.

^ I could be completely wrong about that small Hypothisis though.

[Yog]

I have them all listed in a notebook something like this:

1(16)5(3)X6
1(6)5(5)X6
1(15)1(4)4(3)X30
1(14)1(5)4(3)X30
1(14)2(4)3(3)X60
1(13)1(6)4(3)X30
1(13)1(5)1(4)3(3)X120
.
.
.
Total:7668

Basically I'm trying to figure out every combination of six numbers between 3 and 18. The sums will vary between 18 and 108. So far I've figured out up to 33. The tally so far is:
sum: combinations
18: 1
19: 6
20: 21
21: 56
22: 126
23: 252
24: 462
25: 792
26: 1302
27: 1972
28: 2788
29: 4008
30: 5828
31: 7668
32: 10278
33: 14384

The most will occur at 63. The overall shape should be close to the bell curve. All values equal distance from 63 will have the same total, so the total for the sums of 62 and 64 will be the same.

[madkill]

oh yeah of course, I left out that that small factor.
hah...imagine if it was 7666. Math's is a global language used by all, even satanists!..or...whatever else that people conceive as 'evil'

EDIT: Damn it's confusing of all the combinations with the ones that are like this '7 6 6 5 4 3' etc.
Oh the small factor would be that not all can be timez'd by '6' what is this for by the way?..I know it's for a class but...you learn things for a reason so what reason would this serve that is actually used even by the too-well-educated people for a common daily life or even a very high and expertised job.
All I can assume is like a Physicist or something O_O'

[Yog]

Heehee, I think most people believe that math is evil. ;)

Actually this is just a probability exercise. Here is the basic idea. If you have ever played an RPG then there is a system for generating a character. The old systems used to be roll 3 six sided dice and sum them up. There are six stats so you roll three six sided dice six times. The sum of the roll on 3 six sided dice can be between 3 and 18.

So here's the question? If this was your system of generating numbers then what is the probability of getting this series of rolls: 12 10 9 11 14 7?

The probability of getting all 3s or all 18s is 1 in 101,559,956,668,416. But the figuring the other number strings takes a bit of work. ;)

[madkill]

Ah...I understand it more properly now.

So...I take it you're into or partially into the D&D 'playing-field' ?

[Yog]

Yep, I do. The basic idea was that if you wanted to assign stats to an individual based on how rare you wished them to be then what stats would you assign? If you were to want the individual to be 1 in a 1000 what kind of stats should they have? Eventually I'll be able to answer that question.

The real problem is that apparently there is no "closed form solution" to the problem. This means that you have to crank out the answer the old fashioned way. You can't just plug it into a formula and get a result.