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Woody
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PostPosted: Mon Nov 06, 2006 10:30 pm Reply with quoteBack to top

pipthetroll wrote:

They actually started repeating in sequence. Guardians 125 was a while ago tho, I got 151 and there were no repeats.


That's interesting. It may indicate that there is/was a random seed associated with each account.

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Woody
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PostPosted: Mon Nov 06, 2006 10:40 pm Reply with quoteBack to top

_Indimar_ wrote:
Pulled a good one today.

14-11-12-4-4-14-11 Rolling Eyes


A fine dixie. Wink

15-11-12-10-10-15-11 PL

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MasterBeru
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PostPosted: Tue Nov 07, 2006 1:54 pm Reply with quoteBack to top

In a morbid way, it is funny to read your posts. I am considering returning to the arena after quite a while and I can honestly say that since the late 80s when I first began playing this has always been the complaint. And the suggestions have essentially been the same.

Hmmm... Question Also brings to mind why a number of us kicked around creating a similar game... Confused

Oh Well...
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Deke
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PostPosted: Tue Nov 07, 2006 4:12 pm Reply with quoteBack to top

MasterBeru

Send me a message at DekeYoung@Comcast.net

I have some history for you. Does "Steel Law" ring a bell?

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Woody
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PostPosted: Wed Nov 08, 2006 12:13 pm Reply with quoteBack to top

pipthetroll wrote:
Woody wrote:
Maximillian wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?


Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out! Mad Mad

Oh well, time to dig out the discrete math text... Rolling Eyes


If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.


This (in bold) would be how many 'different sheets' if the same 5 rollups, listed in a different order, were considered a 'different sheet'(a permutation).
The number of different sheets, if the order of list is ignored and redundant rollups are disallowed, is a combination:
(woody's answer)! / 5!(woody's answer - 5)!

with redundant rollups allowed:
(woody's answer + 4)! / 5!(woody's answer - 1)!

Concerning "woody's answer"...
The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,
-with no more than 17(oops! should be 1Cool elements in any container.

(something tells me there's gotta be a better way)

I haven't found any ready-made function to ram this through.

I should be able to find or make a function to do everything before the comma. Concerning the part after, it's kind of a 'negative space' problem to take away the disallowed cases. So far (in my head Smile Sad Confused Mad ), I've got about 18 cases of combo/permu functions which might(I'm probably forgetting something) do the onion-peeling/onion-wrapping. My intiution tells me that this should be reducible to 1 or 2 functions.

If anyone knows any ready-made functions to ram this through, please let me know.

Sigh...an obsessive-compulsive disorder just waiting to happen. Cry

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Last edited by Woody on Mon Nov 13, 2006 8:22 pm; edited 3 times in total
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pipthetroll
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PostPosted: Wed Nov 08, 2006 1:46 pm Reply with quoteBack to top

Woody wrote:
pipthetroll wrote:
Woody wrote:
Maximillian wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?


Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out! Mad Mad

Oh well, time to dig out the discrete math text... Rolling Eyes


If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.


This (in bold) would be how many 'different sheets' if the same 5 rollups, listed in a different order, were considered a 'different sheet'(a permutation).
The number of different sheets, if the order of list is ignored and redundant rollups are disallowed, is a combination:
(woody's answer)! / 5!(woody's answer - 5)!

with redundant rollups allowed:
(woody's answer - 4)! / 5!(woody's answer - 1)!

Concerning "woody's answer"...
The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,
-with no more than 17 elements in any container.

(something tells me there's gotta be a better way)

I haven't found any ready-made function to ram this through.

I should be able to find or make a function to do everything before the comma. Concerning the part after, it's kind of a 'negative space' problem to take away the dissallowed cases. So far (in my head Smile Sad Confused Mad ), I've got about 18 cases of combo/permu functions which might(I'm probably forgetting something) do the onion-peeling/onion-wrapping. My intiution tells me that this should be reducible to 1 or 2 functions.

If anyone knows any ready-made functions to ram this through, please let me know.

Sigh...an obsessive-compulsive disorder just waiting to happen. Cry


I think you'll find yourself playing with a few sets of double integrals there, this is one of the fun project type problems.
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Woody
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PostPosted: Wed Nov 08, 2006 2:22 pm Reply with quoteBack to top

pipthetroll wrote:

I think you'll find yourself playing with a few sets of double integrals there, this is one of the fun project type problems.


...in younger years, I never would have considered doing such things on a day off...but I guess this is what age and boredom does...

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Deke
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PostPosted: Wed Nov 08, 2006 3:16 pm Reply with quoteBack to top

How Deke would generate rollups

Assuming that we would want to have a full assortment of sizes, I would first run a subroutine to determine size. Since I am not Ed Schooner and prefer d100 to 3d6 I would create the distribution I wanted and assign those values to a d100 array (since modern PC can handle data in arrays, and this is not 1983.)

A) Roll a random d100 and assign size as per the following table. We can argue about the distribution of sizes later.

I think Ed rolled 3d8 and rerolled values of “8.” That would be an “Ed Method.”

Size (Weight) [range]
Size 3 (5) [1 to 5]
Size 4 (5) [6 to 10]
Size 5 (5) [11 to 15]
Size 6 (5) [16 to 20]
Size 7 (5) [21 to 25]
Size 8 (5) [26 to 30]
Size 9 (6) [31 to 36]
Size 10 (6) [37 to 42]
Size 11 (6) [43 to 48]
Size 12 (6) [49 to 54]
Size 13 (6) [55 to 60]
Size 14 (5) [61 to 65]
Size 15 (5) [66 to 70]
Size 16 (5) [71 to 75]
Size 17 (5) [76 to 80]
Size 18 (5) [81 to 85]
Size 19 (5) [86 to 90]
Size 20 (5) [91 to 95]
Size 21 (5) [96 to 100]

I would then use a snowfall method for distributing the remaining points.

Distributing 51 to 69 additional points in 6 areas

B) Choose a random 1 in 6 area. If this area is less than 18 then add 4 to the area. If it is 17 or more, choose a different area. Perform this action 5 times (to distribute 20 points.)
C) Choose a random 1 in 6 area. If this area is less than 19 then add 3 to the area. If it is 18 or more, choose a different area. Perform this action 5 times (to distribute 15 points. 35 points total.)
D) Choose a random 1 in 6 area. If this area is less than 20 then add 2 to the area. If it is 19 or more, choose a different area. Perform this action 5 times (to distribute 10 points. 45 points total.
E) If the total of the 6 random area + size is less than 72, Choose a random 1 in 6 area. If this area is less than 21 then add 1 to the area. If it is 21, choose a different area. Perform this action again until total of 6 random areas + size = 72

If Wit + Will < 17 then discard and generate a new candidate. I would argue for < 18 but the group seems to have chosen 17 as the rejection point.

A true snowflake patter would add 1 to random areas many times, however that generates too flat (too boring) a distribution. Adding larger amounts to a single area generates a more varied distribution.

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Woody
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PostPosted: Wed Nov 08, 2006 3:43 pm Reply with quoteBack to top

Deke wrote:
How Deke would generate rollups
I think Ed rolled 3d8 and rerolled values of “8.” That would be an “Ed Method.”

FYI

Rolling 3d8 and re-rolling 8's is isomorphic to rolling 3d7.

3d7 distribution doesn't match the existing data (for size)

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Deke
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PostPosted: Wed Nov 08, 2006 4:06 pm Reply with quoteBack to top

I agree that 3d7 is not representative of our current pool.

I agree with Drake that 2d10 +1 looks good.

What I was trying to point out is that Ed would do something involving dice, that Ed hated percentile dice, and that the computers of the age could not handle arrays and lookup tables.

And I can definetly see Ed rolling dice and re-rolling some of them, on the floor of his apartment, by himself, for hours on end.

So d10 + d10 + 1 is likely.

Regardless, I do not see the rollup pool changing, as much as that would make me happy.

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Caesar
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PostPosted: Thu Nov 09, 2006 1:59 pm Reply with quoteBack to top

Just got a Size 19, 6 Wit, 3 Will "god" back from the Dark Arena. Does anyone want to buy this warrrior? I'll start the bidding @ $100.00...
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Woody
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PostPosted: Sun Nov 12, 2006 10:35 am Reply with quoteBack to top

Woody wrote:
Maximillion wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?


The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,
-with no more than 17 elements in any container.
(something tells me there's gotta be a better way)


I thought about this part some.
Looking at it from a different direction, it's just a combination w/repetition:
55! / 7! (48!)

I pretty sure how to get this part, but it will take more than scribbling on a Subway sandwich napkin to figure out.

note: these calculations involve 7 stats, not 6 as in the original post

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Woody
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PostPosted: Sun Nov 12, 2006 2:23 pm Reply with quoteBack to top

pipthetroll wrote:
Woody wrote:
Maximillian wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?


Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out! Mad Mad

Oh well, time to dig out the discrete math text... Rolling Eyes


If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.


The results are in.

There are: edit...(a big recursive mess) Possible rollups. (Woody's answer)

There are: (woody's answer + 4)! / 5!(woody's answer - 1)! possible team sheets
(if redundant rollups are allowed, and the order of appearance is ignored)

In English:

Possible different rollups = 402,865

Possible different team sheets =
88,435,620,344,256,075,183,711,573


Hmmm...
I think I'll call these
"Woody's Number" and "Woody's insanely huge number".

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Last edited by Woody on Sun Nov 26, 2006 7:29 pm; edited 4 times in total
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Woody
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PostPosted: Sun Nov 12, 2006 8:38 pm Reply with quoteBack to top

Woody wrote:
Possible different team sheets =
88,435,620,344,256,075,183,711,573



I'd like to see some alliance try to log all these. Twisted Evil

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Woody
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PostPosted: Sun Nov 26, 2006 8:51 pm Reply with quoteBack to top

Hopefully, that was the last edit. Rolling Eyes
Eliminating redundant sets and staying under the cap required a nasty little recursive mess.
I'll need to put it into a spreadsheet to verify the arithmetic.

The search did, however, provide some interesting insight and generated an idea.

There may be a way to improve the DA rollup pool without altering the size distribution or doing anything else drastic.

I'll start a new thread explaining the idea.

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