Distribution of bonuses in standard characters

Waldo wrote:
> My group usually generates characters by the standard
> best-three-of-four method. (I offer a point buy, but few of my players
> care to take it.)
>
> Now, according to the PHB, you take the stats you get with this method,
> as long as the character has a net bonus of +1. So, if by some fluke
> you rolled all 11s or lower, you could throw the character out and roll
> again. But if you rolled five 10s and one 12, you'd have to play it.
> (This has actually happened.)
>
<snip>
> A distribution seems scriptable, but the script would be far beyond my
> modest powers. Is there anyone who'd care to try?
>

I wrote a quick Matlab script to generate 100,000 sets of random stats
drawn using 4d6 best 3, then calculated the net character bonuses.
What I find is a distribution of:

Bonus Probability
-6 0.0001
-5 0.0004
-4 0.0010
-3 0.0026
-2 0.0063
-1 0.0127
0 0.0231
1 0.0384
2 0.0589
3 0.0836
4 0.1053
5 0.1189
6 0.1252
7 0.1176
8 0.1004
9 0.0781
10 0.0545
11 0.0353
12 0.0203
13 0.0101
14 0.0045
15 0.0017
16 0.0007
17 0.0002
18 0.0001

So, roughly 2.3% of characters will have net negative bonus, 57% will
have between +4 and +8, and around 0.25% will have a whopping +15 or
greater.
 >> Stay informed about: Distribution of bonuses in standard characters 

Waldo wrote:
> 1) Using best-three-of-four, what proportion of characters will be
> discards (net bonus of 0 or less)? Experience suggests that it's on
> the order of 1%, but I wonder what the correct figure is.
>
> 2) What proportion of characters will be minimally playable -- that
> is, having exactly +1 net bonus?
>
> 3) Is it possible to generate a distribution?

I should have read more carefully before I replied.

I redid my simulation using 1,000,000 random characters....

1) looks like around 4.5% of rolled characters will be discarded due to
having net bonus less than +1.

2) 3.9% of all random characters will be "minimally playable" (net
bonus of exactly +1).

3) see above.

With 1 million random characters, I get a range of net bonus from -9 (4
chars out of 1,000,000) up to +20 (3 chars out of 1,000,000). 56.6% of
the distribution lies between +4 and +8.
 >> Stay informed about: Distribution of bonuses in standard characters 

On Oct 25, 8:19 am, "WDS" <B... DeleteThis @seurer.net> wrote:
> On Oct 25, 6:44 am, "Waldo" <peggolia... DeleteThis @yahoo.com> wrote:
>
> > A distribution seems scriptable, but the script would be far beyond my
> > modest powers. Is there anyone who'd care to try?

> For 24d6 there are 4,738,381,338,321,616,896 (6^24) possible
> combinations so any script/program is going to take a looooooong time
> to run to compute them all. Someone who is good at statistics can
> probably come up with a formula to just compute it.

Since some others posted results of non-complete simulations I thought
I'd take a crack at it. I wrote a C++ program in a few minutes to do
it.

There were 100000000 possible characters and of those 9281478
(9.28148%) were rej
ects (bonuses summed < 1)
-24: 0 0%
-23: 0 0%
-22: 0 0%
-21: 0 0%
-20: 0 0%
-19: 0 0%
-18: 0 0%
-17: 0 0%
-16: 0 0%
-15: 5 5e-006%
-14: 9 9e-006%
-13: 31 3.1e-005%
-12: 133 0.000133%
-11: 495 0.000495%
-10: 1796 0.001796%
-9: 5396 0.005396%
-8: 15544 0.015544%
-7: 40032 0.040032%
-6: 95532 0.095532%
-5: 210658 0.210658%
-4: 434048 0.434048%
-3: 828566 0.828566%
-2: 1472762 1.47276%
-1: 2431192 2.43119%
0: 3745279 3.74528%
1: 5374434 5.37443%
2: 7175761 7.17576%
3: 8927191 8.92719%
4: 10325209 10.3252%
5: 11085649 11.0856%
6: 11035081 11.0351%
7: 10155148 10.1551%
8: 8611797 8.6118%
9: 6707979 6.70798%
10: 4770779 4.77078%
11: 3089134 3.08913%
12: 1804901 1.8049%
13: 946696 0.946696%
14: 440015 0.440015%
15: 180202 0.180202%
16: 63454 0.063454%
17: 19106 0.019106%
18: 4786 0.004786%
19: 1033 0.001033%
20: 146 0.000146%
21: 19 1.9e-005%
22: 2 2e-006%
23: 0 0%
24: 0 0%

Here's the C++ source for it (if you see a bug, let me know):

#include <iostream>
#include <ctime>
#include <cstdlib>

inline unsigned d6() {
return 1+unsigned(6*std::rand()/(RAND_MAX + 1.0));
}


inline unsigned sum(unsigned d1, unsigned d2, unsigned d3, unsigned d4)
{
unsigned lowest = d1;
if (d2 < lowest)
lowest = d2;
if (d3 < lowest)
lowest = d3;
if (d4 < lowest)
lowest = d4;
return d1+d2+d3+d4-lowest;
}

inline int bonus(unsigned attr) {
return attr/2 - 5;
}

int main (int argc, char *argv[])
{
std::srand((unsigned)std::time(0));
unsigned long long rejected=0;
unsigned long long ok=0;

unsigned vals[51]; //-4*6..+4*6 == 51
for (int tot=-24; tot<=24; ++tot) {
vals[tot+24] = 0;
}

for (int cnt=1; cnt <= 100000000; ++cnt) {
unsigned str=sum(d6(),d6(),d6(),d6());
unsigned dex=sum(d6(),d6(),d6(),d6());
unsigned con=sum(d6(),d6(),d6(),d6());
unsigned iq= sum(d6(),d6(),d6(),d6());
unsigned wis=sum(d6(),d6(),d6(),d6());
unsigned chr=sum(d6(),d6(),d6(),d6());
int tot=
bonus(str)+bonus(dex)+bonus(con)+bonus(iq)+bonus(wis)+bonus(chr);
++vals[tot+24];
if (tot < 1)
++rejected;
else
++ok;
}

std::cout << "There were " << rejected+ok << " possible characters and
of those "
<< rejected << " (" << 100.0*rejected/(rejected+ok) << "%)
were rejects (bonuses summed < 1)" << std::endl;
for (int tot=-24; tot<=24; ++tot) {
std::cout << " " << tot << ": " << vals[tot+24] << " " <<
100.0*vals[tot+24]/(rejected+ok) << "%" << std::endl;
}
return 0;
}
 >> Stay informed about: Distribution of bonuses in standard characters 

On Oct 25, 10:04 am, tom.merr....RemoveThis@gmail.com wrote:
> 1) looks like around 4.5% of rolled characters will be discarded due to
> having net bonus less than +1.
>
> 2) 3.9% of all random characters will be "minimally playable" (net
> bonus of exactly +1).

You sure about those? I got different results (9.3% with < +1 total
bonus and 5.4% exactly +1 bonus). I posted the code I ran so maybe
someone can see if I made a mistake.
 >> Stay informed about: Distribution of bonuses in standard characters 

Hmm, I see a bunch of Monte-Carlo's, but no one seems to have noticed
that if you calculate the number of possible bonuses for ONE ability it
requires only O(1,296) opperations, that all bonuses are themselves IID
variables, and that all possible combinations of the nine possible
bonuses is only O(531,441) opperations.

Doesn't ANYONE optimize code anymore? That's less than a second.

Anyway, ways for each bonus to come up: (Works best in a monospaced
font)
-24 1.
-23 84.
-22 3294.
-21 80578.
-20 1388175.
-19 18050268.
-18 185541917.
-17 1559605314.
-16 11002015533.
-15 66478746170.
-14 349728877185.
-13 1623079713930.
-12 6716876468443.
-11 25005427327026.
-10 84350762241015.
-9 259379137929744.
-8 730681209430596.
-7 1893432711980094.
-6 4528633089038838.
-5 10024772116581400.
-4 20583986375146544.
-3 39271575791615784.
-2 69706866594860736.
-1 115212068313686160.
0 177400108223399744.
1 254500213100555072.
2 340081685699284928.
3 423015557179232448.
4 489274276278359104.
5 525451624872764608.
6 522948395041784384.
7 481138310079673664.
8 407994082439260160.
9 317693931831752320.
10 226149103290415200.
11 146376010445986592.
12 85587360645401376.
13 44854171301711776.
14 20869746026205772.
15 8521692413722290.
16 3010813601530293.
17 904558812093702.
18 226169010828433.
19 45810830830716.
20 7261965555075.
21 860029006914.
22 71029517706.
23 3626681688.
24 85766121.

Odds for each bonus:
-24 0.2110425439E-18
-23 0.1772757369E-16
-22 0.6951741396E-15
-21 0.1700538610E-13
-20 0.2929639834E-12
-19 0.3809374477E-11
-18 0.3915723816E-10
-17 0.3291430729E-09
-16 0.2321893346E-08
-15 0.1402984371E-07
-14 0.7380767191E-07
-13 0.3425388718E-06
-12 0.1417546697E-05
-11 0.5277208994E-05
-10 0.1780159944E-04
-9 0.5474003310E-04
-8 0.1542048212E-03
-7 0.3995948562E-03
-6 0.9557342475E-03
-5 0.2115653409E-02
-4 0.4344096848E-02
-3 0.8287973258E-02
-2 0.1471111445E-01
-1 0.2431464798E-01
0 0.3743897013E-01
1 0.5371037239E-01
2 0.7177170408E-01
3 0.8927427929E-01
4 0.1032576879
5 0.1108926476
6 0.1103643596
7 0.1015406529
8 0.8610410905E-01
9 0.6704693555E-01
10 0.4772708206E-01
11 0.3089156561E-01
12 0.1806257432E-01
13 0.9466138416E-02
14 0.4404404292E-02
15 0.1798439645E-02
16 0.6354097616E-03
17 0.1909003928E-03
18 0.4773128339E-04
19 0.9668034276E-05
20 0.1532583684E-05
21 0.1815027094E-06
22 0.1499025011E-07
23 0.7653841293E-09
24 0.1810030036E-10

So non-positive comes up ~9.28% of the time, +1 is about 5.37%, and +10
or more is about 11.32%.

DougL

On Oct 25, 6:44 am, "Waldo" <peggolia....DeleteThis@yahoo.com> wrote:
> My group usually generates characters by the standard
> best-three-of-four method. (I offer a point buy, but few of my players
> care to take it.)
>
> Now, according to the PHB, you take the stats you get with this method,
> as long as the character has a net bonus of +1. So, if by some fluke
> you rolled all 11s or lower, you could throw the character out and roll
> again. But if you rolled five 10s and one 12, you'd have to play it.
> (This has actually happened.)
>
> So, a couple of questions occur.
>
> 1) Using best-three-of-four, what proportion of characters will be
> discards (net bonus of 0 or less)? Experience suggests that it's on
> the order of 1%, but I wonder what the correct figure is.
>
> 2) What proportion of characters will be minimally playable -- that
> is, having exactly +1 net bonus?
>
> 3) Is it possible to generate a distribution?
>
> The possible range of character bonuses runs from -24 (straight 3s) to
> +24 (straight 18s). However, these are somewhat unlikely. (1) The
> average character will be about +6, and almost all characters will be
> between -2 and +15.
>
> A distribution seems scriptable, but the script would be far beyond my
> modest powers. Is there anyone who'd care to try?
>
> Thanks in advance,
>
> Waldo
>
> (1) For straight 3s, you'd have to role four 1s, six times in a row.
> Odds: 1/6^24, or about one in a hundred quintillion. Straight 18s is
> much easier: 5^6/6^24, or about one in ten quadrillion. If you rolled
> a new character every second, it would take you around a million years.
> (As opposed to the straight-3s character, for which you'd still be
> rolling when the sun left the main sequence 6 billion years from now.)
 >> Stay informed about: Distribution of bonuses in standard characters 

On Oct 25, 10:44 am, "WDS" <B... RemoveThis @seurer.net> wrote:
> On Oct 25, 10:04 am, tom.merr... RemoveThis @gmail.com wrote:
>
> > 1) looks like around 4.5% of rolled characters will be discarded due to
> > having net bonus less than +1.
>
> > 2) 3.9% of all random characters will be "minimally playable" (net
> > bonus of exactly +1).

> You sure about those? I got different results (9.3% with < +1 total
> bonus and 5.4% exactly +1 bonus). I posted the code I ran so maybe
> someone can see if I made a mistake.

I tried running an exact solution. And within roundoff my results match
yours.

DougL
 >> Stay informed about: Distribution of bonuses in standard characters 

WDS wrote:
> On Oct 25, 10:04 am, tom.merr....DeleteThis@gmail.com wrote:
> > 1) looks like around 4.5% of rolled characters will be discarded due to
> > having net bonus less than +1.
> >
> > 2) 3.9% of all random characters will be "minimally playable" (net
> > bonus of exactly +1).
>
> You sure about those? I got different results (9.3% with < +1 total
> bonus and 5.4% exactly +1 bonus). I posted the code I ran so maybe
> someone can see if I made a mistake.

Nope. You were right.

I messed up the bonus truncation for stats < 8. All fixed now, and, my
results agree with yours.
 >> Stay informed about: Distribution of bonuses in standard characters 

Waldo wrote:
> My group usually generates characters by the standard
> best-three-of-four method. (I offer a point buy, but few of my
> players care to take it.)
>
> Now, according to the PHB, you take the stats you get with this
> method, as long as the character has a net bonus of +1. So, if by
> some fluke you rolled all 11s or lower, you could throw the character
> out and roll again. But if you rolled five 10s and one 12, you'd
> have to play it. (This has actually happened.)

Actually, according to the PHB you can also re-roll if your highest score is
13 or lower.

--
Mark Blunden.
 >> Stay informed about: Distribution of bonuses in standard characters 

"Waldo" <peggoliathy.DeleteThis@yahoo.com> wrote in news:1161776680.490048.246890
@m7g2000cwm.googlegroups.com:

> 3) Is it possible to generate a distribution?
>
> The possible range of character bonuses runs from -24 (straight 3s) to
> +24 (straight 18s). However, these are somewhat unlikely. (1) The
> average character will be about +6, and almost all characters will be
> between -2 and +15.
>
> A distribution seems scriptable, but the script would be far beyond my
> modest powers. Is there anyone who'd care to try?

May not be exactly what you are looking for, but I found this online:

http://www.math.uah.edu/stat/objects/experiments/DiceExperiment.xhtml

It will roll dice for you, and generate a whole bunch of statistics. I
will let it run for a couple of hours and post the results. Preliminary
data with almost 11k runs gave the following distributions:



3_____0.00463_____0.00654
4_____0.01389_____0.01318
5_____0.02778_____0.02811
6_____0.04630_____0.0483
7_____0.06944_____0.07125
8_____0.09722_____0.09254
9_____0.11574_____0.11531
10____0.12500_____0.13006
11____0.12500_____0.12324
12____0.11574_____0.11568
13____0.09722_____0.09946
14____0.06944_____0.06609
15____0.04630_____0.04609
16____0.02778_____0.02701
17____0.01389_____0.01272
18____0.00463_____0.00442


The first column is, of course, the number rolled. The second is the
percentage of the time that the number will be rolled according to
theory, while the third is how often it actually came up. Multiply the
last two columns by 100 to get it in %.

The experimental results will only approach the theoretical results as
the number of die rolls are increased.
 >> Stay informed about: Distribution of bonuses in standard characters 

WDS wrote:
> On Oct 25, 8:19 am, "WDS" <B....TakeThisOut@seurer.net> wrote:
>> On Oct 25, 6:44 am, "Waldo" <peggolia....TakeThisOut@yahoo.com> wrote:
>>
>>> A distribution seems scriptable, but the script would be far beyond
>>> my modest powers. Is there anyone who'd care to try?
>
>> For 24d6 there are 4,738,381,338,321,616,896 (6^24) possible
>> combinations so any script/program is going to take a looooooong time
>> to run to compute them all. Someone who is good at statistics can
>> probably come up with a formula to just compute it.
>
> inline unsigned d6() {
> return 1+unsigned(6*std::rand()/(RAND_MAX + 1.0));
> }

Minor FYI: std::rand isn't really good enough for this large a data set
where you're trying to extract statistical numbers... often you'll get funny
interactions between two numbers generated which throws the generation off.

Since you're using C++, let me point you over to a wonderful, and very easy,
library that has better generators for this purpose:

http://www.boost.org/

Specifically this:

http://www.boost.org/libs/random/index.html

And the mersenne twister is generally considered "good enough" for
statistical analysis, but there are several you can choose from, and they're
"drop in replaceable" which means it's easy to try one, then another,
without having to rewrite huge chunks of code.

Enjoy!

--
Reginald Blue
"I have always wished that my computer would be as easy to use as my
telephone. My wish has come true. I no longer know how to use my
telephone."
- Bjarne Stroustrup (originator of C++) [quoted at the 2003
International Conference on Intelligent User Interfaces]
 >> Stay informed about: Distribution of bonuses in standard characters 

"DougL" <lampert.doug.TakeThisOut@gmail.com> wrote in message
> Hmm, I see a bunch of Monte-Carlo's, but no one seems to have noticed
> that if you calculate the number of possible bonuses for ONE ability it
> requires only O(1,296) opperations, that all bonuses are themselves IID
> variables, and that all possible combinations of the nine possible
> bonuses is only O(531,441) opperations.
>
> Doesn't ANYONE optimize code anymore? That's less than a second.

I built the same thing. Took longer to run, though, because I was running
it through an interpreter.

I built it previously, years ago, to figure out how many of every possible
class there would be. Long since lost the code, though.
 >> Stay informed about: Distribution of bonuses in standard characters 

Marcel Beaudoin wrote:
> May not be exactly what you are looking for, but I found this online:
>
> http://www.math.uah.edu/stat/objects/experiments/DiceExperiment.xhtml
>
> It will roll dice for you, and generate a whole bunch of statistics. I
> will let it run for a couple of hours and post the results. Preliminary
> data with almost 11k runs gave the following distributions:

There is also Torben Mogensen's "Roll" programming language. I have not
used it myself, but Torben often uses it to help others on the
RPG-Create mailing list.

< http://www.diku.dk/~torbenm/ >

--
Peter Knutsen
sagatafl.org
 >> Stay informed about: Distribution of bonuses in standard characters 

Wow, that was great. Thank you, DougL, Tom, WDS, and everyone else who
responded.

Some thoughts.

1) Almost 10% of rolled characters are discards! Who knew? I would
have sworn it was more like 1%, but I guess the brain just skips over
the rerolls.

2) So, the most common single character type is +5 (11.1%), but the
mean and median are +6 (11%). That makes sense.

3) Strong characters of up to +12 are not too unusual, but after that
the numbers drop off fast.

My general philosophy is "I may believe you if you say you rolled a 1
on a d20. If you say you rolled a 1 on a d100, I want to see the
roll." Following that, if someone brings me a +10 character (4.7%
chance) I might roll my eyes and accept it, but if they bring me a +13
character (0.95% chance) I'm going to say, sorry, have to see that.

4) One thing that strikes me is what a wide range is possible here.
It's really not hard to roll a +2 or lower character. After accounting
for discards, these are almost 14% of all valid characters. On the
other hand, it's not hard to roll a +8 or higher character; about a
quarter of all characters will be this good.

So, if you have a party of four, and each rolls best-three-of-four,
there's a very good chance that the best character will be +5 or more
better than the worst character. Bob rolls 7, 8, 10, 12, 13, 14 while
Adrian rolls 10, 11, 12, 13, 15, 17: this is perfectly likely.
Adrian's character is at least a full ECL better than Bob's, but the
RAW says they march into the dungeon side by side.

If Bob is a real grognard he may rise to the challenge and even have
fun, but it would be easy to see him getting a little unhappy. I might
get a little unhappy myself. I've never liked point buy -- IMO it
tends to produce bland, cookie-cutter optimized characters -- but I see
the logic of it more clearly now. Best-three-of-four gives you a
pleasant randomness, but the not-so-pleasant side effect is that you
may have a PC who is one or even two ECLs behind another.


Waldo
 >> Stay informed about: Distribution of bonuses in standard characters