Probability in Char. Gen.

[Charwoman Gene]

This is quite technical. I think about this stuff. Sue me.

I am curious about researching the expected DMG point value of a "Standard" "4d6, drop the lowest"(4d6dl) D&D character. The trick is, I don't want to use statistics, I want to count up, using integer arithmetic, the actual point values for the roll of the dice, and divide by the number of possible ways the dice can be thrown.

Now, calculating the expected value for a single 4d6dl roll is fairly simple. I've done an Excel spreadsheet that does it for me. 12.24 is the result I get. Things get complicated because:

Point Value vs. Ability Score is nonlinear.
D&D skews scores higher by rejecting "worthless" characters.​

The definition of "worthless" is +0 or lower total modifier, or no ability score above 14.

This complicates matters as now we have to look at rolls for entire characters, not just individual 4d6dl rolls. Knwing the individual expected value is useless.
I used a spreadsheet with 1296 lines to calculate the Expected Value given above, using an increasing index, MOD 6, DIV 6, to calculate what the dice needed to show. [6 positions per die with 24 independent dice, 6^24 possibilities.] I'd need a 4,738,381,338,321,616,896 line spreadsheet to calculate this. (Excel stops at 65,536.) That is 4 quintillion possibilities. Yay.


How to solve this problem?

I don't know enough hard-core (Calculus) statistics to be comfortable with them in this case.

Okay, lets say I make a java program to do this brute force. and assume we can calculate the values once out of every 20 CPU actual cycles. (I don't know how good a number that is, I'm not that much of a hardware guy.) Crap. That's like 15 million hours. Which is like 1700 years.

I guess I should look into distributed parallel computing, huh?

Donate your CPU cycles now!

Anyone see a way to get me out of this trap?

 

[dcollins]

The trick is, I don't want to use statistics, I want to count up, using integer arithmetic, the actual point values for the roll of the dice, and divide by the number of possible ways the dice can be thrown.

Amusingly, there's no difference. Counting up indepedent outcomes and dividing by the total is the first element of statistics.

I have a 4d6 probability curve on my website, with counts of each score. Possibly you can use that as beginning source data (multiply each option by 6, reject options below the "worthless threshold", average the remaining options):
http://superdan.net.home.comcast.net/dndmisc/4d6curve.html

 

[Charwoman Gene]

Amusingly, there's no difference. Counting up indepedent outcomes and dividing by the total is the first element of statistics. ]

Oh, here's the rub.

The 6 4d6dl rolls are NOT independent in aggregate stats, because ALL are rejected based off of holistic properties comparing them all.

I need to work out how to you assign point values to low scores?

I think dcollins, while wrong about his curve being truly accurate for D&D due to the reroll rules, may have pointed me in the right direction for my brain to figure it out myself.

I need to do some work to back up my points though.

The page linked above is well done for the issue it addresses. Can I use your data for the individual probabilities for each number in my solution?

 

[dcollins]

Yes, the probabilities are accurate for any given 4d6 roll.

 

[Thanee]

Calculating it would be fairly easy. You just take all possible results (3-18) and calculate the probability to get it (taken from dcollins' table) and then weigh it with the PB value of that score, but I think you are already doing that, given that other thread of yours (for this purpose it's probably best to use a linear negative PB score, btw).

So you get (16x21+13x54+10x94+...-5x1)/1296 ~ average PB value of one 4d6dl roll.

The numbers in the sum are PB value of the score (16 for 18; 13 for 17; 10 for 16; ...; -5 for 3) multiplied by the number of combinations, which result in that score (21 for 18; 54 for 17; 94 for 16; ...; 1 for 3).

Multiply that number by 6 and you got the approximated average PB value of the 4d6dl method.

The result is 6293/1296*6=29.134 (if all scores of 8 or lower count as 0 PB) and 6166/1296*6=28.546 (if scores below 8 count as -1, -2, -3, -4 and -5 PB respectively).

(Hopeless characters and re-rolls not figured in yet, but they only increase those numbers and certainly not by much. And a bit higher for 4d6dl is ok, anyways, since you do not have the same flexibility in distributing the points among the abilities.)

Therefore PB 28 is most close to 4d6dl. PB 25 is more like low-powered, not average.

Bye
Thanee

 

[Charwoman Gene]

My hypothesis is that 4d6dl with rerolling worthless characters is roughly equivalent to 30 or 32 point buy. I do understand that I could be wrong, and it doesn;t matter, but I am academically fascinated by the challenge of figuring it out. Unfortunately, I need some time to work out all my details to test my hypothesis.

 

[Crothian]

are you using the same defintions of "worthless character" as the PHB?

 

[Thanee]

Yep, filtering out the "worthless" characters (@Crothian: yes, obviously by the book) is a bit more of a hassle.

You need to figure out the probability to roll up a worthless character, once you have that, it's about done.

Bye
Thanee

 

[CRGreathouse]

I need to work out how to you assign point values to low scores?

I'd work this out twice, once counting all scores under 8 as 0 points and once counting them as
(score - 8) points. These should set up reasonable bounds, as their actual value should be between these two.

I used a spreadsheet with 1296 lines to calculate the Expected Value given above, using an increasing index, MOD 6, DIV 6, to calculate what the dice needed to show. [6 positions per die with 24 independent dice, 6^24 possibilities.] I'd need a 4,738,381,338,321,616,896 line spreadsheet to calculate this. (Excel stops at 65,536.) That is 4 quintillion possibilities. Yay.

Using table lookup from the 1296 problem above, you can reduce the brute force calculation to from 6^24 = 1296^6 to 16^6 (16 ability scores from 3 to 18). 16^6=16,777,216 isn't too bad to work with. (Of course, each array will need to hold large numbers---probably enough that long variables rather than normal ints would need to be used.)

 

[Crothian]

Yep, filtering out the "worthless" characters (@Crothian: yes, obviously by the book) is a bit more of a hassle.

I wouldn't say obvious, I've had people on these baords provlaim all manner of character worthless