Post/Author/DateTime	Post
Eliza_Stormwhisper 07-18-05, 05:06 AM	So I punched this code into my TI-83: :ClrHome :{0,0,0,0}->L1 :{0,0,0,0,0,0}->L2 :{0}->L3 :For(C,1,500) :Output(1,1,C) :For(B,1,6) :For(A,1,4) :randInt(1,6)->L1(A) :End :SortA(L1) :sum(L1)-L1(1)->L2(B) :End :mean(L2)->L3(C) :End :ClrHome :Fix 2 :Disp mean(L3) :Float :Stop I ran the program a total of four times, for a total of 2,000 sets of six rolls, or 12,000 rolls, or 48,000 dice being rolled (wow). For each four sets of 500, I got average results of approximately 12.31, 12.25, 12.19, and 12.25, for an overall average of 12.25. Interesting, huh?
J Milburn 07-18-05, 06:58 AM	I am in the progress of working out the odds of all the ability scores coming up, sadly just on pencil and paper. I think it is odd that the average of the roll should just be 11 as far as i can see, but that, as you said, the actualresults always end up higher. The average should be 3, 3, 4, 4, surely?
Eliza_Stormwhisper 07-18-05, 07:45 AM	I am in the progress of working out the odds of all the ability scores coming up, sadly just on pencil and paper. I think it is odd that the average of the roll should just be 11 as far as i can see, but that, as you said, the actualresults always end up higher. The average should be 3, 3, 4, 4, surely? There is a huge difference between theoretical probability and experimental probability. The theoretical average on a six-sided die is 3.5. It doesn't matter how many dice you roll, the theoretical probability will stay the same. So theoretically, on a 4d6 drop lowest, you will end up with an average result of 10.5. On the other hand, this isn't true in practicality, as I noted. There are 1,296 possible results when rolling 4d6. Here are the probabilities (expressed in percentages rounded to the nearest tenth) of getting a particular number (source: my head): 3: 0.1% 4: 0.3% 5: 0.8% 6: 1.6% 7: 2.9% 8: 4.8% 9: 7.0% 10: 9.4% 11: 11.4% 12: 12.9% 13: 13.3% 14: 12.3% 15: 10.1% 16: 7.3% 17: 4.2% 18: 1.6%
Nurgan_the_drunked 07-18-05, 07:47 AM	Unfortunately the average isn't as easy to work out, because the lower end of the scale per dice is often ignored, and the higher end is almost always used (a 1 is unlikely to impact the average, as it will be dropped unless there's another 1 rolled in the same set) so it will skew in favour of higher than the average expected. When applying this to characters, there's another factor: if the rolls are below a certain threshhold (ie unplayable), then they will get rerolled, so the average ability for a character using this method get skewed upwards in that respect as well. You do realise you could have got a more exact result by working through all the possibilities on each dice, which is only about 1,000 combinations, don't you?
Solaris 07-18-05, 08:14 AM	The mean result for the best three of four dice is exactly 15869/1296, or about 12.24. And the previous poster is right: there's not much point in simulating 12,000 rolls to get an incorrect answer when you could more easily iterate through all 1296 possibilities to get the correct one: Die Probability Percent roll (exact) (rounded) ---------------------------- 3 1/1296 0.08 4 1/324 0.31 5 5/648 0.77 6 7/432 1.62 7 19/648 2.93 8 31/648 4.78 9 91/1296 7.02 10 61/648 9.41 11 37/324 11.42 12 167/1296 12.89 13 43/324 13.27 14 10/81 12.35 15 131/1296 10.11 16 47/648 7.25 17 1/24 4.17 18 7/432 1.62
cvazi 07-18-05, 09:19 AM	here is a slightly longer htread on the same topic: http://boards1.wizards.com/showthread.php?t=463429
J Milburn 07-18-05, 09:27 AM	When applying this to characters, there's another factor: if the rolls are below a certain threshhold (ie unplayable), then they will get rerolled, so the average ability for a character using this method get skewed upwards in that respect as well. I personally egnored this when i tried to work it out, and Solaris, where on earth did you get/work them out? That is what i spent hours with a pencil and paper trying to work out (and they match my answers reasonably well by the look of them!)
Solaris 07-18-05, 09:59 AM	Solaris, where on earth did you get/work them out? That is what i spent hours with a pencil and paper trying to work out I wrote a little Lisp program to do it.
celtredleg 07-18-05, 10:50 AM	Non-sarcastic "And??" The numbers are interesting, but I knew the average would be higher than 3d6. I keep waiting for a reason this is important. If it is just for information, then thanks for enlightening me. If it is going somewhere, where is the rest of it???
BadCatMan 07-18-05, 11:39 AM	I generated 1 million stats, and got an average/mean of 12.24. The median is 12. The standard deviation is 2.85, meaning that most scores, ~69% IIRC, will lie between 9.39 (9) and 15.09 (15). I didn't include any other factors like rerolling if the stats are too low. These seem rather subjective, and I don't have time to fiddle (I hate statistics). Hence the averaged average adventurer should have stats of: 12, 12, 12, 12, 13 and 13. Slightly above the average commoner, assuming a simple 3d6, who has an average stat of 10.49, which would be: 10, 10, 10, 11, 11 and 11.
Nurgan_the_drunked 07-18-05, 11:54 AM	Hence the averaged average adventurer should have stats of: 12, 12, 12, 12, 13 and 13. Gonna have to agree with Centredleg here. And? The concept of "average adventurer" is fairly nonsensical, especially if you . The spread of classes is remarkable, and what you have there is something that could be only slightly less immensely unlikely than any other set of statistics, and given the set of duplicates, I wouldn't be surprised if it was more unlikely than a reasonable spread. As my point earlier made, by the rules in the DMG, this is rated as not powerful enough to be a character (no stat higher than +2, I believe), and I've certainly not seen someone with these sort of stats. It is nice to know you've done the exercise (good practise for you, in terms of maths and programming, and the like), but not really too enlightening on its own.
SodaGuy 07-18-05, 11:58 AM	We should all just roll them on real dice. Then we'll see what the dice gods want the average to be.
weasel fierce 07-18-05, 01:01 PM	pretty nice to know. Thanks a bunch.
Mightyflapjack 07-18-05, 01:09 PM	Yet when the players are left to roll the dice themselves, mine seem to have enchanted dice that roll between 14-18 each stat.
Sulaco 07-18-05, 02:10 PM	Well I am a math dunce an I could've told you the result would be higher. The mean of a single d6 is 3.5, so teh average of a 3d6 roll is 10.5 (3.5x3). However, you are adding a 4th die into the equation. There's what, a 50/50 chance of the result of the 4th die being higher than one fo the other three so if you add 50% of 3.5 (which is 1.75) you get 12.25. I'm certain the math is screwed by the result seems self-evident to me.
Dave Stebbins 07-18-05, 07:25 PM	I personally egnored this when i tried to work it out, and Solaris, where on earth did you get/work them out? That is what i spent hours with a pencil and paper trying to work out (and they match my answers reasonably well by the look of them!) I used a spreadsheet, with liberal use of fill down and copy/paste, to list all 1296 equally probable rolls and figure the totals. It took me about 10 minutes about 10 years ago to some up with the exact probablilities, which are as Solaris presented (at least from memory that looks right). I had started out by digging out my old college probability and stats textbooks and fooled around with them for about a half hour before I realized that the key was that every single different, unique, combination of four dice was equally probable and that 6^4 meant there were ony 1,296 combos. That's when I turned to the spreadsheet and had the stats in no time at all. -Dave
Joni-san 07-18-05, 08:43 PM	We should all just roll them on real dice. Then we'll see what the dice gods want the average to be. Quite Right :clap: In related news: One of my friends has a perfect die rolling technique. With 4d6 drop lowest he can get 6's on all of the dice. It is, as they say, all in the wrist.... :twitch: Works (nearly) every time. :blink:
Elzaban the Ethereal 07-18-05, 09:37 PM	I'd like to thank Eliza_Stormwhisper for posting this little bit of code: :For(B,1,6) :For(A,1,4) :randInt(1,6)->L1(A) :End :SortA(L1) :sum(L1)-L1(1)->L2(B) :End That just shaved off several bytes from my various D&D-related programs. I can't believe I never thought of using lists with the sort function.
J Milburn 07-19-05, 06:50 AM	However, you are adding a 4th die into the equation. There's what, a 50/50 chance of the result of the 4th die being higher than one fo the other three so if you add 50% of 3.5 (which is 1.75) you get 12.25. I dont know whether this works, but i looked at it that if the average roll was 3.5, we could take that four dice to have the average results of 3, 3, 4 and 4, drop one of the threes for a result of eleven. I knew this had to be wrong after a few dice rolls, but i still cant see why.
Solaris 07-19-05, 07:54 AM	I knew this had to be wrong after a few dice rolls, but i still cant see why. The mean is just an average. For example, if you wanted to find the mean of x,y,y,z,z,z, you'd simply compute (x + 2y + 3z) / 6. With ability score rolls, it's the same thing. Here's a table showing the number of ways to get each possible outcome with the best three out of four dice: Roll Ways 3 1 4 4 5 10 6 21 7 38 8 62 9 91 10 122 11 148 12 167 13 172 14 160 15 131 16 94 17 54 18 21 Here we have one 3, four 4s, ten 5s, and so on up to twenty-one 18s. To find the mean, we have to add all of these together and then divide by the total number of ways: (1*3 + 4*4 + 10*5 + ... + 21*18) / 1296. The result is about 12.24.
Sulaco 07-19-05, 10:56 AM	Here we have one 3, four 4s, ten 5s, and so on up to twenty-one 18s. To find the mean, we have to add all of these together and then divide by the total number of ways: (1*3 + 4*4 + 10*5 + ... + 21*18) / 1296. The result is about 12.24. What I find so interesting is that I got essentially the same answer (12.25) just by looking at it and seeing what seemed the obvious and self-evident answer. Thing is, I suck at math. I've often wondered if I actually was born with a "math brain" but that school cured me of it. For example, I failed 10th grade math cos I could give the correct results for equations but I could never show my work. I don't know how I got the results, they just seemed "right" (kinda like the example above). If they did cure me they did a damned good job of it. While I can still sometimes look at simple stuff like this and "see" the right answer - even if I still can't explain it right and "show my work" - it is now a rare and increasingly infrequent occurence. Oh well. Guess I'll stick to words...
Solaris 07-19-05, 04:38 PM	Unfortunately, it was just a near-coincidence. If we change the number of sides, or the number of dice, then your results diverge from the actual means. [EDIT: Table slightly off due to programming error. New table follows.] Actual Sulaco Roll mean mean ---------------------- 3/4d4 8.61 8.75 3/4d6 12.24 12.25 3/4d8 15.85 15.75 3/4d10 19.46 19.25 3/4d12 23.07 22.75 3/4d20 37.48 36.75
Sulaco 07-19-05, 04:54 PM	Unfortunately, it was just a near-coincidence. If we change the number of sides, or the number of dice, then your results diverge from the actual means: Actual Sulaco Roll mean mean ---------------------- 3/4d4 8.6379 8.7500 3/4d6 12.2469 12.2500 <--- close 3/4d8 15.8591 15.7500 3/4d10 19.4670 19.2500 3/4d12 23.0724 22.7500 3/4d20 37.4834 36.7500 Umm...what? I made no prediction at all what the mean of any of those apart from the 4d6 one might be. Why are you suggesting otherwise? I would appreciate you not attributing to me something that I did not say or do.
Solaris 07-19-05, 05:00 PM	You presented a method: The mean of a single d6 is 3.5, so teh average of a 3d6 roll is 10.5 (3.5x3). However, you are adding a 4th die into the equation. There's what, a 50/50 chance of the result of the 4th die being higher than one fo the other three so if you add 50% of 3.5 (which is 1.75) you get 12.25. The validity of your method is not dependent upon the number of sides of the dice. If you believe that it holds for d6, then there's absolutely no reason for you not to believe it would hold for other dice. There's nothing d6-specific in the logic. If we apply the argument to d20, we get identical reasoning: The mean of a single d20 is 10.5, so the average of a 3d20 roll is 31.5 (10.5x3). However, you are adding a 4th die into the equation. There's, what, a 50/50 chance of the result of the 4th die being higher than one of the other three so if you add 50% of 10.5 (which is 5.25) you get 36.75. Unfortunately the reasoning itself is not valid, even for d6, even though it is close in that case.
Vanamir 07-20-05, 04:03 AM	There is a huge difference between theoretical probability and experimental probability. The theoretical average on a six-sided die is 3.5. It doesn't matter how many dice you roll, the theoretical probability will stay the same. So theoretically, on a 4d6 drop lowest, you will end up with an average result of 10.5. Sorry if i am nitpicky but this is an entirely wrong statement. The 'theoretical' average of 4d6 drop lowest is not 10.5 since you are throwing away the lowest die not a random die. This way you use only the high results in your statistic and thus get a higher average than 10.5. Besides that a theory is to predict reality or an expperiment. If it does not then the theory is wrong and has to replaced with a refined theory that does predict experimental results.
Sulaco 07-20-05, 09:37 AM	You presented a method: The validity of your method is not dependent upon the number of sides of the dice. If you believe that it holds for d6, then there's absolutely no reason for you not to believe it would hold for other dice. There's nothing d6-specific in the logic. If we apply the argument to d20, we get identical reasoning: As I said, I didn't do the math I just sussed out the result and then tried to bodge some math explain it, math that I pointed out was probably wrong. Really, I don't give a hairy rat's ass what the mean is and, for me, close enough is good enough.