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## Comments

6,4563,009Nothing more to see here so just move on.

2,2693,009111, 112, 113, 114, 115, 116, 122, 123, 124, 125, 126, 133, 134, 135, 136, 144, 145, 146, 155, 156, 166, 222, 223, 224, 225, 226, 233, 234, 235, 236, 244, 245, 246, 255, 256, 266, 333, 334, 335, 336, 344, 345, 346, 355, 356, 366, 444, 445, 456, 455, 456, 466, 555, 556, 566, and 666.

Interestingly, this is the sum of the first 6 perfect numbers. There are 6+5+4+3+2+1=21 results beginning with 1, 5+4+3+2+1=15 results beginning with 2, 4+3+2+1=10 results beginning with 3, etc. until we have only one result beginning with 6. 21+10+15+10+6+3+1=56, the total set of possible results when rolling 3d6 *if we consider only the sum total and not the individual die results*. That is the key to what I have been saying--ignore the individual results and look only at the sum. Based on this reasoning, p(rolling an 18) = 1/56.

1683,009If you want a more difficult challenge at some point try analyzing the results of 5d4-2, which will also give you the range of results 3 to 18. The expected value is still 10.5 but the distribution is skewed towards the mean even more than 3d6.

2,029I'll give an example using 2d6, because it's simpler, and trying to get an 11. You say there's only one way to do that, because it's always going to be a 5/6 set. That's wrong, and I'll explain why. So, how can we get that 5/6 set? Well, let's do this sequentially. It's the same thing, since the events are independent. It's definitionally impossible for rolling one die first to influence the probability at the end (and I can demonstrate why it's the same in more explicit detail, if you'd like). When you roll the first die, think about what you can possibly roll that will let us get this 5/6 spread when we roll the second. It's pretty obvious. We have to roll a 5 or a 6. Otherwise there's no way we can get a 5 and a 6. So we have to roll a 5 or a 6 on the first die, or in other words we have a 1/3 chance of getting a roll that allows us to potentially roll an 11 once we add the second die. Now we roll the second die. No matter what the first die rolled, there's precisely one roll that will add up to 11. If we rolled a 5 before, we need a 6 now. If we rolled a 6 before, we need a 5 now. So either way, we have a 1/6 chance of getting out 11. Now, using the definition of sequential probability, we have a 1/3 chance and a 1/6 chance, so the chance of both of them happening is 1/3 * 1/6 = 1/18, which is consistent with the 2/36 predicted by the idea that there are two possible ways to make 11. In other words, this sequential model makes it obvious that 5/6 and 6/5 must be unique events with unique probabilities that must be summed to get the total probability of rolling an 11.

Now, if that doesn't convince you, I urge you to find a pair of d6s and roll them a couple hundred times. See how many 11s you roll, and how many 12s. I don't like using observed frequencies to argue the definition of probability, because that's a remarkably frequentist way of thinking, but it is nonetheless illustrative of the fact that you tend to get roughly twice as many 11s as 12s.

3,009I never knew that discussing probability analysis could make some people so riled up. At least they take it seriously, which is a good thing. This is almost as good as the time on a different board we were discussing the Monty Hall problem, a classic. If you don't like my way of analyzing dice throws then don't use it.

6,4562,029So, @Mathsorcerer, my problem with your analysis is that, while I understand how you got it, I don't understand what definition of probability you think it works under. I know you have a formula that yields that as an answer, and that's cool, but what does it mean? Granted, certain subjectivist sub-camps would accept it as a correct measure of your confidence of rolling an 11 (for example), but they'd also accept any other number you gave between 0 and 1, so that's not terribly useful. I think you have something more concrete in mind, and I simply have no idea what.

Also, if you only care about the sums, why isn't everything on 2d6 just 1/11? If 4/3/1 is the same as 1/4/3, why is 2/2/4 different?

3,0096,4562,029But that's a secondary concern. What I'd really like to know is not what method you're using to get your results, or even the logic behind that method, but rather what your results actually mean. Does your 1/56 answer reflect anything we can expect to see in the real world, or is it just the output of your formula? Because, to be frank, if it's just the output of your formula I don't really see the point.

2,2696 18s is well beyond a 2 standard deviation cutoff though. That being the normal standard in human populations. I honestly can't think of a character who would need it. For any class. Even for a paladin, who have crazy requirements, what would 18 Int add to the character?

Given that I've rolled at least a couple of thousand times, and so have a lot of other people, it would not surprise me had someone rolled an 18/18/18/18/18/18 but it would be vanishingly rare.

*edit* I can contribute to stats discussions, but probably shouldn't do so while drunk. My points stand.

3,009@Jarrakul, the way I am collapsing the results would count both 4/3/1 and 4/2/2 as an 8, which they are, so those 6+3=9 results would all go into the "8" bucket. There isn't anything "real world" about it; rather, it was an exercise in intellectual curiosity that struck me back when I was playing PNP games and I wondered what the results would be if I analyzed 3d6 and cared only about the sum rather than the individual rolls. The results don't necessarily "mean" anything but they do make the case that extreme results--3s, 4s, 17s, 18s-- are *more* likely to happen than traditional analysis would state, given that 1/56 is much more likely than 1/216. I suppose I should put together a short program and have the computer give me 1,000,000 rolls of 3d6 over the weekend to get some experimental data for comparison.

6,4562,029I guarantee you that if you run that experiment you will get roughly the distribution predicted by traditional probability. I guarantee this because current traditional probability was made up by frequentists who used exactly that sort of experiment to define what they thought probability was. I happen to disagree with their definition, but the Bayesian camp would still predict exactly the same results and use the same math to do it.

3,009Either way, at least I have helped make the day more interesting.

251Unfortunately, without knowing the relative likelihood of those results, it's not very illuminating. It's wrong to call this the 'probability' of that outcome.

I'd also disagree with the way that the outcomes are grouped, as I think that if you are going to call (eg) 1-1-3 and 1-3-1 the same result, you might as well go whole hog and say 1-2-2 is as well, as that also adds up to five, which is the final outcome.

So actually we have reduced the outcome from 1/216 to 1/15. Pointlessly, admittedly, but we've done it.

3,0092,029I remain confused by much of what you're saying. Oh well. Life goes on. And for the record, I have no problem with you having an ultimately pointless hobby. I just feel like your bringing it up here added unnecessary confusion to the thread, since it doesn't really answer the question at hand. Of course, then I had to go exacerbate the problem by calling you out on it, thereby derailing the thread, so perhaps I shouldn't talk. :P

3,009I'll have to look up the book's info; it may or may not be in print anymore even though I have a copy--he gave me one himself.

I just ran 50,000 test rolls--a good start--and the experimental results match what we would expect from the standard 216 results, not my 56. *shrug* It is what it is--a wonderful hypothesis ruined by an ugly fact. I don't expect those results to change if I run 500,000 or the full million.

2516,456EDIT again error in formatting

EDIT 3 here's the file https://us.v-cdn.net/5019558/uploads/FileUpload/b6/2f3a2fbd53d916a0a369d5da0722be.7z

@Dee please remove it if sharing such files is not allowed.

On the topic you can simulate all 18s by running the program with 18d6 (i.e. 6 times 3d6).

1,433*click, click, click, click*

"74... 77... 78... 79... 97... 82... Oh,

**** me!"288I recently did a playthrough with all 19 to my stats.. I figured that was proper for the prophesied demigod. GREAT FUN! (I could have gone up to 25 but that felt too high).

42,4385124d6 and drops the lowest die

reference:

http://web.fisher.cx/robert/infogami/Classic_D&D_house_rules

http://catlikecoding.com/blog/post:4d6_drop_lowest

http://prestonpoulter.com/2010/10/19/the-mathematics-behind-4d6-drop-the-lowest/

http://www.sosmath.com/CBB/viewtopic.php?t=49843

http://kill-0.com/duplo/2008/06/02/dd-dice-probabilities/

Under that, the result for one single ability, e.g. strength, comes from a summation of three order statistics. Someone claims that the game simply use "discrete uniform" to generate 3~18, but I do not think this is true. The reason is simply because from my experience, the generated value is more likely to stay "in the middle", not evenly distributed from 3 to 18.

If you would like to calculate the probability that sum=108, it is actually something like "4D6 and get at least 3 dice as 6" and repeat this process 6 times if assuming independency. There are 21 combinations to generate 18, one for (6,6,6,6), and four for each (6,6,6,i), i=1,2,3,4,5, so 1+4*5=21

Here I would like to extend the problem a little bit. Suppose we call the final result as: big3sum, then first let us calculate the mean.

E(big3sum)=E(x1+x2+x3+x4)-E(min)

where min is the smallest one, or we can write it as x(1) as textbook names the order statistics. The above equation holds due to the nature of expectation nomatter whether the variables are independent.

It is easy to see that E(x1+x2+x3+x4)=4*3.5=14. So thenext step is to calculate E(min)

Let us look closely at the probability mass function of min.

(1) P(min < k)=1-P(min>=k)=1-P(x1>=k,x2>=k,x3>=k,x4>=k)=1-((7-k)/6)^4

(2) P(min <= k)=1-P(min>k)=1-P(x1>k,x2>k,x3>k,x4>k)=1-((6-k)/6)^4

Use (2)-(1), we get:

P(min = k)=P(min<=k)-P(min<k)=((7-k)/6)^4-((6-k)/6)^4

Then by definition,

E(min)=sum(k=1 to 6) k*P(min=k)

The result is 1.7554. Thus, E(big3sum)=14-1.7554=12.2446

However, the calculation of variance will be very difficult. We have to consider the covariance.

V(x1+x2+x3+x4-min)=V(x1+x2+x3+x4)+V(min)-2*Cov(x1+x2+x3+x4,min)

V(x1+x2+x3+x4) is again easy to calculate. The answer is 11.66667. V(min) is also easy, since we have got the mass function. The V(min) is 0.91.

Next step is really dirty. Cov(x1+x2+x3+x4,min)=Cov(x(1),x(1))+Cov(x(1),x(2))+Cov(x(1),x(3))+Cov(x(1),x(4)). Here I call min as x(1) in order to make it clearer.

Cov(x(1),x(1)) is exactly V(min), which we have got. In order to calculate other numbers, you have to deal with the joint distribution of order statistics.

But let us do it in a "cheating" way. It is well known that, for n continuous uniform variables u's in [0,1], if we denote u(i) as the ith value in increasing order, then the distribution of u(i) is Beta(i,n+1-i). In our problem, n=4. If we just assume that x(i)=6*u(i), then by calculate the covariance of u(i) and times that by 36 we can "estimate" the covariance from x(i)'s. This is kind of "non-rigorous" but workable. By textbook,

cov(u(i),u(j))=j(n+1-i)/(n+1)^2/(n+2)

If we plug n=4, i=1, j=2,3,4 in to the formula, and times 36, we will get:

cov(x(1),x(2))=0.72

cov(x(1),x(3))=0.48

cov(x(1),x(4))=0.24

Thus, Cov(x1+x2+x3+x4,min)=Cov(x(1),x(1))+Cov(x(1),x(2))+Cov(x(1),x(3))+Cov(x(1),x(4))=2.35

Next we combine everything together, V(x1+x2+x3+x4-min)=V(x1+x2+x3+x4)+V(min)-2*Cov(x1+x2+x3+x4,min)=11.66667+0.9100789-2*2.35=7.9

The big3sum follows some distribution with mean 12.2446 and variance 7.9, although the probability mass function is not easy to get, but luckily we are more interested in the distribution of the sum of "six independent big3sum". By the i.i.d condition and central limit theorem, we may assume that the sum of a character's ability follows a normal asymptotically. The normal has mean 73.4676 and variance 47.4

However, it is worth noting that, the tail probability, such as P(sum>=107) may not be approximated well due to the nature of asymptotic theory, so this normal model is just provided as a general mechanism for this problem. i also wrote some R codes for MCMC study:

generateone<-function(min,add,n)

{

temp=sample(1:6,4*n,replace=TRUE)

temp=matrix(temp,ncol=4)

temp2=rowSums(temp)-apply(temp,1,min)

temp2=temp2+add

(temp2<min)*min+(temp2>=min)*temp2

}

This function generate one of the character's ability. the min set the lower bound. The add is modifier for extra bonus (for elf, the dex has add = +1)

Here is a simple example. We get 100000 observations, do the kernel density estimation and then numerical integral. The result is for P(sum>=94)

N=100000

str=generateone(9,1,N) #9-19

dex=generateone(3,0,N) #3-18

con=generateone(4,1,N) #4-19

int=generateone(1,-2,N) #1-16

wis=generateone(3,0,N) #3-18

cha=generateone(3,0,N) #3-18

total=str+dex+con+int+wis+cha

pdf=density(total)

apdf<- approxfun(pdf$x, pdf$y, yleft=0, yright=0)

integrate(apdf, 93.5, 108)$value

There could be something not correct in the paper, I appreciate any other suggestions. Also, could the programmer gives some hints for how the game actually generates those numbers?