TRPG General Chat - 2018-10-04 (page 2 of 2)

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7:01 PM

@DavidCoffron This a discrete statistics problem. What is the probability that observed outcome happens if each options truely has the same probability of selection?

So the "surprising" part of the original inquiry is essentially asking, "if all classes were equally likely to be picked, what is t]e probability that this many people selected a class from [paladin,fighter,wizard] ?"

I think the answer is "No. It's not surprising that some classes are more popular than others."

@ColinGross Especially given that Paladin and Wizard are OP, while Fighter is easy to build and play (and performs well in early levels, where most play is done).

@GreySage That makes sense.

11

Q: What does a red left eye mean?

In the "When Will You Rage?" anthology (classic WoD fiction), background characters are referred to as having a red left eye. I also seem to remember this particular bit being brought up in other pieces of White Wolf fiction, although I haven't read them recently. However, I can't find any inform...

world-of-darkness werewolf-the-apocalypse

Thoughts on this question? There seem to be a few issues with it

Copying my comment:
"Given that the answers are basically just speculating, it seems like either this is a "designer-reasons" question... Or the current answers should be edited to mention that this has not been officially clarified in the lore. (Then again, it seems like this question isn't even about the RPG, but rather about a book series based in the same setting... Not sure how to handle this.)"

@V2Blast I have supreme confidence in your editing ability to get the question to an answerable state.

That's a lot of confidence

If I were more familiar with the system/setting, maybe

7:11 PM

@V2Blast I VtC for offtopic because its about a series of novels, not the RPG

@GreySage I think you've got it - but I also know nothing about the RPG and whether or not a red left eye is/is not part of it.

user image

@NautArch Neither do I, but he's specifically asking about passages of books. Even if it had an explanation in the RPG, it's not about the RPG

@GreySage very true

voting

Welp my desktop crashed because the program ate all the RAM in the universe

7:13 PM

@MikeQ chrome?

@MikeQ Chrome does that to me, but only at work for some weird reason

My question was indeed reopened

@ColinGross Mike wrote up our people choosing classes question in matlab, so I'd guess that

@Delioth Gotcha. Probably tried to generate all the set members. Combinations tend to blow up like that.

@DavidCoffron I just noticed that you mention a probability in your comment as "so finite" which seems funny to me

7:15 PM

@GreySage Didn't vote...still unsure. I don't think I know enough to say either way.

@Delioth 'So minute' was what I meant. Too late to edit

Finite has a specific mathematical definition different than the colloquial "small"

Yeah, I figured as much; I even read it as what you meant for a while. It just stuck out to me when I clicked the link again

So I should've used minute

Made me chuckle a little so I figured I'd share

@DavidCoffron minuscule

7:17 PM

@Delioth Thank you for sharing. It made me facepalm when I came back and read it

@ColinGross Teeny-tiny is the proper mathematical term

@Delioth insie weensie

@ColinGross Yes. Also I think I figured out the actual formula but I need to double check

@MikeQ Cool. I sorted out the discrete probability calcs, but I think it's six of one half dozen of the other for the approaches.

The discrete stats approach is basically, if equal, there is a 3/12 chance to select the "popular" choices. What is the probability that X people select popular? That's (3/12)^X.

What is the probability that X people select the popular and Y people select the not popular? (3/12)^X * (9/12)^Y

@GreySage Really not sure what to do with that question now. I'm just gonna look over there now...

7:24 PM

That's to get the exact combination. Asking what is the probability that at least X people select the popular group is marginally more complicated, but essentially the same.

@ColinGross Just look for the possibilty that n-X people select the 9 least popular.

And subtract it from 1

Ok I got it, I tihnk

let f(n,k) = number of permutations of n objects into k buckets

@GreySage That's the find out the probability that X don't select the popular and subtract from one. If the question is at least X, then you have to sum the probabilities. The prob that all pick popular, all but one pick popular, all but two pick popular... etc.

probability that 11 of 22 will be in the first 3 buckets = (number of permutations of 11 into 3) * (number of permutations of the other 11 in the other 9) / (total number of permutations of 22 into 12) = f(11,3)*f(11,9)/f(22,12)

since we're looking for the probability of at least 11, we'll need a summation

@MikeQ I think buckets is not the way to go for that. It's a discrete choice for each event. Choose one of twelve.

What is the probability that a particular combination comes up if all choices are equally probable?

7:30 PM

@MikeQ Problem: this overcounts significantly

@Delioth Really? I got 8.1%

Just use Monte Carlo methods and be done with it x)

I think there's a way to simplify your math a whole lot. Imagine a 12-sided die, where 3 sides have '1', and the other 9 sides have '0'. Roll 22d(12*), what are the odds you roll at least 11?

Because you've distributed 11 into 3 and then 11 into 9, there's not a guarantee that the 3 are actually the most popular 3

@MikeQ That's what the answer that got deleted basically did. (except it added some factor to take the probability that the first 3 are the greatest 3)

7:31 PM

Then take that basic logic, and make it generic to search for other possible outcomes.

You can simplify down to a stats textbook example. The classes are balls. Three of them are green. Nine are purple. What is the chance of 11 of 22 people getting green by pulling them out of a bag.

@ColinGross That's what I started with, but any of the purple balls can actually be green

@DavidCoffron A fighter can be a cleric?

Since determining which is "most popular" can't be done until the end

I really love the Monty Hall problem because I can trip up freshman mathematicians with it regardless of whether they know the classical correct solution or not :>

7:32 PM

You might have 11 in the "most popular" set distributed 3/4/4, and then all 11 in the "less popular" group are 11/0...

@Delioth If you have 11 objects in 3 partitions, doesn't that make them the 3 most popular partitions by definition?

@DavidCoffron Oh yeah. Hmph.

@ColinGross THe probably that 3 specific classes get chosen is about 1%.

BUt we need the 3 most popular

@MikeQ No, because as you distribute the rest you may end up with a more popular partition

how?

22 - 11 = 11

cant be more than 11

7:33 PM

@MikeQ What if those 11 all end up in one of the others

@DavidCoffron So that's a different question. That question is a distribution question. Luckily, you've still go discrete bins.

Is the question then, what is the likelihood that 3 classes account for 50% or more of the choices?

@ColinGross I think thats it

@ColinGross Yes. Because if 3 do than the 3 highest necessarily do

aka the chances that 9 bins contain less than 50%

Yeah, if you distribute separately, you have 11 in 3, something like 5,5,1... and then you distribute into the other 9; which could be distributed also 5/5/1/0..., or 11/0/... - thus your "most popular" group is no longer the most popular

7:35 PM

@SirCinnamon I think that's the safer question

@DavidCoffron ...ah, okay. that's my error.

@SirCinnamon That any set of 9 bins do, yes

(including the set of the 9 smallest bins)

This is a skewness question. For continuous I think it's a kurtosis calculation... for discrete statistics... I have to think or look it up.

Mike's method will give you the probability that 3 specific classes have 50% of the group, but not necessarily the 3 most popular classes

@Delioth yes. You could have 1 bin with 50%, 3 with 50% and 8 empty

7:36 PM

@Delioth No 3 specific classes is a similar binomial distribution with a chance of 25%

It comes to 1%

yesterday, by David Coffron

Using this calculator. Prob of success is 0.25, 22 trials, more than 11 successes

Still ends up being similar to the discrete choice if we can apriori specify the number of bins.

(22,0,0), (21,1,0),(20,2,0),(20,1,1)... until you reach (4,3,3) right?

@SirCinnamon 4,3,3 wouldnt have 50%

@DavidCoffron Right, I think that ordering, its the first case that wouldnt

And it still runs into the issue that the 3 you've got aren't guaranteed to be the 3 most popular

7:41 PM

@SirCinnamon Does (4,4,2) come before it?

For 9/12 of the classes to contain 50%... we don't actually need to define which classes are popular. Only that exactly 3 will be.

@Delioth Well in any set where 3 possible bins have 50%, the most popular bins will also have 50%

Ok, ok, what about

@Delioth assume those 3 are top 3 after sorting

Generate all permutations of 11 into 3

For each permutation p, count the number of ways to permute min(p) into 9

7:42 PM

@DavidCoffron Ah yeah... hmmmm

Sum that formula from 11 to 22. Then divide by the number of ways to permute 22 into 12.. Would that do it?

@SirCinnamon So how many permutations did we discard?

@Delioth the total possibles minus the sum

Is it correct to assume this is not an official wiki?

@NautArch What constitutes an "official" wiki?

7:47 PM

@Rubiksmoose I...don't know?

@MikeQ Whatever method you come up with, at least check yourself on this spreadsheet. Just change the number of classes to 4 and the number of players to 6 (where we need at least 3 to reach 50%) and use the top 2 buckets for the first sheet and the top 1 bucket for the second sheet. Those are calculated mechanically

@Rubiksmoose But trying to help this guy out.

@NautArch Fan wikias are usually good enough sources for non-controversial topics

But I see no way of verifying whether it is moderated or curated by game developers

@NautArch If by official you mean run by the company I highly doubt it. Looks like a fan wiki, but as David said, it is probably good enough.

@Rubiksmoose Is it legit enough to use in an answer? If someone made an answer utilizing an equivalent for 5e, I'd probably downvote.

7:52 PM

@NautArch I don't know. Not every game has the same sense of "official" being required as D&D. I would think as long as you noted the source and its potential fan nature that would avoid some downvotes? I'm really not sure.

@NautArch Would you downvote if somewhere answered a lore question based on the Forgotten Realms wiki? (the hypothetical question accepts all editions of lore)

@DavidCoffron I probably wouldn't vote, I don't know the wiki. But my gut says (for 5e) use only official sources.

8:05 PM

@DavidCoffron @Rubiksmoose I also feel really weird answering with absolutely zero knowledge of the books or rpg.

@NautArch Yeah I get that. I answered a couple Pathfinder questions without knowing the game very well and it felt weird

Okay, third attempt:
for n = 11 to 22
> p1 = all permutations of n into 3
> p2 = all permutations of 22-n into 9
> for each p1, count the number of p2 such that max(p2) <= min(p1)
sum up the counts
divide by the number of permutations of 22 into 12

Because it's saying for each way to fit the majority into 3, count the number of ways to fit the rest into the other 9, such that none of the other 9 are larger than any of the 3

I count 98/905 sorted distributions where the top 3 do not add to 11 or more

@DavidCoffron Upon further reading of the wikia...i don't think it's related.

807/905 sorted distributions have the top 3 adding to 11 or more

8:18 PM

@MikeQ Yes that would do it

Result is 0.7766%

How did you count the "number of p2 such that..."

Also, I'm pretty sure the answer should be greater than 1%

(or is that just the probability of exactly 11, not at least 11)

here's my analysis ghostbin.com/paste/zss7b

Since the probability of any specific 3 being greater than 11 is 1% and the probability of any specific 3 being exactly 11 is 0.2%

I'm getting 89% chance that the top 3 add to 11 or greater

instinctively that feels high... but I cant see any flaw in my math...

8:23 PM

@SirCinnamon Hm....

it would be a 100 percent chance that the top 6 are greater than 11 though

@SirCinnamon Yeah. When I did this analysis I had a similar revelation. Since everyone has to choose a class inevitably you reach a point where the classes are forced into high enough numbers where it works

With 22 people the minimum is already 6 out of 11 in the top 3

99.55% that the top 5 are, 97.9% that the top 4 are

For you to get a top 3 that DON'T result in half of them seems less likely when you think about it that way

(since they have to distribute themselves a specific way)

I've got a simulation I just finished programming that is calculating a 48.86% chance that the top 3 classes represent at least half the total number of players.

So far it has simulated about 87 Million samples of 22 players across 12 classes.

8:27 PM

@Xirema Umm... Now we have four numbers lol. 89%, 48.86%, 0.776%, and 2.2%

@DavidCoffron I cant follow that link from my work computer but im glad we reached the same ballpark

@DavidCoffron Yeah, I see that.

@SirCinnamon Your work blocks google drive!? That's weird

@SirCinnamon Basically I did a spreadsheet for all permutations for 4 classes and 6 players with different "top X" numbers

for the bottom 9 to sum to more than 11, while still being the bottom nine, you have to evenly distribute across all 9 with non being greater than a 3

(I couldn't think of an excel formula for all combinations instead of permutations, but permutations was pretty easy)

8:29 PM

@DavidCoffron Mine does a weird thing where it doesn't "officially" block it, but everytime I try to access Google Drive (and some other sites), it times out without being officially blocked.

If I change it to track the top 2 classes, the odds drop to 1.74%.

@Xirema Yeah its the same. it's (i believe) something to do with google thinking all the requests from your whole work network are coming from the same machine, so it thinks youre doing some kind of attack

@Xirema What do you get for 6? Because it HAS to be 100%

the top 6 must add to more than half the group

@SirCinnamon 100.00% exactly. It agrees with your assessment.

And the top 5 should almost always (only 2 combinations allow it)

Where 10 classes have 2 and the last 2 classes are 1,1 or 2,0

@DavidCoffron I get 99.99% for 5.

@Xirema Hm...

@SirCinnamon Now I'm trying to figure out what combinations you are missing

Xirema's simulation should be close to accurate

8:34 PM

@Xirema That sounds like roughly the expected, so that's good

Incidentally, all my simulation is doing is just performing the test over and over, and tabulating results as I let it run.

@Xirema So it's just making a bunch of 22 player campaigns?

Yeah, assuming the randomness is well-distributed, it should be good

@DavidCoffron Yup. Randomly assigning players to a class.

Then it sorts the classes by most popular→least popular, and counts the top X.

There's always the possibility that it skews the results due to randomness being able to cluster results; but as the # of trials goes up that becomes less and less likely

8:36 PM

If the number is greater than or equal to the threshold (11), it counts it as a success, if it's less than the threshold, it counts it as a failure.

8 mins ago, by Xirema

So far it has simulated about 87 Million samples of 22 players across 12 classes.

Yeah, I'd caught that. I'd guess that after a million samples it should be as close as possible to the actual percentage.... probably.

At the very least, I'd trust it after a million samples for a problem set of this size

@SirCinnamon How did you generate the combinations because it should be more than 905, no?

22 players choosing 12 options is <sup>(k+n-1)<\sup>C<sub>(n-1)<\sub>

careerbless.com/aptitude/qa/permutations_combinations_imp7.php

k is 12, n is 22 (it is equivalent to sorting 22 balls into 12 boxes)

The boxes aren't distinct

@Delioth you right. We don't care what classes they are, only if they are the most picked

8:44 PM

@DavidCoffron You have N players. Place M players in class 0, such that M<=N. Place L players in class 1 such that L<=M and L<=(N-M)

repeat for all possible values

until youve placed 22 players or run out of classes

So you are assuming they are sorted by size and then just filling in the values

yes, exactly

it should be every possible sorted array

if I'm not crazy

Another way to generate is to start with the first case and then move 1 from the last non-0 cell to the each of the earlier cells. So you have with 4 players and 3 classes: 2,1,1 -> 2,2,0 -> 3,1,0 -> 4,0,0

does that give you the same set?

I think so, but in a different order

My program is working recursively

It is essentially generating all the sets of 12 non-negative integers that add up to 22

48

Q: Number of ways to write n as a sum of k nonnegative integers

How many ways can I write a positive integer $n$ as a sum of $k$ nonnegative integers up to commutativity? For example, I can write $4$ as $0+0+4$, $0+1+3$, $0+2+2$, and $1+1+2$. I know how to find the number of noncommutative ways to form the sum: Imagine a line of $n+k-1$ positions, where e...

If this doesn't give us 905 then you are missing something

9:03 PM

Well its giving me 7 which i cant imagine is correct

@SirCinnamon Is this what your program does

@DavidCoffron Yes, I think so

backwards, but yeah

Then I don't know. It should work

Is there a way to post images directly to chat, or do I need to find an image uploading site?

@Xirema Chat has an upload button

user image

Imgur has a partnership with StackExchange and it stores all images uploaded in this way there

9:07 PM

user image

@Anaphory Here is your answer experimentally ^ (i.e. 48.84%)

@Xirema What happens if you change the participants to 28 (that was another question Anaphory had since he still had a 50% margin after 28 people)

[If you don't mind checking]

@DavidCoffron As they say on my favorite game show: let's find out!

@Xirema We're evaluating different things here - by distributing randombly you only have 1 case where its [22,0,0,0...] and millions of cases where it's even or close to even

@DavidCoffron I didn't know that

@DavidCoffron (Am I bumping the threshold up to 14, or keeping it at 11?)

9:10 PM

48

Q: Any details about stack.imgur.com?

Since May 2011, images uploaded through SE sites are uploaded to stack.imgur.com. It seems that removing that stack part does not get one (the same) image, so it seems to be a standalone installation? Any details on that new domain? And: any new way to circumvent blocks of imgur.com? And this ni...

discussion images imgur

With a nice fancy interface like that it might be worth throwing it on GitHub? Could be worth making it generic (e.g. classes kept in a config file or something)

@Xirema to 14

He got 3 fighters in that +6 sample (bumping it to 14 for Paladin, Fighter, Wizard)

user image

@DavidCoffron The 0.7766% was wrong, it was only checking a specific 3 rather than any 3.

Huh, the 5 result sticks out to me in that sim

9:18 PM

@Delioth Yeah, I'm not sure what to think of that.

@Xirema It's very well possible that there isn't a way to add 12 numbers to equal 28 that doesn't involve 5 numbers being >= half

28 spread over 12 classes has a minimum of 2 per class, with at least 4 needing to have 3

his program hasnt hit the single distribution where every class has 2 and the exact minimum have 3

I mean, the flattest distribution that would work best would be 3,3,3,3,2,2,2,2,2,2,2,2, which gives us 14 in the first 5

oh right its looking for 14 not 15

so yeah, guaranteed

So that's a breakpoint where those specific numbers can't result in a single (needs 6 to be half) combination

If you go up to e.g. 32 & 16, it works again (since you have the combination of e.g. ten 3's and two 1's, where the first 5 add up to 15, which means we'd need 1 more digit to be at our halfway point)

9:26 PM

It appears github doesn't let you upload directly from your webbrowser?

Xirema's program is working with the classes being distinct, mine considers only the distribution of numbers

@Xirema gist.github.com

gist.github.com/sircinnamon/236cfc45561593772fdbfad83f415ae8

thats what I wrote. I think that works for counting the number of distributions, if you treat all sets as sorted

gist.github.com/Xirema/744ba7a0c00ddf87320405942af5d651

Mine is a little overcomplicated to give a nice interface.

ahhhhh java! run away!

Incidentally, my code is a perfect case-study either in why Java should have a built-in Pair type, or why having a builtin Pair type is a mistake regardless of the language, depending on your perspective.

I think we can all agree that an atrocity like List<Pair<Integer, Integer>> is evidence you've fornicated up, somewhere along the line, in writing your code.

@Xirema Ah Java, that takes me back

9:41 PM

@Xirema I think its funny that only 2 in 40 million sets of 28 players has 14 of one class

I'm imagining a Warlock cult infiltrating the 28 player party

What about 28 monks, all drunken master style

@MikeQ Do they say ni?

@DavidCoffron No, you're thinking knights

@MikeQ Oh right. Here they are

Monks - Monty Python and The Holy Grail

But 28 of them

I'd be willing to DM it for 1 round

9:49 PM

@MikeQ 1 round? I'd give em at least 3

With all the dice, we're looking about ~30 seconds per turn, minimum. So that's at least 14 minutes per round, for the players alone.

Then you have to double that because it's D&D and people are inevitably gonna waffle on decisions or need to look stuff up

I like waffles.

@MikeQ I've done larger group before. I answered a question based on it, one sec

@BESW Sure, but 28 waffles per round is gonna be too much

I want 12 of them!

9:53 PM

Hm, how many ways to distribute 28 waffles such that @DavidCoffron gets exactly 12 and @BESW gets at least 1

3

A: How to design a combat encounter for 20 characters?

Break it into multiple sectors When working with large groups (especially groups of substantially different levels), each encounter should be subdivided or the waiting time becomes excessive for the players. This requires a bit of work and is better managed with the help of a co-GMPC for each gr...

This is what I've done in the past

It works well sometimes, but you have to design the encounters specificallly for the group

Square One TV: "Combo Jombo" by Regina

@MikeQ Are those the only 2 waffle eaters? If so, there is exaclty 1 way, because BESW must have 16

@DavidCoffron I don't know. I'm still confused about the previous problem.

1 hour later…

11:26 PM

hey there @MarkWells, welcome to the RPG.SE lair :)

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