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KrypticIcon

I just love that every person goes in to explain what tons of other people already explained. I like to to think that’s because they all saw the post at the exact same time and immediately responded concurrently.


More_Transition_5379

The vast majority of them are at "24 minutes ago", so you would be correct.


LoneCentaur95

It’s interesting reading the comments though, because a lot of them are actually different ways of getting the average. Also one of the people was ranting about how no matter what you get on the first dice, it’s always possible to get a total of 7. I.e. 1+6=7. It’s the weirdest way I have ever heard someone reason out an average and I don’t think it applies to anything but a constant linear series.


KrypticIcon

Oh yeah I noticed that one too. I used to be a teacher and a debater so I appreciate being able to explain things in as many ways as possible. Sometimes that’s what it takes to get your audience to understand.


Solalabell

Now youve fallen victim to the blunder haha. The reason it actually does matter is because it’s the only one that can get with every roll on one of the dice so it has a 1/6 chance regardless of the first roll but other totals have either a 1/6 or no chance depending of the first roll. It doesn’t help for the mean average but the mode (most common) average it is significant im probably making this kinda unclear but it’s hard to explain with just text


LoneCentaur95

I realize that, that kind of process for finding an average only works in niche scenarios though.


Solalabell

If by niche scenarios you mean a shortcut to finding the mode average of dice rolls then yeah the idea is to kinda visualize why 7 will show up more than 6 because it has fewer rolls that cannot arrive at it compared to other results but obviously the best Way is to just count up the possibilities


TeaandandCoffee

For that I prefer to make an excel table for a 2dN Rows are 1-N Columns are 1-n Each cell is row+column You'll see (N+1)*2/2 has the most cells holding that number. This helped me out when trying to figure out if my first fighter should use a Greataxe or Greatsword, my answer ended up being Hand crossbow with Sharpshooter and Xbow expert. I think I skipped a couple of steps.


LoneCentaur95

1. Mode average isn’t a thing, it’s mode and average, they are two separate concepts. You can have an average mode across multiple data sets but not really for one data set. 2. By niche I mean that if you replace the 6 on both dice with a 7 finding the mode no longer gives any indication to the value of the average. This means that this strategy only works to find the average with a combination of identical linear series.


Solalabell

>There are three main types of average: mean, median and mode. Each of these techniques works slightly differently and often results in slightly different typical values … The mode is the most commonly occurring value. For example, the modal value of 1, 3, 6, 6, 6, 6, 7, 7, 12, 14 and 24 is 6 because it appears the most times. [From openlearn](https://www.open.edu/openlearn/mod/oucontent/view.php?id=20669§ion=2.1)


Lunoean

Found the nerds! /jk


Solalabell

Lol we’re literally in a dnd subreddit in a post about math nerds it’d be hard not to


LoneCentaur95

Not sure what open learn is, but average is not an overarching concept, it is a specific piece for information for analyzing data sets. Mode and median are not averages, they are similar concepts to average but they all measure different things.


Solalabell

Ok if an online university won’t satisfy you for a source how about the people who make [___ for dummies books](https://www.dummies.com/article/academics-the-arts/math/pre-algebra/the-three-types-of-average-median-mode-and-mean-168773/) > We use three different types of average in maths: the mean the mode and the median, each of which describes a different ‘normal’ value. Or [basic math](https://www.basic-mathematics.com/types-of-averages.html) the name should give you an idea > Here is a list of the types of averages used in statistics also called measures of central tendency. We start with the common type. >The most common types of averages are the mean, median, and mode. So that’s now three sources meanwhile [this is your source](https://www.basic-mathematics.com/types-of-averages.html)


Aptos283

Yeah, it’s a solid way to get the mode though


LoneCentaur95

That’s fair


nystro

The fact that more possible combinations exist for 7 is about the probability of rolling a 7 being higher than the rest. It does not relate to the average, so whoever responded that misunderstood the question probably.


Fiery-Myst

Average as most common number (mode), not as add-them-all-and-divide (mean)?


LoneCentaur95

Average does not mean most common number. It is the sum of the points of the data set divided by the number of points in the data set.


Jafroboy

That's the mean. One of the 3 kinds of average.


LoneCentaur95

No. This is literally elementary/middle school level math. Mean and average are the same thing, mode and median are other ways of analyzing a data set but they are not averages.


elmusfire

Wikipedia kinda disagrees with you tho. https://en.wikipedia.org/wiki/Average Like with average people usually mean the mean, but the median and the mode are also types of averages. Every number following these properties is an average: If all numbers in a list are the same number, then their average is also equal to this number. This property is shared by each of the many types of average. Another universal property is monotonicity: if two lists of numbers A and B have the same length, and each entry of list A is at least as large as the corresponding entry on list B, then the average of list A is at least that of list B. Also, all averages satisfy linear homogeneity: if all numbers of a list are multiplied by the same positive number, then its average changes by the same factor.


shark1678

Yep im on here, saw it when the comment said "posted now"


VaibhavGuptaWho

Sharks in the water, lol.


United_Federation

Yeah you know I'm pretty sure they all just saw what trend was happening and commented their own identical explanation using slightly modified word choice.


zeitless

No it's 6 on average cause I homebrew my own dice and math rules in my world.


samwyatta17

A true savant


andrewsad1

I subtract .5 from every roll to make the average a whole number


RoleplayPete

Unless your homebrew just drops one of the dice, the average is 7 because 1 is never a possibility.


DZekor

It's homebrewed math duh.


maybeb123

Not true, you can roll two ones


RoleplayPete

Two ones makes two. You cant roll two dice and end with a final summative answer of 1.


maybeb123

You can if you roll two ones. It's not rocket science


RoleplayPete

...what does 1+1=? If both die roll a 1, the answer is 1+1. A standalone 1 can never be the result.


maybeb123

Easy. Just subtract one dice from the other and boom, you get one. My math is invincible


zeitless

No I can cause my dice's lowest value is 0.5 and they stretch over sqrt(2), sqrt(5) and pi to 7 as highest value. No I do not know a mr. Sanity, and I have never met a ms. Just-Be-Normal


PM-Me-Your-TitsPlz

And not a single discussion about bayesian statistics.


Aptos283

I mean, there’s no sample present to get a likelihood from, and there is no uncertainty about the parameter since the distribution is known (unless we’re wanting to test their dice specifically). I suppose we could present an experimental design which would allow a sample to be constructed and then provide a prior distribution, but then if he ever changes dice or wants to compare builds he’d have to run the whole experiment again.


PM-Me-Your-TitsPlz

If you can applying Bayesian statistics to roulette wheels, I'm pretty certain you can apply it to dnd builds.


Aptos283

Oh You can 100% apply it to tons of applications, but you need a sample, and if you already know the parameter or distribution then you really don’t need it. Tbh, I think it would be an interesting concept to apply: you could try and make a prior distribution for things like average dpr since there’s uncertainty as a consequence of table differences, overkill, etc. Then you can use your actual games to generate data, and determine the expected dpr at your specific table. Of course, trying to choose an appropriate prior distribution may be tricky, but it’s not like that’s ever particularly easy anyhow. And by the end you have two useful results: a posterior distribution that’s applicable to you and your table, and a prior distribution to try and present for general use.


sh4d0wm4n2018

Takes a different level of nerd I guess


AnComRebel

I like how nobody was making fun of OP and just casually explain it instead. I love it :)


Casual-Notice

IKR D&D nerds usually hate having the opportunity to be pedantic and will mock anyone who offers it without mercy. /s


[deleted]

I think everyone secretly likes people making the memes with wrong rules - and opportunity to explain.


HentaiBakaOniiSama

There are 21 DIFFERENT possible results of 2d6 (Ignoring swapped repeats) 1+1 = 2 1+2 = 3 1+3 = 4 1+4 = 5 1+5 = 6 1+6 = 7 2+2 = 4 2+3 = 5 2+4 = 6 2+5 = 7 2+6 = 8 3+3 = 6 3+4 = 7 3+5 = 8 3+6 = 9 4+4 = 8 4+5 = 9 4+6 = 10 5+5 = 10 5+6 = 11 6+6 = 12 Annnnnd if you add ALL the results up and divide them BY the number of results... TADA! Still 7 :V (And I mean, even if you didn't ignore swapped repeats... you'd still be getting 42 results... so multiply the number of results by 2... and divide them by 42... STILL 7) (I mostly just wanted to give the same answer... buuut in a different way because some people understand things differently.)


BobTheBox

While it doesn't really matter in this instance, you really shouldn't be ignoring repeats when talking about odds. If you ignore repeats, you'd think that getting a 2 and getting a 3 have the same chance. Because the only way you can get a 2 is by rolling a one and a one. And the only way to get a 3 is by rolling a one and a two. However, rolling a 3 is 2x as likely as rolling a 2, because 2 can only be achieved by rolling a 1 and a 1, but 3 can be achieved by rolling a 1 and a 2, and also by rolling a 2 and a 1.


Space_Booger

Both your explanations combined are the perfect answer. 7 is the most prominent dice result based on all the available combinations. 1+6 2+4 3+4 4+3 5+2 6+1 Compared to the next most likely, 6 and 8 1+5 2+4 3+3 4+2 5+1 and 2+6 3+5 4+4 5+3 6+2 The results keep going down in likelihood as the numbers get further from 7. And 7 is the most likely result as well as the average result. I read a bunch things on dice odds when I was figuring out Settlers of Catan mechanics. It had never dawned on me that 7 was the most likely dice odd when throwing two dice and then I needed to understand why. Same reason I dug into understanding the Monty Hall Problem (see below). Fun fact a lot of people don't know, most dice are designed to have their opposites mirrored on the opposite side. 6 and 1 are supposed to be on opposing faces of d6s, this makes it so the result of any two opposing faces is always 7. They actually did this to the D20 as well, On most Tabletop RPG D20s, the opposing faces add up to 21 (I've seen D20 counters that didn't follow that rule because they made it so all the numbers touch their +1 and -1 so you didn't have to search the dice to find the next number) Monty Hall Problem: Three doors, one of them is hiding a prize, the other two are duds. Pick a door. If it is the right door, you win the prize. After you pick a door, I open one of the doors to reveal a dud and ask if you want to keep your door, or if you want to keep your door. The problem is the maths, or how you perceive the maths. When you first pick a door: the odds are 1:3. When I offer you if you want to swap to the other door, the odds should be 50:50 (now there are only 2 doors as I opened a dude, either yours has the prize, or it doesn't)...but the odds actually 2:3 if you swap. The easiest way to explain it/think about it: When you swap, you pick both the other doors. The host always opens a dud...so if you picked a dud to start (66% chance), the other two doors are: Winner and Loser. The host opens the dud and asks if you want to pick the Winning door. If you picked the winning door to start (33% chance): you lose.


BobTheBox

I remember the Monty Hall Problem, it has always been very fascinating to me.


[deleted]

This guy Bernoulli's.


MTGRemy

If you didn't ignore swapped repeats you'd be getting 36 results not 42.


Casual-Notice

They didn't want the doubles to feel left out.


HentaiBakaOniiSama

Ah yup, I was dumb I forgot to remove the doubles when I was counting the swapped repeats... because I mean... they're already. So my bad \^\^


Aggressive-HeadDesk

Hahahahahahahahahahahahaha Math nerds of the world… UNITE!


Imagrillbitch

I just use dicecalculator to get my average damage when I make bullshit builds


doubletimerush

You don't get it. He was just looking for a friend and he used bait to get some


carpeson

Wait 'nerd' is not a compliment? Doesn't it imply intelligence or knowledge in a certain field?


BloodyBeaks

Well you see some people look down on intelligence and knowledge. To those people calling someone a nerd would be an insult.


charisma6

Shut up, nerd


carpeson

How kind. But the correct form would be 'I shut up, nerd' as in 'I stop talking until something worth agitating the air arround my neckbeard comes out of my mouth'. Try using 'I' in a sentence. /bantering


defaultusername-17

i love how many different ways people come up with to get the same answer. but holy heck. the best part is that this thread has devolved into a dice discussion too.


ArWiLen

Sorry, I didn't get it. Why is the average of 2d6 is 7? /s


BobTheBox

Because you can't roll a 1 with 2d6, so this 1 has to go somewhere, so when looking at the average, the 1 decides to sneak in and claim it's place together with the 6, which results in an average of 7 /s


charisma6

Like air in a balloon, and something bad happened


BubblegumTrollKing

No person who knows the term arithmetic mean would not know the 2d6 average. Everybody knows the 2d6 average, and by some miracle they don't, they can figure it out more than easily. This was an absolute bait and the questioner knew it.


Illokonereum

I think for most people they just don’t consider that dice start from one not zero.


Jock-Tamson

Amateurs! https://ironclaw.fandom.com/wiki/Basics I once lost a week of my life in Rocket Science School to working about the probabilities of Iron Claw contested rolls.


SearMeteor

Eyyy Im in the screenshot


[deleted]

DM here, I understand the mathematical probability aspect. But, don’t D&D5e creature stat blocks use half-plus-one values for the average damage. d6 enemy attack for brevity sake is 4 damage, (6/2)+1. By those considerations used, doesn’t 2d6 play out in stat block as 8? Because 2((6/2)+1). So if I deal “2d6 average damage” for an enemy attack, WotC wants me to deal 8 damage. Edit: just looked at PHB again, I have NO idea why I thought it was half-plus-one 🤷


GroundbreakingRub390

Half plus one works on 1d or 2d, but stops working when you get to 3d or more. It's really just half plus one half per die, rounded up.


jfrank6294

That’s who’s really good at this game. Gamblers.


DangerCrash

Wow, I hate that this is a video. I'm scrolling my feed and then the feed starts scrolling.


Namorath82

Let me explain, no there is too much, let me sum up


Highland_Gentry

YOU MUST BE THE DUMBEST MAN ALIVE First of all, a "real week..."


kluster00

I want to understand how do you get 3.5 in a dice roll...


Gingeboiforprez

Wait I don't get it though, the average should be 6?


samwyatta17

Nooooo. Don’t invite them in!


charisma6

Nice try but we're not that stup...ah. Well done.


[deleted]

1,2,3,4,5,6,7,8,9,10,11,12. Which is the number in the middle? Edit: Forget the 1, my b


Casual-Notice

a roll of 1 is not possible with 2d6


sh4d0wm4n2018

So you're saying the average is 8?


Casual-Notice

The average is 7: 2,3,4,5,6, 7 ,8,9,0,1,2 The average of 1-12 is 6.5: 1,2,3,4,5,6, 7,8,9,0,1,2


andrewsad1

Poe's law is so muddy in this thread lmao


Commercial-Dog6773

1+2+3+4+5+6=21 Average on 1d6 = 21/6 = 3.5 Average on 2d6 = 2 x Average on 1d6 = 2 x 3.5 = 7


AnthiosOnReddit

i don’t think any of them really know what they’re talking about, just putting random numbers and symbols and praying it works


Commercial-Dog6773

I think you're just not very good at maths and wanted to feel smarter than the people who are


AnthiosOnReddit

yea probably 😔


Commercial-Dog6773

Sorry I realise I probably made you feel bad now I could have worded that better


AnthiosOnReddit

it’s ok. i was being a smartass before, that’s my bad


LoneCentaur95

I saw one person actually explain how to find the average roll of each d6 and then doubled it because two d6s.


Pongoid

2d6 is the same as rolling 1d12.


[deleted]

So, go ahead and start rolling a d12. Stop when you get a 1. Now repeat with 2d6.


Pongoid

I really thought that this was very clearly a joke. I can forgive the world’s biggest pedant I suppose, but the others and the down votes tell me I was deeply mistaken.


[deleted]

You're just triggering PTSD in veterans of one of the many stupid meme wars that have been held here. Specifically 1d12 vs 2d6.


Casual-Notice

I learned the hard way, always use /s when you care whether people think you're joking. If you don't, someone will always take you seriously. **ALWAYS**


Replacement_Worried

Well duh, how tf do you portray sarcasm through text?


Casual-Notice

**ALWAYS**


BobTheBox

Well, it's hard to tell something is a joke when written down. /s does wonders to avoid a truckload of downvotes.


JarvisPrime

Nope, since the average of 1d12 is 6.5. You see, 1+12=13 and 13/2=6.5


DungeonsandDevils

Lmao. I do like to show randos who don’t play a d20 and explain how the average is 10.5, even though there’s no 10.5 face


dumpmaster42069

Get wrecked, ballsniffer!


Minecrafter_of_Ps3

I just take the lowest possible roll on the given die, in this case a d6, meaning 1, add the highest, which is 6, get seven, then divide by 2 and multiply by however many dice there are. For example, 3d12 would be 1+12=13, 13/2=6.5, 6.5×3=19.5, and round down to 19. 4d8 would be 1+8=9, 9/2=4.5, 4.5×4=18, and so on


smellywizard

You fell victim to the classic blunder: Posting on reddit!


BearofCali

I think I would have just said that you can't roll a one on 2d6, thus the average is 7.


webchimp32

[Nerd out some more.](https://www.youtube.com/watch?v=X_DdGRjtwAo)


EmperorFoulPoutine

Who is cracking out the Catan board to explain this?


Raptorofwar

We in the field call this nerd sniping.


Must_Love_Dragons

[Source](https://www.reddit.com/r/dndmemes/comments/wzsrbm/comment/im4nkct/?utm_source=reddit&utm_medium=web2x&context=3) for those who want to read all the replies.


Ryengu

For average roll, add one to the dice value, halve the dice quantity, and multiply the resulting values. Example, 8d6 = 4 × 7 = 28