Please teach me probability so I don't have to keep asking

[zeruslord]

Standard deviation is a really big deal for bonus size - the smallest bonus that people are really going to remember and feel in play is one that shows up on at least one roll in 10 when you're near the center of your game's bell curve. The smaller your standard deviation relative to the full range of the bell curve, the more bonuses of that size you can fit in before getting to the point where one player is auto-hitting and one player is auto-missing. +1 bonuses on a d20 are pretty tiny - any particular one will show up maybe once a night, but going from needing an 11+ to a 10+ on 3d6 is a swing of 12.5%., equivalent to +2 or +3 bonus. On a d20, eight of those will push you to the point where you always succeed at something you used to always fail at, while it takes 15 +1 bonuses to do that on 3d6. Essentially, the number of bonuses that feel significant that you can include while keeping players close enough together that there are target numbers where every player actually needs to roll is bigger when the standard deviation is small.

[PhoneLobster]

What I wonder about is the effect standard deviation has on die mechanics.

Yeah... that's gonna be a bad one to ask around the "3d6 can have more +1s in it before you go from 3 to 18 than d20 can have +1s in before you go from 1 to 20" gaming den.

As... zerus pretty much demonstrates.

Your question is a bit flawed on causality, standard deviation is an important number to know, but outside of it's actual mathematical definitions has next to no "magic" meaning.

The "favourite" gaming den interpretation, an attempt to justify "curved" mechanics as having something, anything, positive to say for themselves, as put forward by zerus lord, is a gigantic pile of crap that relies very heavily on a number of flawed assumptions.

The most blatant of which of course is, that your target number before bonuses is always at the center of the "curve" and that there is then one and only one modifier applying to the roll.

The reality is that anyone with any sort of maths background looking at something like "what is the value of a +3 in a 3d6 roll mechanic for an RPG" is going to look at the value of the +3 over the range of different target numbers/other modifiers it will be applied on top of, which typically is "all of them" and give you an average of that.

The gaming den methodology of "measure it at the biggest point only" is... insane. And it results in things like zerus claiming a +1 worth about 5.55% (or maybe 6.25%, depending on... stuff) on a 3d6 roll is worth totes big more than a +1 worth about 5% on a d20 roll, because, at it's biggest ever it's worth 12.5%, and cough cough never ever mention that sometimes it's only worth 0.46% and for 8 out of 16 results it's worth less than 5%.

As a general rule I would suggest if anyone starts telling you that you can fit more of a fixed linear bonus onto a smaller total range of numbers, that person is either innumerate or dishonest.

[momothefiddler]

Yeah, and if your TN is 47 on a d20, the difference between a +1 and a +10 is ZERO PERCENT why aren't we accounting for THAT.

Probability is a FRAUD and math is USELESS.

[codeGlaze]

Probability is a FRAUD and math is USELESS.

Man, I KNEW the power was in the bones and tea leaves all along!
... That's it, I'm releasing a game with a dart board as it's base mechanic.

[PhoneLobster]

Yeah, and if your TN is 47 on a d20, the difference between a +1 and a +10 is ZERO PERCENT why aren't we accounting for THAT.

Depending on what you are trying to learn about the system, and whether you know that the 47 TN is possible and how likely it is, you actually DO want to, and CAN, account for that.

It's just generally information we don't want to know for a TN that generally doesn't occur, with a frequency we cannot easily predict.

Now, "the difference between +1 and +10 at TN of 18" is much more useful and should be accounted for more often, and frequently isn't, but, whatever. That's not what your stupid "point" is about.

You post is about defending "a single bonus ever, only vs a TN at the mid point of the RNG" as information you do want to know... and it is... but it isn't actually particularly useful on it's own and if you pretend it's how you get the value of a bonus in general you get the wrong number and that WILL fuck your system up.

Compared to you know "determining the bonus vs the actual range of base TNs/other modifiers we are likely to encounter". Ideally you'd include some accounting for the frequency of net TNs, but if you don't know that just the range they cover WILL provide you with some useful information.

Pretty much NO ONE is writing an RPG where there is only one bonus and it only applies vs a fixed TN at the mid point of the RNG. If they are, it likely sucks, but they aren't.

Hell you are lucky if TNs before applying any single given bonus are even "mostly" at the mid point of the RNG.

So, if you are trying to determine "how much is this skill I am writing that gives a +1 bonus to a 3d6 roll in my typically complex RPG system worth?" the answer, sadly is NOT "12.5%". And any analysis of your system that either claims that it is, or revolves around that claim, is flawed.

[momothefiddler]

Depending on what you are trying to learn about the system, and whether you know that the 47 TN is possible and how likely it is, you actually DO want to, and CAN, account for that.

And yet somehow you still go around shitting yourself about +1 on a d20 being 5%.

Now, "the difference between +1 and +10 at TN of 18" is much more useful and should be accounted for more often

Are you seriously already telling me that different spots on the curve are of different analytical worth?

Jesus christ man.

The fact that the same linear bonus has different probabilistic effects at different places on a curved system as opposed to a linear one is not only something i'm not ignoring, it is literally the advantage.

---

PL has had this argument a lot of times, and he still argues shit like "16<20 so you can't have more +1s" as though they're relevant, so he's obviously not worth arguing with at this point, but seeing as this is a probability education request thread, here's the situation for the rest of you.

Start off by assuming the base, unmodified situation is a coinflip. Not only is that a valid approach from the standpoint of a default probability, it also gives you the most play for bonuses and penalties before things fall off. So, DC 10.

Now say you have four different ways you want to improve something. An extra point of skill, a magical item, a favorable circumstance, and... idk, a piece of mundane gear. If these are +1 each, then on 1d20 taking any one of them gives you +5% and on 3d6 taking any one of them gives you +12.5% - that is, the 3d6 incentivizes tossing in a piece of something to help. Meanwhile, the d20 option is significantly less useful and you end up having to grab 2 or 3 of them just to keep up, giving you a bunch of bonuses that you don't care about but have to track because that's how you get your odds of success. If you end up with all four, that's +41% on 3d6 and +20% on 1d20.

"But momo!" you say, "That's not a fair comparison! Obviously each of those things should be a +3 in the d20 system to provide the same result!"

Sounds good. Now we have four different ways of getting +1/+3 on 3d6/1d20. Grabbing any one of them gives you roughly +15% success. This is a sort of thing you can feel pretty immediately. It's a satisfying thing. It makes it worthwhile to be like "I brace my gun on the top of the barrel before firing". You've succeeded at making it a thing that matters rather than just a string of fiddly bits.

But... what if someone wants all four? Now they're at +12 on 1d20! They're off the RNG. How do you solve this? Do you go back the basic bonus being too small to give a shit about so that more of them can be stacked? Do you tell people they can't have more than one or two bonuses? Do you have a giant mess of bonuses that may or may not stack with each other depending on type and source? If you want to have a bunch of different ways for people to get a reasonable, notable bit of an edge without even a few of them pushing them off the edge, you might be tempted to have some sort of "your first bonus counts as the full +3, the second is only +2, and all the rest are +1" or something. But that's exactly what the curved rng does for you. Your first +1 is +12.5%, your next is +11.6%, your next is +9.7%, next is 6.9%, etc.

If your goal is to have a bunch of ways to increase your odds but to organically have people not take more than one or two of them, the curve's smaller StDev is exactly the sort of thing you're looking for. You don't have to worry about the order of the stacking, or about making everything shitty, or about not letting people pick the options that work for them. The first +1 bonus just is ~3x as big as the fifth.

Now, I'm not here arguing for curved RNGs, as a matter of fact. It's not anywhere near strictly better. But the StDev thing is this, and that's the main benefit.

[PhoneLobster]

Your first +1 is +12.5%, your next is +11.6%, your next is +9.7%, next is 6.9%, etc.

Notice how that always just... stops there from posts like yours. No one ever talks about the 0.42% ends. You certainly don't and your "math" actually deliberately excludes and ignores them, rounds in a deeply dishonest manner exceptionally favorable to your insane claims and makes laughably unfair comparisons under the cloak of "yeah I'm totes going to be more fair by actually getting this even more wrong..."

Seriously your "more fair" scenario is a god damn fraudulent difference engine. You cherry pick the big end of your curve, you add the total chances from it, you demand the individual chances for the 1d20 system be a very generously rounded 15% for your arbitrarily selected biggest single chance of 12.5%, instead of your average 10% for no god damn reason, then pretend that the total 60% chance from that is TOTES the same as your 40% chance from the other one, and that using up 60% of the entire range of a 1d20 for a 60% odds of success chance is totally the same as using up a specific 25% of a 3d6 range for only 40% change in odds of success.

And this is why you are so fucking stupidly wrong and dishonest.

To what god damn degree curved systems provide the "desirable" behaviour of having different segments of the range provide different changes in chances of success, even within the part of the range where the behaviour is MOST evident and "desirable" YOUR MATH IS STILL FUCKING WRONG.

Even when you selected your favorable starting point, truncated your math at a favorable end point, you STILL couldn't bring yourself to honestly compare a +40% total to a +40% total or a +10% average to a +10% average, you STILL needed to constantly cherry pick the god damn fat end and then use that to make a dishonest comparison of the god damn thin end.

Because even under your own stupid argument, if you compared apples to apples, and actually stuck to the same damn total or averages EVEN for the limited cherry picked truncated section of the ranges you are talking about, the actual differences aren't nearly as "obviously totes cool!".

The entire methodology is flawed and dishonest.

If someone actually used your last post as any part of the methodology for designing how bonuses worked in their 3d6 system or their 1d20 system. Their fucking system's maths would break.

[Grek]

Could we please not derail this thread with yet another 3d6 vs 1d20 argument?

[momothefiddler]

Could we please not derail this thread with yet another 3d6 vs 1d20 argument?

Eh, a reasonable request.

[tussock]

The thing is, probability theory is a proper degree you can get at proper universities. Professionals do disagree with each other about how to use it in the real world.

It's also all counter-intuitive. Humans seem to naturally assume a bunch of stuff about it that isn't at all true, as some sort of internal story about how the world is fair. Which works well for warm months coming back after cold months, but really poorly for 20's coming after 1's. Seasons are seasonal, roulette wheels are not.

Not to mention iterative probabilities, or that Bayesian stuff where you sort of carefully redefine all your questions until they match your results, in a way that isn't instantly stupid and broken as doing that normally would be. Because it's always about asking the right questions, and it turns out there's valid systems for figuring out what they should be, if you start somewhere close by.

---

For instance, figuring out the Fighter has 65% chance to hit, you might ask what his average damage, is, or the quartiles over four rounds, or the mode, or the chance of not having any successes after 5 rounds, or 10 rounds, and how many players do you need testing the thing before you're 95% likely to have someone suffer those fates.

8+ to hit, (1 - 7/20) = 0.65
Average damage = 0.65 * damage per hit.
In four rounds you get 0-4 hits at
0 hits, 02%: 1*0.35^4
1 hits, 11%: 4*0.35^3*0.65
2 hits, 31%: 6*0.35^2*0.65^2
3 hits, 38%: 4*0.35*0.65^3
4 hits, 18%: 1*0.65^4

And from there you can map out, say 4d8+16 for four hits, 3d8+12 for 3, and so on, multiply each by it's chance of getting that number of hits, and make a nice graph. Maybe notice the rare events (which are more memorable) are mostly negative, so add some rare critical damage functions to produce memorable positive events too.

After 5 rounds you've got 0.35^5, or half a percent. LN(0.05)/LN(0.995) ~= 600 trials will almost certainly show up that event.
After 10 rounds ... forget it. 100,000+ trials to be almost certain of seeing it.

At 0.015 for the 0 hits in four rounds, you'd need LN(0.05)/LN(0.985) ~= 200 trails to have a 5% chance to still not hit that result. Or 45 trails for a 50% chance. So at 13 fights per level, after 3 levels around half your Fighters would never have had a swish fight. By adding attacks, lengthening battles, you make that much more rare.

Logarithms are handy for stuff like iteration
X^Y = Z
=> Y*LN(X) = LN(Z)
=> Y = LN(Z)/LN(X)

Number of trials needed to almost certainly see something is pretty important in tabletop. Saving throws until you almost certainly die, for instance. We don't actually roll very many dice, maybe a few thousand hit rolls from 1-20 in D&D, maybe a hundred saves, but only a few hundred hit rolls and just tens of saves on the way to 7th level. If you can roll hundreds of trials with some chance of never seeing a certain result, you might want to reconsider the existence of that result, at least at low levels.

---

But at a basic level for RPG design, it's easy. Work out what results you want to have occur. Like, really occur, not some bullshit where Demogorgon ports in and eats your face for casting CLW one time in ten thousand. Lay them out, work out an acceptable range for each result in your head and note it down (so between 0-2 times over 20 levels, or between 2-5 times per fight, or whatever). Then plug in the simplest possible randomiser for each that produce those ranges of result.

Then take whatever randomiser (or set thereof) is used most of the time, and try to make it fit all the other events as well. Make the thing simpler to learn. Maybe chuck out some result sets that aren't common and don't fit the system anyway. If you're finding it hard to actually see if your randomiser produces certain results in the first place: STOP. Do not do that. It's hard enough for players when the chances are obvious to everyone, do not obfuscate it any further with crap that's hard to understand.

Like, 4th edition D&D gave us fights where the averages looked OK, but the outliers were where too many daily and encounter powers missed, there was more HP worth of monsters and four hours later you were all still rolling dice and desperately wanting the game to end. You can't only consider averages in design, so the RNG has to be simple enough to work through the probability space. Spreadsheets and their graphing functions are handy there.

[Orion]

This is not new information, but perhaps you will find it helpful. Let's talk about:

The chance of rolling "12" on 2d6
The chance of rolling "doubles" on 2d6
The chance of rolling "8" on 2d6
The chance of rolling "At least 10" on 2d6
The chance of rolling "at least one 6" on 4 dice
The chance of rolling "at least one 5+" on 4 dice
The chance of rolling "exactly 1" 5+ on 4 dice
The chance of rolling "at least 2 5+" on 4 dice

Bonus Round: The chance of getting the same number twice in a row on 2d6.

If two events don't affect each other, multiply the chance of each event together.

To get 12 on 2d6, you need to roll 6, 6. Each die has a 1/6 chance of rolling 6. The chance that both dice roll 6 is 1/6*1/6, or 1/36.

If more than one event is acceptable, add the chances of each.

Rolling 12 is one of 6 equally-likely ways to get "doubles" so the chance of doubles is 6 times greater, or 1/6. (Another way to think about this one: no matter what number you rolled on your first die, the next die has a 1 in 6 chance to match; we'll come back to this technique)

When you add two dice together, there's more than one way to get the same total. You add up the chance of each way (often called a "case").

With dice, each case is equally likely. Start by listing off the ways to roll 8. If you got "2,6; 3,5; 4,4" you done fucked up because you forgot about "5,3; and 6, 2". Even if you don't care about the order events happen in, you have to include every order in your calculation. There are 5 ways to get 8 and 36 ways to roll 2 dice, so your odds are 5/36.

(This means that if you flip a coin twice, you're twice as likely to get one heads and one tails as to get 2 tails. As a child, I found this brain-breakingly weird, but it's true. In gaming terms, your W/U magic deck gets 1 island, 1 plains half the time)

Checking for 10+ is another case of adding multiple events.

You just add (chance of 10) + (chance of 11) + (chance of 12) = (chance of 10+). When you add multiple dice, you get a "bell curve" where events get less likely at the extremes. 2d6 actually has a smooth and memorable distribution. 6/36 at the center (7), -1/36 per step to the outside. You can get 10 3 ways (4,6; 5,5; 6,4) 11 2 ways (5,6; 6,5) and 12 one way (6,6) so the odds of a 10+ are 3+2+1/36, or 1/6 again.

So much for adding dice. On to dice pools. This is weirder, because the dice aren't really independent in the same way as before. No die influences the number that comes up on another die, but it influences the chance that we care.

--------------------------------------------------------------------------------------

It's Arkham Horror and you're Cursed, which means you only hit on a 6. You have 4 dice and you need 1 hit to succeed.

What's your chance? Each die has a 1 in 6 chance, and you get 4 of them, so you might think you have a 4 in 6 chance, but you would be tragically mistaken. You will get an average of 4/6 hits, but sometimes you'll get multiple hits on the same roll, which is a waste. To solve this kind of problem, you often need to think backwards. Instead of asking for your chance to succeed, ask for your chance to fail. That's easier, because succeeding required you to roll a "6" at least once, but failing requires you to roll "not 6" every time. We're back to multiplying independent events. The chance of "not 6" on one die is 5/6. the chance of "not 6" 4 times in a row is. 5/6*5/6*5/6*5/6. That's equivalent to 5*5*5*5/6*6*6*6, or 625/1296. Your chance to succeed is the the reverse of that -- 671/1296 or about 52%.

If you hit on 5+, the rules are the same, you're just rolling for "4 or less" instead of "not 6." The fail chance is 2/3 to the 4th power, which is 16/81, or 19.7%. The success chance is 80.3%.

The chance of exactly one hit is a huge pain to calculate.

When the dice all affect each other in a complicated way, you need to look at "branches" of a "probability tree" (like branching parallel worlds.) Another way to think about it: make assumptions to simplify it, but then go back and adjust for the chance your assumption holds. Let's assume that the 1st die is a hit. If the first die hits, you need all 3 of the other dice to miss. By now you should know how to do this one: there's 8/27 chance of 3 misses in a row. Now, go back and remember that the first die only hits 1/3 times. So the chance of a hit (1/3) followed by 3 misses (8/27) is (8*1)/27*3), or 8/81. (That's 9.9%).

To put it another way, "There is a 9.9% chance that the first die, and only the first die, is a hit).

But! You are rolling 4 dice, and any one of them could be your "only hit." You also need to consider the chance of "only the second die hits" "only the third die hits" and "only the fourth die hits." The good news is that these cases never overlap. You can't have 2 "only hits," so you just add them together. You get exactly one hit 32/81, or 39.5% of the time.

If you need multiple hits, calculate the chance of each number of hits (chance for 0 hits, chance for 1 hit, chance for 2 hits, etc.) and add the relevant ones.

We already know the chance of zero hits is 19.7%. We also know the chance of exactly 1 hit is 39.5%. The chance of "0 or 1 hits" is 19.7+39.5, or 59.2%. The chance of "two, three, or four hits" is 40.8%.

Bonus Round: What are the odds of getting the same number twice in a row on 2d6? This belonged in the 2d6 sections, but it's harder than the rest so I moved it. Also, it makes dice interact in similar ways to what we just saw.

This would be a really easy question if we cared about the order dice rolled in. If "3, then 5" were different from "5, then 3", each result would have a 1/36 chance. No matter what result you got on the first roll, there would be a 1/36 chance of getting it again, so a 1/36 chance of getting the same result twice. This is like the doubles question.

However, because we didn't track order, we saw that you were way more likely to get a 7 than an 11. This makes a big difference, and you can see why if you tell yourself a story (use a probability tree). Suppose you've already made your first roll, and it's a 7. Now, your chance of a repeat is pretty decent. Your next roll needs to be a 7, and that's the most likely number. On the other hand, suppose you already made your first roll was a 12. Now you're screwed, because you're making your next roll hoping for 12, and that hardly ever happens. Now, remember that the first story (I already rolled a 7, and I feel okay about trying for another one) happens way more often than the second story (I already rolled a 12, and I feel terrible about trying for another one.)

At this point, you're gonna have to add up the chance of each kind of "repeat." "7, 7" happens 1 in 36 times, because you need a 1 in 6, followed by another 1 in 6. "8, 8" happens (5/36)*(5/36) times (because the chance of "8" was 5/36) "9, 9", "10, 10", "11, 11", and "12,12" come up "(4/36)(4/36), "(3/36)(3/36)", "(2/36)(2/36)", and (1/36)(1/36) respectively. 2-6 is a mirror image of 8-12. Add it all up and you get

(1+4+9+16+25+36+25+16+9+4+1)/1296 (which is 36X36), or 8.8% of the time.

You may notice that the calculations for probabilities can get really painful, really fast. The good news is that there are general formulas you can plug stuff into, like the 2kN "combinatorial" formula that computes odds for an arbitrary number of rolls of an arbitrary size of dice, but if you don't think of yourself as a math person, you might stare at it for a while but not quite get it. Hopefully, you can use the stories provided here to understand why the formula works, and for a sanity check if it seems to go wrong.

There is also a formula for the results of multiple draws from a deck of cards. This is called "selection without replacement" because you can't get the same card twice, while rolling dice is "selection with replacement" because you can get the same roll repeatedly.

You should be able to reason out the probabilities for cards in theory by using the probability branch stuff. Like, what's the chance of drawing all 4 aces in 4 draws? The first time, you have a 4/52 chance of pulling an ace, but the second time you only have 3/51, and then 2/50 and 1/49. If you're dealing with "at least" problems, though, it gets horrible real fast. If you want to know the odds of drawing 3 or more hearts in a hand of 7, you've gotta cope with the fact that your first draw has a 1/4 chance of hearts, but your second draw has either a 12/51 11/51 chance, and those cases are themselves not equally likely.

Basically, anything with cards is not computable by hand, and you should just learn the formulas if you need to do it.[/spoiler]

[OgreBattle]

How do I find the standard variation of a flat RNG (d20, d6, d10 )?
How do I find the standard variation of a curved RNG (2d10, 3d6)?

[Grek]

Same way!

Since the probability space for results is small and enumerable you can calculate the ideal standard deviation. Do the following steps in order:
1. Figure out the mean result and all possible results rolled.
2. From each result you could roll, subtract the mean.
3. Square each difference.
4. Add up all the squared differences.
5. Divide that sum by the number of possible results.
6. Take the square root of that number.

Example: 1d6.

The mean is 3.5 and your possible results are (1, 2, 3, 4, 5, 6).
The differences are (-2.5, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5).
The squared differences are (6.25, 2.25, 0.25, 0.25, 2.25, 6.25).
Those sum to 17.5.
Divided by 6, that is 2.92.
The square root of that is 1.71.

For curved rolls, just remember that both order and specific dice rolled both matter. On 2d10, 1 and then 10 is a separate result from 10 then 1 and from 2 then 9.

Example: 2d2

The mean is 3 and your possible results are (1+1=2, 1+2=3, 2+1=3, 2+2=4).
The differences are (-1, 0, 0, 1).
The squared differences are (1, 0, 0, 1)
Those sum to 2.
Divided by 4, that is 0.5.
The square root of that is 0.707.

This is of course very time consuming to do by hand, and prone to errors (I made a basic arithmetic error in the 1d6 example my first time through it!). So I prefer using http://www.anydice.com/ and the Summary function.

[Morat]

Technically you're multiplying the square of each difference by its probability. But with flat RNGs, of course, you can just multiply by the probability of any of the results after you sum them all up, since it's all the same. That's where the "divide by 6" step comes in for 1d6. Each outcome has a 1/6 probability, so you can divide the sum by 6 and get the same result.

But for curved RNGs, it's faster to multiply each outcome by its probability than it is to add the difference of every possible result.

So for 2d6, the probability of a 2 is 1/36 (you must roll 1,1), of a 3 is 2/36 (1,2 or 2,1), etc. So you go (2-7)^2 * 1/36 + (3-7)^2 * 2/36 + (4-7)^2 * 3/36 + ... all the way to + (12-2)^2 * 1/36 for the variance, and the square root is the standard deviation.

But yes, it is vastly faster to just use anydice.

[Username17]

Note that for 1d20 the deviation is 5.77 while for 3d6 the deviation is 2.96. This means that you can fit physically more sigmas into a 3d6 roll than you can for a d20 roll. And of course, that Phone Lobster's rants on the subject are completely innumerate or perhaps deliberate falsehoods. Six of one, half dozen of the other.

Standard deviations implicitly expect data to be normally distributed. Curved die rolls are, while flat die rolls are not. Which means that when you do the standard deviation calculations on 3d6 you get useful data, while when you do the standard deviation calculations for 1d20 you kinda get gibberish.

The fact that you move one standard deviation on 3d6 about every 3 is why the standard target numbers in HERO system are 8, 11, and 14. Those numbers feel adequately different from each other because they are about one standard deviation apart. It would have been just as good to choose target numbers of 7, 10, and 13, because 3d6 is normally distributed in its outputs and the same absolute shift in either direction will be a standard deviation no matter where you start. Again and still: remember that literally everything Phone Lobster says about math is insane and wrong.

-Username17

[tussock]

Note that for dice, you can solve for standard deviation.

It's SQRT(n(x-1)(x+1)/12). So 1d6 is SQRT(1*5*7/12), or 1.71, and 2d2 is SQRT(2*1*3/12), or 0.707, as above.

[Judging__Eagle]

Found this probability calculator via a FB RPG Design group

http://topps.diku.dk/torbenm/troll.msp