Another shitty blog

Exactly What It Says On The Tin

[blignick]

[orcboxer]

his worldview is utterly divorced from reality and brother those alimony payments are brutal

[teathattast]

[someoldjokes]

From The New Anecdota Americana, 1944

[princessparfaiti]

diversity win! this farmer is poly

[polyamorousmood]

from NINETEEN FORTY FOUR

[summonfish]

[shutyourmoustache]

Someone said “loved to the point of invention,” and I’m personally in shambles.

[glacecakes]

[sniperct]

tumblr waiting for news on mitch mcconnell (image source)

[fellahimedia]

I think since rules dont matter anymore that they should bring back cabo verde into the finals like when x factor used to bring back eliminated contestants

[blckgirlacid]

I need my weird alone time or I will explode

[deslizada]

man sometimes friendship really is just "I saw this and knew it would give you psychic damage. please respond with agony" and then they do. and it's great

[weirdmageddon]

i know most people have seen it but i cant emphasize how much this is literally my favorite breath of the wild clip of all time. also i can never fucking find this clip when i need it especially in high definition so here it is

[eikonoklasms]

i cant stop thinking about this post so bad. cat named daniel

[onlytiktoks]

[piedude]

[y2k-internetexplorer]

[prokopetz]

How do you calculate odds of successive rolls in "x of outcome 1 before y of outcome 2" scenarios? I'm specifically trying to figure out how much effect certain adjustments to the death save mechanic have on dnd5e's lethality, but it would also be useful for estimating things like montage tests in Draw Steel.

Fortunately, death saving throws are a lot simpler than the general case of "X of outcome 1 before Y of outcome 2" because (probability of outcome 1 + probability of outcome 2) equals 1; the presence of a "neither" case can throw a monkey into the wrench.

Since there's no "neither" outcome, the number of rolls you can possibly make is bounded; "three successes before three failures", for example, will never exceed five rolls, because after the fifth roll you're guaranteed to have at least three of one outcome or the other. That makes it possible to enumerate every possible sequence of outcomes:

SSS SSFS SFSS FSSS SSFFS SFSFS FSSFS SFFSS FSFSS FFSSS FFSSF FSFSF SFFSF FSSFF SFSFF SSFFF FFSF FSFF SFFF FFF

Let's assume that our S (success) result is rolling a 4 on 1d4, and our F (failure) result is rolling anything else. We can then work out the odds of each case, like so:

SSS = 1/64 = 16/1024 SSFS = 3/256 = 12/1024 SFSS = 3/256 = 12/1024 FSSS = 3/256 = 12/1024 SSFFS = 9/1024 SFSFS = 9/1024 FSSFS = 9/1024 SFFSS = 9/1024 FSFSS = 9/1024 FFSSS = 9/1024 FFSSF = 27/1024 FSFSF = 27/1024 SFFSF = 27/1024 FSSFF = 27/1024 SFSFF = 27/1024 SSFFF = 27/1024 FFSF = 27/256 = 108/1024 FSFF = 27/256 = 108/1024 SFFF = 27/256 = 108/1024 FFF = 27/64 = 432/1024

Adding up our target outcomes, we get 106/1024 odds of three successes before three failures, and 918/1024 odds of three failures before three successes. (If these don't add up to 1, you know you missed a case somewhere.)

Once we've got the symmetrical case (i.e., X successes before X failures), we can deal with asymmetrical cases simply by changing up which rows in the probability tree we've already constructed count as which outcome – we needn't start from scratch. For example, the odds of two successes before three failures moves all the 27/1024 rows in the above table into the "overall success" bucket, and we arrive at 268/1024 odds. Similarly, "three successes before two failures" would flip the 9/1024 rows over to the "overall failure" bucket.

(Now that we know what's going on "under the hood", the obvious follow-up question is "is there a general formula I can use to skip constructing the probability tree and get the answer directly?". The answer is "probably, but I don't recall it off the top of my head". If we're lucky, one of the several mathematicians following this blog will be able to correct my omission!)

[rlyehtaxidermist]

Modelling this mathematically is pretty straightforward. My first thought is an absorbing-state Markov chain. This can model pretty much any sequential roll system, but in this particular case, it's simplified significantly because this system is acyclic.

To build a full Markov model, label your states by ordered pairs (S,F). The starting state is (0,0), and the absorbing states are (X,*) and (*,Y). Each non-absorbing state transitions to either (S+1,F) with probability P(S) or to (S,F+1) with probability 1-P(S). You can order your states and make a huge matrix and do all the math, but there's also software to do absorbing-state Markov calculations for all kinds of complicated systems.

With that said, we can hella cheat at this specific one using combinatorics. Breaking it up into cases makes it more work! "Roll X successes before Y failures" is the same as "Roll X or more successes on X+Y-1 dice", which is a simple sum of a binomial distribution.

To use your earlier example, we have p = 1/4, X=Y=3, which gives

The general formula for X successes before Y failures with probability p of success is then

If you don't want to write all that out in Wolfram-Alpha or the like, you can also use a binomial CDF calculator. For the linked one, X+Y-1 is your number of trials, X is your number of successes, and you want the line that says Cumulative probability: P(X≥)

[rlyehtaxidermist]

I'll go into a bit more detail on using the Markov model for a more complex model that allows neutral results, critical successes, and critical failures (as are often of interest). This means that each "normal" state (S,F) has the following transitions

to (S+1,F) with probability P(S)

to (S+2,F) with probability P(CS)

to (S,F+1) with probability P(F)

to (S,F+2) with probability P(CF)

to (S,F) with probability 1-P(S)-P(CS)-P(F)-P(CF)

The way to solve this is to construct a matrix where each row and column matches a state, with row i and column j being the probability of going from state i to state j. You then begin with a row vector of 1 followed by a bunch of zeroes and multiply it by this matrix multiple times, with each time representing one roll.

Having spend this morning watching GDQ without any other low-intensity work to occupy my hands with, I have written a crude calculator that should be able to figure out your odds of a success or failure after a given number of rolls given a few parameters. Make a copy and please tell me how much I fucking broke this thing.

[infamousbloodevoker]

Y'all are a little bit killing me here overcomplicating the simple case. Without incorporating crits, if your outcome can only be a binary Yes/No (Save/Fail), then the number of rolls R before S saves follows a Negative Binomial Distribution with probability P=[the probability of a Success/Save for any roll]. Use any online calculator for the NB distribution. Do not use a Binomial distribution calculator. Now if you are incorporating crits? Or a neither case? Ignore all that. I'm too sleep deprived to hash it out a better solution but traditional probability distributions won't save you.

[raisinchallah]

really humiliating trying to write horror like they went into the creepy house and there was a creepy ghost and the creepy guy with a creepy knife and everything was very creepy are you scared yet and thats like literally not how suspense and tension actually work but like all u can do is say well maybe something else was creepy?