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@rlyehtaxidermist
There's a very funny thing that happens in transhuman science fiction where sometimes an author will turn out to be a reactionary conservative who thought he was writing dystopian fiction and intended all the sickass radical body modification stuff to be scary, but he assumed that was obvious and forgot to put it in the actual text.
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if you're going to um actually me you should probably take a moment and make sure you're not just restating all of the things i already said with less explanations
so it turns out that chemical runoff from ai datacenters is causing chromosomal defects in sheep populations, causing something like 95% of affected sheep to be born female
that's right: ai is to blame for the ram shortage
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!)
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≥)
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.
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.
You still need to do a summation if you're using the negative binomial distribution; otherwise you're only finding the probability of rolling X successes and a fixed number of failures, not at most that many failures. The formula and calculator instructions I posted is equivalent to the CDF of a negative binomial distribution, because the negative binomial distribution is just a special case of the binomial distribution.
"incorporating crits? or a neither case?" is the specific case I called out as requiring the Markov chain model (or, if solvable combinatorially, not being doable as a Monday morning pastime).
The big reason it gets me when folks act like it's unlikely that the author was considering the esoteric symbolism of the colour of the curtains is that, like, there are people out there who consider the esoteric symbolism of their home decor in real life. They're not even rare – Pinterest is full of them.
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as per my last straw,
Reading the dev notes for Mina the Hollower has been a pretty stark reminder to consider the practical reasons for creative decisions. I was all "okay, but what are we really saying by making the protagonist a mouse", and then the devs are like "yeah, the main reason she's a mouse is because GBC-style sprites are tiny and her bigass ears make it easy for the player to tell which direction she's facing".
any system where clicking a block of text automatically opens that text in an editor (Atlassian stuff loves to do this) is my enemy. get out of my way you piece of shit, I'm trying to copy-paste
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!)
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≥)
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.
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!)
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≥)
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
Free to watch • No registration required • HD streaming
Op turned off reblogs but I MUST
"no more gnc men erasure!!!" ok but you losers don't even keep this energy for gender non conforming black boys because black people, especially dark skinned black people regardless of their gender, gotta fight to be seen with an ounce of femininity
it reminds me of the whole "black trans men pass easier" but the dipshits who say this don't even wanna admit *WHY* they assume this.
black people are hyper-masculinized no matter what we do it fucking sucks.
from my own experiences people assumed I'm on testosterone or assumed I'm trying to pass despite my goal being fem/gnc and the fact I'm not going on hormones it's just annoying.
so many misguided metaphors around violence and desire. if the open maw of a panting beast fills you with the want to be devoured, that does not make you prey. while the rabbit trembles in fear, its deepest desire is to run. evolution demands it. in fact, the desire to be eaten does not make you any small animal at all.
it makes you a fruit.
drupe Fruit perfec t size for take food from to e\at! inside very Sweet and Juicy mammal eat tasty put drupe in Mammal Mouth. Put Drupe In Mammal Mouth. no problems ever with druppe fruit because good Soft and Chewy for mammal teeth weak of tough leafy stalks. Adrupe Fruit yes a food for mammal put drupe in mammal mouth can trust drupe for giveing good food to mammal. friend drupe
there’s this term i coined in my friendgroup i call “the charizard effect” and it can apply to anything and everything, but it was born from me explaining my feelings about the pokemon charizard. the term is basically about how overexposure to something be it by corporate shilling or fandom prominence drives me away from really enjoying something bc i’m exposed to it so much against my will i become tired of it. it came to me bc i was ranting about how tpci does not, and cannot stop reinventing charizard, and how it is popular and obtusely included in almost every region, merch, etc in every way possible and it’s highly commodified.
i dont dislike the pokemon charizard, in fact i really like its X form, but i am exposed to so much charizard in my pokemon consumption that i cant be bothered to care for it in any more than in passing. this applies to a bunch of other stuff i’d otherwise be ok with, but i always just call this aversion phenomena “the charizard effect”
making this term has done numbers for me being able to concisely express how i feel abt something. like. its not charizard’s fault i feel this way, im sure i’d feel normal abt it if it was stripped of all this over commodification, but i cannot. hence the name
Hey, did y'all see this?
I saw this when running newpipe. But wait, it gets deeper. I clicked on the details buttons and it said as of today, we have 83 days left until Google rolls out this new requirement for apps inside and outside of the google play store. If any developer disagrees with their new terms and fees, they will be blocked!
I'll share some of the info below:
Looks like they're trying to nuke the remaining privacy and freedoms we have left on the internet.
What to do?
-Get your developer friends to not comply to their new guides
- Sign the open letter on the site and take action by checking out the full resources list on their website as well!
To summarize, this is all daunting especially when you feel all alone with unfair and inhumane regulations comming out faster than improvements but we got this working together!
Share the link with your friends, family and anyone who will listen!
Your phone is about to stop being yours. In September 2026, Google will block every Android app whose developer hasn't registered with them.
If you're in the US, I created a petition to make it easier to contact senators and congressmen.
Join 1 people. Google is trying to make people hand over government id in order to make an Android app. If they don't, then that app can't b
If you're not in the US, see if your country is listed here for whom to contact.
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
Free to watch • No registration required • HD streaming
men and women are not opposites. men and women are not enemies. men and women are two parts of a broad coalition which fights against a mutual enemy: inkjet printers
I'm a big fan (enthusiast) of big fans (devices for creating air currents), but Big Fan (collective of large, prosperous fan merchants) is putting bad clicky noise in all of them to prevent me from understanding the dialogue in Big Fan (2009 film starring Patton Oswalt)
