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Okay so every time you attack there's a 50% chance of making another attack, which can continue to trigger additional attacks. So the first is guaranteed, likelihood of the second is 50%, likelihood of the third is half that again so 25%, fourth is half that again so 12.5%, and so forth.
100% is 1/2^0 , 50% is 1/2^1 , 25% is 1/2^2 , so clearly the formula is 1/2^(n-1) where n is the number of attacks. This means there's a 1/2^7 = 1/128 chance of making at least eight attacks with one crossbow shot.
And since half the time you continue shooting after that, the odds of *exactly* eight shots are 1/256.
Edit: fixed an inaccuracy
If he doesn't understand a diagram that I understood when I was probably in the single digits, then what the fuck am I doing with my life - should I be doing better, help me
Honestly though they didn't do the best at addressing the issue though I don't think? At no point did they explicitly state what I would have said, and I genuinely think what I would have said would have done more to help the incorrect person than what "op" said.
There is a difference between the number of possible outcomes and the chance at which they occur. Yes there are 8 possible outcomes but from the very beginning, not at the time a shot is taken, these possible outcomes are weighted differently.
That should have been explicitly stated. But it wasn't. The problem was that the incorrect person was partially correct in their assumption that there were 8 possible outcomes so it should have been attempted to explain that the outcomes are weighted differently rather than to explain that they are flat out wrong.
Exactly the different outcomes just have a different probability. This is a Laplace experiment so every outcome does have the same probability if you don’t stop once you miss.
But the outcomes relevant for the game that he states have different probabilities because the hidden permutations have to be accounted for with this approach.
Omega approach for this problem instead of decision tree is stupid but it works.
Nah he's saying, because you can have cases like WW,WL, LL, but not LW, since you can't get a trigger AFTER you didn't trigger before, so it's not 2^n ...
What he needs to hear/understand is that has no impact on the probability of the previous trigger, and you can only use (favourable outcomes)÷(all possible outcomes) in certain cases. Here the outcomes are not independent, you can't just count them all and divide.
Minor knit-pick from a statistician here. 1/128 is the probability of getting *at least* 8 attacks. Getting 8 attacks (exactly) has a probability of 1/256. You need to multiply by the 1/2 which ends the sequence.
An easy way to verify this is to note that the probabilities given in the first slide do not add up to 1, but multiplying each of them an extra 1/2 gives you a geometric series that will add to 1.
That's not correct, you're missing a factor 1/2 in all your probabilities. Probabilities have to sum to 1 but yours does sum to 2. See [my answer here](https://reddit.com/r/confidentlyincorrect/s/jp1bWOsUau)
That's what I was thinking. May as well say it's a 50% chance of 8 shots. You either get 8 shots, or you don't. Same way your chances of dying tomorrow when a plane crashes into you while being struck by lightning during a threesome with Taylor Swift and Beyonce are 50/50.
That's why I buy lottery tickets. You either win millions or you don't. 50/50, baby.
It literally is. Dude's logic is "Every outcome has an equal chance, there are 8 possible outcomes, therefore the chance is 1/8, in order to have a 1/128 chance you'd have to have 128 outcomes."
I think the easiest way to break the logic is to ask them what the odds are of hitting 9 shots. If all the other previous outcomes were 1/8, then hitting a 9th shot has to be impossible, right? The biggest obvious weakness in this thinking is that there's not 8 outcomes, there's infinity outcomes, nothing forces a stop at 8.
That’s always bugged me, every time I’ve heard it in movies and tv shows (Aussie, so we say “couldn’t” but we get a lot of American media, obviously)
“I could care less” = I care to at least some extent aka the opposite to the sentiment generally being expressed
“I couldn’t care less” = I have zero care regarding this, therefore it is impossible for me to care any less than I already do aka the exact sentiment generally being expressed.
I always assumed it was a truncated phrase whose ending was lost in time. "I could care less *but I'd have to be dead*" or something along those lines.
Similar to phrases like "I can't even" or when people will use just part of a well known phrase "well a bird in the hand..."
The origins are “I couldn’t care less” and etymologists believe “I could care less” is a Yiddish-English sarcastic variant (in the same spirit as the sarcastic Yiddish English phrase “I should be so lucky”). Except the sarcastic tone is disappeared and the phrase is now said, generally, in a rather earnest tone.
I knew “couldn’t” came first, I just couldn’t remember how it changed in (primarily) American English so if you Google, you’ll probably find the same comparison (I added the earnest tone part though, because yeah, it’s typically said earnestly, which is the opposite of how it should be said).
ETA: just found another thing where “could” was used in an advice column in 1966, which would make the sarcastic origin less likely. Maybe it originated from a typo? Who knows, all I know is that somehow, the vernacular in American English was changed for the phrase to, when taken literally, mean the opposite, and often, if not mostly, used in a somewhat earnest tone of voice.
In plain language, Occam’s razor states that the simplest explanation is preferable to one that is more complex. On the basis of this principle I'm going to assume it's probably just Americans fucking it up.
But if you couldn't care less you wouldn't find the need to say it. The act of saying "I couldn't care less" means you care enough to say it. Therefor, you could care less.
Incorrect.
I couldn’t care less about the price of fish in China. I *do*, however, care (in a negative way) about someone *telling* me, in minute detail, about the price of fish in China. Therefore, I say “I couldn’t care less about the price of fish in China” to inform said party about my lack of care about the topic and, therefore, ending that topic of discussion.
For example…
I would just like to point out that I said I *don't* give 1/128 of a fuck, which technically only excludes that specific amount and leaves it ambiguous as to whether I actually give a greater or lesser proportion of a fuck.
When I say “I could care less,” what I mean is, “I care so little about this that I don’t even care how much I care about it. In fact, if someone asked me to, I would be perfect happy to care about it even less than I already do.”
Wrong. The fact that you commented means you give at least 1/2 of a fuck, and when I aim my crossbow at you you're going to give a lot of fucks, unless I miss my first shot and the percentage of fucks you give will be 50.5%, which can be rounded out to one, so you clearly give a fuck.
Flip coin 8 times. Chance of heads all 8 times in a row = 1/128. Confidently incorrect person claims chance of heads all 8 times in a row is 1/8. The end.
They mean a chance of 8 or more obviously even if not specified. They don't care about the possibility that a no th shot could've happened or not they would've been equally impressed.
Yeah the 1/8 logic assumes that each possible outcome is of equal probability, which is very wrong.
I'm not great at math so I can't tell you for sure if 1/128 is correct, but I can easily tell you 1/8 is stupid lol.
Assuming it *could* theoretically fire infinite attacks if it continues to hit the 50/50, then there are infinite possibilities for how many shots it may fire.
Therefore the probability of it hitting *any* amount of attacks is 1/∞
/s if it wasn't obvious
Any time you need thingA *and* thingB to happen, you multiply the two probabilities together. Because probability is always a number between 0 and 1 (say 0.5 in the case of a 50/50) this always gives you a lower number.
0.5 x 0.5 = 1/4
Do that 8 times and you get there
The context is in the post, the first is indeed guaranteed. Also that first paragraph makes no sense, it doesn't even look like you understand what you're talking about there lmao.
I had a similar argument with my friend yesterday lol I had just done something where I hit a 5% chance and wanted to do another which I said was still a 5% chance he said that’s not his probability works and the next one had a lower chance and I disagreed. Idk if I was right or not but I said it’s a smaller chance to hit them back to back if I hadn’t hit the first one but since it isn’t compounding chance it’s the usual 5%
You're right - your friend is committing the [gambler's fallacy](https://en.wikipedia.org/wiki/Gambler%27s_fallacy).
The chance of *succeeding on both rolls* is .25% (I think, somebody check my math), but *each roll* remains 5%.
Assuming you mean "or" inclusively (succeeding on both still counts): the easiest way is to take the probability of both being a failure and subtracting that from 1. The opposite of at least 1 success is all failures.
1-.05 = .95 <-- the probability of a failure
.95^2 = .9025 <-- the probability of 2 fails in a row
1-.9025 = .0975 <-- the probability of succeeding on at least 1 roll
The long way is to list all the outcomes and add together the acceptable ones.
P(S)\*P(S) = .05^2 = .0025
P(S)\*P(F) = .05\*.95 = .0475
P(F)\*P(S) = .95\*.05 = .0475
P(F)\*P(F) = .95^2 = .9025
We add together every outcome except the last one:
.0025+.0475+.0475=.0975
If you want to use the exclusive "or" and want the chance of only succeeding on a single roll but not both then you do .0475+.0475=.095.
Some games (and other programs) do have coding that reduces the pure randomness of stuff, so without knowing the specifics, your friend very well could be right here. But assuming that's not relevant, it sounds like a discrepancy between "unlikely to do the same thing consecutively" vs. "the previous results don't affect later ones." The odds of succeeding at a 5% chance twice in a row are 1 in 400. But if you've succeeded at that 5% chance 99 times in a row, the odds you'll succeed on the 100th attempt are 5%; the previous successes or failures, no matter how improbable, don't have any bearing on the very next roll of the die.
> Some games (and other programs) do have coding that reduces the pure randomness of stuff
If you're interested in how pseudorandom number generators work (in a very basic sense), what they do is they take an initial seed (some value) and generate a deterministic sequence of numbers from it. This means if you know the seed, and you know how a sequence was generated from it, AND you know which element from the sequence you're about to pull, you will always be able to predict what the outcome will be.
On top of this, the sequence may have biases for certain outcomes due to some values appearing more often in the sequence. If you know the seed and you know what algorithm was used to generate the sequence from the seed, then you can perform analysis on the sequence and find that you're more likely to have one outcome than another, when they're supposed to be an equal probability.
Going a tiny bit into CPU architecture (mega nerd shit that I may or may not be wrong about):
Intel x86 CPU architecture provides two Assembly language instructions for RNG: RDSEED (read seed) and RDRAND (read random). As far as I know, RDSEED is true random number generation (TRNG) that uses a source of noise from the hardware to achieve this, but takes a lot of clock cycles to perform, so relying on it will incur large performance penalties. I think RDRAND is pseudorandom number generation (PRNG) relying on an initial seed, but it also re-seeds periodically? So if you're generating random numbers very frequently it can be predicted, but if you're doing it infrequently it cannot.
Because of the performance hit, I believe RDSEED is generally only used to generate a seed, rather than generating ALL of your random numbers.
NOTE: I'm a complete gobshite at the best of times so if someone is better at this sort of stuff definitely feel free to correct me
I know most computer programs aren't technically capable of what we consider "true" random. The processes they use to simulate it can be pretty interesting for sure. But some make deliberate efforts to skew towards (or away from) certain results. For example, Diablo 3 will gradually increase your chance to find a legendary item the longer it's been since you've seen one. And I once read that iTunes users complained that shuffle wasn't random enough since, in true random fashion, you would sometimes get 2 or more consecutive songs from the same artist (or even the same album). So they implemented a feature that made that less likely to occur, technically reducing randomness but increasing the illusion of it to its users.
Probability is weird and you were both kinda right.
Assuming the events are independent then yes the chance does not change the second time around.
However it is also true that hitting both 5% would be much rarer than hitting only one.
You had already hit the first 5% that makes it now irrelevant to the second opportunity (If they're independent) thus making him wrong that it would lower your chances of hitting it again.
But if before hitting that first 5% you asked the probability of hitting two 5% chances in a row, it would be 0.25%
> grabbing a random number generator with 1 to 2, or a dice and pick 1-3 or 4-6
It's a real shame that people don't carry around an artifact in their pockets that would help them choose randomly between two outcomes. That would really change the world. You'd probably save at least a quarter hour a day with something like that. So much extra time to coin new phrases. You'd be able to hold your head high and not slink away with your tail between your legs.
OP didn’t really give context and thinks that this subreddit is r/IHadAnOnlineFight.
People posting themselves and their boring arguments should be banned.
They seem to be looking at each 50/50 chance as independent of each other rather than calculating the probability of a string of binary options dependent on each preceding success or failure all succeeding.
If shot one is guaranteed, then shot two is the only one that is actually 50/50 because shot three's potential is dependent on shot two's outcome and so on and so forth.
Depending on what you mean exactly you might be technically correct, but your reasoning is wrong.
After the first 1 attack that's guaranteed, you can say that:
The probability of 7 **or more** subsequent consecutive attacks is the infinite sum 1/2^(7) × 1/2 + 1/2^(8) × 1/2 + … = 1/2^6 × 1/2 = 1/128.
The probability of **exactly** 7 subsequent consecutive attacks is 1/2^(7) × 1/2 = 1/2^(8) = 1/256.
If you don't believe me, do a sanity check: sum all the probability of doing 0, 1, 2, 3, … consecutive attacks and you should end up with 1, because probabilities must sum to 1. You will get 1/2^(0) × 1/2 + 1/2^(1) × 1/2 + … = 1/(1-1/2) × 1/2 = 1.
To be more precise, the number of subsequent consecutive attacks is a random variable following geometric law of probability 1/2. The expected value here is 2 subsequent attacks.
In general the geometric law of probability p Is measuring the number of consecutive successes for repeated Bernoulli trials of probability p. The probability to have k consecutive successes is p^(k)(1-p) and the expected value is 1/(1-p).
He fell victim to one of the classic blunders! The most famous of which is, 'never get involved in a land war in Asia,' but only slightly less well-known is this: 'Never assume all outcomes have the same probability!”
It’s one of the most common mistakes you see when kids start learning probability, and one of the reasons that probability is one of the least-intuitive branches of mathematics that most people will come across.
I mean i can understand to some degree of what this person is thinking, their thought process follows the basic outcomes rather than the % overall of the highest outcome being the chosen one. That’s how they got the 1/8 instead of the 1/128, out of the 8 choices of an outcome(following their logic) it would make sense it can only happen 1/8th of the time but percentage wise it’s a different story
Yes. Think of a coin flipped three times and you want tails-tails-tails. There’s a 1/2 chance of getting tails. If I flip it again, there’s again a 1/2 chance of getting tails. Do it one more time, it’ll either be heads or tails.
The possible permutations (similar to combinations; the order matters in perms) are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Eight combinations.
TTT is one possibility: 1/8. 1/2 * 1/2 * 1/2 = 8.
The funniest thing about this is that you're both incorrect. The probability to get exactly 8 shots is neither 1/8, nor 1/128, but it is 1/256.
You have a 1/2^7 probability to get up to 8 shots. After that, you have 0.5 probability to either shoot an additional shot, getting you up to 9 shots, or to stop there. To get exactly 8 shots, you need to stop, which means you multiply the total probability by another 0.5, getting 1/2^8.
If you want to see how this works out in practice, assume you conduct m (=128) experiments. You would expect to get only one shot in m/2 (64) of those, two shots in m/2^2 (32) of those, etc. The general formula is that for s shots, you would expect m/2^s occurrences where you get exactly that many shots. If you conduct 128 experiments, then you expect to get exactly 8 shots m/2^s = 128/2^8 = 0.5 times, meaning once in 256 experiments (thus 1/256).
Technically correct, but in reality if someone feels like 8 shots is a sufficiently rare event, he's going to think of anything at least as rare as 8 shots to be significant as well. So we should really calculate for ≥8.
While that's a good point, they were discussing the probability of an event that already happened. It's not correct to say that that probability is 1/2^7.
There are only 8 outcomes in that particular sequence, so I get where they're coming from, but they clearly don't get even basic probability if that's their conclusion.
It's far more probable that the xbow gets just a second shot because it has a 50% chance of happening.
I would've liked to see their reply if they were directly asked "Do you think that there's the same 12.5% chance of getting 8 shots as 2 shots?" or point out that if someone was lucky enough to get 100 consecutive shots, by his logic there'd only be a 1% chance of any particular outcome, including only 2 shots after 1 hit, and getting 100 consecutive shots? They have no clue what they're saying.
So, then presumably the odds of 7 shots is 1/7, and 6 shots is 1/6, and so on. I wonder what the sum 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8 will be....(don't even need to argue the infinite sum, going to 4 gets into a problem).
Can we ban maths posts? If it's not the order of operations ones where the original is written in a stupid way with the intention of being misunderstood, it's ones like this where no one is good at explaining themselves and it's less that 1 person is "confidently incorrect", and more that at least 2 people don't know how to communicate clearly
Edit: I know that the 2nd person is incorrect. But the people arguing in the pictures have phrased their (correct) arguments in an incredibly confusing way - if I wasn't already familiar with the maths involved, I wouldn't have a clue what they were talking about and their arguments wouldn't sway me. Which is why I don't like posts like this, because although it does show a person being incorrect, and they seem confident about their false assumption, it doesn't really give the incorrect person an opportunity to realise their mistake and it's understandable to dismiss counterarguments if it sounds like they're a load of nonsense.
I know the 2nd guy is consistently incorrect, but all the explanations given to him are shit. I understand the maths involved, but it's like everyone has chosen the most confusing way possible to explain it to him, at which point I have a lot more sympathy for him being consistently incorrect. There are some much clearer explanations in the comments on this post.
That’s very true. I hated OP’s approach to explaining and kept saying “why not explain that there’s 128 combinations and hitting all eight times is only one?”
your explanations are not as clear as you think they are.
But, also, I think you are arguing about game mechanics, not probability. I take it you have a weapon that can take up to 8 shots. If you shoot and you miss, you get a 50% chance of getting another shot. So, assuming you miss every shot, you're right that you have a 1/128 chance of taking 8 shots.
But isn't the other guy saying that you ALSO get to take another shot if you hit on your shot? So to actually calculate the % chance of taking 8 shots requires you to factor in the chance that you will hit? If that's the case, then I'm pretty sure you're both wrong.
I got the explanation of the setting fine.
The crossbow fires once, this is guaranteed, it then has a 50% chance to fire again. If it does fire again, then there is another 50% chance it may fire once more, repeating until it fails to fire.
Thus the question is, "what is the probability of the crossbow firing 8 times in total?".
>The crossbow fires once, this is guaranteed, it then has a 50% chance to fire again. If it does fire again, then there is another 50% chance it may fire once more, repeating until it fails to fire.
Okay, yes, that's what OP is saying. But the guy OP thinks is wrong seems to be saying that it has a 50% chance to fire again if you miss but it always fires again if you hit.
>. But the guy OP thinks is wrong seems to be saying that it has a 50% chance to fire again if you miss but it always fires again if you hit.
"On a failed shot, it 100% never shoots again this round" is a quote from "the wrong guy". He's re-defining a little bit, as OP said "every attack there is a 50% chance" and he's saying "every attack, there's a 50% chance to hit, and if you hit, you attack again". They're calculating the same thing, "the wrong guy" is just flat-out wrong about it.
>every attack, there's a 50% chance to hit, and if you hit, you attack again".
So in this version:
you hit: you get another shot
you miss: you get a 50% chance at another shot.
So, if this really is the fact pattern, the Confidently Incorrect guy is right and OP is wrong.
In this situation you get:
shot 1: 75% chance of a second shot
shot 2: 56.25%
shot 3: 42%
shot 4: 32%
shot 5: 24%
shot 6: 17%
shot 7: 13% chance of taking the 8th shot -- or roughly 1/8.
I'm not sure that's actually the fact pattern, but it's almost certainly the rationale the incorrect guy is using. Hence my point: they disagree on the assumptions, not on the math.
>you miss: you get a 50% chance at another shot.
This is incompatible with the
> On a failed shot, it 100% never shoots again this round
from "the wrong guy".
Otherwise, with your set of assumptions that nobody here or on the original thread holds, you do get .75^7 chance to get at least 8 shots, and it IS pretty close to 1/8.
> If it was 1/128, there would be 128 possible situations that could have happened.
Except, "the wrong guy" strikes again and makes it abundantly clear that they think it's 1/8 because there are 8 distinct end results that can happen (?) and that makes it a 1/8 chance (??)
Which, admittedly on little thought, I highly doubt comes to 1/8 anyway.
Also, I don't think that's the logic they're using. They listed 8 possibilities, assumed they each had equal probability and therefore arrived at 1/8. Which is comically wrong.
Also, I assume this is a game of some sort, in which the item would have the specific description OP describes. If we call that onto question then there's not much point in the whole thing since OP could be making shit up to seem right, but that's not the most productive thought to entertain.
>Which, admittedly on little thought, I highly doubt comes to 1/8 anyway.
It turns out it comes out to about 13%, which close to 1/8. The wrong guy's approach would give you .75\^7.
No matter how you state it, probabilities always sound like bullshit. Besides, when is reality predictable? or, would you use a formula to calculate your chances in playing "Russian Roulette?"
So….
If the math works out….
Starting with 128 chances to successfully hit a target firing a crossbow without out failure 8 times in a row….
I should see success one time.
I will not hit the 8th shot one time and I will not get to the 8th shot 126 times.
Is this the proper understanding of the maths?
I suppose the underlying conceptual issue here was mixing up elementary states with macrostates. It is true any shooting combination falls in the 8 categories they mention, but it is crucial that there are multiple ways to do so.
In cases like this (like for example in the Monty Hall problem), it woulf be so much nicer to very early on see people test their intuition/minitheory with a statistical simulation before taking all the time to argue.
It gives them a 50% chance of shooting again after each successful shot.
Assuming the first shot had a 100% chance of success, the probability of getting 7 more shot is 1 × ½ × ½ × ½ × ½ × ½ × ½ × ½\
= (½)⁷\
¹/128
1/128 is probability of outcome (note that this probability is eight, not EXACTLY and EXCLUSIVELY eight)
1/8 is which outcome occurs, not the probability of said outcome
Dude’s just an idiot
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Okay so every time you attack there's a 50% chance of making another attack, which can continue to trigger additional attacks. So the first is guaranteed, likelihood of the second is 50%, likelihood of the third is half that again so 25%, fourth is half that again so 12.5%, and so forth. 100% is 1/2^0 , 50% is 1/2^1 , 25% is 1/2^2 , so clearly the formula is 1/2^(n-1) where n is the number of attacks. This means there's a 1/2^7 = 1/128 chance of making at least eight attacks with one crossbow shot. And since half the time you continue shooting after that, the odds of *exactly* eight shots are 1/256. Edit: fixed an inaccuracy
Exactly
The real incorrect was continuing to trade comments instead of drawing that teapot a probability tree
The real incorrect is trying to correct him in the first place, he'll never accept it so why waste your time
If he doesn't understand a diagram that I understood when I was probably in the single digits, then what the fuck am I doing with my life - should I be doing better, help me
Honestly though they didn't do the best at addressing the issue though I don't think? At no point did they explicitly state what I would have said, and I genuinely think what I would have said would have done more to help the incorrect person than what "op" said. There is a difference between the number of possible outcomes and the chance at which they occur. Yes there are 8 possible outcomes but from the very beginning, not at the time a shot is taken, these possible outcomes are weighted differently. That should have been explicitly stated. But it wasn't. The problem was that the incorrect person was partially correct in their assumption that there were 8 possible outcomes so it should have been attempted to explain that the outcomes are weighted differently rather than to explain that they are flat out wrong.
Exactly the different outcomes just have a different probability. This is a Laplace experiment so every outcome does have the same probability if you don’t stop once you miss. But the outcomes relevant for the game that he states have different probabilities because the hidden permutations have to be accounted for with this approach. Omega approach for this problem instead of decision tree is stupid but it works.
> 50/50, bro. > What, 50% we live, 50% he still doesn’t get it?
Probability trees are hard to draw in Reddit comment formats.
Nah he's saying, because you can have cases like WW,WL, LL, but not LW, since you can't get a trigger AFTER you didn't trigger before, so it's not 2^n ... What he needs to hear/understand is that has no impact on the probability of the previous trigger, and you can only use (favourable outcomes)÷(all possible outcomes) in certain cases. Here the outcomes are not independent, you can't just count them all and divide.
Right, the other guy didn't really explain it very well
You can test this with a coin. Not sure why you went with a d6. Seems to needlessly complicate the problem
Minor knit-pick from a statistician here. 1/128 is the probability of getting *at least* 8 attacks. Getting 8 attacks (exactly) has a probability of 1/256. You need to multiply by the 1/2 which ends the sequence. An easy way to verify this is to note that the probabilities given in the first slide do not add up to 1, but multiplying each of them an extra 1/2 gives you a geometric series that will add to 1.
Ok I far more appreciated the way you explained it.
That's not correct, you're missing a factor 1/2 in all your probabilities. Probabilities have to sum to 1 but yours does sum to 2. See [my answer here](https://reddit.com/r/confidentlyincorrect/s/jp1bWOsUau)
What about the likely hood of the pi*(10 * limit(n~>0){1/n})th shot?
This is just "It's 50% chance. Either it happens or it doesn't" taken to extremes.
Me, waiting for my Thai fried rice to arrive: “there is a 50% chance I’m having pizza tonight”.
There is a 50% chance a shrimp fried this rice
And a 50% chance he threw his dad in it.
and a 50% chance said dad rose out of it like a zombie in an old horror movie
That's what I was thinking. May as well say it's a 50% chance of 8 shots. You either get 8 shots, or you don't. Same way your chances of dying tomorrow when a plane crashes into you while being struck by lightning during a threesome with Taylor Swift and Beyonce are 50/50. That's why I buy lottery tickets. You either win millions or you don't. 50/50, baby.
"You can't crit if you can't hit" *Cries in vanilla wow*
Ever played XCOM?
It literally is. Dude's logic is "Every outcome has an equal chance, there are 8 possible outcomes, therefore the chance is 1/8, in order to have a 1/128 chance you'd have to have 128 outcomes." I think the easiest way to break the logic is to ask them what the odds are of hitting 9 shots. If all the other previous outcomes were 1/8, then hitting a 9th shot has to be impossible, right? The biggest obvious weakness in this thinking is that there's not 8 outcomes, there's infinity outcomes, nothing forces a stop at 8.
Exactly. My chances of winning the lottery are 50%. I either win or I don’t. /s
I'll be honest, I don't give 1/128 of a fuck about this argument.
So you could care less, mathematically.
True, which is also why that phrasing to convey indifference is faulty. Logical speakers of English say "I couldn't care less".
That’s always bugged me, every time I’ve heard it in movies and tv shows (Aussie, so we say “couldn’t” but we get a lot of American media, obviously) “I could care less” = I care to at least some extent aka the opposite to the sentiment generally being expressed “I couldn’t care less” = I have zero care regarding this, therefore it is impossible for me to care any less than I already do aka the exact sentiment generally being expressed.
Kiwi here. We may not agree on much, but this is one area where we are in complete alignment.
I always assumed it was a truncated phrase whose ending was lost in time. "I could care less *but I'd have to be dead*" or something along those lines. Similar to phrases like "I can't even" or when people will use just part of a well known phrase "well a bird in the hand..."
The origins are “I couldn’t care less” and etymologists believe “I could care less” is a Yiddish-English sarcastic variant (in the same spirit as the sarcastic Yiddish English phrase “I should be so lucky”). Except the sarcastic tone is disappeared and the phrase is now said, generally, in a rather earnest tone. I knew “couldn’t” came first, I just couldn’t remember how it changed in (primarily) American English so if you Google, you’ll probably find the same comparison (I added the earnest tone part though, because yeah, it’s typically said earnestly, which is the opposite of how it should be said). ETA: just found another thing where “could” was used in an advice column in 1966, which would make the sarcastic origin less likely. Maybe it originated from a typo? Who knows, all I know is that somehow, the vernacular in American English was changed for the phrase to, when taken literally, mean the opposite, and often, if not mostly, used in a somewhat earnest tone of voice.
In plain language, Occam’s razor states that the simplest explanation is preferable to one that is more complex. On the basis of this principle I'm going to assume it's probably just Americans fucking it up.
American here. That's a safe bet.
when in rome...
But if you couldn't care less you wouldn't find the need to say it. The act of saying "I couldn't care less" means you care enough to say it. Therefor, you could care less.
Incorrect. I couldn’t care less about the price of fish in China. I *do*, however, care (in a negative way) about someone *telling* me, in minute detail, about the price of fish in China. Therefore, I say “I couldn’t care less about the price of fish in China” to inform said party about my lack of care about the topic and, therefore, ending that topic of discussion. For example…
You cared enough to blabber on about it. Therefor there's less to care about. Also, sarcasm is a thing in idioms. Get over it
But…. He *COULD*
Obviously because 0=“care less” and 1/128 of care is >0, so he can care 1/128 in the direction of “could”.
I would just like to point out that I said I *don't* give 1/128 of a fuck, which technically only excludes that specific amount and leaves it ambiguous as to whether I actually give a greater or lesser proportion of a fuck.
This is the best comment in this whole thread.
My math is correct, actually.
Congratulations.
(I thought we were just riffing here)
We were, sorry, it was very late, that's all I had...
When I say “I could care less,” what I mean is, “I care so little about this that I don’t even care how much I care about it. In fact, if someone asked me to, I would be perfect happy to care about it even less than I already do.”
But what about 1/8 of a fuck?
Nope!
Wrong. The fact that you commented means you give at least 1/2 of a fuck, and when I aim my crossbow at you you're going to give a lot of fucks, unless I miss my first shot and the percentage of fucks you give will be 50.5%, which can be rounded out to one, so you clearly give a fuck.
Ehhh...I don't have enough maths to make a counterargument, but I *feel* you are at least 73.2% full of shit.
Wrong! You got it backwards. I am 26.8 fos. 73.2 is the number that proves I'm right. 1/128x73.2× fucks to give (no fucks left)/my crossbow.
You 100% can suck my dick. QED.
Quod Erect Dickmunchum
Nomnomnom. (& username checks out)
what did I just witness?
Lesbian. Does. Not.Compute.
Wrong! You can't compute because you're a *woman* , not because you're a lesbian. /s
[Indeed](https://youtu.be/4F-lYM1YzBs?si=SuUqQGZTCggB6lRb)
Good comment lol
1/250.
My head hurts
Flip coin 8 times. Chance of heads all 8 times in a row = 1/128. Confidently incorrect person claims chance of heads all 8 times in a row is 1/8. The end.
This is still not true. Chances of 8 heads is 1/256, the example in the OP is a scenario where the first occurance is a gimmie.
Some of us here have never had to look at binary/positional notation and it shows.
They mean a chance of 8 or more obviously even if not specified. They don't care about the possibility that a no th shot could've happened or not they would've been equally impressed.
it’s 2^8. 1/256
Yeah the 1/8 logic assumes that each possible outcome is of equal probability, which is very wrong. I'm not great at math so I can't tell you for sure if 1/128 is correct, but I can easily tell you 1/8 is stupid lol.
Assuming it *could* theoretically fire infinite attacks if it continues to hit the 50/50, then there are infinite possibilities for how many shots it may fire. Therefore the probability of it hitting *any* amount of attacks is 1/∞ /s if it wasn't obvious
You can't divide nothing by s. Checkmate.
Brilliant showcase on why the logic is flawed lol.
Any time you need thingA *and* thingB to happen, you multiply the two probabilities together. Because probability is always a number between 0 and 1 (say 0.5 in the case of a 50/50) this always gives you a lower number. 0.5 x 0.5 = 1/4 Do that 8 times and you get there
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The context is in the post, the first is indeed guaranteed. Also that first paragraph makes no sense, it doesn't even look like you understand what you're talking about there lmao.
I had a similar argument with my friend yesterday lol I had just done something where I hit a 5% chance and wanted to do another which I said was still a 5% chance he said that’s not his probability works and the next one had a lower chance and I disagreed. Idk if I was right or not but I said it’s a smaller chance to hit them back to back if I hadn’t hit the first one but since it isn’t compounding chance it’s the usual 5%
You're right - your friend is committing the [gambler's fallacy](https://en.wikipedia.org/wiki/Gambler%27s_fallacy). The chance of *succeeding on both rolls* is .25% (I think, somebody check my math), but *each roll* remains 5%.
0.25% is correct. 5% = 1/20 (5%)^2 = (1/20)^2 = 1/400 1/400 = (1/100)/4 = 1%/4 = 0.25%
Thank you.
Never seen anyone do percent like this and tbh after seeing it im surprised I haven't
but the chance that they’ll succeed on either roll 1 or roll 2 is calculated how? Addition?
Assuming you mean "or" inclusively (succeeding on both still counts): the easiest way is to take the probability of both being a failure and subtracting that from 1. The opposite of at least 1 success is all failures. 1-.05 = .95 <-- the probability of a failure .95^2 = .9025 <-- the probability of 2 fails in a row 1-.9025 = .0975 <-- the probability of succeeding on at least 1 roll The long way is to list all the outcomes and add together the acceptable ones. P(S)\*P(S) = .05^2 = .0025 P(S)\*P(F) = .05\*.95 = .0475 P(F)\*P(S) = .95\*.05 = .0475 P(F)\*P(F) = .95^2 = .9025 We add together every outcome except the last one: .0025+.0475+.0475=.0975 If you want to use the exclusive "or" and want the chance of only succeeding on a single roll but not both then you do .0475+.0475=.095.
I actually don't remember that, so I'll leave it to someone who does. Heh.
Some games (and other programs) do have coding that reduces the pure randomness of stuff, so without knowing the specifics, your friend very well could be right here. But assuming that's not relevant, it sounds like a discrepancy between "unlikely to do the same thing consecutively" vs. "the previous results don't affect later ones." The odds of succeeding at a 5% chance twice in a row are 1 in 400. But if you've succeeded at that 5% chance 99 times in a row, the odds you'll succeed on the 100th attempt are 5%; the previous successes or failures, no matter how improbable, don't have any bearing on the very next roll of the die.
> Some games (and other programs) do have coding that reduces the pure randomness of stuff If you're interested in how pseudorandom number generators work (in a very basic sense), what they do is they take an initial seed (some value) and generate a deterministic sequence of numbers from it. This means if you know the seed, and you know how a sequence was generated from it, AND you know which element from the sequence you're about to pull, you will always be able to predict what the outcome will be. On top of this, the sequence may have biases for certain outcomes due to some values appearing more often in the sequence. If you know the seed and you know what algorithm was used to generate the sequence from the seed, then you can perform analysis on the sequence and find that you're more likely to have one outcome than another, when they're supposed to be an equal probability. Going a tiny bit into CPU architecture (mega nerd shit that I may or may not be wrong about): Intel x86 CPU architecture provides two Assembly language instructions for RNG: RDSEED (read seed) and RDRAND (read random). As far as I know, RDSEED is true random number generation (TRNG) that uses a source of noise from the hardware to achieve this, but takes a lot of clock cycles to perform, so relying on it will incur large performance penalties. I think RDRAND is pseudorandom number generation (PRNG) relying on an initial seed, but it also re-seeds periodically? So if you're generating random numbers very frequently it can be predicted, but if you're doing it infrequently it cannot. Because of the performance hit, I believe RDSEED is generally only used to generate a seed, rather than generating ALL of your random numbers. NOTE: I'm a complete gobshite at the best of times so if someone is better at this sort of stuff definitely feel free to correct me
I know most computer programs aren't technically capable of what we consider "true" random. The processes they use to simulate it can be pretty interesting for sure. But some make deliberate efforts to skew towards (or away from) certain results. For example, Diablo 3 will gradually increase your chance to find a legendary item the longer it's been since you've seen one. And I once read that iTunes users complained that shuffle wasn't random enough since, in true random fashion, you would sometimes get 2 or more consecutive songs from the same artist (or even the same album). So they implemented a feature that made that less likely to occur, technically reducing randomness but increasing the illusion of it to its users.
Very good point!
Probability is weird and you were both kinda right. Assuming the events are independent then yes the chance does not change the second time around. However it is also true that hitting both 5% would be much rarer than hitting only one. You had already hit the first 5% that makes it now irrelevant to the second opportunity (If they're independent) thus making him wrong that it would lower your chances of hitting it again. But if before hitting that first 5% you asked the probability of hitting two 5% chances in a row, it would be 0.25%
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Me too. Then I shoved them all up my nose. Some of them are still up there.
Thank you for your service
> grabbing a random number generator with 1 to 2, or a dice and pick 1-3 or 4-6 It's a real shame that people don't carry around an artifact in their pockets that would help them choose randomly between two outcomes. That would really change the world. You'd probably save at least a quarter hour a day with something like that. So much extra time to coin new phrases. You'd be able to hold your head high and not slink away with your tail between your legs.
This is the same as flipping a coin, but the game ends once you get a tails? Just tell him that and hopefully he can logically see the mistake
Practically yes that’s how it works
/) Unrelated, but can the crossbow hit a maximum of 8 times or is that just when it missed?
Just when it missed
So you mean that crossbow could *technically* do #INFINITE DAMAGE?
/) I kind of figured that was the case because the original comment said 1/256, but the wording of all of the replies made it seem like 8 was the max.
I'd like to introduce these guys to my casino. New customers are always welcome! The odds are very good. 😉😉
what
The person saying 1/8 doesn’t understand how probability works
It wasnt the explanation I was missing, more like the point 😆
OP didn’t really give context and thinks that this subreddit is r/IHadAnOnlineFight. People posting themselves and their boring arguments should be banned.
Pointless, yes, but he did at least show the context
WTF just happened? I have a nose bleed and a head ache.
makes you wonder how some people managed to graduate from high school
I'm so bad at math I have no idea who's wrong.
At least you know. That's half the battle.
I am just ignorantly going to blame the first guy who used fractions. Fuck fractions
this happens a lot in retos comments it feels
They seem to be looking at each 50/50 chance as independent of each other rather than calculating the probability of a string of binary options dependent on each preceding success or failure all succeeding. If shot one is guaranteed, then shot two is the only one that is actually 50/50 because shot three's potential is dependent on shot two's outcome and so on and so forth.
You can't understand how to use different colors to show who said what. Takes 1/128th the effort as my comment
Ahh, the good old "You either win the lottery or you don't. It's 50/50". Or in this case 1/8 apparently
Depending on what you mean exactly you might be technically correct, but your reasoning is wrong. After the first 1 attack that's guaranteed, you can say that: The probability of 7 **or more** subsequent consecutive attacks is the infinite sum 1/2^(7) × 1/2 + 1/2^(8) × 1/2 + … = 1/2^6 × 1/2 = 1/128. The probability of **exactly** 7 subsequent consecutive attacks is 1/2^(7) × 1/2 = 1/2^(8) = 1/256. If you don't believe me, do a sanity check: sum all the probability of doing 0, 1, 2, 3, … consecutive attacks and you should end up with 1, because probabilities must sum to 1. You will get 1/2^(0) × 1/2 + 1/2^(1) × 1/2 + … = 1/(1-1/2) × 1/2 = 1. To be more precise, the number of subsequent consecutive attacks is a random variable following geometric law of probability 1/2. The expected value here is 2 subsequent attacks. In general the geometric law of probability p Is measuring the number of consecutive successes for repeated Bernoulli trials of probability p. The probability to have k consecutive successes is p^(k)(1-p) and the expected value is 1/(1-p).
Nah man. Either you hit all 8, or you don’t. That’s a goddamn 50/50 right there /s
Even if yes it’s still not the 1/8 lol
Don’t miss at all. 1/1 chance 😎
He fell victim to one of the classic blunders! The most famous of which is, 'never get involved in a land war in Asia,' but only slightly less well-known is this: 'Never assume all outcomes have the same probability!” It’s one of the most common mistakes you see when kids start learning probability, and one of the reasons that probability is one of the least-intuitive branches of mathematics that most people will come across.
These people will never realize how wrong they are :/
No they won’t
Idk why people are always arguing about the chances, its 50/50. It either happens or it doesn't.
The chances of it happening 7 times in a row are not 50/50
It happens50/50it doesnt
The argument happening in op is the chance of 8 in a row, which is not 50/50 even if each individual shot is
r/woooosh
I mean i can understand to some degree of what this person is thinking, their thought process follows the basic outcomes rather than the % overall of the highest outcome being the chosen one. That’s how they got the 1/8 instead of the 1/128, out of the 8 choices of an outcome(following their logic) it would make sense it can only happen 1/8th of the time but percentage wise it’s a different story
When I have no clue about the math that is being talked about online I just shut the hell up.
[Looks like you got 141 & 2/3% chance of proving your point to me.](https://youtu.be/WFoC3TR5rzI?si=lfbuapgsl8eRksE6) You see, the numbers don’t lie…
If I’m understanding this correctly, you multiply the odds?
Yes. Think of a coin flipped three times and you want tails-tails-tails. There’s a 1/2 chance of getting tails. If I flip it again, there’s again a 1/2 chance of getting tails. Do it one more time, it’ll either be heads or tails. The possible permutations (similar to combinations; the order matters in perms) are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Eight combinations. TTT is one possibility: 1/8. 1/2 * 1/2 * 1/2 = 8.
lol is that retromation?
Yes lol
I've read the comments and have come to the conclusion that I am tired
I feel like you should’ve just compared it to flipping a coin and landing it heads 7 times in a row and he might’ve understood a bit better.
I actually wasn’t one of the people talking
Ah my bad then
No problem I just stumbled across this and thought it was interesting and funny
Got maybe 4 pages in before I realized I don't give a fuck about any of this
You really only need 4 pages
The funniest thing about this is that you're both incorrect. The probability to get exactly 8 shots is neither 1/8, nor 1/128, but it is 1/256. You have a 1/2^7 probability to get up to 8 shots. After that, you have 0.5 probability to either shoot an additional shot, getting you up to 9 shots, or to stop there. To get exactly 8 shots, you need to stop, which means you multiply the total probability by another 0.5, getting 1/2^8. If you want to see how this works out in practice, assume you conduct m (=128) experiments. You would expect to get only one shot in m/2 (64) of those, two shots in m/2^2 (32) of those, etc. The general formula is that for s shots, you would expect m/2^s occurrences where you get exactly that many shots. If you conduct 128 experiments, then you expect to get exactly 8 shots m/2^s = 128/2^8 = 0.5 times, meaning once in 256 experiments (thus 1/256).
Technically correct, but in reality if someone feels like 8 shots is a sufficiently rare event, he's going to think of anything at least as rare as 8 shots to be significant as well. So we should really calculate for ≥8.
While that's a good point, they were discussing the probability of an event that already happened. It's not correct to say that that probability is 1/2^7.
A retromation watcher, nice
Sounds like the 1/8 dude could have tried it real quick to see if his theory was at least feasible
I don’t need Monty Hall ruining my place of work when Monty Hall has already ruined my home life
Your wife chose the third door and walked out?
what game is this?
It’s called endgame of devil retromation has a good series on it
I can understand why people would think it's still 50-50. How the heck did he get to 1/8?
I do not know whatsoever
There are only 8 outcomes in that particular sequence, so I get where they're coming from, but they clearly don't get even basic probability if that's their conclusion. It's far more probable that the xbow gets just a second shot because it has a 50% chance of happening. I would've liked to see their reply if they were directly asked "Do you think that there's the same 12.5% chance of getting 8 shots as 2 shots?" or point out that if someone was lucky enough to get 100 consecutive shots, by his logic there'd only be a 1% chance of any particular outcome, including only 2 shots after 1 hit, and getting 100 consecutive shots? They have no clue what they're saying.
Yeah I kind sorta see what they thought it’s just that they were so confident and argued so much
Ain’t reading allat
I miss the days when this subreddit meant actual arguments between right n wrong ☹️
I’m sorry if this isn’t what you wanted it’s what I had❤️❤️
Sorry buddy I didn’t mean to diss your post I was just reminiscing, my bad
No you’re fine I was joking I miss that too❤️
I love this subreddit tbh I just see way too many incorrect math or grammar posts, at least you avoided that 🤣
omg im famous
Even my brain hurts now
The chance of making 8 consecutive crossbow shots is just a 1/2… either you made 8 consecutive shots or you didn’t.
So, then presumably the odds of 7 shots is 1/7, and 6 shots is 1/6, and so on. I wonder what the sum 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8 will be....(don't even need to argue the infinite sum, going to 4 gets into a problem).
\- 1/12
Nerd
Can we ban maths posts? If it's not the order of operations ones where the original is written in a stupid way with the intention of being misunderstood, it's ones like this where no one is good at explaining themselves and it's less that 1 person is "confidently incorrect", and more that at least 2 people don't know how to communicate clearly Edit: I know that the 2nd person is incorrect. But the people arguing in the pictures have phrased their (correct) arguments in an incredibly confusing way - if I wasn't already familiar with the maths involved, I wouldn't have a clue what they were talking about and their arguments wouldn't sway me. Which is why I don't like posts like this, because although it does show a person being incorrect, and they seem confident about their false assumption, it doesn't really give the incorrect person an opportunity to realise their mistake and it's understandable to dismiss counterarguments if it sounds like they're a load of nonsense.
No, the second guy is confidently, disappointingly incorrect.
I know the 2nd guy is consistently incorrect, but all the explanations given to him are shit. I understand the maths involved, but it's like everyone has chosen the most confusing way possible to explain it to him, at which point I have a lot more sympathy for him being consistently incorrect. There are some much clearer explanations in the comments on this post.
That’s very true. I hated OP’s approach to explaining and kept saying “why not explain that there’s 128 combinations and hitting all eight times is only one?”
Ugh. Who *cares*
your explanations are not as clear as you think they are. But, also, I think you are arguing about game mechanics, not probability. I take it you have a weapon that can take up to 8 shots. If you shoot and you miss, you get a 50% chance of getting another shot. So, assuming you miss every shot, you're right that you have a 1/128 chance of taking 8 shots. But isn't the other guy saying that you ALSO get to take another shot if you hit on your shot? So to actually calculate the % chance of taking 8 shots requires you to factor in the chance that you will hit? If that's the case, then I'm pretty sure you're both wrong.
I got the explanation of the setting fine. The crossbow fires once, this is guaranteed, it then has a 50% chance to fire again. If it does fire again, then there is another 50% chance it may fire once more, repeating until it fails to fire. Thus the question is, "what is the probability of the crossbow firing 8 times in total?".
>The crossbow fires once, this is guaranteed, it then has a 50% chance to fire again. If it does fire again, then there is another 50% chance it may fire once more, repeating until it fails to fire. Okay, yes, that's what OP is saying. But the guy OP thinks is wrong seems to be saying that it has a 50% chance to fire again if you miss but it always fires again if you hit.
>. But the guy OP thinks is wrong seems to be saying that it has a 50% chance to fire again if you miss but it always fires again if you hit. "On a failed shot, it 100% never shoots again this round" is a quote from "the wrong guy". He's re-defining a little bit, as OP said "every attack there is a 50% chance" and he's saying "every attack, there's a 50% chance to hit, and if you hit, you attack again". They're calculating the same thing, "the wrong guy" is just flat-out wrong about it.
>every attack, there's a 50% chance to hit, and if you hit, you attack again". So in this version: you hit: you get another shot you miss: you get a 50% chance at another shot. So, if this really is the fact pattern, the Confidently Incorrect guy is right and OP is wrong. In this situation you get: shot 1: 75% chance of a second shot shot 2: 56.25% shot 3: 42% shot 4: 32% shot 5: 24% shot 6: 17% shot 7: 13% chance of taking the 8th shot -- or roughly 1/8. I'm not sure that's actually the fact pattern, but it's almost certainly the rationale the incorrect guy is using. Hence my point: they disagree on the assumptions, not on the math.
>you miss: you get a 50% chance at another shot. This is incompatible with the > On a failed shot, it 100% never shoots again this round from "the wrong guy". Otherwise, with your set of assumptions that nobody here or on the original thread holds, you do get .75^7 chance to get at least 8 shots, and it IS pretty close to 1/8. > If it was 1/128, there would be 128 possible situations that could have happened. Except, "the wrong guy" strikes again and makes it abundantly clear that they think it's 1/8 because there are 8 distinct end results that can happen (?) and that makes it a 1/8 chance (??)
Which, admittedly on little thought, I highly doubt comes to 1/8 anyway. Also, I don't think that's the logic they're using. They listed 8 possibilities, assumed they each had equal probability and therefore arrived at 1/8. Which is comically wrong. Also, I assume this is a game of some sort, in which the item would have the specific description OP describes. If we call that onto question then there's not much point in the whole thing since OP could be making shit up to seem right, but that's not the most productive thought to entertain.
>Which, admittedly on little thought, I highly doubt comes to 1/8 anyway. It turns out it comes out to about 13%, which close to 1/8. The wrong guy's approach would give you .75\^7.
Fair enough, I didn't feel like mathing that part out. My bad.
I doubted it to! But some other comment prompted me to do the math.
No matter how you state it, probabilities always sound like bullshit. Besides, when is reality predictable? or, would you use a formula to calculate your chances in playing "Russian Roulette?"
Omg I’m famous
Wow
So…. If the math works out…. Starting with 128 chances to successfully hit a target firing a crossbow without out failure 8 times in a row…. I should see success one time. I will not hit the 8th shot one time and I will not get to the 8th shot 126 times. Is this the proper understanding of the maths?
Didn't she hide the crossbows?
I suppose the underlying conceptual issue here was mixing up elementary states with macrostates. It is true any shooting combination falls in the 8 categories they mention, but it is crucial that there are multiple ways to do so. In cases like this (like for example in the Monty Hall problem), it woulf be so much nicer to very early on see people test their intuition/minitheory with a statistical simulation before taking all the time to argue.
It gives them a 50% chance of shooting again after each successful shot. Assuming the first shot had a 100% chance of success, the probability of getting 7 more shot is 1 × ½ × ½ × ½ × ½ × ½ × ½ × ½\ = (½)⁷\ ¹/128
Pretty sure the OG comment is just a reference to the 1/256 glitch from Pokémon… idk if it wasn’t intended to be mathematically accurate
Dm: Crossbow breaks after 4 shots. (problem solved)
1/128 is probability of outcome (note that this probability is eight, not EXACTLY and EXCLUSIVELY eight) 1/8 is which outcome occurs, not the probability of said outcome Dude’s just an idiot
🤓
Do people just not understand extremely basic math
God damn it dude why is there so much math in this subreddit
Lol true tho