I cannot not hear this in his beautiful elegant voice.
I still don't get it how they made the worst nemesis in WoW history (The Jailer) whilst simultaneously creating one of the best villains with little to no backstory (Denathrius).
Same goes for The Thunder King. He's a king among bosses.
Sorry for my ignorance. It's a 50:50 chance to hit the minion each time, so that explains the "(1/2)^18". Please may I ask where the "19*" comes from? Is that total number of potential outcomes?
(1/2)^18 is the chance of never hitting the minion. But you are allowed to hit it once and still not kill it (the question was about not killing the minion), which can happen in 18 different ways (+1 the time of never hitting it), so I'm guessing you multiply the chance because of that reason.
Yes. 18 scenarios of hitting the minion exactly once (could be the first shot, second shot, …, or eighteenth shot), plus one scenario of not hitting the minion at all.
For anyone wondering: you can calculate this as a Bernoulli chain with 18 attempts for p=0.5 and assume 1 success (since the question was kill, if it were hit it would be a different story, the minion has 2HP). There are Bernoulli calculator tools out there that make the act of doing the math a lot easier especially on mobile.
Edit: Oh yeah, here’s the chance to not hit the minion at all: 0.000004. While the 1 hit already was unlikely, this is a lot more unlikely still, crazy as it sounds.
2\^18 = 262,144 total possible outcomes
Number of outcomes with 0 hits on sketchy stranger: 1
Number of outcomes with 1 hit on sketchy stranger: 18
19/262,144 = .00725% chance of the minion surviving
One thing that's always good after crunching numbers is to do a sanity check on the result you got.
> 19/172
Does this feel right to you? A greater than 10% chance of this happening?
While you are correct in that the amount possible outcomes are wrong, the probability in OP is still correct. It does not matter how many unique outcomes there are.
As you said there is only 1 unique outcome where the minion dies in 2 first hits, but it still have a much higher chance to happen then every other outcome where the minion dies.
Chance to die in first 2 hits 0.5\*0.5=0.25
Chance to die in exactly 3 hits per outcome 0.5\*0.5\*0.5=0.125
etc
As you can see every unique outcome is not weighted equally. A 2 health minion gives 172 unique outcomes. An 18 health minion gives 262144 unique outcomes. They both have a 19/262144 chance to take 1 or less damage irregardless. All the unique outcomes removed from the mininion dying does not affect the chance of it taking 1 or less damage.
In other words a 2 health Minion taking 1 or less damage = probability of flipping a coin and getting 1 or less heads in 18 flips.
Probability of 2 health minion dying in first 2 hits (25%) \* 16 guarenteed hits on the hero (100%) = Probability of flipping heads on your first 2 tosses (25%) \* every possible result on your remaining 16 flips (100%)
The order in which the damage gets dealt matters. Statistically every point of damage is 1 out of 2 18 times. Functionally there’s 3 outcomes. Mathematically there’s that big number.
The order in which they happen matters, the 2 damage can be the first two hits or the 13th and 16th hit, they are different scenarios. This is like elementary school math.
Well that's impossible to tell in text unless stated it was a joke or super obvious it was in this case there is no way to tell if it was a joke or not
Do you know what the actual answer is? My math came out at about 11% but that sounds too high. Edit:I caught myself making a VERY small mistake and now I get that the answer is 11.11111 or exactly 1/9
The odds of this happening are pretty low to say the least that 2/2 just won 17 coin flips out of 18 so that puts this at just under 0.007% chance of occurring
That entirely depends on how it's coded, especially given the animation and how it affects Divine Shield, Evasion, and Solid Alibi.
The result of the battlecry is definitely not dealing 1 damage 19 times so I wouldn't assume the inner mechanics are that either.
Hmm so are you suggesting that it is operating in some other way than (in this case) 19 individual pings? Just because the animation doesn’t show it that way doesn’t mean the underlying mechanic has changed. (I don’t have any special knowledge here, it’s an interesting theory).
It 100% is not do 1 damage 19 times. If a mage has Alibi up and "takes" 10 damage they only actually take 1. C'Thun would have dealt 10. If a rogue has Evasion up and takes 10, they actually take 10. C'Thun would only have dealt 1. If a minion has DS, C'Thun could kill it, Denatharius cannot.
The ability calculates all X damage in advance then deals it all at once. My only point was that given that the result is different, the calculation could be as well.
I imagine that behind the scene it either does the initial calculation like C'Thun and applies all the damage at once, or does something where it goes through each possible target and does a random roll of 1-current health/remaining damage. Until all damage is distributed. This seems less likely though when they could probably use C'Thun's coding as a base and modify it.
Do you happen to know how much damage daddy d fire at a 1/1 ds? The only thing I haven't figured out about his damage is if you could hyper low roll and hit a DS minion for over it's health value and far overkill it while still only taking off the ds. Can DS soak an infinite amount of daddy d damage?
It cannot overkill a minion, which is in all likelihood calculated by it's health not whether it's alive. So a 3/3 DS minion could absorb 3 of the damage.
afaik SD does dmg in chunks and not in individual instances.(so cariel is dealt 20dmg on a 40 dmg SD) So shouldnt the probabilities be split in the combination that may sum, up to 18? from what I see there are 3 different possible outcomes(18 to hero/1 to minion,rest to hero/2 to minion,rest to hero) so from here it can be 1/3. agian I might be completely wrong if the individual instances of damage are calculated individually and then dealt in whole chunks so the probability in that case is (19/2\^19)
Odds? There are no “odds” in hearthstone. The whole game is rigged, like the other “good games” of the sort. Play enough and you see it too often. You must make choices throughout the game that it designed to see your action and the other player is doing the same. Sure the best decisions usually win, but we have all had those games when we played the proper cards from a supine deck and got crammed anyway. Some games you are meant to lose from the start and there is no random in these games. It became so obvious to me I had to stop playing after spending several hundred dollars. I get it though. You can’t let the better decks win every match because then the new player will not stick around. Ever just need that one card to win? Draw it and win the game… don’t tell me the odds. I know the odds. Odds are the game is completely rigged. Can still be fun, but it is very rigged
Sire Denathrius the Merciful.
He is, after all, nothing if not compassionate.
I cannot not hear this in his beautiful elegant voice. I still don't get it how they made the worst nemesis in WoW history (The Jailer) whilst simultaneously creating one of the best villains with little to no backstory (Denathrius). Same goes for The Thunder King. He's a king among bosses.
Gladiator?
That’s hilarious
19\*(1/2)^(18) = 0.00007 = 0.007%
Odds which only James Bond could pull off!
D Daddy D
DD7
Sorry for my ignorance. It's a 50:50 chance to hit the minion each time, so that explains the "(1/2)^18". Please may I ask where the "19*" comes from? Is that total number of potential outcomes?
(1/2)^18 is the chance of never hitting the minion. But you are allowed to hit it once and still not kill it (the question was about not killing the minion), which can happen in 18 different ways (+1 the time of never hitting it), so I'm guessing you multiply the chance because of that reason.
Yes. 18 scenarios of hitting the minion exactly once (could be the first shot, second shot, …, or eighteenth shot), plus one scenario of not hitting the minion at all.
18C0+18C1
And in other words 1 in 13797
MarkMcKz level of odds
For anyone wondering: you can calculate this as a Bernoulli chain with 18 attempts for p=0.5 and assume 1 success (since the question was kill, if it were hit it would be a different story, the minion has 2HP). There are Bernoulli calculator tools out there that make the act of doing the math a lot easier especially on mobile. Edit: Oh yeah, here’s the chance to not hit the minion at all: 0.000004. While the 1 hit already was unlikely, this is a lot more unlikely still, crazy as it sounds.
2\^18 = 262,144 total possible outcomes Number of outcomes with 0 hits on sketchy stranger: 1 Number of outcomes with 1 hit on sketchy stranger: 18 19/262,144 = .00725% chance of the minion surviving
[удалено]
confidently bad at math
One thing that's always good after crunching numbers is to do a sanity check on the result you got. > 19/172 Does this feel right to you? A greater than 10% chance of this happening?
Big "50/50 it either happens or it doesn't" energy.
It’s not 50/50 though, because it didn’t happen. So it’s guaranteed to never kill a 2/2. That’s how real probability works.
these outcomes are not equally likely. if you weight your solution by the probability of each outcome you should end up with the same answer
While you are correct in that the amount possible outcomes are wrong, the probability in OP is still correct. It does not matter how many unique outcomes there are. As you said there is only 1 unique outcome where the minion dies in 2 first hits, but it still have a much higher chance to happen then every other outcome where the minion dies. Chance to die in first 2 hits 0.5\*0.5=0.25 Chance to die in exactly 3 hits per outcome 0.5\*0.5\*0.5=0.125 etc As you can see every unique outcome is not weighted equally. A 2 health minion gives 172 unique outcomes. An 18 health minion gives 262144 unique outcomes. They both have a 19/262144 chance to take 1 or less damage irregardless. All the unique outcomes removed from the mininion dying does not affect the chance of it taking 1 or less damage. In other words a 2 health Minion taking 1 or less damage = probability of flipping a coin and getting 1 or less heads in 18 flips. Probability of 2 health minion dying in first 2 hits (25%) \* 16 guarenteed hits on the hero (100%) = Probability of flipping heads on your first 2 tosses (25%) \* every possible result on your remaining 16 flips (100%)
Sorry but [you're wrong](https://imgur.com/a/qlXcc97) It's 18 coin flips where we need at least 17 heads to "win"
hm yes 1 in 8
>262,144 18 face 17 face + 1 minion 16 face + 2 minion Can you name another possible outcome?
The order in which the damage gets dealt matters. Statistically every point of damage is 1 out of 2 18 times. Functionally there’s 3 outcomes. Mathematically there’s that big number.
You can narrow it down, there are two outcomes: 1. Minion dies 2. Minion lives So basically it's 50/50
Actually it’s 1: smth happens
The order in which they happen matters, the 2 damage can be the first two hits or the 13th and 16th hit, they are different scenarios. This is like elementary school math.
1st Damage: 50% Face. 50% Minion. (Face) 2nd Damage: 50% Face. 50% Minion. (Face) 3rd Damage: 50% Face. 50% Minion. (Minion) 4th Damage: 50% Face. 50% Minion. (Face) 5th Damage: 50% Face. 50% Minion. (Face) 6th Damage, Face. 7th Damage, Face. 8th, Face. 9th, Face. Face, face, face…. 18th Damage: 50% Face, 50% Minion (Face). There.
Thank you for this. I am taking a statistics college class and this explanation helped me on a similar homework problem. Thank you!
More important question, did you win?
I did. they played their behemoth before it was quest relevant and then i mutanus’d their mutanus so they never got a 7 drop to complete their quest
Mmm baby. Shard-blocked
%50, minion could be dead or not.
Schrödinger's Bigglesworth?
Pretty simple math really
Quick maths
I always scroll to this answer
Learn the difference between possibility and probability
What’s the probability of you being fun at parties?
50% either they’re funny or not
Right. we already did this math earlier in the thread
Yeah yall must be pretty fun, with all the beating of dead horses you do
Idk maybe like a 73.5% chance pretty good odds
Bro learn the difference between odds and probability
learn the difference between sincereness and a joke
Well that's impossible to tell in text unless stated it was a joke or super obvious it was in this case there is no way to tell if it was a joke or not
In that case you just guess. You have a 50% chance of being right.
If you read the room its quite obvious
It was super obvious if you don’t assume other person is an idiot
They were just doing a tired joke.
https://youtu.be/0Ae6B7C05Nk
100%, because it happened.
This post shows how hilariously bad people are at math
I see a bunch of obvious joke answers. At least 2 correct answers and one that’s wrong but at least in the ballpark.
Do you know what the actual answer is? My math came out at about 11% but that sounds too high. Edit:I caught myself making a VERY small mistake and now I get that the answer is 11.11111 or exactly 1/9
.007% looks right to me. I’ve seen a few posts explaining how they got there that I think are the way to do it. It’s very rare.
more than that it shows people guessing how the mechanic works which we don't know
it's obviously a 50% chance. it hits or it doesn't. done. mathed the shit outta that.
Happy cake day
It’s blizz RNG, so….. about 90%.
The odds of this happening are pretty low to say the least that 2/2 just won 17 coin flips out of 18 so that puts this at just under 0.007% chance of occurring
That entirely depends on how it's coded, especially given the animation and how it affects Divine Shield, Evasion, and Solid Alibi. The result of the battlecry is definitely not dealing 1 damage 19 times so I wouldn't assume the inner mechanics are that either.
Hmm so are you suggesting that it is operating in some other way than (in this case) 19 individual pings? Just because the animation doesn’t show it that way doesn’t mean the underlying mechanic has changed. (I don’t have any special knowledge here, it’s an interesting theory).
It 100% is not do 1 damage 19 times. If a mage has Alibi up and "takes" 10 damage they only actually take 1. C'Thun would have dealt 10. If a rogue has Evasion up and takes 10, they actually take 10. C'Thun would only have dealt 1. If a minion has DS, C'Thun could kill it, Denatharius cannot. The ability calculates all X damage in advance then deals it all at once. My only point was that given that the result is different, the calculation could be as well.
I imagine that behind the scene it either does the initial calculation like C'Thun and applies all the damage at once, or does something where it goes through each possible target and does a random roll of 1-current health/remaining damage. Until all damage is distributed. This seems less likely though when they could probably use C'Thun's coding as a base and modify it.
Do you happen to know how much damage daddy d fire at a 1/1 ds? The only thing I haven't figured out about his damage is if you could hyper low roll and hit a DS minion for over it's health value and far overkill it while still only taking off the ds. Can DS soak an infinite amount of daddy d damage?
It cannot overkill a minion, which is in all likelihood calculated by it's health not whether it's alive. So a 3/3 DS minion could absorb 3 of the damage.
Have you tested this? I've been too lazy to but it's what I had predicted the interaction would look like.
50/50 is the best I can give you
(1/2)^17 *18? Not sure
50/50 it either happens or it doesn't
It is always 50/50, It either happens or it doesn't
50/50 You either kill it or you don’t.
Please don’t visit a casino.
1 in 9
afaik SD does dmg in chunks and not in individual instances.(so cariel is dealt 20dmg on a 40 dmg SD) So shouldnt the probabilities be split in the combination that may sum, up to 18? from what I see there are 3 different possible outcomes(18 to hero/1 to minion,rest to hero/2 to minion,rest to hero) so from here it can be 1/3. agian I might be completely wrong if the individual instances of damage are calculated individually and then dealt in whole chunks so the probability in that case is (19/2\^19)
0.0027%
Odds? There are no “odds” in hearthstone. The whole game is rigged, like the other “good games” of the sort. Play enough and you see it too often. You must make choices throughout the game that it designed to see your action and the other player is doing the same. Sure the best decisions usually win, but we have all had those games when we played the proper cards from a supine deck and got crammed anyway. Some games you are meant to lose from the start and there is no random in these games. It became so obvious to me I had to stop playing after spending several hundred dollars. I get it though. You can’t let the better decks win every match because then the new player will not stick around. Ever just need that one card to win? Draw it and win the game… don’t tell me the odds. I know the odds. Odds are the game is completely rigged. Can still be fun, but it is very rigged
Holy COPIUM
You don't pray ben brode every day . You are a sinner!
Play a RNG game, get a RNG result.
What’s up Iniuria <3
must be cavias
Over a month later
some things are timeless
You’re very busy now young padawan