You should be able to solve this.
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You should be able to solve this.
You should be able to solve this.
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50%, we're assuming at least one of them is a crit
50%
If one has to be a crit, it's only a 50/50 chance to get the second, right?
everything is 50/50
happens or it doesn't
But it said one of them had to hit, so it is 100% on the one of the two.
50%
00
01
10
11
3 have at least 1 crit, so 33% for 2 crits.
>probability memes when the context is a srpg video game
yikes
Crit-Crit 25%
Crit-No 25%
No-Crit 50%
0 crit hits because my luck is dog shit
https://en.m.wikipedia.org/wiki/Boy_or_Girl_paradox
Go frick yourself
I wish i could.
I like how they refuse to unambiguously point out the correct answer of 1/3 in the summary, and instead say
>The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,[3] and Raymond S. Nickerson.[4]
That's not talking about the probabilities, that's talking about the difference in answers based on how the question was presented. And that's the big problem, it's not that it's hard to understand, it's that the question is poorly asked.
>It's ambiguous, these mathematicians that changed the question told me so!
if you have to stop and ask the same question again, it was a badly asked question
based
lmao @ all the pseuds posting actual replies to this bait thread every time
1/2 * 1/2 = 1/4
>You hit an enemy twice
In one combat or not? I.e. are you doubling?
>At least one of the hits is a crit. Assuming a 50% crit chance
Is the assumed crit considered guaranteed, like when a story animation shows a crit? Then the premise of the 50% crit falls apart since in that case, the chance is 100%. If not, then it's 25%, since there's no way to know if a crit will occur or not if it isn't guaranteed, so all 4 possible outcomes (0/0, 0/1, 1/0, 1/1) are still relevant.
And I am able to solve it.
Mathgays, explain to me in moron terms why it matters that it's 33% or whatever when the question is (for all practical purposes)
>Do you do a crit and a normal hit or two crits?
In a practical sense, why does the sequence matter?
The end result of crit + normal is the same as normal + crit.
Same with the Monty Hall problem, I am legitimately too smooth brain to understand it.
>3 choices
>Choose 2
>Host says 1 is wrong
>Now choose between 2 and 3
>Better to choose 3
For all practical purposes, how is this different from a 50/50 chance?
>"Because the initial odds haven't changed"
>"The answer is being curated"
From my moron brain's perspective, the choices went from 3 choices to 2 choices, one is correct, the other is wrong.
Statistically, whether or not the crit comes first or second matters because it is a different outcome. This may not be immediately obvious but in context it's natural. Let me demonstrate:
>you're playing Darkest Dungeon
>your vestal is poisoned and on Death's Door
>vestal has a chance of dying when they take the tick of poison damage at the start of their turn
>before your vestal can take her action, you have your graverobber's turn
>there is one enemy left that is able to be killed in one crit or two attacks
>if your grave robber can crit and kill the enemy before the vestal's turn begins, you can give her an antidote and stop the poison damage to save her life
>if your grave robber doesn't crit, the enemy lives and your vestal might die to DoT before they could take their action, meaning even if they had a guarenteed crit, it wouldn't matter
In Monty Hall a wrong choice is guaranteed to be removed so you are either switching from right to wrong or wrong to right. Your initial choice will be wrong 2/3 if the time so in those 2 cases you'd be switching to the right answer.
You must surely have read the Monty Hall problem scaled up to 99 goat doors and one car door, with 98 goat doors being revealed after your initial choice. I'm not sure how the problem could still seem unintuitive when asked if you should switch from an initial choice which only had a 1% chance of being correct.
I watched a video of the scaled up version, but choosing door 48 and being told doors 1-47, 49-66, and 68-99 are all wrong, makes me think that door 67 is pretty suspicious, because why would the host pick 48 and 67 specifically out of 99 doors? Obviously, in the context of the problem, I wouldn't be immediately told I was wrong, so to my moron brain, of course I'm wrong because so many options have been declared wrong EXCEPT the one I chose and this weirdly specific door out of the other 98.
>of course I'm wrong because so many options have been declared wrong EXCEPT the one I chose and this weirdly specific door out of the other 98.
Exactly, and this logic still applies when scaled back down to three doors.
"at least one" is not the same as "the first" or "the second". this means from the total choices of 4 outcomes from 2 rolls is down to 3, because "at least one" means "no crits" is eliminated. in those 3 outcomes, only one of them is a "both crits", hence why it's a 1/3 chance of getting both.
monty hall problem is different entirely, and will never be 50% outcome. that is because your odds of guessing are 1/3 chance of picking the prize, and 2/3 chance of picking the goat. in this scenario, there will always be a wrong door which the host knows. that wrong door is the one not chosen, so it's revealed. when he asks to change your answer and this ends up changing the odds inversely.
the answer depends on whether the guaranteed crit was guaranteed independently of the crit chance or not
if it was a guaranted crit outside of the crit chance and you're asking the chance that the second one happens to roll a crit, it's 50%
if the guaranteed crit in this hypothetical was only guaranteed in the sense that it successfully passed the crit chance check, then it would be a 25% chance that both would have been crits.
we know one hit crits though so you only need one split. 50/50
>there are 3 red balls, 3 blue balls, 3 yellow balls, 1 green ball in a box
>therefore, the chance of getting a green ball is 25% because there are only 4 possible outcomes
I’ve gotten hit by a 3% crit at the end if a level once. Not happy.
The absolute chance that two crits happen in the aforementioned scenario is 25%
Assuming we are in the same scenario, with the expectation that one hit is a crit, the chance of two critical hits is still 25%, but since the requirement reduces the space in which our unity lies, the chance is now p / (1-q) where q is our negated requirement chance (neither hit is a crit, or 25%) and p is our expected chance (both hits are a crit, or 25%)
This simpliefies to 1/3 chance of both hit being a crit, with the requirements that at least one is a guaranteed crit.
This is akin to rolling the dice until at least one of the two is a crit, and only then resolving both attacks. This would mean that for 5 attacks, with 4 guaranteed crit, you'd have to roll A LOT of dice before an attack is done.
This is obviously neither efficient or effective.
The way to fix this is to roll dice, and then turn any hit into a crit until the minimum number of crits required is met.
>The absolute chance that two crits happen in the aforementioned scenario is 25%
that's not what was asked tho nerd-kun.
it's the 100% of the outcomes as described, it either crits or not, and it has a 50% chance to happen for independent rolls
Once the rolls are declared not-independent (one has to crit out of the two, causing them to become dependent on eachother's outcome) then you are reducing the 100% of the outcomes, without changing the absolute chance that two independent crits happen (still 25%)
It's this chance that is taken into account, along with the dependency of the outcomes, that leads to a 1/3 chance.
You're not asked "If the previous/next hit is a crit, what is the chance that the next/previous hit is a crit" (50% - the roll is indipendent)
You're asked "Out of two dependent rolls, what are the chances that both of them are a crit if both of them cannot be non-crit" (33% - the rolls are dependent without chance fixture)
You can't eliminate NoCrit-NoCrit without tampering with the odds somewhere, which would mean that the Crit odds are not always 50%. This violates one of the premises in the question itself.
>NoCrit-NoCrit without tampering with the odds somewhere
You can, if you just consider the nocrit-nocrit roll invalid, literally eliminating it from the possibilities.
p/(1-q) is literally doing that, the absolute chances of each roll are the same, and only the combined dependent outcomes are influenced, as both rolls have the same independent chance of rolling a crit by themselves, but if they both roll a nocrit they literally aren't counted as a valid outcome, and discarded.
This is the main difference between making the rolls dependent (at least one has to crit out of 2 -> 2 dependent rolls -> 33%) instead of keeping them independent (forgoing one of the rolls, and making it a 100% crit, the rolling only once -> 1 independent roll -> 50%) or forcing the outcome (turning one of the nocrit into a crit if they happen to be both nocrit -> 2 independent rolls -> 25%)
This is babby tier probability study
100% of course