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/v/ couldn't agree on this.
Is /sci/ smarter than a bunch of vidya virgins?
>>
>>10188804

50%, it either happens or it doesn't
>>
>>10188804
I like how you used the same font and tone of white in the background. It's a nice touch.
But you could have actually posted it on /v/ first, so 6/10.
>>
>>10188804

1/3
>>
>>10188804
Taking the second hit individually, 50%.
>>
>>10188809
>But you could have actually posted it on /v/ first, so 6/10
>>/v/441575871
>>
>>10188814
nvm I don't know how to cross-board link
>>
>>10188817
>>>/ /
>>
>>10188814
>>10188817
>>10188822
>>>/v/441575871
>>
>>10188814
Weird, didn't show up in google image search.
>>
>>10188811
what if the first hit is a non-crit
>>
If they are independent:
A B %
Miss Miss 0.25 <-- This is excluded by the condition
Miss Hit 0.25
Hit Miss 0.25
Hit Hit 0.25

Renormalising gives a probability of 1/3.
>>
>>10188829
Then it's still 50%. The chances of individual hits being crits don't change.
>>
it's like the central limit theorem. the crit chances are uniform but there are two of them, so there are more ways to make the middle than the ends.
>>
>>10188836
then you have to assume 3 possibilities, 0/1, 1/0 and 1/1
>>
>>10188836
If you know that at least one hit is crit, and that the first hit is not crit, then the second hit being crit is 100%
>>
>>10188840
You forgot 0/1, so it's 50%.
>>
>>10188804
no_crit -1,0-crit
0,5 /
start crit
0,5 \ /0,5
crit
\0,5
no_crit

0,5 - only last one crits
0,25 - both crit
0,25 - only first one crits
>>
50%. Easy. We know one was a crit. Crit chance is 50%.

Probability of two crits in two hits without any conditional information is 25%. This is intro stats stuff. You guys are supposed to be doing physics and very advanced calculus
>>
>>10188835
I guess they never miss, huh..
>>
>>10188824
This is causes mental pain to me. I'll pretend at least 50% (heh) of those guys are baiting
>>
>>10188867
so what's the correct answer according to you
>>
>>10188869
3^(-0.9999...)
>>
>>10188804
This is not a mathematically unambiguous formulation because of this condition:
>At least one of the hits is a crit.
There are three possibilities unless OP clarifies this.
1. Both rolls are simultaneous. If no crits are rolled, one of them is converted into a crit.
2. Attacks are rolled consecutively; if the first isn't a crit, the second is an automatic crit.
3. Attacks are rolled consecutively; the first one is an automatic crit (although the problem is formulated in such a way that it is unlikely this is what was meant).

Interestingly, whether it is 1) or 2) doesn't matter for the OP question, the answer is 0.5*0.5=0.25.
For 3), the answer is 1*0.5=0.5.
>>
>>10188804
[math] p_{othercrit} = p(\mathrm{2 crit|1 guaranteed crit})p(\mathrm{1 guaranteed crit}) + p(\mathrm{1 crit|2 guaranteed crit}) p(\mathrm{2 guaranteed crit}) [/math]

From the fact that the 'other'attack has a 0.5 crit chance, [math] p(\mathrm{2 crit|1 guaranteed crit})=p(\mathrm{1 crit|2 guaranteed crit})=0.5[/math] Also noting [math] p(\mathrm{1 guaranteed crit}) = 1p(\mathrm{1 guaranteed crit})[/math]

[math] p_{othercrit} = 0.5 \times (p(\mathrm{1 guaranteed crit}) + (1-p(\mathrm{1 guaranteed crit}))) = 0.5[/math] QED
>>
>>10188804
Crit, Crit
Crit, No Crit
No Crit, Crit
No Crit, No Crit

1/3

Btw wtf is a crit?
>>
>>10188861
>math
>very advanced calculus
>no other math exists
>>
>>10188955
>Btw wtf is a crit?
At least one of the two hits
>>
1/4. The chance of a critical hit is 1/2 for both hits. The fact that we know at least one is a crit does not change the fact that is still a 1/4 chance they are both crits.
>>
It's 1/3.
>>
75%
>>
>>10188840
>>10188843
You meant
0/0?
>>
>>10188942
Epic [math]fail[/math]
>>
>>10188804
Minimum is said to be one hit is a crit. The other hit is not guaranteed to be crit, only 50%. So 50% chance
>>
>>10188955
A critical hit. Comminly dealing 150-200% the damage of a normal hit. A dated game mechanic coming from dungeons and dragons that games should have no business using.
t. turbonerd
>>
>>10189039
Thank you, based game-anon
>>
>>10188804
Ambiguous bait question. Answer can be both 1/3 and 1/2 depending on interpretation.
>>
>>10189050
Also
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

Now stop arguing, brainlets.
>>
>>10188835
brainlet here, what is that "renormalisation"?
>>
>>10189143
Making sure the remaining probabilities sum to 1.
>>
>>10189143
It's just renormalization but done by a posh Brit.
>>
BAYES RULE YOU FUCKING RETARDS
>>
>>10188955
50%, as atleast one hit was a crit out of 2
Thus
c=crit
n=non-crit

S=cc, cn, nc
But order doesn't matter in this case. Either it's a crit and a noncrit or a crit and a crit. So 1/2th or 50%
>>
>>10189163
Cannot be applied given the conditions from OP. How do you translate “at least one crit” into bayes?
>>
>>10189172
Rethink your logic. You only know that a (one) hit was crit. You don't know which one (first or second)
>>
>>10189163
>>10189172
are there any incorrect high view CS fag gaymer videos on YT that attempt to sloppily work this out? Also, why are you all incapable of working this out?
>>
>>10188804
1/3
>>
depends on how much damage you've already done, if we're talking about tf2
>>
>>10188804
U can find the answer here:
>>10187135
>>
It depends on whether OP is a fag
>>
>>10188835
you're not measuring hits and misses you dumb fuck. the combinations are: (noncrit, noncrit), (noncrit, crit), (crit, noncrit), (crit, crit). Since there is only one out of four options with two crits, the probability is .25. Fuck off with renormalizing, don't read shit you don't understand pleb.
>>
>>10189264
Hows that 90 IQ treating you
>>
>>10189191
why would order matter in this case?
noncrit-crit and crit-noncrit are the same thing.
If you are specified that your previous strike wasn't a crit, it would still fall under the same case that you did an crit and next case would not be a crit.
Question doesn't imply that order matters. All we know is that atleast one was a crit, which means it falls under both nc and cn no matter if it was the first strike or the second strike
>>
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>>10188804
Come on...
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>>10189350
Why do both middle two have to be listed? The order doesn’t matter. You already know for certain one is a crit, so it’s not even worth pondering the “if non crit, then guaranteed crit” scenario
>>
>>10189302
They aren’t the same; they’re two different possible events.
>>
>>10189302
If you took a random family off the streets with 2 children and at least 1 boy, then there's a 33% chance that they have 2 boys.
The order doesn't matter, what matters is that having 1 boy and 1 girl is twice as likely than having 2 boys or 2 girls.
>>
>>10189369
Statistically you're twice as likely to have 1 crit and 1 non-crit than you are to have 2 crits.
>>
>>10189379
If order doesn’t matter than boy/girl and girl/boy should not be counted as two separate outcomes
>>
>>10189400
It's one outcome that's twice as likely to happen than having 2 boys
>>
25%
>>
>>10189373
>>10189379
SHUT THE FUCK UP YOU FUCKING VIRGIN FAGGOTS FUCK OFF SUCK MY DICK SUCK MY DICK I DON'T GIVE A SHIT MY ANSWER IS CORRECT AND YOU ARE FUCKING RETARDED FUCKING CUNT GO FUCKING KILL YOURSELF YOU MOTHERFUCKERS FUCKING 1/3RD FAGS INSERT A 12 INCH DILDO IN YOUR ASS AND GET AIDS AND DIE CUNT FUCK YOU FUCK YOU
SUCK MY DICK
SUCK MY DICK
SUCK MY DICK
>>
>>10189428
So this... is the power... of the 50% poster... whoah...
>>
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>>10189369
>>10189428
>>10189302
Proven wrong numerically.
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>>10188804
Now I've never taken a probability or logic class, but how is it not 1/2?

If one is guaranteed to be a crit hit, then how is the question not asking for the probability of the other being crit?

Assuming order doesn't matter,

crit - non crit = non crit - crit
The four possibilites are
crit - non crit
crit - crit
non crit - crit
non crit - non crit
but since order doesn't matter, the list shortens to
crit - non crit
or
crit - crit

how is it not 1/2?
>>
>>10189440
You are so dumb
>>
>>10189451
If we’re given order it’s 50%, but we’re not, so we have to consider both ways of crit/no crit happening
>>
>>10189451
>since order doesn't matter
People keep saying this, but the reason people write it with order is because the prior probabilities are uniform over the possibilities. You can write it without the order, but then you have to say crit-nocrit is twice as likely as crit-crit, and also twice as likely as nocrit-nocrit.
>>
>>10189456
Heh yeah, you sure told that empiricist.
>>
>>10188804
>/v/
Please stay there.
>>
Let's say the first one isn't a crit. Then, is the next one a crit "by default" or does the game roll to see if it's a crit, and if it's not it makes it a crit?

0 - 0 (1) (can this happen?)
0 - 1
1 - 0
1 - 1

Or maybe, the game rolls a number, and if it's not a crit then it makes it a crit. Then it rolls the second number

0 (1) - 0
1 - 1
1 - 0
1 - 1

This scenario is not against OP's rules.

It's an ambiguous problem which doesn't have a single answer unless you know which algorithm the game is using to ensure you crit at least once
>>
>>10189494
>or does the game roll to see if it's a crit, and if it's not it makes it a crit?
That makes no sense. Why "roll", only to throw away the result? That's the same as not rolling and just make it a crit
>It's an ambiguous problem which doesn't have a single answer unless you know which algorithm the game is using to ensure you crit at least once
The game doesn't ensure you crit at least once. It's a simple conditional probability question
>>
I tried the following:

P(A|B)=P(A^B)/P(B)
P(B)=3/4
P(A^B)=1/4
P(A|B)=(1/4)/(3/4)=1/3

But then there's also the factor of crit chance which is 50%. Where does this come in the calculation?
>>
>>10189523
^ means AND in this case btw
>>
You're fighting a god damn dragon, and you've almost killed it. The dragon is really close to dying: it won't quite die in two hits, nor will it die in a single crit, but a crit plus a hit is enough to kill it. Thankfully your built is really crit heavy so you have a 50% crit rate.

Your thirst overcomes you, so you turn towards your Mountain Dew and take a nice gulp, and you click twice to launch two more attacks. After doing the Dew, you turn back towards the screen and the dragon was killed by the two attacks, which means that there was definitely at least one crit.

What is the probability that both of your attacks were crits?
>>
>>10189523
How did you get numbers like 1/4 and 3/4 if you didn't use the 1/2 crit rate already?
>>
>>10189541
doesn't fuckin matter, the dragon's dead
>>
>>10189494
0 - 0 is impossible, but that doesn't meant that the first roll being 0 will automatically make the 2nd role a 1.
>>
>>10189561
1/4th because there's only one case out of the following 4: "CC", NC, CN, NN
and 3/4ths because of the 4 there's 3 that has a crit in it: "CC". "NC", "CN", NN

Thing is, the crit strike could have been say 30% or so, thus there would be a greater chance of getting NoCrit and NN probably being the more common occurence. That's why I think that the 50% crit chance must have some sort of influence
>>
>>10189593
what I mean is you don't get a 50% chance of the first roll being a 0
>>
There is only one crit roll. The text essentially gives you one of those [first hit is a critical attack] buffs sometimes found in games.

100% chance for 1 critical hit
50% chance for 2 critical hits
0% chance for 0 critical hits

Not sure why something so simple has /sci/ buttblasted.
/v/ here btw.
>>
>>10189614
wew
>>
>>10189594
You implicitly assumed a 50% crit chance when you counted the possibilities.

When it's a 50% crit chance, all the possibilities you listed have a 25% chance, so you can solve everything just by counting. Otherwise you have to weight the possibilities according to crit chance.

So your numbers are correct, but only for the 50% crit rate, as I imagine you probably expected.
>>
>>10189614
Brainlet.
>>
>>10188998
If 0 represents crits, that 0/0 option is not possible. Some of the initial info was at least one of the hits are crits.
>>
>>10189636
No, he's right. The problem is ambiguous. It could very well be that the player has a flatline 50% crit chance and got a buff that makes the next hit a critical strike (vanilla wow Cold Blood, for example). As long as the problem doesn't state how the guaranteed crit is handled, all calculations are bogus
>>
>>10189643
There's nothing saying there was a guaranteed crit. It just says one of them was a crit. It didn't have to be a crit, it just was by chance a crit.
>>
>>10189648
"at least one of the hits is a crit" is semantically isomorphic to "guaranteed crit"
>>
>>10189651
If I flip two coins, look at the results, and then say "at least one is a heads", nothing I said implied there was a guaranteed heads.
>>
>>10189648
It could be that, as a way to ensure you get at least one crit. Or it could be that you have a guaranteed crit. The latter is more common in games than the former. And the way the problem is frames both are valid
>>
>>10189658
You’re making an assumption because you think it makes more sense. But the problem can be interpreted in more than one way.
>>
>>10189651
This is the text that made me think its guaranteed crit. Otherwise it's pointless to mention it, and without it the problem becomes a simple matter of throwing coins twice and calculating the chance they are both tails.
>>
>>10189665
I'm reading the problem as written.
>>
>>10189264
But we know that at least one of the hits is a crit, so there are only three possibilities (all of equal probability)
>>
Its 1/3, jesus christ I hope y'all are just shitposting
>>
does the answer change depending on how i know "at least one hit is a crit?" it seems that either a)i observed that one hit was a crit, and now have to guess the next hit on a 50% chance or b)someone is telling me that in a future scenario, the probability of the hits will adjust based upon the first one, where if the first hit is a crit, the next hit will be a normal 50% chance, but if the first hit is not, then the next hit will be a 100% chance
>>
>>10188804
this is the boy girl problem just reworded
>>
we LITERALLY had this thread yesterday with boys and girls and the same exact argument was had
>>
actually, it wasn't even yesterday, it's right up at the top of the catalog right now >>10187135
>>
>>10188899
This seems like the most well-rounded analysis, why is everyone saying the answer is 1/3 or 0.5 unconditionally? Please explain, brainlet here.
>>
>>10188804
Kinda poorly worded, seen a bunch of times in variations that can promote a response from one of the other variations.

Considering there is a 50% chance to crit the possible outcomes of the cycle is (Crit, Nocrit)
NN 25%
NC 25%
CN 25%
CC 25%

We know that at least one is a crit, so the NN option is eliminated, leaving a 1/3 probability that both are crits.
But you could argue some semantics on this.
>>
>>10188899
Eh, you made up the automatic and converted part.
It doesn't say that one will ALWAYS be a crit. It's just giving you one instance where there were two attacks and one happened to be a crit.
>>
>>10190722
The problem is the way it’s phrased. Some people are analyzing it like this. 0 being a hit and 1 a crit
0-0
0-1
1-0
1-1
But because the problem states that you crit at least once, they remove the first possibility and the odds are 1/3.
But, you don’t know how it works. Perhaps it works like the post you quoted said. Basically, it’s all up to whether the possibility of rolling 2 hits in a row is allowed or not. The people saying 1/3 are saying that because the problem states that you crit at least once then that possibility doesn’t exist. The ones saying 0.25 are saying that the game would roll twice anyway, and the possibility of getting two hits exists (and if that happens then the game turns your last hit into a crit)

tl;dr do you allow for 0-0 to happen or not? Depending on that you get .25 or 1/3.
>>
>>10190722
Another way of thinking about it is if you allow artificial crits. Meaning, a 0-0 roll gets artificially turned by the game into a 0-1 roll. If you only allow natural crits (that is, once it’s rolled whatever the result remains) then you know for a fact that 0-0 is impossible.
People saying 0.5 and 0.25 probably play games and have an idea how rng works
>>
>>10190791
Almost everyone here is treating the four outcomes as equiprobable when in truth vidya RNG is loath to give you two crits in a row.
But yes, given the information in OP and using only that, the probability is 1/3.
>>
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>>10188804
>Probability for double crit = 0.250008
>Number of double crits = 250008
>Number of non-double crits = 749992
>>
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>>10190813
>>10188804
Its 1/4.
>>
>>10190816
How did you reach the conclusion that the probability of only the second hit being a crit is 50%, my man?
>>
>>10190833
>the first is 50/50
>if crit then second event is 50/50
>if non crit then second event is 100% crit
There is always 50% crit chance unless the first was noncrit since the problem states their MUST be at least one crit out of the two
>>
>>10188804
>>10190816
its not a quarter, weve been assured a crit in the set of 2, so the sets probability is changed.
it remains at 50%
balls dropping on pins confirms this!

For 10 rows (n=10) and probability of bouncing left of 0.5 (p=0.5), we can calculate the probability of being in the 3rd bin from the right (k=3) as:

f(3;10,0.5) = (10 choose 3) 0.5^3 (1-0.5)^(10-3)
also:

(10 choose 3) = 10! / 3!(10-3)! = 120
(This means there are 120 different paths that
can end with the ball in the 3rd bin from the right.)

So we get:
f(3;10,0.5) = 120 times 0.5^3 (1-0.5)^(10-3) = 0.117

just walked you through it boys.
who likes a bell curve!
distribution motherfuckers.
>>
>>10190841
this guy gets it!
bell curve distribution.
one crit is assured
one is 50%
so assuming that you miss the first your second is guaranteed.
always ends 50/50
>>
>>10190841
OP doesn't make it sound like a rule enforced at roll-time. You simply have some knowledge of the outcome that helps you prune the sample space (by a quarter). It's a simple matter of calculating the conditional posterior probability.
>>
P(2crit | 1+ crit)
= P(2crit and 1+ crit) / P(1+ crit)
= P(2crit) / P(1crit or 2crit)
= P(2crit) /(P(1crit) + P(2crit))
= p^2 /(2*p*(1-p) + p^2)
= p^2 /(2p - p^2)
= p /(2 - p)
for p = 0.5
= 0.5 / 1.5
= 1 / 3
>>
>>10190841
>>10190851
You are interpreting the problem differently. You are assuming that the "game" never dishes out two non-crits in a row. Not that in hasn't on this particular occasion, you're assuming it never happens, as a rule.
>>
>>10190857
He states and I quote "At least one of the hits is a crit."
do we not take that into account as rule in the following?
>>
>>10190857
Well then the question isn't specific enough if it allows two different interpretations. More information is needed to only have one possible answer
>>
>>10190860
At least one of two specific hits is a crit. He never says that every non-crit following a non-crit is automatically turned into a crit.
>>
>>10190813
your script is wrong.
The distribution should have (0, 1), (1, 1) amd (1, 0) in equal amounts while yours is 50% (0, 1) and the rest equal.
>>
>>10190865
then its as simple as checking the possible outcomes in the set.
with 4 possibilities its so simple!?
we want the probability as a percentage maybe?
1/4 or 25%
is this better?
>>
>>10190782
>do you allow for 0-0 to happen or not?
No, because it says "At least one of the hits is a crit."

>>10190791
This is more the issue. That assumption should not be made, because it also assumes you always attack twice which is, again, not in the problem. This could just as easily be two single attacks and if you had a rule where you always get a crit then obviously you'd always crit, 100%.

>People saying 0.5 and 0.25 probably play games and have an idea how rng works
No. Yes, game RNG calculations are usually not as simple as this problem, but they're also usually less well known

It doesn't say the two attacks are connected. It doesn't say they're a combo or as single attack cycle. It doesn't say you are always guaranteed a crit.
You need to remove these assumptions from your consideration.
>>
>>10190870
>the possible outcomes in the set
It's a possible outcome in general, but not for our case. We are told (non-crit, non-crit) is not the case for our two hits. The possibilities are 3, so 1/3.
>>
>>10190867
he says we are guaranteed a hit one at least one so the set shrinks accordingly to remove the possibility of two non crits
so it actually becomes 1/3 or 33.3%
>>
>>10190860
Because it doesn't say "At least one of the hits will always crit".
>>
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I've got the definitive answer here folks:
Depending on how the hits are determined your probability is 50% or 25%.
-If the order of hits is unspecified or unimportant (ex. You accomplish both hits simultaneously) then you get 50%. You can think of it as having probabilities Pa and Pb of criting between your two attacks. You know that there is a 100% crit between the two of them and therefore the other must be 50%. Pa * Pb = 50%.
-If order does matter (Pa -> Pb) then you have to follow a decision tree like pic related. It is intuitive to say that since you get 3 final states that your probability is 1/3 but that assumes that all states are equally probable. You have a 50% chance of not criting on the 1st hit, and in this scenario you get a 100% crit for the 2nd. If you crit on you're first hit, then you're left with a 50% chance of criting again. To be in that situation though, you need to crit once which is an initial 50%, therefore Pa * Pb = 50% * 50% = 25%.

Thanks OP, my gf and I had a fun time figuring this one out
>>
>>10190867
Its not wrong. I implemented it with the interpretation of "at least one of the two hits is crit" as a hardcoded game mechanic rather than as a hint of the results for a one time roll.
>>
>>10190875

im pretty sure thats how he wanted it interoperated tbqh

>>10190877
>>10190874
>>10190872
>>
>>10190877
You and your gf are retarded.
>>
>>10190879
no look at the first if clause, if hit1 is 0 then hit2 is immediately 1.
hit1 is 0 50% of the time so your distribution consists of (0, 1) for 50% of the samples. This is already wrong, it should be a third of each, (0, 1), (1, 1), (1, 0) because these all have the same probability in the original distribution when you dont hahe the filter (at least one crit)
>>
>>10190877
just try creating a dataset by flipping 2 coins and setting them aside.
Then look at all your coin pairs and take out the ones that have 2 tails.
youll end up with 1/3 of the remaining coins as 2 heads and the rest mixed
>>
>>10190883
Please explain.
>>
>>10190877
>-If order does matter (Pa -> Pb) then you have to follow a decision tree like pic related. It is intuitive to say that since you get 3 final states that your probability is 1/3 but that assumes that all states are equally probable.
They are all equally probable.
>You have a 50% chance of not criting on the 1st hit, and in this scenario you get a 100% crit for the 2nd.
Incorrect. The probability is not changing.
The language is kinda important here. It states "At least one of the hits is a crit".
There is no order to this. If the first one wasn't a crit it doesn't become a 100% chance the second one WILL crit. The thing that changes is OUR KNOWLEDGE of which branch would have been followed, not the PROBABILITY of that branch being followed.

The probabilities remain the same. The only thing we can rule out is the possible outcomes. That does not affect the probabilities.
>>
>>10190896
Rolling for crits generally and then removing all the no-crit cases is different from starting out knowing that there is at least one crit.
>>
>>10190905
Doesn't change the probabilities, though.
>>
>>10190905
Ok Ill try again.
A person draws one out of for cards, named aa, ab, ba, bb.
They then look at it and if they see aa, ab or ba they say to you "one plus" If they see bb they say "zero".
It is clear that when running this experiment you will hear one plus three out of four times.
Now imagine yourself running a trial of this experiment and your opposite says one plus to you.
the chance for bb is now obviously zero.
the three remaining possibilities were drawn with EQUAL probability and this can't change now. The aa card gets the exact same treatment as the ab card. They are indistinguishable to the person looling at them to give you the hint.
There are three possibilities with equal chance remaining, so aa as a 1/3 chance.
This is how you have to interpret the problem because you dont start put knowing there will be at least one crit. The crits aren't generated by a gnome which takes care that if the first hit wasnt a crit at least the second one is, no they are randomly generated and afterwards we argue about the outcome given certain hints about what happened.
>>
>>10190896
>>10190902
>>10190907

There is a difference in probabilities between these two scenarios. If we take the frequentist approach to OP's problem then we can assume that we have it an enemy with 100 pairs of attacks. What you are implying is that we perform these attacks with the possibility of getting a no-crit combo and then removing those possibilities later. The problem is that you've already observed these no-hit combos occur so they are inherently a part of your sample and can't be removed. Alternatively, if you view it in terms of routes that can be taken, you are implying that there are four equally likely routes and we're just not going down one of them. The problem is that there is no fourth route to go down. Because each point where routes diverge is split symmetrically, the end probability of each route cannot be symmetric. We can only have 100% knowledge of an outcome after it's been observed so if we have 100% knowledge of a crit after a no-crit has occurred, it is because there is no further splitting down the line and a final state has already been observed. If this final state has a 50% chance of occurring, then the other remaining final states have a 25% chance each.
>>
>>10190939
>If we take the frequentist approach to OP's problem then we can assume that we have it an enemy with 100 pairs of attacks. What you are implying is that we perform these attacks with the possibility of getting a no-crit combo and then removing those possibilities later.
No. What I am implying is that 99 of those pairs of attacks have no relation to this situation. 25 of them could have been no-crit combos and it DOESN'T MATTER.
All we know is that in THIS "combo" there was definitely at least one crit.

>Alternatively, if you view it in terms of routes that can be taken, you are implying that there are four equally likely routes and we're just not going down one of them.
Exactly.
>The problem is that there is no fourth route to go down.
No, the route is STILL THERE. We just know it WASN'T taken.

It's like your chance of winning Lotto is one in hundreds of millions (or worse). If you win Lotto does your chance of winning Lotto become 100%? No. It's STILL one in hundreds of millions.

The fact that you know that some branch of the probability tree HAS NOT BEEN TRAVELED, does NOT change the probabilities. There are no special rules in this case that will rework the probabilities.

>We can only have 100% knowledge of an outcome after it's been observed so if we have 100% knowledge of a crit after a no-crit has occurred, it is because there is no further splitting down the line and a final state has already been observed.
Yes, the outcome is already known. Which is why you can say "At least one of the hits is a crit". Again, that doesn't change the probabilities. It's like if I throw a pair of dice right now and tell you "At least one of the dies is a six. What is the probability that both are sixes?"
Nearly 50% of the possible outcomes of a two die roll are not possible, but it's not going to still be 1/36 and it's not going to be 1/6 either.
>>
Stats Major here, agree with this anon >>10188899. I also enjoy theorycrafting in Path of Exile where there are many interactions like "lucky" attacks (roll twice fir crit) with diamond flasks, etc.
>>
>>10188804
Mutually exclusive events so therefore 50%.
>>
lol
crit crit
crit no crit
no crit crit
epic 1/3
>>
Flip two coins in a row. Do this hundreds of times. Now, only take into account the pairs with at least one head. You will see that head/tails combo is twice as likely as two heads.

The answer is 1/3.
>>
It's such a bait question.

In a real game example, it really depends on the constitutionality of the second crit.

Is the game scripted to ALWAYS give a crit after the first hit? Because then it wouldn't be true 50%, it'd be much higher unless it weighted in the other direction of the average crit rate is over 50%.

It just depends if we're talking about "for every two hits, there's at least one crit" or "from a set of hits, we randomly derived two and at least one was a crit".

The answer would be 50% or 33% depending on how the crit function was programmed, and the "50% crit chance with at least one crit" is a misrepresented question, since it wouldn't really be a 50% crit chance if it was forcing a crit.
>>
I do not understand how the question of the boy-girl paradox is ambiguous. Wikipedia gives two possible intepretations:

1. From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3
2. From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2

Whats the difference here? In both scenarios all you learn is that the family has at least one boy.
.
>>
>>10192566
In the second scenario you also have to account for the probability of selecting a family with at least one boy in the first place, and also the probability of selecting the boy in the families with only one boy, which would mean that, using this selection process, the odds of being told there is at least one boy is significantly higher (that is, 100%) for families with two boys, 50% for families with 1 boy, and 0% for one with two girls.
>>
>>10192566
In scenario 1, you are selecting from ALL families. The chances of selecting a family with at least one boy is 3/4. Because they then disregard all the other families, this is how they arrive at 1/3

In scenario 2, you start the selection process from families that already have at least one boy. Because you know that one is already a boy you just have to determine the gender of the other child.
>>
>>10188804
Because one hit is already defined, it is outside of the range for the probability, thus it shouldn't factor into it. Assuming order doesn't matter, as we're looking at an objective sum, the chance the second hit is a crit is 50%

|C+N|
|C+C|
|N+N| <-- outside probability range.
>>
>>10188942
[math]quantumelectrodynamics[/math]
>>
>>10193072
>Assuming order doesn't matter
>the second hit
Wow.
>>
>>10193086
Semantics, kys
>>
>>10193072
We're not given an order, so you need to consider both ways of crit/no crit happening.
This gives us
Crit, no crit
No crit, crit
Crit, crit
All with equal probability, the probability of getting two crits is 1/3. You can't just say order doesn't matter and then immediately apply an order to it, non sequitur.
>>
>>10192566
1) Selects all families with at least one boy.

2) Will miss about half of families that have one boy and one girl since you could still select the girl and exclude that family
>>
Crit, no crit & no crit, crit are equal and redundant. You assume order, therefore artificially increasing the probability of this outcome, from 1/2 to 2/3.
>>
>>10193106
Meant for
>>10193097
>>
This is absolutely hilarious watching people argue this across the boards.
>>
>>10193106
You are wrong.
>>
>>10188804
Why are people still discussing this shit that gets posted everyday?
>>
>>10193088
It's a straight contradictory statement. Yes, if you know the state of the first strike, the chance of a critical on the second is 50%. If order doesn't matter it's 1 third. Reformulate the problem as so:

In response to market demand, m&ms now only come in two colors, red and yellow. They are also now packaged in packets of two m&ms only. The process that puts the m&ms into bags is random, depositing a red m&m 50% of the time and a yellow one 50% of the time. In packets with at least one yellow m&m, what are the chances that both are yellow?

So yeah, you have three options again: two red (not included in the above probability), one red one yellow, and two yellow, but in the above process the packets with one red and one yellow are twice as common as the all yellow packets, accounting for 50% of packets. That makes it 1/3
>>
>>10193112
The moment you started analyzing a predefined 'pool', you assume order.

This question is not asking to analyze a pool, but to analyze the probability of an individual set of hits.
>>
>>10192808
>>10193056
>>10193100
Thank you for the help but I am still extremely confused. The scenarios seem identical to me. Is this what it is?

The answer is 1/3 when: You are trying to determine the probability of both children being boys, given the family has at least one boy.
The answer is 1/2: when: You are trying to determine the probability of both children being boys, given you see the family has a boy but you don't see the other child

Is this essentially the difference? Because I don't see much of a difference. In both cases you know they have at least one boy.
>>
Any sort of "guaranteed crit" scheme is contradicted explicitly by the statement that crit chances are 50%
>>
>>10193257
>Because I don't see much of a difference.
There isn't that much of a difference. The 1/2 answer is more like: the probability that the other child is a boy where you find that one child is a boy. If you were to see one child randomly from the mixed families, half the time you'd see a girl and that means that automatically half of the mixed families are removed.

Alternatively you ask the question "Do you have a boy?" of the 2 children families, out of that group 1/3 will have 2 boys.
>>
>>10193117
But aren't we only selecting from packages we KNOW have at least one yellow M&M?
>>
>>10193257
There's a subtle difference between the construction of the 1/3 case and the 1/2 case, it has to do with precisely how many Bernoulli trials must be run, due to uncertainty (or lack thereof).

In the 1/3rd case, we know at least one is a boy but to pick one and state what it is a priori is to defy what else you're told, which is the probability is 50% - certainties can't have probabilities. In other words we'd be assigning a different, degenerate probability mass function to one of the Bernoulli trials but to act like we're allowed to do that is add assumptions to the problem that do not already exist.

On the other hand, that's exactly what we're allowed to do in the 1/2 case, since we'd be told explicitly which child is a boy. At that point we only have one Bernoulli trial to run, so the probability of both being boys is equal to the probability of just the unknown one being a boy, 50%.
>>
>>10193296
There is no functional difference from "at least one of the kids is a boy" to simply seeing a boy and not knowing the gender of the other kid.
>>
>>10193327
There is a difference between peeking at one of the kids and seeing that it happens to be a boy and being told that there is at least one boy.
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>>10193336
How so? In both cases, the info is identical. You know 100% that one of the kids is the boy, and you dont know the gender of the other child. I dont not see how the information differs in either scenario.
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>>10193353
The chance of "there's at least one boy" given that it's actually BG is 100%. The chance of "I peeked at a random kid and it was a boy" given that it's actually BG is only 50%.
>>
Arent there just two cases? One crit or two, so it would be 50%?
>>
>>10188840
O/1 and 1/0 are the same thing
>>
>>10193368
Do you think if you flipped a coin 500 times the chances of them all being heads in a row is 50%?
>>
>>10193649
They either are or they aren't.
>>
>>10193649
No. Probability of getting at least one crit is 1, probability of getting another crit is .5
1 x .5 = .5
>>
>>10193664
Or to put it another way, if you flipped one coin that only had heads and you flipped another coin that had heads or tails, what's the probability of you getting two heads?
>>
>>10193649
>>10193658
>>10193664
This shows why the problem is stupid.
First:
Probabilities only work with LARGE data sets. They can not tell you anything about any single crit, or flip. They can tell you that out of 10,000 50% will be crits.
But to use that to predict the next crit is junk science, especially with toy data sets
Same reason the “chances “Nucleotide evolutioning in two places in the dna code by chance at the same time is longer than the age of the universe” is wrong too.
And second: probabilities only really work with fixed, I.e. in the past data sets. Thier predictive value is zero if you are dealing with anything of complexity or over a long enough time scale.
These facts will enrage the averge /sci/ poster
>>
Every single one of you drooling homunculi is a complete DOLT and DOESNT UNDERSTAND ELEMENTARY MATH

50%*50%=25%
We know one was already a crit so thats a 25% added on to begin with
Three options: first is crit, second is crit, or both are crit. We're not looking at two of those cases, so 2/3 is the factor
25%*25%=1/8
1/8+2/3= 17/24 chance of getting both crits
>>
>>10193745
What is this called?
>>
1) If the guaranteed crit is hardcoded to happen at least once every hit pair (which the premise implies), it's 50%.

2) If the guaranteed crit is not hardcoded to happen at least once every two hits, and you are simply omitting all of the double noncrits from the total of possible outcomes, then it's 1/3nd.

Since the premise is about a computer game and is phrased in the present tense, one should assume that the crit is hardcoded to happen at least once every hit pair.
Making it 50%.
>>
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>>10193863
The same logic applies to the similar problem with the children btw, see pic.

Except in this case, it's more reasonable to assume the "guaranteed boy" is not "hardcoded" because
1) these are children and it's harder to imagine a magic entity preventing double girls from happening
2) the phrasing (male "has" children) implies the children have already long been born

So in the children version of the problem, it's more likely to be 1/3 since the odds of a family having two boys are 25%, which becomes 33.3333% if you omit the double-girl families from the "population".

So if you saw the video game version first, you're likely to get the children version wrong, and vice versa.
>>
>>10193646
You're so retarded leave this board
>>
>>10193646
anon they’re not
>>
people who say 1/3 seem to use some weird logic where the probability of the events will CHANGE based upon the first child. I just think its a very strange, non-intuitive way of arriving at the answer. I don't see how it makes sense that the probability of a boy goes to *100%* just because the first would be a girl. It feels like you're just forcing that into the scenario. When people argue 1/2 they already know a boy exists so they don't have to argue any change in probability, the probability of the unknown child is also 50%
>>
>>10193997
This is true if you imagine a godlike creature actively stopping girl-girl combos from happening, in order to fit the premise.
Since you're talking about children, that's not really the reasonable thing to do.

However, this is the reasonable way of thinking if you consider the scenario in OP, whereby it's very easy to hardcode a two noncrit combo from happening.
Here it's more reasonable to say 50% chance of double crit.

See >>10193863
and >>10193876
>>
>>10193863
>Since the premise is about a computer game and is phrased in the present tense, one should assume that the crit is hardcoded to happen at least once every hit pair.
No. You're just a dumbfuck.
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>>10194008
If one crit is hardcoded per hit pair, odds are 50%.
And the framing of the problem definitely hints that way.
Deal with it.
>>
>>10194005
The op question is basically just a rephrase of the boy girl question
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>>10194017
Not really, since in the child version it's more reasonable to assume girl-girl combos did happen, but were simply omitted (giving 1/3 as the answer), while in the gaming version it's more reasonable to assume the "at least one crit" part is hardcoded into the game (giving 50% as the answer).
>>
How does actually seeing a boy raise the chances to 1/2? There is still a 2/3 chance it belongs to a boy girl family and not a boy boy one since there are twice as many boy girl families.
>>
>>10189440
The number is false.
>>
>>10194037
If they show you the first child and it's a boy, then you already have the one guaranteed boy, so the next one could be either a girl or a boy. So 50/50.
However, if they show you the first child and it's a girl, then the next one can only be a boy or it wouldn't fit the premise. Meaning 100% a boy.
Combine the two and you get 1/3rd odds of boy-boy.

However, this implies that the children are "revealed", and this is done one after the other. Which is not a requirement under the premise.

For the real solution(s), see >>10193863
and >>10193876
>>
>>10189264
89IQ
>>
>>10194043
>If they show you the first child and it's a boy, then you already have the one guaranteed boy
But you have the guaranteed boy in any case. You know at least one is 100% a boy
>>
>>10194096
And if they show you one child of the two and it's a boy, then the next could be either a boy or a girl.
>>
>>10194110
But you already know one is guaranteed, and you know one is a 50/50 boy girl. How does actually seeing the boy first change this? Even from just the phrasing of the problem we know 100% one is a boy and it doesn’t matter if t came first or second
>>
>>10194123
>How does actually seeing the boy first change this?
Because if you see a boy first, you know the next child can be either a boy or a girl.
>>
>>10188808
based probabilityquestionsniper
>>10188863
underrated
>>
>>10189614
read the thread
>>
>>10194126
But the child you DONT know about can either be a boy or girl. You already KNOW one is a boy. Actually seeing this boy doesn’t change any of the information
>>
>>10194164
But if you see a boy first, you know the next child is just as likely to be a boy or a girl.
>>
>>10194167
But seeing the boy first doesn’t tell you if it was born first or second, so there’s still two possible boy/girl or girl/boy outcomes
>>
>>10193876
>1) these are children and it's harder to imagine a magic entity preventing double girls from happening
Chinese One Child Policy
>>
>>10194185
It doesn't matter if it was born first, it was shown first.
You can apply the same logic to births: if the firstborn is a boy, the next child has a 50/50 chance of being either boy or girl.
If the first to be shown/born is a girl however, the next has to be a boy.

However, all of this implies that the children are "revealed" (shown, born, ...), and this is done one after the other. Which is not a requirement under the premise.

For the real solution(s), see >>10193863 (You)
and >>10193876 (You)
>>
Mr. Smith has 50 children. At least 49 of them are boys. What is the probability that all his children are boys?
>>
>>10194226
1/2 because 49 are guaranteed and one is a 50/50
>>
>>10194226
49.5/50
or 99/100
>>
>>10194226
The correct answer is 1/51. It’s not even ambiguous.
>>
I fucking swear stats is /sci/'s weakest topic. Goddamn post after post about algebraic topology and ya'll can barely string together some basic statistics.
>>
>>10193257
Correct, that's the difference. Keep in mind the second scenario, they can have a boy, but you don't see him since they show you the girl.
>>
>>10193257
Conditional probability vs. the probability of a single event. Yes.
>>
>>10188835
where does it say anything about missing
>>
>>10194249
But most are saying 1/3 and it’s the correct answer
>>
>>10194275
1/2 is the most likely correct answer in the gaming version.
1/3 is the most likely correct answer in the child version.
>>
>>10194279
No even in the game version it’s 1/3 because the probability would be programmed to change based on the condition

If the first hit crits, the second hit is 50%
If the first hit non-crits, the second hit is 100%
>>
>>10194288
What about simultaneous hits?
>>
>>10194288
Why would it be programmed, it doesn't say that in the question, if the crit chance is 50% you can roll two misses in a row
>>
>>10194294
The question says at least one is definitely a crit.
If this were programmed, it would have to randomly pick one hit (first or second, left hand or right hand, fist or knee, ...) per hit pair to make into a crit.

Doing it like this: >>10194288
wouldn't fly, since that means simultaneous hits (e.g. dual wielding) can't be programmed.

>if the crit chance is 50% you can roll two misses in a row
Not according to the premise.
>>
I don’t understand this in a video game scenario. If you are guaranteed at least one critical hit in two hits, how is the critical hit rate 50%? Seems more accurate to say for every 2 hits there is a 100% critical hit chance, and a 50% chance they are both crits.
>>
>>10194370
>how is the critical hit rate 50%?
It's only 50% for one of the two hits.
The other hit is always 100% crit.
>>
Let A be: At least one crit will hit
Let Be be: Two crits will hit
We're looking for P(B|A) (P(B) knowing A)
We use Bayes Theorem:
P(B|A)=(P(A|B)*P(B))/P(A)
P(A|B)=1 since if two crits hit, at least one hits
P(B)=0,5*0,5=0,25
P(A)=0,25*3=0,75

thus P(B|A)=1/3
>>
>>10194374
So there isn’t a constant critical hit probability? In that case you must state the probability of both at once, otherwise it doesn’t make sense
>>
>>10194304
The premise just says that you hit twice and one is a crit, it doesn't say it will have a crit every time you take two hits
>>
The answer is 1/3 if this happened in the past and each hit had a 50% chance of critting

The answer is 1/2 if the game is programmed to guarantee you at least one crit in every 2 hits

Is this correct?
>>
>>10194378
>So there isn’t a constant critical hit probability?
Well no, because the premise says one of the two has to be crit.

>In that case you must state the probability of both at once, otherwise it doesn’t make sense
Indeed.
The probability of one of them is 100%, the probability of the other is 50%. Even in case of a simultaneous hit.

>>10194379
Nobody's talking about "taking" hits.
>>
>>10194380
That's my understanding.

It would still be 1/3 if the game is programmed like this: >>10194288
but this doesn't work if the hits land at the exact same time, something the premise doesn't exclude.
>>
>>10194382
So are you arguing it’s 1/2 or 1/3
>>
>>10194383
No it would still be 1/2 in that case because it means non-crit/non-crit would not even be a theoretical possibility. It would mean that both hits were NOT 50% chance of critting which is what makes it differ from the boy girl problem.
>>
>>10194388
I'm saying 1/2 makes more sense in the video game version since one crit per hit pair would be hardcoded into the game (even in case of simultaneous hits). So noncrit+noncrit combos simply never happen.

And 1/3 makes more sense in the child version, since here it's more reasonable to assume that girl+girl combos do happen, but you simply disregard them.
>>
>>10194382
I meant hit twice, not native speaker
>>
>>10194391
In that case, you'd have a 50/50 chance of the first hit being crit or noncrit.

If first hit is crit => next hit has a 50/50 chance of being crit or noncrit.
If first hit is noncrit => next hit is 100% crit.

This yields 1/3.

But it does not take into account simultaneous hits, so it's not theoretically sound.
>>
>>10194396
>>10194379
But you have to consider all possible scenarios if you want to calculate probability.
>>
>>10194397
But unlike the boy girl problem, in this case you are 100% guaranteed a crit even before the hits occurred. You are going to get 2 crits 1/2 the time because there’s a 100% chance of one and a 50% chance of the other
>>
>>10194405
>But unlike the boy girl problem, in this case you are 100% guaranteed a crit even before the hits occurre
That's my view, yes.

But if you program the game like this: >>10194288
you still get 1/3.
However, I feel that this manner of programming is flawed, since it does not account for simultaneous hits.
>>
If the game was programmed for simultaneous hits, the only proper way to phrase it would be 50% chance of double crit.
>>
>>10194412
Wrong.

Your outcomes would be:
Left hand crit - right hand no crit
Left hand no crit - right hand crit
Left hand crit - right hand crit
>>
>>10194418
This implies one hand comes before the other.
>>
>>10194418
But if they occur at the same time how could a program possibly alter the probability of one hit based on the other?
>>
>>10194419
>>10194421
Program is already programmed to output those 3 outcomes. Then it rolls the dice
>>
>>10194429
You’re right . Didn’t even think of that
>>
>>10194429
That would be artificial.

If you follow the theory of the premise, you'd program a dice roll on which of the two hits to make the guaranteed crit.
>>
>>10194429
But this is wrong because it then means the critical hit chance is not 50%
>>
>>10194437
This.

One of the two hits has to have a 50% crit chance, the other is the "at least one".
>>
Doesn’t the wordage “assuming a 50% critical hit chance” imply this is a past scenario where each hit had a 1/2 chance, and thus the answer is 1/3?
>>
>>10194446
The present tense seems to imply that things are ongoing or about to happen.
The "assuming a 50% crit chance" could simply apply to the one hit that isn't the "at least one crit".
>>
>>10194370
Think of it as a double attack skill with one guaranteed to crit and the other one affected by your crit chance, which is 50%.
>>
>>10194437
>>10194436
Yes, it is artificial, but to the player if the ability was described as what it was in the OP in an item description, it would look like the item's ability is working as intended.
>>
>>10194457
>it would look like the item's ability is working as intended.
No because there wouldn't be 50% crit chance on one of the two hits.
>>
>>10194464
But if the chance of a double crit is 1/3 then the other two options are also a chance of 1/3 and 1/3.

You should have the same chance to get a double crit as you do to only get one crit and a miss. So the answer has to be 50% then, is what you're saying?
>>
>>10194483
>then the other two options are also a chance of 1/3 and 1/3
That's not the same as "50% crit chance".

One hit will always be crit, and then the other hit only has a 1/3 chance of being crit.
>>
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>>10188804
Brainlet here:
Can we even assume 50% crit chance with the knowledge that one of two hits is a crit? It seems to me that it has non-constant probability due to being reliant on past scores. Or are we talking about average probability? Or just a crit chance without looking at the sure crit factor? If so I think it's 25% for double crit (see picture)
>>
>>10194497
You're right, it's not the same, which means 1/3rd fags are btfo.
>>
You guys are all wrong, Jesus fuck.

The possibilities are

Don't crit / Don't crit
this is ruled out by the description

Crit / Crit
Nice

No crit / crit or crit / no crit
You're a dumb fucking idiot that can't hit properly and kill yourself from shame

So the chance that both are crits is 100%.
>>
>>10194742
Guess they never miss, huh?
>>
>guessing someone's bank credentials
>Two possible outcomes: 1. I guess them correct. 2. I guess incorrectly.
>Therefore I have a 50% chance of guessing your bank info.
>>
>>10194545
The question just says that you tried to hit twice and one of them was a hit. It could've happened that none was a hit. For any case the chance of a crit is 50%. In your picture you'd need another no crit in the right branch. Since (not, not) doesn't count, because the question tells you landed 1 crit, only the other three matter and the probability is 1/3
>>
>>10195003
What if the premise is how the game is programmed?
i.e. one hit has to be crit, and the other hit has a 50% crit chance.
>>
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The probability isn't split evenly. If you roll a nocrit first, the next must be a crit. If you roll a crit first, the next is a 50% chance. The rolls are individual.
>>
>>10188986
The only correct answer ITT
>>
>>10195129
What if the hits are simultaneous?
>>
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Our sample space S = {(1,0), (0,1), (1,1)} where 1 is a crit and 0 is a non-crit gives us the set of pairs we're considering for this specific two hit scenario. And the question explicitly tells us to assume a 50% crit rate. It should be natural and obvious that 1/3rd of these cases results in a pair of crits.

P((1,1)) = N(1,1)/N(S) = 1/3.

If we were to assume that the game is hardcoded, as people have been discussing, such that "of every two hits one must be a crit" And this is true for every single hit, not just this pair, (complicating the problem) then our sample space for pairs of hits would still always be one of the pairs in our sample space S. Therefore in the hardcoded scenario every single hit draws from (1,1,0) without replacement and the second hit draws from either (1,1) if first draw is a non-crit or (1,0) if first draw is a crit. Third and fourth and beyond simply repeats. But this is not a 1/2 crit rate this is a 2/3 crit rate which contradicts the premise.

Any given hit selected at random will have a 2/3 chance to be a crit. Clearly greater than 50%. Which is expected since we've entirely removed the possibility of sequential non-crits. We can calculate the chance of pairs easily using conditional probability. P(a intersection b) = p(a|b)*p(b) = .25. So we'd predict 25% double crit rate in the hard coding scenario. Not the 1/2 number people are throwing around. This is somewhat non-intuitive, but it makes sense because we would expect half of the pairs that have a crit on the first hit to have a crit on the second hit and no contribution from the pairs that don't have a crit on the first hit. We're expecting 50% of all hits to crit on the first hit, and we're expecting 50% of those to have a crit on the second.

I wrote up a quick test simulation just to confirm it, and the numbers are close enough for me. I'll do one with simultaneous hits later.
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>>10188804
big brain
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>>10188804
There are four possible outcomes; each attack is assumed to be independent:

No crit, no crit
No crit, crit
Crit, no crit
Crit, crit

Each of these outcomes has a 25% chance of occurring.

Since there is a condition that at least one crit occurs, the sample space is reduced to:
No crit, crit
Crit, no crit
Crit, crit

Dividing each probability of 25% by the sample space sum of 75% (as the probabilities must add up to 100%), the new probability is 33% for each event.

The probability of there being two grits, given that there is at least one crit, is 33%.

This is the objectively correct answer, and anyone who says otherwise is a brainlet.

t. Actuarial student
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>>10189369
>Why do both middle two have to be listed?
Because they're different events. Binomial probability you fucking brainlet.
>>
>one of the hits [has to be] a crit
1/2
>one of the hits [happens to be] a crit
1/3
/thread
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Can someone honestly conceive of a situation in which you learn that at least one is a boy without learning about the specific boy? In almost any conceivable situation the probability of this problem will be 1/2.
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>>10195480
It’s twice as likely for there to be one crit than two crits
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>>10195003
I skipped the right branch because of zero probability (guaranteed second crit). To be honest the question should have a better definition of a guaranteed hit and 50% crit chance. I think that since it's from /v/ it's closer to a game setting where my answer is probably the closest to how it would be coded.
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>>10188804
100%, ur ar welcom
>tfw 30 iq
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>>10189428
capped and saved for future use
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>>10195480
You're still wrong. I'll try to break this down in a way you can understand.

If one has to be a crit then it's reasonable to expect that half of the first hits will be a crit and half of the first hits will be a non-crit.

If it's a crit on first hit then we can reasonably expect the second hit to be a crit 50% of the time. If it's a non-crit on the first hit then there's no way we can achieve a double crit so those events can be discarded.

Therefore the number of events in which we score a double crit is a subset of events where the first hit is a crit. So it'/
a subset of only half of all possible events. In fact it's one half of one half. Or 25%.
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>>10195790
Not that guy, but that is exactly how i thought about it.
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>>10195790
>If it's a non-crit on the first hit then there's no way we can achieve a double crit so those events can be discarded.
They can't be discarded since they satisfy the initial condition of the problem.

>Therefore the number of events in which we score a double crit is a subset of events where the first hit is a crit. So it's a subset of only half of all possible events. In fact it's one half of one half. Or 25%.
That's just the chance of getting a double crit without conditions. If you discarded events then what is the point of calculating the probability from all possible events??? The point is that the initial condition restricts the possible events. But you've both restricted the wrong events and then failed to restrict any events. It's baffling to me why people like you try to randomly guess the method to answer a question.
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>>10188804

It's 1/3

4 possibilites
Hit Hit

But because we assume one is a crit, we're working with theee
Hit Crit
Crit Hit
Crit Crit

As you can see,ASSUMING ONE IS A CRIT, you have a 1/3 chance of both being crits (all possibile combinations equally likely)
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>>10196086
This kind of assumes that you can order the hits, what if they strike simultaneously and there is no visual cue, how would you distinguish hit/crit and crit/hit?
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>>10195929
Are we both talking about the same problem here? I'm speaking in regards to the "hardcoded" problem that previous anons were discussing. Which I assumed is what the previous post meant by "[has to be] a crit."

It's easily checked numerically, so I can only assume you're talking about a different problem.
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>>10195480
This is a very good summary.
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>>10195790
You're making two assumptions:
1) You're thinking sequentially, i.e. one hit comes after the other.
2) Plus, you're thinking conditionally, i.e. "IF the first is X" (i.e. conditional probability)

Regarding assumption 1), there's nothing in the premise that excludes two perfectly simultaneous hits
Regarding assumption 2), it's also possible to code the game into randomly selecting one hit per hit pair to be crit, meaning the first hit could be a 50/50 crit (which is a more "pure" probability than your assumption of conditional probability)
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>>10196141
See >>10196175
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>>10195444
>The probability of there being two grits, given that there is at least one crit, is 33%.
What if the one "guaranteed" crit is hardcoded into the game?
Whereby the game randomly selects one hit per hit pair to become crit, meaning the first crit could be a 50/50 crit.
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>>10196175
The hits could be sequential or they could not be sequential. Either is a reasonable assumption given the vagueness of the question. We could program the game to randomly select pairs of ordered hits from a bag, or pairs of unordered hits. We can also program the game to decide the outcome of the first hit by coin flip then determine the outcome of the second hit conditionally -- as was a previous anon's premise. 1/2, 1/3, or 1/4 are all solutions to different problems that still satisfy the initial premise in one way or another.

Again I think everyone is just confused about who is talking about what as the conditional case is specifically what I was referring to as that's what I was talking to another anon about way up many posts before. I also think I misfired on the wrong anon.
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>>10196285
>The hits could be sequential or they could not be sequential.
Not according to the conditional approach.
Because under the conditional approach, you have to first look at one hit to determine the odds of the other hit.

For instance: "if the first hit is crit, then the next hit is 50/50, but if the first hit is not crit, then the next hit has to be 100% crit".

You can't do that with two perfectly simultaneous hits.

The only way to account for both consecutive and simultaneous hits, is to make the game hard select one of either hits as the 100% crit in advance.
In which case the odds of double crit are 50/50.
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>>10196104
it doesn't assume anything like that.

It's just that there are four, equally likely, possible outcomes.

Hit Hit, Hit Crit, Crit Hit, Crit Crit. These are the ONLY THINGS that can happen in the space of 2 possible attacks, and each has a 25 percent chance of happening.

Next, the problem informs us that we know at least one of the hits is a crit, doesn't matter the order. The NEW possibilities we're operating on:

Hit Crit
Crit Hit
Crit Crit

Again, all equally likely. So now we look at Crit Crit and we know it's a 1/3 chance.
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>>10196316
>there are four, equally likely, possible outcomes.
>Hit Hit, Hit Crit, Crit Hit, Crit Crit.
Not if you hardcode one crit into every hit pair.

That not only keeps the "hit hit" combo from ever happening, but also allows for two possible "crit crit" combinations:
1) non-guaranteed crit + guaranteed crit
and
2) "guaranteed crit + non-guaranteed crit
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>>10196326

There's absolutely zero reason to do that though. That's not what the problem asks. Also I'm not sure that matters. By imposing the conditional "You know at least one of them is a crit" you're already guaranteeing a crit.

I think if you do what you're saying, you change the probabilities:

First hit has a 50/50 chance of being a hit or a crit.

If it's a hit, we know 100 percent that hit 2 will be a crit. So Hit-Crit has a 50 percent chance.

If first hit is a Crit, then the second hit can be either a Crit or a hit, with 50/50 chance. Crit-Hit has a 25 percent chance, Crit Crit has a 25 percent chance.

So by doing what YOU said, you actually change the probabilities so they are not equally weighted.

Hit-Crit = 50 percent chance
Crit-Hit = 25 percent chance
Crit-Crit = 25 percent chance
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>>10196347
>First hit has a 50/50 chance of being a hit or a crit.
>If it's a hit, we know 100 percent that hit 2 will be a crit. So Hit-Crit has a 50 percent chance.
Again you're assuming the hits are sequential and cannot be simultaneous.
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>>10189421
WOW
>>
>>10189440
that's actually not even wrong
>>
>normies argue over black and blue dress
>come here see this thread, /sci/ arguing about critical chances
>come back 2 days later
>still arguing, 278 responses
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>>10189541
I-I'm not upset about nobody seriously answering my formulation of the question. Really, I'm n-not.
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>>10188863
bet he doesn't kiss ya
>>
Hit 1: normal hit 2: must be crit. 50% of the time this will happen.

Hit 1: crit hit 2: 50 50 crit or normal. 25% chance of 1 crit

The probability of 1 normal hit and 1 crit is 75% assuming 2 normal hits is not a possibility
>>
Just because there are only 3 possible scenarios crit-hit, hit-crit and crit-crit doesn't mean that they all have an equal chance of happening so the 1/3 is wrong
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>>10197737
In this case, because there is a 50% hit chance, the 1/3 is right
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>>10197526
Mwah!
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Jesus christ I thought this was the smart board.
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>>10188804
>>10189541
1/3 feels about right
Q.E.D
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>>10188804
crit,crit
crit,dead
crit,no crit
no crit, dead
no crit, crit
no crit, no crit (not possible)

1/5
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>>10188804
Its 50% chance
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>>10188808
50% like this guy said. One hit is 100% because it's a crit. The second one is 50% because there's a 50% chance. So 1/1 x 0.5= 0.5
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>>10199020
what is the probability of tossing 2 tails if 1 tail is always guaranteed
OR
what is the probability of tossing 2 tails if 1 coin is rigged to be always tails

are they the same?
>>
>>10188804
1/3
anyone who says otherwise should go back to elementary school
>>
Use Bayes Law. Let A denote the first strike is a crit, let B denote the second strike is a crit. I’ll use and to denote intersection and or to denote union. Then P( A and B|A or B) = P(( A and B) and (A or B))/P(A or B) = P(A and B)/(A or B) = .5
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>>10199067
>first strike is a crit
Try again. The problem doesn’t say this.
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>>10199067
>using Bayes law for simple conditional probabilities
this board is more retarded than I thought
>>
>>10199067
self-thought
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>>10188804
One crit is equal to 50%(1/2)
The probability of two succesive crits is then 50% again, meaning the probabilty is (1/2) * (1/2), which is 1/4, or 25%
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>>10199075
>one of the hits has to be critical
read OPs question more carefully
>>
crit+hit: 0
hit+crit: 0
crit+crit: 1
1 out of 3
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>>10199069
You treat each hit independently because they are independent events. Think flipping a coin
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>>10199072
Just trying to show it using airtight logic, somebody doesn’t write proofs : ^)
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>>10199091
but flip two coins at the same time and 1 is always tails, so it is 1/2
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>>10189541
This actually makes the problem clearer. Answer: 1/3.
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>>10199091
They are not independent events.
For example if the first event is a non-crit, then the next event must be a crit
If the first event is a crit then the next event can be both a crit or a non-crit
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>>10188804
1/3.

There are only three possible outcomes, each are equally likely.

1/0
0/1
1/1
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>>10189373
According to binomial probability the order wouldn’t matter, just look at the formula. It uses combinations not permutations.
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>>10199121
In the problem OP gave us the order does matter, one event changes the possibilities of the next event.
>>
computer decides which hit to fix 1st or 2nd
if it fixes 1st, it is either "crit,hit" or "crit, crit"
if it fixes 2nd, it is either "hit,crit" or "crit, crit"
1/2

computer corrects after player
player hit 1st, computer makes 2nd crit
player crit 1st, computer makes no move, 2nd is either crit or hit
1/3

[spoiler]I can probably bullshit my way to 3/7 but that would be hard[/spoiler]
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>>10199110
Incorrect, they are independent events that have already occurred and we are examining data seeing that one event was success. The we determine the probability that both events are successes given we know that one was a success.
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>>10199163
Nice circular argument.
You did nothing to argue against what I said, and the fact that we are analyzing data has nothing to do with whether the events are independent or dependent.
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>>10199169
Yes it does. Consider a scenario, I flip two coins and keep the results to myself, later I tell you that one of them was heads. Then what is the probability that both were heads?
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>>10199128
The critical strike chance is fixed at .5. Just because the first strike is not a crit doesn’t change the probability the second strike is a crit to 1.
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>>10199199
Oh I see. Yeah that actually makes more sense.
I was given the impression by OP that the events were undetermined.
But given what you said, and the past tense of OP this interpretation makes much more sense
>>
The enemy is dead because the first hit was critical and thenquestion is wrong.



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