/sci/ - Science & Math

>>7317442
How the fuck would it not be 50%?

>>7317454
3 possible outcomes 1 is when they both crit

>>7317454
well there are three outcomes
noncrit+crit*
crit+noncrit*
crit+crit

*distinct outcomes with same total result

1 out of 3 has 2 crits. so 1/3 chance.

should the two distinct outcomes be counted as the same? then it's only two outcomes, which makes 1/2 chance.

>>7317456
The outcomes are Crit-NoCrit and Crit-Crit. One of the hits will ALWAYS crit.

How come /sci/ can pretty much solve everything except simple probability theory questions?

>>7317461
well neither can /int/ >>>/int/42896524

>>7317457
This. Save yourselves the embarrassment of arguing about this. It depends on whether or not you discount distinct outcomes or not, which all depends on your interpretation of the question.

Propability space (all possible outcomes):
Ω={FF, TF, FT, TT}

Event "at least one is critical":
B={TF, FT, TT}
Probability: P(B)=|B|/|Ω|=3/4 (|*| means the number of elements in the set *)

Event "two crits"
A={TT}
P(A)=1/4

Event "at least on crit" AND "two crits"
A intersection B = {TT}
P(A intersection B)=1/4

Probability two crits having known that B has happened
P(A|B)=P(A intersection B)/P(B)=(1/4)/(3/4)=1/3

>>7317458
"One of the hits will always be a crit" and "50% crit chance" are mutually exclusive.

You can either read the problem as a paradox, or accept that we somehow a priori have knowledge about the results and can eliminate any possibility that doesn't include at least one crit once we get to the end.

It's like rounding but with shitty metaphysics.

>>7317470
You are clearly the Dutch guy from the /int/ thread and equally as incorrect.

>>7317458
If you crit first, there is a 50% of critting again. So then we have two outcomes.
If you don't crit first, there's 100% of next being a crit. So we have one outcome from that.

Three outcomes in total, one being the wanted outcome. So 1/3.

>>7317468
Very fun read

>>7317475
Did my accent give me away?

I'm still right, btw.

See >>7317473

>>7317458
If one hits always a crit, then it isn't a flat 50% chance, is it? The question is asking for the chance of two crits given that you get at least one. Using the conditional probability formula, you get 1/3

>>7317442
this one is actually easy to understand: it's 33%

but to this day I still don't understand the Monty Hall Problem http://betterexplained.com/articles/understanding-the-monty-hall-problem/

Can somebody explain it to me?

>>7317458
Well, no.
Consider all four possibilities, ignoring the "one hit is a crit" requirement:

>25% chance of no crit
>50% chance of 1 crit
>25% chance of 2 crits

Do you agree that this is the case? If you don't, try flipping two coins many times, and you will clearly see that one-heads is twice as common as two-heads.

Since 1-crit (the combination of Crit-NoCrit and NoCrit-Crit) is twice as likely a 2-crits, then restricting our probability space to only these by requiring one or more crit gives us:

>66% 1 crit
>33% 2 crits

Therefore, there's a 33% probability that both hits are crits, given that at least one is.

>>7317473
this

>>7317484
What's more likely- that the host opened the non-prize door that he planned to from the beginning of the game, or that he had to pick another door to open because you selected the one he was planning on opening? 2/3 and 1/3, respectively

The host has to know where the prize is, otherwise he would risk opening the wrong door and spoiling the game. That knowledge is partially revealed to you by the fact that he had a 2/3 chance of opening that door anyway, because he knows there's a non-prize behind it.

>>7317482
>I'm still right
Nah, you're still wrong.

>>7317484
instead of using 3 doors think the problem as if it were 100 doors and you realise why if you don't change you are stupid

>>7317484
2/3 chance of picking the wrong door and since he always opens the other wrong door you will get the car 100% if you picked the wrong door first. Since there is a bigger chance you will pick the wrong door first since 2/3 are wrong you should switch.

>>7317496
Prove this infinitely more intelligent and eloquent than me guy wrong, then: >>7317473

>>7317473
In a way, the problem is ill-defined.

>There's a 50% crit chance.
>Assuming at least one of the hits is a crit,
This rewording would make it a lot better.

>>7317442
>At least one of the hits is a crit
>Assuming a 50% crit chance

Does that 50% crit chance apply to that one hit you say is always a crit?
Then 33%.

Does that 50% crit chance not apply to that one hit you say is always a crit, and therefore this one hit doesn't have a 50% crit chance but a 100% crit chance?
Then 50%.

The first option is a much more logical interpretation. I always like to interpret things however the fuck I want because fuck da police, though.

>>7317503
see
>>7317483
>Using the conditional probability formula, you get 1/3
Just try it for your self.

>>7317520
And where does it say in the question you should use the conditional probability formula?

>>7317522
which other formula do you prefer?
protip: there are none.

>>7317503
The thing is, you dont know which hit will be critical hit. So there are 3 probabilities. 1/3 33%
Guess its hard.

Can someone explain what's wrong with this diagram? Especially the 50% people, since they're furthest away from my calculation.

If you perform the first hit, there's a 50% chance to get a crit.
If it is, there's a 50% chance to get another crit, so that's 50%*50%=25%.
It it isn't a crit, then the second hit must be a crit, because we're guaranteed of a crit.

So there's a 25% chance to get a double crit and a 75% chance to get a single crit.

>>7317542
Statistics troll pls

>>7317542
CC
NC
CN
NN
1/3
THE PROBLEM WITH YOUR DIAGRAM IS THAT IT IS WRONG.

>>7317552
>Statistics troll pls
Of course, but I'm unable to debunk it given the premises of the exercise.

Question, are "chance" and "probability" (of something happening) the same?

Ambiguous. It doesn't explicitly state the experimental conditions, and there's no obvious (for some value of that word) interpretation of them based upon what was asked.

The fact that there's even discussion of this should be proof enough.

It doesn't matter the order of the crit-nocrit, it's still only one possibility

If I put my Toshiba laptop to the left of my Asus laptop that doesn't mean I have more laptops if I then put it to the right of my Asus.

One is crit, one is hit

Both are crit

Those are the only two options

>>7317584
how is there no obvious interpretation when
>hey <person who needs to solve this problem>, I observed two hits, with 50% crit and at least one of those hits was a crit
is more obvious than russian troops
giving equally obvious 33% as the answer

>>7317542
Your diagram is supposing that, a priori, there was no possibility that neither hit would be a critical. Said another way, you are assuming that the first roll can affect the second. That assumption is not obvious from the statement of the problem (and because it's dice, it's reasonable to think that a lot of people wouldn't interpret that to be the way that the rules of a game would work).

Your diagram is correct under your set of assumptions, but you have to recognize that that is not the only set of assumptions that is consistent with what is stated in the OP. (So do the people claiming 1/3.)

>>7317604
What were the conditions before the rolls were made? What was the probability of rolling two crits then, before knowing anything about the dice, and does the price of Eskimo snot in Zimbabwe have any effect on that?

>>7317609
Made irrelevant by the fact that it's state that the chance of a crit is 50%

One is definitively a crit, one is up in the air

If one coin has to be heads, what are the odds of the next one, which is not set, being heads or tails?

>>7317611
stated*

>>7317605
>there was no possibility that neither hit would be a critical
Yes. As the problem states: "at least one of the hits is a crit."

Of course what is meant is that the result of a TRIAL has one or two crits, whereas some people (those without a healthy dose of statistics classes) will interpret it as a property of the system.

>>7317611
No, it is not irrelevant. That's the whole point. You are assuming some rules/conditions which are not explicitly stated, and which are not even obviously implied.

If you assume that, prior to the rolls taking place, those rolls will be completely independent, then there is a 75% chance of rolling a crit, and a 25% chance of rolling two crits. Can we agree on that? Good. Note that in this scenario, knowing that one hit was a crit doesn't change a damn thing about the a priori probability.

Now, if you further assume that this question is only asked when at least one of the rolls is a crit, then you can cut it down to 1/3. And that's what you are doing.

However, THAT IS TWO BIG ASSUMPTIONS.

Let's go back to the first one. Let's say that you were guaranteed before rolling that at least one of your rolls would be a hit. Now you are back to the diagram posted >>7317542. That's a valid and reasonable interpretation of the game, wholly different from yours.

But it gets even worse, because we are not even assured that this question was the only one that could be asked. That's your second assumption. What happens when no crits are rolled? Do we then ask
> "At least one of the hits is not a crit. Assuming a 50% crit chance, what is the probability that none are crits?"
Or maybe we ask the first question when the first roll is a crit, and the second is not a crit. That would change the odds again.

The moral of the story is that your experimental conditions have to be established rigorously before examining the data, and your assumptions about them must be clearly stated. None of this is true, so the question is ambiguous.

>>7317485
You can't just ignore that part of the problem.

>>7317669
It's entirely irrelevant.

Name a configuration other than


One is crit, one is hit

Both are crit

There are none, that's it

>>7317442
nice bait, Destiny
>tfw this took up like 2 hours of discussion on stream

>>7317457
There are only two outcomes you put the same thing twice. I'm sure you probably noticed by now.

There is only one variable here and that is the second hit which has a 50% chance you don't need to factor in the other hit making the probability 1/2.

Why does /sci/ find this so hard? if you can't do basic probability you should be ashamed

>>7317542
But you always get one free crit no matter what

there you go, it's 1/3

>>7318204
Seems like studying engineering is useful after all

>>7318225
tbh i knew it would be 1/3 all a long, I'm just practising my terrible programming skills

>>7317442
Co tam na karaczanie kolezko?

>>7317442
(>>7317442
(2C2)(1/2)^2(1/2)^0=1*.25*1=.25

25% chance both or none are hits, 50% chance only one is. Order matters.

>>7318204
Robby please stop this. Very embarrassing to tutor you and have you do this. Please, don't drop out of school.

>>7317442
Let B represent "Both hits are crits"
Let E represent "At least one of the hits is a crit"

Let S be the sample space of two hits, each being regular (R) or crit (C): {RR, RC, CR, CC}

P(E) = |{RC, CR, CC}|/|S| = 3/4
P(B) = |{CC}|/|S| = 1/4
P(~B) = 1 - P(B) = 3/4
P(E|~B) = |{RC, CR}|/|{RR, RC, CR}| = 2/3

P(B|E) = P(E|B)*P(B)/P(E)
= P(E|B)*P(B)/[P(E|B)*P(B) + P(E|~B)*P(~B)]
= (1)*(1/4)/[(1)*(1/4) + (2/3)*(3/4)]
= (1/4)/[(1/4) + (6/12)]
= (1/4)/[(3/12) + (6/12)]
= (1/4)/(9/12)
= 12/36
= 1/3

It's 1/2, you fags are overcomplicating it

There are two states

one is a hit one is a crit

or

both are crit

Order doesn't matter with the former as one has to be crit regardless

>>7317442

One third, assuming you discount entirely double-non-crits.

>>7319547
see
>>7319527


>You flip two coins. One is heads. What are the odds the other is heads?

>>7317557
NN is impossibru since at least one hit will be crit

1/2 confirmed, /thread

>>7319527
It's 1/8 you faggot, your overcomplicating it

There are eight states

one is a hit one is a crit one is a hit one is a crit

or

both are crit both are crit both are crit both are crit

Order doesn't matter with the former as one has to be crit regardless because swinging and missing the first time is a combination of being a retard and messing up. That makes the 8 crit states a fact.

>>7317454
ikr. they either are or they aren't

>>7317473
the question should be; with a 50% crit chance, what is the probability that there will be two crits given that there will be at least one crit

>>7317442
PART 1/2
Okay, we will propably have the exact same thread 2 to 3 days from now and Im just wasting my time, but I will try to answer this anyway.

Imagine you had a friend come over (hard to imagine, I know) and he is playing WoW on your PC while you ... idk sit in a corner with your head facing the wall.

After a while your friend turns around and says:
> Hey Anon, I just hit the training dummy twice
Lets pause here for a moment

there are two possible outcomes to a hit:
a crit (C) and a noncrit (N), both equally likely (50%)
Any given set of 2 hits consequently has 4 possible outcomes

NN
NC
CN
CC

again all equally likely (25%)
The summed up propabilities of all possible outcomes must always be 100% (Because there has to be an outcome 100% of the times)
At this point in time, given only what we know about the crit chance, and the information our friend has just provided us with, we can therefor reason that the propability that he had 2 crits is 25%, but hes not done yet

unpause
> and at least one of them was a crit.
Lets pause again
Our previous calculations were made under the assumption that there are 4 possible outcomes, however given this new piece of data we have to adjust this number a bit.
Lets start with something simpler.
Lets pretend that our friend said something entirly different, like for example
> and both of them were crits
So what is the propability that he had 2 crits ? Well assuming he didnt lie its exactly 100% (duh) it seems painfully obvious, BUT how do you calculate it ?

>>7319725
PART 2/2
Well first of all you look at your original set of possible outcomes (reminder NN ,CN, NC, CC all equally likely, SUM MUST BE 100%) and you modify it to match the new conditions, namely 2 crits
leaving us with only (... ... ... CC, all equally likely, SUM MUST BE 100%).
There is only one possible outcome matching all given criteria and if only one outcome is possible the propability of that outcome is... well 100%
Now we do the exact same steps again, only this time with a different condition

> and at least one of them was a crit
Again we have 4 possible outcomes (NN ,CN, NC, CC all equally likely, SUM MUST BE 100%) and again we sort out the outcomes that match the given condition (at least one crit), but this time we are left with 3 possible outcomes
(..., CN, NC, CC all equally likely, SUM MUST BE 100%). 3 possible outcomes, all equally likely, you dont need a masters degree in mathematics to figure out that thats a chance of 1/3 or 33,333...%.

Alright almost there
We were asked for the propability of him having 2 crits.
take a look at the set of possible outcomes again and see if you can figure out which of them match this condition, then add up their propabilities.
33,33% CN --> Nope
33,33% NC --> Nope
33,33% CC --> yes
-------------------
33,33% or exactly 1/3

there are 3 possible outcomes matching the condition "at least 1 crit", all equally likely (1/3) and within this set of 3 possible outcomes there is only one outcome matching the condition "2 crits", therefore
the propability of him having 2 crits is and will always be exactly 1/3

congratulations Anon !! you have now reached the mathematical understanding of a 5th grader

>>7317442
PART 1/2
Okay, we will probably have the exact same thread 2 to 3 days from now and Im just wasting my time, but I will try to answer this anyway.

Imagine you had a friend come over (hard to imagine, I know) and he is playing WoW on your PC while you ... idk sit in a corner with your head facing the wall.

After a while your friend turns around and says:
> Hey Anon, I just hit the training dummy twice
Lets pause here for a moment

there are two possible outcomes to a hit:
a crit (C) and a noncrit (N), both equally likely (50%)
Any given set of 2 hits consequently has 4 possible outcomes

NN
NC
CN
CC

again all equally likely (25%)
The summed up probabilities of all possible outcomes must always be 100% (Because there has to be an outcome 100% of the times)
At this point in time, given only what we know about the crit chance, and the information our friend has just provided us with, we can therefor reason that the probability that he had 2 crits is 25%, but hes not done yet

unpause
> and at least one of them was a crit.
Lets pause again
Our previous calculations were made under the assumption that there are 4 possible outcomes, however given this new piece of data we have to adjust this number a bit.
Lets start with something simpler.
Lets pretend that our friend said something entirly different, like for example
> and both of them were crits
So what is the probability that he had 2 crits ? Well assuming he didnt lie its exactly 100% (duh) it seems painfully obvious, BUT how do you calculate it ?

>>7319738
PART 2/2
Well first of all you look at your original set of possible outcomes (reminder NN ,CN, NC, CC all equally likely, SUM MUST BE 100%) and you modify it to match the new conditions, namely 2 crits
leaving us with only (... ... ... CC, all equally likely, SUM MUST BE 100%).
There is only one possible outcome matching all given criteria and if only one outcome is possible the probability of that outcome is... well 100%
Now we do the exact same steps again, only this time with a different condition

> and at least one of them was a crit
Again we have 4 possible outcomes (NN ,CN, NC, CC all equally likely, SUM MUST BE 100%) and again we sort out the outcomes that match the given condition (at least one crit), but this time we are left with 3 possible outcomes
(..., CN, NC, CC all equally likely, SUM MUST BE 100%). 3 possible outcomes, all equally likely, you dont need a masters degree in mathematics to figure out that thats a chance of 1/3 or 33,333...%.

Alright almost there
We were asked for the probability of him having 2 crits.
take a look at the set of possible outcomes again and see if you can figure out which of them match this condition, then add up their probabilities.
33,33% CN --> Nope
33,33% NC --> Nope
33,33% CC --> yes
-------------------
33,33% or exactly 1/3

there are 3 possible outcomes matching the condition "at least 1 crit", all equally likely (1/3) and within this set of 3 possible outcomes there is only one outcome matching the condition "2 crits", therefore
the probability of him having 2 crits is and will always be exactly 1/3

Congratulations Anon !! you have now reached the mathematical understanding of a 5th grader

>>7317484
Numberphile has a great video about it.
It basically says what you have already been told.
https://www.youtube.com/watch?v=4Lb-6rxZxx0

>>7317454
>>7317458
>>7317688
>>7317470
>>7318157
>>7319691
>>7317542
>>7319554
>>7317605
>>7317577
>>7317584
>>7317609
>>7317594
>>7317733
>>7318161
>>7318321
>>7319527
>>7319552
/sci/ would be better off if you just left forever

>>7317484
You might be thinking about too much in math terms. The key fact is the MC knows exactly which door the car is hidden, and he's not going to open the door that has the car.

Secondly this puzzle relies less on you choosing the right door and more on you choosing the WRONG door. Which ends up confusing people.
So we have 3 doors
Door 1---> Nothing
Door 2---> Car
Door 3---> Nothing

No matter which door you choose, he's going to open another door. But the door he opens you know for a fact WON'T have the car behind it.

Door 1---> Nothing
Door 2---> Car
Door 3---> Nothing

Let's ignore initial probability and focus on the pathway. Assume you pick a door with nothing behind it. The remaining door combinations are now

Door 1---> Nothing
Door 2---> Car

and


Door 2---> Car
Door 3---> Nothing

The MC can't open door two. If he does the games over. So he opens the door that doesn't have the car. Thus the possible options are such.

Door 2---> Car
or
Door 2---> Car

As we can see, assuming you picked a door with nothing behind it, you are guaranteed a car if you switch.

Now let's go back to the start. To pick your doors, you pick at random. You have a 2/3rds chance to pick an empty door. Which means the above pathways are valid, and you have a 100% percent chance of getting the car if you switch.

>>7318157
They're not indistinguishable bosons. They're swords. Christ.

>>7318204
You have made unwarranted assumptions in setting up your code. Zero marks.

do any of you faggots even bayes theorem
P( 2 heads|atleast 1 heads)=P(atleast 1 heads| 2heads)*P(2 heads)/P(atleast 1 heads)
P(atleast 1 heads| 2heads)=1
P(2 heads)=1/4
P(atleast 1 heads)=1/3
(1/4)/(1/3)=3/4
theres a 75% chance of two heads given atleast 1 head

>>7319843
>do any of you faggots even bayes theorem
>>7317471
>>7318372

>P(atleast 1 heads)=1/3
You mean 3/4. The prior probability of at least one head on two coin flips is 3/4.

So you then get (1/4)/(3/4)=1/3
theres a ~33% chance of two heads given atleast 1 head

>>7319858
oh fug
i fucked up big time
goodbye /sci/ i must an hero

There's a ton of debate about the monty hall problem, and both sides assume the other is retarded.
The 50% crit chance question is a bad example though.

The monty hall problem can be explained thusly:
doors 1, 2, and 3:
x-x-x
Two of these have nothing behind them.
One has your brand new car.
Say you pick door 2.
Monty then tells you door 1 is empty.
Now, you had a two in three chance of picking the wrong door before, so the likelyhood is that you picked a wrong door, and monty identified the other.
The door with the highest probability of having the car behind it is door 3, logically.

>>7319866
Just realized I misspelled likelihood.

>>7319836
You still have to consider the quantum mechanical effects.
https://www.youtube.com/watch?v=0NkdEfvmr5g

>>7317542
>Pictured: A line that says 100%
>Can someone explain what's wrong with this diagram?

Easy: OP specified 50% crit chance so you need to get that shit sorted.

>>7317611
>If one coin has to be heads, what are the odds of the next one, which is not set, being heads or tails?

>the next one

OP didn't say it was the first one that was known to be a crit.

>>7319527
>It's 1/2, you fags are overcomplicating it
>There are two states
>one is a hit one is a crit
>or
>both are crit
>Order doesn't matter with the former as one has to be crit regardless

>It's 1/2, you fags are overcomplicating it
>There are two states
>one is that you win the lottery
>or
>you don't
>I don't understand likelihood ratios

>>7319552

1/2

Now:

>You flip two coins. At least one is heads. What are the odds that both are heads?

1/3

Oh look, if you ask different questions you get different answers WHO COULD HAVE KNOWN.

>>7317442
1/3 is correct, and the question is entirely unambiguous.

>>7320901
>>7320890

You mean 1/2

If one is already known to be heads then there is only one 50/50 variable.

Try it yourself

Sit two quarters on your desk and assign them values 1 and 2. Put one heads up and flip two. What are the odds you're going to have two heads? Put 2 heads up and flip 1. What are the odds you're going to have two heads?

If one state is known and the only remaining state is 50/50 it can never be anything other than 50/50

>>7319742
>Congratulations Anon !! you have now reached the mathematical understanding of a 5th grader

X ~ Binomial(2, 0.5) => n° of crits
P(both | at least one) = P(X = 2 | X >= 1) = P(X = 2)/(1 - P(X = 0)) = (1/4)/(3/4) = 1/3.

>>7321867
I hate you

>>7317442
50%

1 of the hits has a 100% chance of criting
The other has a 50% chance of criting

The combined chance of both of them criting is 1 * .50 or 50%

>>7321914
1/(2^100 - 1)

>>7319738
>>7319742
You keep in mind the first and second hit possibilities (first letter first hit, second letter second hit)
>NN, NC, CN, CC
and have previous knowledge that one of the hits is confirmed C, this leads you to disregard NN based on there's no chance of this happening (0%) something that you know before hand.
So the point is why don't you also disregard NC, keeping in mind that out of the three possible outcomes left (NC, CN, CC) with only the first hit of this outcome (N) you already have 100% of not being CC

>>7321888
I love you (:

Instead let's solve why /sci keeps trying to solve these purposefully ambiguous questions.

>>7322014
Because it's not ambiguous, retards just don't know that "at least one is" is different than "the first is"

>>7317484
Monty hall switches cars with goats and goats with cars. If you pick a goat, your force monty hall to eliminate the only goat left. If you pick a car, then both doors will be goats and switching makes you lose 33% of the time

>>7321822
>If one is already known to be heads then there is only one 50/50 variable.
Knowing that at least one is heads is not the same as knowing that one particular coin is heads.

I think a more reasonable interpretation of the situation would call for flipping two quarters, recording how many times both are heads as well as how many times at least one is a head, and then dividing the former by the latter.

Try it yourself.

>>7322048
I think you can argue room for ambiguity in "at least one of the hits is a crit".

If you flip two coins, look at one of them at random, and see a head, you could describe this as learning that "at least one of the coins is a head". If you flip two coins, ask a trustworthy friend to look at the coins and tell you if at least one is a head, and your friend does so and says "yes", you could also describe that as learning that "at least one of the coins is a head".

But these are not the same pieces of information. With reasonable assumptions in the former case the probability that both coins are heads is 1/2. In the latter case it is 1/3.

>>7322750
No. In both cases, the probability is 1/3. Try it yourself.

It really is just as straightforward as counting the cases, eliminating those that don't fit the assumptions, and then determining the probability from the remaining cases.

>>7317442
Either we know which one is a sure crit or we don't.

If we know which, then it's 50% for the one left (just one independent event)

If we don't, then the table

NC
CN
CC <--- *

with 1/3 becomes relevant.

So if there is anything we should learn about all this fuss, it is that it is incredibly important to be precise when formulating mathematics questions.

>>7322886
>No. In both cases, the probability is 1/3. Try it yourself.
Here's a simple program that models the first case:

https://repl.it/sCg

I can also show it with math:

Let H represent "Both coins are heads"
Let E represent "You looked at a random coin and it was a head"

Let S1 be the sample space of a single coin flip: {H, T}
Let S2 be the sample space of two coin flips: {HH, HT, TH, TT}

# from the definition of probability
P(H) = |{HH}|/|S| = 1/4

# from the definition of probability. Presumably the coin flips are independent and identically distributed, so it doesn't matter which one you pick
P(E) = |{H}|/|S1| = 1/2

# if both coins are heads, looking at a random coin will always find a head
P(E|H) = 1

# bayes rule
P(H|E) = P(E|H)*P(H)/P(E)
= (1)*(1/4)/(1/2)
= 2/4
= 1/2

>>7323066
And here's a program for the second case:

https://repl.it/sCx

And here's math for the second case:

Let H represent "Both coins are heads"
Let E represent "You ask a trustworthy friend if at least one coin is a head and they say 'yes'"

Let S be the sample space of two coin flips: {HH, HT, TH, TT}

# from the definition of probability
P(H) = |{HH}|/|S| = 1/4
P(E) = |HH, HT, TH|/|S| = 3/4

# if both coins are heads, your trustworthy friend will always say "yes" when asked if at least one coin is a head
P(E|H) = 1

# bayes rule
P(H|E) = P(E|H)*P(H)/P(E)
= (1)*(1/4)/(3/4)
= 1/3

>>7317442
>statistical question written in one or two sentences
>on /b/
>Is the question ambiguous?
what the fuck do you think
these kind of shit-tier bait pictures have been posted so many times it's not even remotely funny anymore
>LE PLANE ON LE TREADMILL, WIL IT TAEK OF :'D???
>MUH 2 COINS ONE IZ HEDZ!!!!
>LOLOL AT LEASED ONE HIT IZ CRITZ, WAT DO??
just fuck off already

>>7321822
>Sit two quarters on your desk and assign them values 1 and 2. Put one heads up and flip two. What are the odds you're going to have two heads? Put 2 heads up and flip 1. What are the odds you're going to have two heads?

Now: Flip both of them and if you don't have at least one head, discard the result as out of line with OP's specification.

If there are three equally likely states (HT, TH, HH) it is one third, it can never be anything other than one third.

>>7323626
Why is the chance of HH not 25%?

>>7323382
>>LE PLANE ON LE TREADMILL, WIL IT TAEK OF :'D???
>>MUH 2 COINS ONE IZ HEDZ!!!!
>>LOLOL AT LEASED ONE HIT IZ CRITZ, WAT DO??

Monty Hall, the cupboard and the two coins work because stats are unintuitive, they aren't actually ambiguous. Plane on Treadmill is kind of ambiguous but both readings have the plane taking off once you draw the free body diagram. The only one of the regular troll threads that is actually ambiguous is the one with a+b/c*e where it makes a huge difference whether you read it a+(b/c)*e or a+b/(c*e) and there are stacks of idiots who learned PEMDAS and think it applies to this question.

>>7323636

When all states are equally likely, the odds are 1/N, where N is the number of states.

So if the states are TT, HT, TH and HH, your N is 4 and the odds are, yes, 1/4

When you learn that "at least one is heads," you remove TT and N is now three so the odds are 1/3

To all those saying that is 1/2, I'm afraid to say you're fucking wrong and that you should kill yourselves.
Suppose and ideal situation in which you make the experiment 10 times. Then things will be like this:
25 times two hits
50 times one hit one crit
25 times two crits
"At least one of the hits is a crit" >>100-25=75 is E
>>25 is A
P(A)=25/75=1/3
Order does fucking matter, as you don't know which one was the crit. Get over it. BTW this is a recurrent thread, stop posting this shit 100000 times.

>>7321940
>>7319738 here, sry for the late reply
PART 1/2
I think you slightly misinterpreted the experiment specified in the OP, you will probably notice it if you carefully read it again but ill try to explain it anyway:

The OP describes a classic probability experiment ( similar to the toss of a coin or multiple coins ) and asks for the probability of one certain outcome under a given condition.
In this context "outcome" can be defined as the final state of the experiment AFTER its over.
Notice that in this particular experiment the outcomes NC and CN are identical, because we are only interested in the number of crits, not their order.
We could also use a different notation to make this a lot clearer "0" for no crit (formerly NN), "1" for 1 crit (formerly NC and CN) and "2" for 2 crits (formerly CC), so we would end up with only 3 possible outcomes with
the probabilities
0 = 25%
1 = 50%
2 = 25%
This becomes a lot easier to understand if you imagine another experiment where we have 3 hits and we calculate the probability for 2 crits. Even though NCC, CNC and NCC are 3 different possible outcomes
they are identical in their number of crits.
IMPORTANT !!!1!
The individual steps during the experiment are of no importance to us, nor do we have any knowledge about them, before the experiment is over. We are only working with the theoretical results.

>>7321940
>>7323672
PART 2/2
The experiment described in the OP has 4 possible outcomes (NN, NC, CN, CC) and askes for the probability of the outcome CC under the condition that at least one of the 2 hits was a crit.
When calculating the probabilities for the possible outcomes we can set NN to 0 because NN is not possible under this condition.
If at least one of the hits was a crit we can say with 100% certainty that the final outcome is NOT NN, however we can NOT eliminate NC ( nor CN or CC ) because this outcome is possible under the given condition.
If we were given the condition that the FIRST hit was a crit we could throw NC out the window, but under the current condition, which only specifies the minimum number of crits in the outcome and not their order NC is possible and
happens in 33% of the times.

>>7321940
>>7323672
PART 2/2
The experiment described in the OP has 4 possible outcomes (NN, NC, CN, CC) and asks for the probability of the outcome CC under the condition that at least one of the 2 hits was a crit.
When calculating the probabilities for the possible outcomes we can set NN to 0 because NN is not possible under this condition.
If at least one of the hits was a crit we can say with 100% certainty that the final outcome is NOT NN, however we can NOT eliminate NC ( nor CN or CC ) because this outcome is possible under the given condition.
If we were given the condition that the FIRST hit was a crit we could throw NC out the window, but under the current condition, which only specifies the minimum number of crits in the outcome and not their order NC is possible and
happens in 33% of the times.

>>7323672
>>7323681
But you are disregarding NN based on previous knowledge that it can't be CC since one of the hits is confirmed C, and then proceed to treat NC, CN and CC as if you didn't have this previous knowledge and had to actually run the experiment to know the outcome, when in fact you could previously disregard every outcome but CC, since this one is the only one that meets the requirements of being two crits leaving only 2 options, either it is (50%) or it isn't (50%). If we continue down this logic knowing that we are after a CC we know that at least one of the hits is Crit, this leaves only one variable that we don't know, the other (not that I don't say first or second) hit, and this pending variable has only 2 possible outcomes C (50%) or N (50%).
> we are only interested in the number of crits, not their order.
False, well atleast false for you, since you consider NC and CN as different outcomes when both are the same with the only possible difference is the order of the hits
Now if you consider NC and CN as different based on this order of hits then we have to run the experiment with this in mind from the start, which leads us to recognise only 2 possible outcomes on the first hit C (50%) or N (50%), this possibility of having 2 different outcomes on the first hit can only send us on one path; we use the previous knowledge we have of one hit being confirmed C and evaluate that, since our first hit presents 2 different possible outcomes the next hit presents only one, C.

We only have 2 hits, and whether we know one of them is confirmed C or not makes no difference on the fact that just one of them is a variable with only 2 possible outcomes C (50%) or N (50%)

>>7324057
>(note* that I don't say first or second)*

>>7317442
>ITT autism thread

>>7323066
You have defined P(E) wrongly. You've already flipped two coins so you cannot define your sample space as a single coin flip.

>>7317484
2 goats 1 car. 2/3 you pick goat. you switch. you get car.

>>7324057
PART 1/2
Hey >>7319738 here again, it seems to me that I made the critical mistake of overcomplicating things in my explanations.
I thought that a longer, more detailed explanation including a demonstrative example would make the problem much easier to understand but apparently that didnt work out.
Time for another approach:
You seem quite interested in this topic, maybe you just want to right but ill give you the benefit of the doubt and assume that you are actually interested in finding the right answers to mathematical problems.
If thats the case then I strongly advise you to rethink your argumentation, because it is fundamentally flawed. And trust me (Im speaking from experience here) once you realise that you were wrong -and you will eventually-
you will feel incredibly stupid, it will almost physically hurt

I will try to simplify this problem as much as possible:

Essentially this is a two coin toss experiment, 2 coins, 2 possible values per coin (Heads = H and Tails = T), 50% chance for each value.

>>7324403
PART 2/2
Time for some cointossing: Im flipping coins, you cant see the results so you have to calculate probabilities.
I flip coin #1, and one of two things happens: its either Heads or Tails
I flip coin #2, and once again one of two things happens: its either Heads or Tails
There are 4 possible outcomes to this scenario: (coin#1,coin#2) HH,HT,TH,TT see pic
On the desk in front of me lies now one of these 4 outcomes, all equally likely (25%) and you have no way of knowing which one it is
So ill give you a hint
>at least one of the coins is Heads up
which is just another way of saying
>the outcome is not double Tails
think about that for a second. If at least one coin is Heads up that means the outcome on my desk can not be double Tails.
So now you can eliminate the outcome TT and are left with 3 outcomes that could potentially lie on my desk: HT, TH, and HH ALL EQUALLY LIKELY ( see pic )
You might say that thats only 2 outcomes, because HT and TH are identical and you would be (half) right because there are 2 ways in which you can reach the outcome one tail + one head (see pic) which means the
probabilities for the combined outcome "one Head + one Tail" would have to be the combined probability of the two outcomes HT and TH and would therefor be bigger than the probability of HH.
They would NOT be equal. They would NOT be 50:50
Again:
There is one of 4 possible outcomes on my desk. all equally likely (25% see Pic). we eliminated one (because I just told you its not TT). the remaining three are STILL equally likely, just no longer 25% each.
Because when there are 3 possible outcomes -all equally likely- the probability for each of them must be 1/3.
Now there is one of 3 possible outcomes on my desk: HT,TH or HH and there is no way for you to tell which outcome it is. But can you tell me the probability for double Heads ?

>>7324342
>You've already flipped two coins so you cannot define your sample space as a single coin flip.
I explained this. Presumably the coins in two coin flips are independent and identically distributed. Do you disagree?

If you want to ignore that it's tedious but simple to work out the same thing like this:

Here is the sample space for the two coin flips: {HH, HT, TH, TT}

# each outcome in the two flips sample space has a 1/4 probability. In the first outcome a randomly selected coin is always heads. In the second and third outcome a randomly selected coin has heads half of the time. In the fourth outcome a randomly selected coin never has heads
P(E) = (1/4)*(1) + (1/4)*(1/2) + (1/4)*(1/2) + (1/4)*(0)
= 1/4 + 1/8 + 1/8
= 2/8 + 1/8 + 1/8
= 4/8
= 1/2

Do you disagree with any of that?

>>7323669
>>7322750

As I argued here, there is some ambiguity in "at least one of the hits is a crit". In real life the manner in which you learn that "at least one of the hits is a crit" can matter and lead to you having different information.

I think most people who say 1/2 aren't saying it for good reasons though.

>>7324534

That's not the chance of 2 heads, that's the chance that a randomly selected coin from the sample space is heads. I cannot even comprehend why you think those would be the same chance.

>>7317442

Obviously this is an attack function for a game and must be considered as it might be implemented.

Here is my sudo code.

int crits (int &attk_1, int &attk_2){

int attk_1 = quantam_rand (0,1);
int attk_2 = quantam_rand (0,1);

/****************************************************************************
This will produce the following truth table each being equally likely
0 0
0 1
1 0
1 1
*****************************************************************************/

if (attk_1 + attk_2 == 0) {
if (quantam_rand (0,1)) attk_1 = 1;
else attk_2 = 1;
// in the event that both attacks fail to hit a critical strike assign a critical strike to one of them randomly.
}
}

The answer is 25% chance both hits are crits. If you dispute this then code up an alternate function.

>>7324720
>That's not the chance of 2 heads, that's the chance that a randomly selected coin from the sample space is heads. I cannot even comprehend why you think those would be the same chance.

Ah. I think the confusion here is because you misread >>7323066 . "P(E)" is *supposed* to be the probability that a randomly selected coin is heads. I defined "E" as the observation of looking at a random coin from the two and seeing a head. The prior probability of getting two heads is "P(H)". They are not the same chance, one is 1/4, the other is 1/2.

>>7323066
>Let H represent "Both coins are heads"
>Let E represent "You looked at a random coin and it was a head"
>P(H) = |{HH}|/|S| = 1/4
>P(E) = |{H}|/|S1| = 1/2

# Actually, the formula for P(H) should read "P(H) = |{HH}|/|S2| = 1/4"

>>7324749
I dispute that.

I coded two interpretations (using coins rather than crits) here: >>7323066 >>7323074 .

The issue with yours is the part where if you get no crits you randomly assign a crit to one of the hits. This is favouring outcomes with only one crit above outcomes with two crits without justification. Instead you need to randomly assign a crit to one hit, a crit to the other hit, or a crit to both.

Your answer will then be 1/3.

>>7324749
Nope. Terrible implementation, it does not give 50% crit chance as OP specifies. It's not part of the attack function that you get at least one crit, that's just an observed output of one double-attack.

>>7324945
>I think the confusion here is because you misread >>7323066

It is, I missed the previous post where you stated it was for; I thought you were still on OP's picture rather than the "look at one coin yourself and observe heads" thing you were actually doing - I just saw you write 1/2 and was like "WELL well weLL, SOMEONE is wrong on the internet today!"

>>7325093
No worries. Glad it's cleared up.

>>7317442
It really oughta be 1/4. Given these possible outcomes:

miss - crit (edited by system to avoid the miss-miss outcome)
miss - crit (a fair 50% second crit rolled)
crit - crit
crit - miss

>>7323638
>Monty Hall
monty hall threads are almost always reduced down to a pointless discussion about what kind of ruleset is actually implied

>>7325154
>miss - crit (edited by system to avoid the miss-miss outcome)

The OP does not say that the system edits the second hit to be a crit if both were going to be non-crits, it just says that there is a 50% crit chance. Nothing stopped a non-crit non-crit outcome, it just happened not to be that case.

>>7317442
it's 1/3.

you know that at least one of the hits is a crit but you don't know which one.

the possibilities of two hits:
>both crit
>first hit crits, second hits doesn't crit
>fist hit doesn't crit, second hit crits
>none crit (excluded by the premise of the question)

>Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

>Any time Sleeping Beauty is awakened and interviewed, she is asked, "What is your belief now for the proposition that the coin landed heads?"

should she say it's 1/2? it's a fair coin flip, so the probability of heads is 1/2, just that she gets asked twice if it's tails (BUT SHE HAS NO MEMORY)

or should she say it's 1/3, because out of the times she gets woken up, it's heads 1/3 of the time?

>>7325242
That follows the most intuitive interpretation to me, but as I've argued above there is some ambiguity in "at least one of the hits is a crit". In real life the manner in which you learn that "at least one of the hits is a crit" can matter and lead to you having different information.

>>7325263
>In real life
all you know from the premise of the question is that you hit twice (with 50% crit chance) and crit at least once. it's incorrect to assume that that you have any information about whether it was the first hit or the second hit that critted.

>>7325154
>miss - crit (edited by system to avoid the miss-miss outcome)
>miss - crit (a fair 50% second crit rolled)
>crit - crit
>crit - miss


>8 attacks
>5 crits

OP said 50%, not 62.5%

>>7325253

1/2. "I am currently awake and answering a question" gives you no additional information because you knew that would happen anyway.

>>7325253
1/2. Waking up is something that will happen with probability 1 if the coin landed heads and with probability 1 if the coin landed tails, which means it provides no evidence on the matter. Thus, from the fact that you have been awoken, you should not shift initial probabilities regarding the coin in any way.

>>7325307

I should elaborate:

Your prior is (obviously, I'd say) .5

Your posterior, adjusted for new evidence, is also .5, because you have no new evidence - your current experience is identical for both outcomes, so your current experience is not evidence.

There's a different one that's quite a bit weirder.


1: You know these rules ahead of time
2: You go to sleep.
3: You get perfectly cloned three times (There are four of you)
4: Experimenter flips a coin.
4A: If heads, he puts one of you in a red chamber and three of you in green chamber
4B: If tails, he puts one of you in a green chamber and three of you in a red chamber
5: You wake up in a green chamber. What are the odds of the coin being heads?

>>7325323
3/4. This yields a sum log score of 3*lg(3/4) + 1*lg(1/4) or about -3.24, versus a sum log score of 4 for an estimate of 1/2.

>>7325329
A log score of -4 for 1/2, of course.

>>7325301
>OP said 50%, not 62.5%
So... to make a 50% crit system that conform with all of Op's rules (never miss-miss) we need to lower the chance for rolling a crit bellow 50% to compensate for the auto-crit that sometime must occur after a miss.

How do one go about calculating what that number is?

>>7325307
>>7325308
>>7325316
i (currently) think it's 1/2 but these fags say it's 1/3

http://math.stackexchange.com/questions/487350/sleeping-beauty-paradox-fair-prior

>>7325352
>So... to make a 50% crit system that conform with all of Op's rules (never miss-miss) we need to lower the chance for rolling a crit bellow 50% to compensate for the auto-crit that sometime must occur after a miss.

OP never said you couldn't miss. He just said that in these two specific attacks that you just did, you got at least one crit.

OP's system is super simple: Flip coin. If heads, hit, if tails, crit. Done.

>>7325355
write a program, do 10000 rolls and check the result

>>7317594
One of your laptops is broken.
>Doesn't matter which one hurr durrr
Fuck this I'm out.

>>7325355

Ugh. It might be 1/3. I'm trying to set up an Expected Utility calculation for each value where SB doesn't guess what the odds are (because fuck that) but is rewarded for guessing correctly.

I'm thinking $2 if she wakes and guesses "heads" and $1.5 if she wakes and guesses "tails."

If it's 50%, she should guess heads because her expected utility from saying heads is $1 and from tails $0.75.

If it's 33%, she should guess tails because her expected utility from saying heads is $0.66 and $1 from tails.

So. She believes it is 50% and says "heads" every time she wakes. After three runs, she has $2

So. She believes it is 33% and says "tails" every time she wakes. After three runs, she has $3.

Huh. Might be 33%.

>>7325402
>but is rewarded for guessing correctly.
It's also rewarded for getting to guess at all. What happens if you count the average score instead of the sum?

>>7325402
from the perspective of betting on heads or tails, the 1/3 answer makes sense, but only because she gets to bet twice if tails comes up from the one coin toss.

but she doesn't bet "heads or tails", she is only supposed to say the probability that heads came up. from the perspective of the people conducting the experiment, the 1/2 answer makes sense because there is only one coin toss. imagine if she gets awakened 999 times instead of 2 times if it lands on tails. from the perspective of the people who conduct the experiment, the probability of heads is 1/2, and they would kek if she said 1/1000 just because they ask her 999 times if it's tails.

>>7325443
...but fuck me, now i'm yet again thinking that it might be 1/3 after all. she knows that she got awakened and questioned. this happens more often when it's tails. so 1/3 might still be the probability based on her knowledge (that she got awakened and questioned).

>>7325447
...but the fact that she got awakened and questioned doesn't affect the probability of the actual coin flip

i don't even

>>7325447
>she knows that she got awakened and questioned. this happens more often when it's tails.

No, that doesn't work - because of the amnesiac, she knows she was interviewed at least once, but that occurs in both scenarios and is equal evidence towards both and does not change her prior one whit.

>>7324403
>>7324404
Wrong, look at this graphic, there's only 2 possible outcomes, HT and TH are in fact the same, you have 2 coins
1. Heads only
2. Heads and tails
The only variable is the one from heads and tail coin, and as you can see it only has 2 outcomes leaving a 50/50 chance

Try to find a flaw in the graphic, you can't the one with a flaw is yours, your flaw is that you are not keeping in mind the fixed coin parameter
2 flips
2 coins
1 fixed - Heads only
1 random - Heads/Tails
Once you flip one coin you can't flip that same coin again, I actually prefer the coin scenario because it allows you to flip the two coins at the same time. Not that it would make any difference because, there's always 1 fixed coin and 1 with 2 possible outcomes heads 50% or tails 50%

Before replying look at the graphic again and try to fly a flaw in it.

The answer is 3/4. Find the probability that none of the hits are critical and take the complement. What is the probably of getting no critical hits?

1/2(1/2) = 1/4

What is the complement?

1-1/4 = 3/4

Another way to think about it? There are four possibilities. No crits, the first is a crit, the second is a crit or both are crits. In 3 of the four scenarios, you have at least one critical hit.

>>7325501
woops, im tired and completely misread the question. The answer is 1/3. You can extrapolate that from my previous explanation

>>7325481

>OP says the crit chance is 50%
>I bet he meant one of the chances was 100%

>>>/out/

>>7317442
It's actually 2/3

The answer is 1/3, the question is not ambiguous.
Here's the reasoning:
The sample space is: {NN,CN,NC,CC}
where N is the event of not landing a crit, and C is the event of landing a crit.
If you landed at least a crit, it means that the possible events we are taking into consideration are {NC,CN,CC}, hence, assuming we are talking about an uniform sample space, the probabilith of landing two crits is 1/3.

>>7325507
>OP says we get 2 hits
Hit:
Hit:
>OP says one if the hits is a crit
Hit: Crit
Hit: -
>OP says the crit chance is 50%
Hit: Crit
Hit: Not crit
>or
Hit: Crit (100%)
Hit: Crit/not crit (50%)

In any scenario you only have 2 options leaving a 50/50 chance, think the two hits at the same time, since OP doesn't not specify they happen at different order, different order is a variale you decide to implement on your own

>>7325529

>>OP says one if the hits is a crit
>Hit: Crit
>Hit: -

Or

Hit: -
HIt: Crit

There are two ways to get At Least One Crit.

>>7325535
OP does not specify hits are in different order, one if them is hit the other isn't

>>7325507
>>OP says the crit chance is 50%
>I say it's 33%

>>>/comejoinusyou'llfitrightin/

>>7325542
that should be this, sorry

>>7325542
>You will immediately cease and not continue to access the site if you are under the age of 18.

>>7325546
>you will inmediatly proceed to post irrelevant replies if you run out of arguments

>>7325542

Nope, I generate samples with a 50% chance, then discard any that do not live up to "at least one crit."

That gives me 33% chance of Crit+Crit.

You generate samples with a 100% chance, then a 50% chance. That gives you 50%, sure, but it's not what OP specified so >>>/out/ I say to you.

>>7325549

He didn't "run out of arguments," because he isn't me and I'm the one who's been shoving your shit in these last ten posts.

>>7325552
>>7325557
if you do not interpret "at least one of the hits is a crit" as
>Hit: Crit
>Hit: Crit/Not Crit
and proceed to interpretate as
>Hit: Crit/Not Crit
>Hit: Crit/Not Crit
Which then "at least onf the hits is a crit" and "assuming a 50% crit chance" is the same
and you proceed to imput "order" as another parameter then you cannot disregard the NN previously because it is a valid result of the test, which then gives you a probability of 25% of CC,

in any case it's not 33%

>>7317442
Fuck I hate /sci/ sometimes. This question is just the reincarnation of the coin flipping question that used to be posted on here. It gets so many posts only because people have no idea how to properly answer it.

>>7325572
Also let me add

By disregarding NN on the argument that "at least of the hits is crit", considering is not the same as "assuming a 50% crit chance", you are automatically validating one of the hits is 100% crit proving myself right

>>7325572
>>7325613
I'd love to see your logic if the question changed its
>at least one critical hit
to
>at least two critical hits

>>7325572
you have TWO hits. there are FOUR different scenarios, all with equal probability:
>FIRST: crit, SECOND: crit
>FIRST: crit, SECOND: hit
>FIRST: hit, SECOND: crit
>FIRST: hit, SECOND: hit

we KNOW that AT LEAST one hit was a crit, so that eliminates the fourth scenario. therefore there are THREE scenarios remaining:
>FIRST: crit, SECOND: crit
>FIRST: crit, SECOND: hit
>FIRST: hit, SECOND: crit

the probability that both hits are crits is ONE OF THE THREE remaining scenarios.

>>7325572
>>7325643
yeah so are you going to say that
>at least two critical hits
has the same probability as
>at least one critical hit
or what are you going to come up with now you dim-witted teenager

>>7325613
And I'm going to add this last thing

The reason
>at least one of the hits is a crit
is not the same as
>assuming a 50% crit chance
after we've been told be get 2 hits, is because
>at least one if the hits is a crit
makes a direct reference to the final result of that hit, not the probability of hit getting that result, in which case the statement would say
>at least one of the "outcomes" is crit
and this would be the same as saying
>assuming a 50% crit chance

>>7325643
let's see you want to replace
>at least one of the hits is a crit
with
>at least two of the hits are crits

this would mean the OP is defining the end value as CC, it would be like asking you the probabily of getting CC on a fixed test that can only have CC as final result

>>7325655
Nope. That isn't what I asked. I asked you to carry the logic of the question but change the at least one crit hit to two. Those are two separate questions.

>>7325651
This guy know what he's talking about.

>>7325658
>this would mean the OP is defining the end value as CC, it would be like asking you the probabily of getting CC on a fixed test that can only have CC as final result
Exactly. You were able to get this answer because you discard the other possibilities since they are less that two crits. That makes >>7325572 and >>7325613 garbage because, for some reason, we aren't allowed to discard possibilities in OP's original question. In other words, in the two crits problem we did

NN,CN,NC,CC
at least 2 critical
CC
prob for both hits being critical is 1

For OP's question, it turns into this

NN,CN,NC,CC
at least 1 critical
CN,NC,CC
prob for both hits being critical is 1/3.

>>7325287
>all you know from the premise of the question is that you hit twice (with 50% crit chance) and crit at least once. it's incorrect to assume that that you have any information about whether it was the first hit or the second hit that critted.
That's consistent with a scenario where, say, a trustworthy friend tells you that they saw one of your two hits and it was a crit. You have no information about which hit was a crit, but with reasonable assumptions the chance that you crit twice in this scenario is 1/2.

>>7325518
What do you think of the ambiguity described here?: >>7322750

>>7325658
>>7325672
I was going to write a long post but frankly I'm a little tired so I'll keep it short
using order of hits and getting 4 results
>NN,NC,CN,CC
The only way you can get 4 outcomes if you diregard:
that one round is fixed and no matter the order if the rounds it only has one way , this way (look the graphic >>7325481)
You are giving every hit 50% of being C or NC so NN is a possible outcome, that you proceed to disregard AFTER you runned the test, so you actually get the resut and this inflict in the real probabilty but you decide to ignore it so that you get 3 results, 3 results that are THE SAME other than the fact of its order, the same ORDER that FORCES you to get 4 RESULTS and you arbitrality choose to ignore one of them

>>7325714
this leaves of course a real 25% of CC

>>7325714
>2 results that are THE SAME other than the fact of its order*

>>7325688
the chance in that scenario is still 1/3 even if it seems unintuitive. all the friend knows is that one of these happened, and it could be any one of them with equal probability:
>FIRST: crit, SECOND: crit
>FIRST: crit, SECOND: hit
>FIRST: hit, SECOND: crit

>>7325714
>The only way you can get 4 outcomes if you diregard
I'm not disregarding anything. If you allow the player to strike the target twice then all possible outcomes are

NN,NC,CN,CC

There is nothing special going on here anon.

>>7325752
this gives you 25% not 33% you can't disregard NN and if you do, accepting one hit has 100% of being crit, this forced 100% succes chance makes order irrelevant since you can't get more than one result in each hit/round

of 2 variables that had 2 possible outcomes each (4 in total, 25% of CC), you killed one leaving only 1 variable left, with only 2 possible outcomes (2 in total, 50% of CC)

>>7325728
>the chance in that scenario is still 1/3 even if it seems unintuitive.
It's not. The math and a program demonstrating the result are here (using a scenario with coin flips rather than crits): >>7323066 .

>all the friend knows is that one of these happened, and it could be any one of them with equal probability

That's not correct. The friend knows that one of those happened, but they are not equally probable. FIRST: crit, SECOND: crit is twice as likely as the other two possibilities.

In the other two cases the likelihood of the friend who sees one of the two hits seeing a crit is 1/2. Half the time he would see the hit that is a crit and half the time he would see the hit that is not a crit. In the FIRST: crit SECOND: crit case the likelihood of the friend seeing a crit is 1 because both hits are a crit.

G denotes guaranteed crit. Numbers at bottom represent varying paths.

>>7317442
Can we stop having these threads? The answer is 1/3, because we know that the hits occur at different times. If we didn't know, the question would be ambiguous. See: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

Now please, there's been like ten threads on this. Stop it.

/sage

>>7325779
You clearly didn't read my original post here >>7325672

There are 4 total ways to strike the target but only 3 can be used towards calculating OP's probability problem because of the way the problem is phrased. I am not saying 1/4 is right at all.

>>7325481
>>7319738 Here
PART 1/4
>Wrong, look at this graphic, there's only 2 possible outcomes, HT and TH are in fact the same, you have 2 coins
>1. Heads only
>2. Heads and tails
>The only variable is the one from heads and tail coin
>...
>2 flips
>2 coins
>1 fixed - Heads only
>1 random - Heads/Tails


nope, that is not correct. actually it is wrong on two differnt levels:
1) there are 3 possible outcomes(4-1 we previously eliminated). google "theory of probability" and I guarantee you that every single academic article on this topic will prove me right on this one, most likely on the first page.
http://www.pas.rochester.edu/~stte/phy104-F00/notes-2.html

>>7325481
Suppose you follow this procedure:

1. Draw two columns on a piece of paper
2. Flip two coins
3. Look at both results
4. If you see one head and one tail or two heads, put a mark in the first column. If you see two tails, mark nothing
5. If you see two heads, put a mark in the second column
6. Repeat steps 2-5 a large number of times then divide the number of marks in the second column by the number of marks in the first column

Do you agree that this is a good model of the scenario we are discussing?
Do you think you would get a result close to 1/2 rather than a result close to 1/3?

>>7325815
PART 2/4
Your post cleary shows that you dont understand the difference between a "possible outcome" and an "event" (not sure if that is the proper english term). Possible outcomes are the well defined, mutually exclusive,
final states of our experiment. If I roll a dice there are 6 possible outcomes, no more no less. if I roll 2 dice there are 6x6 possible outcomes. 3 dice 6x6x6 possible outcomes and so on.

An Event is a set of [0...infinity) possible outcomes and it can literally be anything you can think of. Going back to our 3 dice example on such event could be "only even numbers".
Notice that there are multiple possible outcomes that realize this event (222, 224, 226, 242, 244, 246, 266, 422, 424, 426, 442, 444, 446, 466, 622, 624, 626, 642, 644, 646, 662, 664, 666)
This gets even clearer when you think about the following example:
Game show, 3 doors , behind on is a prize, behind two is a pic of Moot dressed as a little girl, you have to pick one and because you dont know whats behind the doors you have to pick one randomly.
There are 3 possible outcomes, you choose door 1, you choose door 2, or you choose door 3 however we are not interested which door you choose, only in the events they realize.
Event 1: you go home with your prize (realized by one door 1/3 chance), or
Event 2: you get a moot pic (realized by the 2 remaining doors 2/3 chance)

25%
both events are separate from each other

>>7325819
PART 3/4

Lets go back to the cointossing experiment. there are 3 possible outcomes TH HT HH, but (in this particular case) only two relevant events, namely
Event 1: one head + one tail (realized by TH and HT, 2/3 chance)
Event 2: both head (realized by HH 1/3 chance)


You can come up with as many events as you like, the possibilities are literally endless for example the event (6 heads, 19 tails) is realized by exactly 0 possible outcomes (because we are only flipping two coins) but and
that is the important part the number of possible outcomes is fixed and (in this particular case) defined by the number of coins we flip minus the number of possible outcomes we can eliminate
When you flip two coins the event "one head + one tail" is twice as likely as the event double tails and it is also twice as likely as the event double heads because one head + one tail is realized by
2 possible outcomes (HT and TH) while the other two events are realized by only one possible outcome (remember all possible outcomes are equally likely)

>>7325792
that program is incorrect because it only checks one of the coins and ignores the fact that if the one that it checked was a head the second coin could still be heads.

>>7325823
PART 4/4
2)
Your post (including the pic) imply that you assume one coin is fixed, meaning 100% head. Once again: you misinterpreted the OP. "at least one of the hits is a crit" translates to "at least on of the coins lands head up"
however that gives no information whatsoever about the state of any individual coin. Both coins can still be either H or T, (that is the reason why we still have the two possible outcomes TH and HT) the only information
we get here is that the outcome double tails is impossible.
We have no knowledge about the state of one specific coin, only about the state of both coins combined (that is what we call the possible outcome).

Taking this statement and iterpreting it as "One coin is H, we remove it from the experiment, which only leaves us with one unknown coin" is just plain wrong. it defies logic.
You are calculating the probability of A ( coin #2 = H ) under the assumption that B (coin #1 = H), something called conditional probability.
however we dont have a conditional probability here. because we have no information about coin #1 or #2.
One last time: the given premise is not a statement about one specific coin, but about the combination of both (possible outcome)

Also FUCK 4chans comment length restrictions

>>7325818
How about this procedure, which is more accurate to the scenario:

1. Draw two columns
2. Flip a coin
3. If it turns out heads, flip another coin, and put a check in the first column.
4. If the second coin is heads, put a check in the second column.
5. If the first coin isn't heads, flip the second coin.
6. If the second coin turns out heads, put a check in the first column.
7. Repeat as many times as you want.

Divide column two by column one.

The answer is 1/3.

>>7325827
I don't understand what you're trying to get at. The program only checks one of the coins because that is what happens in the scenario we are talking about, with you checking one of the coins at random and seeing a head (or your friend telling you they saw one of your hits and saw a crit).

>>7325836
ok maybe the probability is 1/2 in the case of where the friend only looks at one at random. but in OP's question, you KNOW that at least one critted so in OP's question it's unambiguously 1/3.

1/3

>>7325843
My issue is that it seems reasonable to describe the friend looking at one at random scenario as one where you know that "at least one critted". After all, you do in fact know that at least one of the two attacks was a crit. Your friend saw one of them and told you so.

But you could also describe a scenario where your trustworthy friend tells you he saw both hits and at least one was a crit as one where you know that "at least one critted". Again, you do in fact know that at least one of the two attacks was a crit. Your friend saw both of them and told you so.

But you have different information in these two scenarios, and the chance that you crit twice is different in each.

So it seems that there is some ambiguity in the phrase "at least one of the hits is a crit". It can describe two subtly different real life situations which impact the answer to the question.

>>7325830
I don't think that procedure is more accurate to the scenario, but yes, that one would also yield results close to 1/3 rather than 1/2.

>>7325860
>that one would also yield results close to 1/3 rather than 1/2
Assuming you did it a large number of times, that is :)

Made a simulation in java.

It goes to 1/3.

is this right?

>>7325857
good point, but for the purpose of the question it should be assumed that the questioner has perfect knowledge of the system unless it specifies otherwise. if the questioner meant that you only have information about one of the hits then it should have been specified in the question, but here it implies that the information given ("at least one of the hits is a crit") applies to both hits, i.e. it asks for P(not both non-crit).

>>7325869
Almost. Some rows would have a hit in the second but not the first column. The hit that crits could be the second hit.

This is just a simple conditional probability problem. Pic related. Read up on it anons.

>>7325872
>it asks for P(not both non-crit)
i mean it asks for P(both crit) within the probability space of (not both non-crit)

>>7325873
What difference does that make? each try, no matter first or second hit, has a crit

Same shit as the gay coin problem.

Force the same one to be heads every time:
1/2

Any one could be heads;
1/3

Here's my shitty code for it if anyone cares, some might find it more readable than the above guy's code

>>7325878
the difference between 1/2 and 1/3 which is exactly 1/6

>>7325878
CRIT | hit
hit | CRIT
hit | CRIT
hit | CRIT
CRIT | CRIT
CRIT | CRIT
CRIT | CRIT
CRIT | hit
hit | CRIT
hit | CRIT
hit | CRIT
CRIT | hit
CRIT | CRIT
CRIT | hit
CRIT | CRIT
CRIT | CRIT
hit | CRIT
hit | CRIT
CRIT | CRIT
hit | CRIT
CRIT | CRIT
hit | CRIT
CRIT | hit
hit | CRIT
CRIT | CRIT
CRIT | hit
CRIT | hit
hit | CRIT
CRIT | CRIT
hit | CRIT

P(CRIT | CRIT) = 0.3333333333333333

>>7325885
are telling me you a see a difference between these two scenarios?

>>7325897
No, those are the same.

>>7325872
Approaching the question the way you're saying feels more reasonable to me, but I would say there is at least a little ambiguity. If someone gave 1/2 as the answer and justified it by appealing to a scenario like one of the one's I've described I wouldn't say they are wrong, they've just given the correct answer to a different interpretation of the question.

>>7325897
those scenarios aren't equivalent to the premise of the question. see >>7325889 for what it could more realistically look like

>>7325878
>What difference does that make? each try, no matter first or second hit, has a crit

There's a difference between "At least one was a crit" and "the first one was a crit."

In particular, consider the amount of information you get:

If your friend says "At least one was a crit" then... well, duh, the odds of getting at least one crit on two hits are 75% so we already considered it probable that we'd gotten a crit.

If your friend says "The first one was a crit" then... That's information. That was only a 50% chance so we did not consider that probable to start with.

To bring it back to OP's question: OP tells you "At least one was a crit." If you proceed to mark your column as if you know the FIRST one was a crit, you are acting as if you have more information than you actually have.

>>7325902
in a formal setting i wouldn't say they are correct. "at least one of the hits is a crit" is a definite statement that implies godlike knowledge. there is no information in the question that justifies a belief in that the godlike questioner would only have information about one of the two hits.

>>7325915

Do you believe the sentences "One of the hits was a crit" and "At least one of the hits was a crit" convey the same amount of information?

>>7325918
>Do you believe the sentences "One of the hits was a crit" and "At least one of the hits was a crit" convey the same amount of information?

Wait disregard that, I suck cocks am an engineer and also misread posts on the internet.

>>7325918
that's open to interpretation and is just arguing semantics. but in the "friend scenario" you're implying that the questioner would discard a (hit, crit) case if the questioner would only look at the first hit for whatever reason. that doesn't make sense because the question doesn't say anything about that the questioner would himself would only look at one hit. the questioner has to include both (crit, hit) and (hit, crit) along with the (crit, crit) unless it is specified that you only look at one of the hits and see that it was a crit.

>>7325901
so now each empty (blue) space, how many chances of being Crit does it have?

>>7325909
Are you fucking me, the columns have no number, you could literally put a first or second on each try and it would be the same shit, it doesn't affect the 50% probability of the empty spaces

I took the liberty of making my own program. It should be a bit easier to understand, no complicated logic.

>>7325940
>>7325941
I did this for you in particular.

>>7325941
Just to further explain, the overlap between the two lists is due to the fact that a round where both hits are critical is counted in both list A and list B. Line 38 eliminates these double counting scenarios.

>>7325951
Here's one that's a bit simpler.
|| means OR
&& means AND
for those of you who don't know Java.

>>7325941
>>7325946
>>7325951
are you seriously fucking telling me in this two scenarios you see a difference?

>>7325962
Did you read the program?

>>7325980
please answer the fucking question, do you, in those 2 scenarios see a fucking difference?

>>7325982
No. Both of those scenarios are the same. I see absolutely no difference.

What I'm trying to say is that that procedure doesn't model the question. Here's what you are doing: (pic)

>>7325995
The reason it's wrong is that it completely leaves out the scenario where the first shot doesn't crit and the second shot does.

Or you can think of it like this: firstShotCrit is actually "One of the shots crits" and secondShotCrit is actually "The other shot crits". This is the only way you can get a 50% probability. You have to leave out part of the scenario.

>>7326007
>The reason it's wrong is that it completely leaves out the scenario where the first shot doesn't crit and the second shot does.
>first shot doesn't crit and the second shot does.

>>7325940
>Are you fucking me, the columns have no number, you could literally put a first or second on each try and it would be the same shit, it doesn't affect the 50% probability of the empty spaces

No, but it affects the 100% probability of the spaces you've filled in.

Before your friend says anything:

You expect this at 50%: "The first one was a crit."
You expect this at 50%: "The second was a crit."
You expect this at 75%: "At least one was a crit."

So if you tell me that the first one was a crit, or that the second one was a crit (changing it to 100%) you have given me 50% worth of information. If you tell me that at least one was a crit (changing THAT to 100%), you have only given me 25% worth of information.

So if you get told "At least one" and then proceed to do the math as if you know which one, of course you get a wrong answer.

>>7326007
It's okay being wrong. Read this: https://en.wikipedia.org/wiki/Marilyn_vos_Savant#.22Two_boys.22_problem

>>7326017
The programs don't lie. They use raw math and logic. If you find an error in my programs, please, tell me what I've done wrong.
>>7326019

>>7325962
>are you seriously fucking telling me in this two scenarios you see a difference?

They're the same scenario mirrored, and different from the scenario of "at least one is a crit."

>>7323066
>bayes rule
>P(A|B) = P(A∩B)/P(B)
P(A) = 1/4 (chance we crit twice)
P(B) = 3/4 (chance we crit at least once)
P(A∩B) = 1/4 (chance we crit at least once and crit twice at the same time)
P(A∩B)/P(B) = (1/4) / (3/4) = 1/3

>>7326040

Re-read the post chain - he's not solving OP's image, 1/2 is the correct answer to the problem he's actually solving.

>>7326051
oups, my bad

<span class="math">P(A\mid B) = \frac{P(A\cap B)}{P(B)}
\newline\newline
P(Both\; Crit\mid At\; least\; one) = \frac{P(Both\; Crit\cap At\;least\;one)}{P(At\;least\;one)}
\newline\newline
P(At\;least\;one) = 1 - P(None\; crit)
\newline\newline
P(None\; crit) = 1/4
\newline\newline
P(At\;least\;one) = 1-1/4 = 3/4
\newline\newline
P(Both\;Crit)= (1/2)^2 = 1/4\;\;\;\;\leftarrow
\newline\newline
P(Both\;Crit\cap At\;Least\;One) = 1/4
\newline\newline
P(A\mid B)=P(A\;Given\;B) = \frac{1/4}{3/4} = 1/3\;\;\;\;\leftarrow[/spoiler]

The first arrow is what everyone keeps getting wrong.

>>7326067
Crap
<span class="math">P(A\mid B) = \frac{P(A\cap B)}{P(B)}
[/spoiler]
<span class="math">P(Both\; Crit\mid At\; least\; one) = \frac{P(Both\; Crit\cap At\;least\;one)}{P(At\;least\;one)}
[/spoiler]
<span class="math">P(At\;least\;one) = 1 - P(None\; crit)
[/spoiler]
<span class="math">P(None\; crit) = 1/4
[/spoiler]
<span class="math">P(At\;least\;one) = 1-1/4 = 3/4
[/spoiler]
<span class="math">P(Both\;Crit)= (1/2)^2 = 1/4\;\;\;\;\leftarrow
[/spoiler]
<span class="math">P(Both\;Crit\cap At\;Least\;One) = 1/4
[/spoiler]
<span class="math">P(A\mid B)=P(A\;Given\;B) = \frac{1/4}{3/4} = 1/3\;\;\;\;\leftarrow[/spoiler]

>>7326017
top derp

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

>>7326157
>Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
>Its answer could be 1/2, depending on how you found out that one child was a boy
>At least one of them is a boy
>depending on how you found out that one child was a boy
>At least one of them is a boy

memes aside what is the actual answer

>>7326957
You should switch doors

>>7326957
The only real answer is to not post and let this shitposting meme thread die.

>>7326957
if you're an autistic/redneck fuck with no theory of mind who incorrectly assumes that the questioner only ever looks at one of the hits to determine "at least one of the hits is a crit" (and completely ignores cases where the hit that the questioner didn't look at is a crit) then the answer is 1/2.

if you're a rational, competent human who takes the statement "at least one of the hits is a crit" at face value with no far-fetched assumptions then the answer is 1/3.

How is this debate still going?
The only real answer is 1/2

>>7327387
gr8b8m8r88/8

>>7327387
P(A) = 0.5
P(B) = 0.5
P(A∩B) = 0.5 * 0.5 = 0.25
P(~A∩~B) = 0.5 * 0.5 = 0.25
P(both crit when at least one hit) = P(A∩B) / (1 - P(~A∩~B)) = 0.25 / 0.75 = 1/3

>>7317442
3 outcomes with 1 crit
1 with 2 crits
1/3

>>7317484
the chance of the car being behind your picked door is a third, the chance for it being behind the other two doors is 2/3
a door is opened
the chance for the car being behind your door is 1/3, the chance of the car not being behind your door is still 2/3. not being behind your picked dor means behing behind the other door. it has 2/3 chance

>>7327717
>3 outcomes with 1 crit

wat

>>7327745

I think he counts 2 crits as an outcome with one crit (because it is not not an outcome a crit.)

>>7327745
That's 3 outcomes with at least 1 critical hit.
1 outcome with 2 critical hits
1/3

>>7317442
We have

CN
NC
CC

right? so, if all of them are equiprobable the answer is 1/3. But can we be sure that all of them are indeed equiprobable?

>>7327809

>But can we be sure that all of them are indeed equiprobable?

Sure.

This is your P-Space:

CC
NN
CN
NC

Which is definitely equiprobable (they're each .5 * .5)

So when you remove one, even though your change their absolute probabilities (by changing the P-Space) you're not changing their relative probabilities except for their probabilities relative to NN.

>>7327840
I saw it this way to, what is confusing me is the fact that, if your first hit is not critical, then the second one must be critical, since the chance of the first hit being critical is 1/2 then the chance of getting NC is 1/2 and 1/4 for CC and CN.

What am I missing here?

>>7327861
>What am I missing here?
a brain

>>7327868
elavorate

>>7327884
I flip a coin. I observe it landed on heads. What is the probability it landed on heads? Does this mean it HAD to land on heads?

>>7327892
No, but what if you have two coins, and at least one of them has to be tails. If your first coin is H then the second one must be tails, if your first coin toss lands on T then you don't care about the second one because the condition of at least one T is already satisfied

>>7327912
>No, but what if you have two coins, and at least one of them has to be tails.
Do you understand the analogy? You just made the exact same mistake. What you're doing is like saying that if I observe a coin landing on heads, it has to land on heads.

These are not magic coins, they are fair coins. There's a 25% chance that both land on heads, that just happened to not occur in this case. That is the only information we actually have. So if double heads did not occur that means we are left only three equally likely possibilities of what did occur. And 1 out of those 3 is double tails.

>>7327925
I think I get it now