Can Jow Forums solve this...

hit1 hit2 0.5*0.5 = 0.25
hit1 miss2 0.5*0.5 = 0.25
miss1 hit2 0.5*0.5 = 0.25
miss1 miss2 0.5*0.5 = 0.25

At least one crits, so remove the last possibility. We now have three possible outcomes

hit1 hit2 1/3
hit1 miss2 1/3
miss1 hit2 1/3

1/3, 33%

it depends if the rolls are made at the same time or one after another

if they're made at the same time, then it they're made one after another, then if the problem statement does not define which one it is, we can't answer

Possible Outcome 1: 1st hit critical, 2nd hit critical
Possible Outcome 2: 1st hit critical, 2nd hit not-critical
Possible Outcome 3: 1st hit not-critical, 2nd hit critical

1 outcome (Possible Outcome 1)/3 possible outcomes, such that p=1/3

Remember, at least one possible c-hit precludes a 4th possible outcome of no criticals

50%
if 1 is Guaranteed then you have 50% that other is also crit, meaning that the chanse of both being crit is 50%

What the fuck is this school level bullshit and do people unironically consider this hard?

the answer is 25%

for the first hit, 50%:
- 50% for NOT CRIT: if it's not crit, then the "at least one" makes the 2nd hit a crit, but it doesn't matter at all because you only care about both hits being crit.
- 50% for CRIT: if it is crit, the next hit is unaffected by the "at least one" condition, because there's already one.

now, continuing from a CRIT result for the first hit (50% chance):
- 25% for NOT CRIT (i.e. 50% of 50%): fail
- 25% for CRIT: BOTH HITS CRIT.

so the answer is 25%

the trick is realizing that the "at least one" condition will only have effect when you missed the first one, and you only care about a scenario where both hits are crit, so it makes no difference at all.

>all the autistic retards writing novels on this
It is 50% you fucking retards.

Those of you who say 50%
At least one of them is a crit. This doesn't mean that the first one is a guaranteed crit or the second one is a guaranteed crit, right?

this
it's basically asking what is the probability the other hit is a crit

brain pickles like this really brighten my day up!

Attached: gladspede.jpg (333x500, 23K)

It's a deliberately misleading question, don't think too hard on it.

>this leaves three outcomes with equal probabilities
not at all

the probability of a result "one hit crit and the other not crit" is 50%, because it's equal to the probability of missing the first hit, 50%

>missing
not being CRIT*

I think the problem is worded vaguely and throws people off. I thought about it in three different ways and got three different answers.

>"one hit crit and the other not crit"
"first hit not crit and the 2nd crit"*

"at least one of the hits is a crit"
Despite the normal 50% rate, in this particular circumstance one crit is guaranteed. The other hit still has a 50% chance to crit, thus the chance of two crits right now is:

50%

Attached: main_picture_548739d2a63fb.jpg (600x397, 94K)

This is a troll question with contradictions

Don't fall for the bait, it's purposefully ambiguous

based nip

It's 50%. We already know that one hit was a crit; that changes everything. The question is essentially "You hit an enemy. Assuming a 50% crit chance, what is the probability of a crit?"

the problem is clear and simple desu

the 50% answer is wrong because it misunderstands the "at least one" condition to mean "the first hit is a crit"

the 1/3 answer is wrong because there are no "three equal probabilities", just like there are no
five equal probabilities of getting a six in throwing a die just because we say that a five also counts as a six, in this scenario the six would get 2/6, and the rest 1/6.

I like the One Third interpretation best.

(not really relevant but I am also many worlds kind of guy)

>five equal probabilities for each result*

50%

the first hit is always a crit so it doesn't matter, you're only looking at the second hit which has a 50% chance to be a crit

Actually while you can make a poitn for 1/2 and 1/3.(as in the problem doesn't explain enough to know which one is true), 1/4 is straight up wrong and they should be laughed at

It definitely is the one which sounds more intuitive. But it's nto wrong to assume we know which one of the two strikes is a crit. And thus 1/2 is acceptable too.

Your idea of a "The first crit is a hit" interpretation has no relevance or meaning. It would only be a cause of error if the question said "What is the chance the SECOND hit was a crit?" It isn't. In a randomized test such as this, one outcome does not affect any others

50%ers need to tell apart combinations from permutations and apply the probability tree of outcomes

1/3ers and 1/4ers, I brought my popcorn, keep arguing

Three differnet out comes

hit crit
crit crit
crit hit

We need crit, crit so that's 1 out of 3.

50%

crit, crit
crit, no crit/no crit, crit

If X is critical and O is non-critical, and at least one hit is X, the possibilities are:
X O
O X
X X
50% crit chance means that the probability is evenly distributed, because the probability that a hit will not be critical is also 50%. The answer is therefore 1/3.

Good post

you have to have an actual subhuman iq to believe the answer is 1/4

>crit, no crit/no crit, crit
I think these two are not the same

the way the question is worded, they are the same. one has to be a crit and the other has to not be a crit to satisfy the conditions, both of which do

Here is a simulation
Of course one could argue that my interpretation of the problem is wrong. Is it?

Attached: prob.png (944x546, 45K)

i dont think you understand the 50% argument and are assuming an incorrect basis for it

the order doesn't matter. one hit will crit or 100% chance.

the other will still have 50%

This is like those order of operations puzzles. It depends on how you interpret it. What does "at least one of the hits is a crit" mean?

If P(a) and P(b) are 0.5 each, P(a|b) = 0.25. But this holds in an unconstrained set of solutions where all the permutations of a and b are possible. The "one is a crit" statement in the manner most people would interpret it narrows the set down to ab, (-a)b and a(-b). (-a)(-b) is rejected by definition. This means that the old probability values of those 3 options must now add up to 1.

It's really a trick question because that constraint statement means you are no longer dealing with classic conditional probabilities, because of how the question messes with the order of the statements. It implies certain foreknowledge which is not something you get in pure probability tasks.

You fucking monkeys nowhere in the problem does it say the first hit is a crit. All it says is that at least one is a crit.

That’s like assuming my oldest child is a boy if said that I have two children and at least one is a boy.

>no crit, crit
>crit, no crit
>crit, crit

one out of three so it's 33,33...%

brainlets

my reasoning is it doesn't matter if it's the oldest child or not, it only matters if one child or both of them are boys so it's irrelevant whether the older or younger is a boy as long as one is guaranteed to be a boy

however i have since just said fuck it and went about this empirically by rolling 100d2 and tallying the results into 10 cases of 0 crits, 30 cases of 1 crit, and 10 cases of 2 crits, and since we discount all of the cases of 0 crits the actual result is apparently 25%

I agree. You can't have "at least one is a crit" AND "50% crit"

Well I ran a simulation and got 1/3. And the funny thing is it is probably irrelevant, because it is the interpretation of the results that is different. So we may get the same results but apply different formulas according to our understanding of the problem and either get 1/3 or 1/4 or 1/2

1/3

Your reasoning is wrong and you did your experiment wrong dumbass.
The probably that “only one hit is a crit” is double the probability that “both hits are crits.” They are not equally weighted.

here is my data, *shrug*

Attached: 1.png (1788x1006, 39K)

Crit no crit 1/4
No crit crit 1/2
Crit crit 1/4
/thread

H = Heads
T = Tails

All combinations:
HH
TH
HT
TT

1/4
25%

it doesn't matter what i think the probability is since i just read the results m8

for every case of two crits there were 3 cases of a single crit

therefore the chance is 25%

Ill continue to post the answer in every thread
A = at least one hit a crit on any totally random toss
B = both hits are crits on any totally random toss
P(2 crits given at least 1 crit) = the chance of B given that A happens = P(B|A) = P(B and A) / P(A)
P(B and A) = chance of 2 crits on a totally random toss AND at least 1 crit = chance of 2 crits on a totally random toss = 1/2 * 1/2 = 1/4
P(A) = 1 - P(not A) = 1 - chance of 0 crits on a totally random toss = 1 - 1/2 * 1/2 = 3/4
The final answer is (1/4) / (3/4) = 1/3

>inb4 chance of 0 crits is 0% not 25%!!!
Read the fucking text, idiot. The way we do conditional probability is by reducing a problem with constraints (like in this problem) to 2 problems without the constraints, and then combining them. This is the way it's meant to be done. The P(B|A) formula accounts for the fact that there can not be 0 crits on its own.

this is my last attempt to enlighten plebs

I can't upload files here so

fileconvoy.com/dfl.php?id=g4a473707e653d1ba1000134009cee3789404e8c362

It says "at least one if the hits is a crit" so one of those cases is sure not to happen. remove it, you get 1/3.

i removed it and got 25%, see

Here is my solution

Attached: crit.png (929x321, 17K)

Forgive me if I don’t trust your experiment which tells us the probability of rolling 0 crits is 1/5.
You don’t count all four possibilities, since you know there is at least one head.

I’m convinced you either are mentally retarded or trolling if you don’t think it’s 1/3.

That is not very many trials. Lots of room for error between values like 1/3 and 1/4.

Hey look its another monthy hall iteration!

Oh boy I sure do love to call people retards over counterintuitive deductions that I myself made before being exposed to the answer!

i tossed a coin and i got heads therfore the chance of getting heads is 100%

You're supposed to calculate probability based only on observations where at least one of the two was a critical strike.

That would come out to be 50% because that's the probability of a critical strike.

upvoted

They’re retards if they refuse to believe the explanation.

Miss1 hit2 0.5
Hit1 hit2 0.25
Hit1 Miss2 0.25

now do it 100 times

btw the chances of real life coin toss are actually not exactly 50% depending on the coin you're using

>But it's nto wrong to assume we know which one of the two strikes is a crit. And thus 1/2 is acceptable too.
But 1/3 is the answer you get on the basis that you don't know which one of the two strikes is a crit. We had this thread last week

Attached: basic.png (1299x342, 16K)

1/3, those saying 50% are brainlets

If you take a large number of samples of sufficient size, the probability distribution of getting heads will be normal and centred around 0.5.

So which hit is the definite crit? first or second?

You can't do it sequentially like this, that's not a valid probabalistic interpretation of the scenario. I know it sounds ridiculous but probability can be a little ridiculous. It is often contingent on how much you know. If you know the first coin flip, it changes a lot of the probabilities, in a similar way to how knowing where one of the goats is makes it better to switch in Monty Hall. We're never told that, we're flipping coins and not getting to see them until they're both flipped (presumably). In which case all you can do is take the 4 possibilities (CC, CH, HC, HH) eliminate the ones that dont happen (CC, CH, HC) and then divide by the number of events. 1/3.

25%

There are 3 outcomes but that doesn't mean they're even. The first-hit-not-crit is a 50% possibility and continues to be 50% total to the end with second hit being a guaranteed crit. If, however, there is no crit on first hit, the next one splits between two outcomes, so 25/25.

It's 3 outcomes, chances are 25-25-50 with two consecutive crits being 25%. Just draw a flow chart with chances for each outcome.

Why do you keep adding layers of unnecessary semantic clutter? Fuck your "boys" and "kids".

I will rephrase the question, see if you agree with it or not.
1. P(a) = 0.5
2. P(b) = 0.5
3. P'(ab + a*b + ab*) = P(1 - a*b*) * 1/(P(1 - a*b*) = 1
Find: P'(ab)

The actual probability of P is computed as the number of events where the condition holds true from the set of all events. But the question doesn't ask for that, it specifically tells you that the events where neither a nor b is true are not possible. This means that you have to subtract all the cases where a*b* from the total. Then you have to find what part of the remainder is ab.

...

There's no guarantee it's going to reflect the actual probabilty. Chances are that's not going to be 50/50 actually.

>If, however, there is no crit on first hit, the next one splits between two outcomes, so 25/25.
You're supposed to discard observations where there was no critical hit.

go back to pakistan

50%

you guys are idiots

it's 50% you nigs, Law of total probability

The problem doesn't state that.

It may not hit exact 0.5, but it will be very close

en.wikipedia.org/wiki/Law_of_large_numbers

The probability of the second punch depends on the first hit and for this reason it's 1/3

0% probability it happens and nobody can say i'm wrong.

>at least one of the hits is a crit

i made another experiment with 200d2 resulting in the data in the pic

the results are 23 cases of 0 crits, 53 cases of 1 crit, and 24 cases of 2 crits

discounting the 0 crit cases, that adds up to 32%

even if you had a perfectly physically balanced coin and flipped it with perfect technique, it still wouldn't necessarily be strictly 50/50, however it would be very close after enough tosses

Attached: 1.png (341x196, 3K)

I figured it out, it's 1/2401.
We originally have 4 possibilities, HH, CH, HC, CC. But we are told that one of these cannot happen. So we need to add a shadow flip, Y/N, to tell if the event happened or not.
So now we have HHN, HHY, CHN, CHY, HCN, HCY, CCN. CCY.
So far only one of these cannot happen, HHY. Remove it. Now there are 7 possibilities. We want the case where CCY happens and HHN, HCN, CHN also happen. This is 1/7 * 1/7 * 1/7 * 1/7 = 1/2401.

That condition doesn't require you to discard any observations, though. It simply means that you'll get a guaranteed crit on second strike if the first one isn't. I've considered this in my reasoning.

what if the first isn't a crit but the second is?

The crit chances aren't ACTUALLY 50%. If they were 50%, then there would be a 25% chance of no crits. The question specifically states that that can't happen, so you are no longer actually dealin with probabilities. So you have to extrapolate from the given probabilities to a special constrained case.

>0 crits
there's your error you fucking imbecile. The question says AT LEAST ONE OF THE HITS IS A CRIT

Which means that cases where there are 0 crits cannot happen.

>If, however, there is no crit
If there IS* crit, goddamit. Always fucking up in smallest details.

and i was not counting them, i discounted them remember?