Why is Jow Forums so dumb?

So, if at least one of the hits is a crit, it's guaranteed that one of the hits will be a crit. This means there is a 100% chance of 1 crit. The chance of both being crits is 1 * .5. Thus, the answer is 50%.

Retards

Honestly it's more likely cause you asked people at a (great if people got it instantly?) college. Ask the average boomer and they'll get it wrong

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33%

.
/ \
C* N
/ \ / \
C* N C N
It's 25% by calculation.
Some may argue it's always 50% to happen in reality.
The question is retarded

Wait if one of them is already critical then it's not 25%

aw shit it is 25%

You might as well just say "You flipped a coin and it came up heads. You flip it a second time, what's the probability it comes up heads?"

Go on then you prick

75%

No no. Its the probability of getting heads twice in a row.
25%

Pr(a given b) = Pr (a & b) / pr (b)

Pr(a) = 0.5
pr(b) = 1

Pr(a given b) = (0.5 x 1) /1
= 50%

It's a third, it's a very famous problem and at least in the UK you get taught it at 17. (Although I got it first time)

75%

50% insta-crit + 50% chance of crit on the remaining hit.

0% I just suck at hitting

So there are 3 possible scenarios

Crit+ Crit

Hit + Crit

Crit+ Hit

therefore 1/3

Isn't it the probability of getting two heads in a row "when at least one is guaranteed to be heads"?

that's completely different from getting two heads in a row when both coins have a 50% chance to be either-or.

can I change my answer to this? also what coins do you hold?

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Congrats

50% for 2 hits not per

Got it right first time at 17 years ago.

About 10 bitcoin and equal amounts of all the bitcoin forks

And you stupid fucks try to tell me how to invest my money...
For two hits we have 4 different combinations of crit (C) and noncrit (N) :
CC
CN
NC
NN

OP told us that at least one of them (but NOT whether it's the first or second hit) is a C. That's why NN is ruled out by definition. Of the remaining 3 only one has 2 crits, that's why the probability is 1/3 or 33%

cheers for the swift response.

Damn, didn't know the UK was that retarded. Sorry for your loss

correct answer, 33%ers are mid level iq

and what do you currently hold?

Link

brilliant.org/wiki/bayes-theorem/

Basic conditional probability - 50%

Actually no, 2/3

But in that scenario, if you flipped a coin and it came up Tails, that means there is a 0% chance for the next coin to be a Tails since at least one must be Heads; that isn't consistent with the problem presented in this thread since we're looking for a specific side of the coin and not a match of either-or.

Nah man. The question is “at least one of them is a crit” so you remove the N,N branch. You’re only left with 3 possibilities. Leaving you with 1/3 a chance.

THIS IS THE ONLY ANSWER

POSSIBLE SCENARIOS:
>MM
>HM
>MH
>HH

The first choice is not possible as at least one is a hit. THEREFORE, HH occurs in 1/3 total possiblities. The answer is 33%. If you came up with ANY other answer, you are a brainlet and you will NEVER make it unless you go all in on BAT before ads go live at the end of the month

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wow took this long for the right answer.

no one here did discrete math?

50% no way anybody here gets it wrong.

>what is a statistical distribution and what is each random outcome

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This is fucking dumb. You know why?

Because for someone to know that at least one of the hits is a crit, then that means someone has to have seen either the first hit, the second hit, or both.

If he saw only the first one, the probability for the second to be also a crit is 50%.
If he saw only the second one, the probability for the first to be also a crit is 50%.
If he saw both hits, then either he saw one crit or two crits. If he saw one crit, then the probability for the other to be a crit is 0%. If he saw two crits, then the probability for the other to be a crit is 100%. So if he saw both hits, there is 50% probability that there are two crits.

So the overall probability that both hits are crits is (1/3)*(1/2)+(1/3)*(1/2)+(1/3)*(1/2) = 1/2 = 50%

This is correct

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If crit chance is 50% and you are guaranteed one of the two hits to be a crit (assuming hits are successful) shouldn't it be like (100%+50%)/2=75%. Sure if you take the scenarios at face value it seems to be 1/3 but isn't the order irrelevant?

>t. brainlet
please explain

Holy shit thank you

What if it both hits were simultaneous? Does his answer suddenly become correct? That seems silly, his answer is correct either way.

I did say that earlier but no one refuted me. Then again I'm a fence sitting brainlet too
Possibly even NPC tier.

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yeah wanted to give you a (you) but your using 50% twice seemed like wrong phrasing tbqh.

no you brainlet
By your logic, when rolling 2 dice, rolling a 3 is just as likely as rolling a 2. Rolling a 1,2 pair is NOT the same as rolling a 2,1 pair.

In laymen terms, its not 50% because you have 2 chances of hitting that 50% crit. You cant take 1 guaranteed hit at face value, as by getting that 50% crit you may use your first hit (as a miss). This is why its less than 50%

Adding 100% + 50% /2 just makes no sense at all lmao I see the flawed logic for thinking its 50% but not 75%.

If I reworded the question but to non-crits your answer would still be 75% based on that math. So regardless of what you hit first, your calculation would argue that both hitting a crit, and hitting a non-crit would have a 75% chance. That makes no sense

this.

The question is worded in a way that has multiple interpretations. "At least one hit is a crit", if you changed the phrase to "one of these hits is already a crit" then the answer changes. This is based on an English technicality and is designed to teach kids about figuring all possible outcomes. But if you took the sentence at absolute value instead of like an autist: 50% due to being able to discount the first crit. This isn't a question of intelligence, it's more of an English trivia than a math problem

I don’t think theoretical logic and mathematical logic line up here fellas

33%

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if you rephrased it to the opposite i.e. "at least one is a normal hit, what are the chances of both being normal hits" then yes it would be 75% again.
If you rephrased it to "at least one is a crit, what are the chances of the other being a normal hit" then it's 50% because that's the crit chance. There is no "both" factor in the latter case.

by my logic any specific dice pair has a 1/36 chance of being rolled and the order is irrelevant. But I dunno, the closest I ve been to probabilities/statistics the past 6 years has been playing dnd.

this sadly

It's 50%.

One of the two hits is already definitely a crit according to the premise, so basically the question is "Assuming a 50% crit chance, what is the chance one hit will be crit?". And that's 50% because the premise just told us.

Incorrect. You could create an array hitcrit =[ ] , then using 1 for crit and 0 for noncrit with index 0 for hit 1 and index 2 for hit 2, write to the array. Return a value for hitcrit[0] or hitcrit[1] using a random number generator without knowing which index is called

How do you double down on that first point? You are saying in both instances their is a 75% chance of the same result occuring. That is just clearly impossible.
Also, I like that you say 1/36 chance, yes there aren't 36 possibliities by your logic:
>1,1
>1,2
>1,3
>1,4
>1,5
>1,6
>2,2
>2,3
>2,4
>2,5
>2,6
>3,3
>3,4
>3,5
>3,6
>4,4
>4,5
>4,6
>5,5
>5,6
>6,6
Notice there are 21 pairs if order is irrelevant. 36 comes from the fact that reversing the order for every pair, excluding doubles (21 - 6 = 15 pairs to reverse), is every possibility.

Also here is proof its 1/3 and for all the other brainlets in here

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You have to completely disregard that first crit hit from the premise you douche.
Regardless of how many crit or non-crits you had before, if you have a 50% chance of crit, then you have a 50% of crit.

Like flipping a coin, every single time you flip it, the chances are ALWAYS 50/50. Regardless of how it flipped the million times before that.

>disregard the first crit
why? no one said it had to be the first. Explicitly point out how that script conflicts with the OP. OP does not specify which hit is the crit. It is either the first or the second. This is what I tested.

>why?
Because every single time you go for a hit, the chances of it being crit are 50%.
Previous hits do not count.

Like flipping a coin, it's always always always 50/50.
Even if you just had 5 heads in a row, the chances of the next flip being heads are 50%.

the outcome of hit+crit is the same as crit+hit. shit question imho. but yeah thats the thing with catch question like these i guess

It's a one third chance of both being critical hits

>shit question imho
It's a bait question in a way.
It's testing whether you are aware of a very basic concept in probability.

No.

One hit is already crit according to the premise, so it all comes down to that one hit which has a 50% crit chance according to the premise.
So it's 50%.

Is this bait or do you really not understand? Not trying to be a dick.

Lmao, please tell me how I'm wrong.
This oughtta be good.

I know past results do impact current ones. Everyone older than 10 understands this. You are just fucking stupid lmao

Where in the posted script do I conflict with OP's post? Answer this question explicitly or kill yourself

math.stackexchange.com/questions/991060/flip-two-coins-if-at-least-one-is-heads-what-is-the-probability-of-both-being?rq=1

If you don't answer where the posted script conflict with op its confirmed b8

>I know past results do impact current ones

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> I know past results do impact current ones.
> You are just fucking stupid lmao

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you got me there

There are three scenarios. Model them using binary. 1 is a crit, 0 is not.
>1 0
>0 1
>1 1
You are guaranteed that either 1 0 or 0 1 will be an outcome.
If the first hit is 1, then there's a 50% chance that the second hit will also be 1. That's where you are confusing it as being a 50% chance.
However, if the first hit is 0, then the second hit is guaranteed to be 1. This is what you're not thinking about. There's no telling which will be the first hit, because the first hit isn't guaranteed to be a crit, so the odds cannot be 50%. It's one third.

>I know past results do impact current ones.
Not how probability works.

You flip a coin five times, by some miracle it's 5 times heads.
What are the odds of the next coin flip being heads again?
50%

According to the premise in OP, we're talking about two hits:
> hit number (1) is pre-established as being crit
> hit number (2) has a 50% chance of being crit
(hit numbers are not necessarily chronological)

With me so far?

Now the question is "what are the odds of both being crit?"

There are only TWO possible scenarios:
(I dare you, double dare you, triple dog dare you to add another viable scenario)

1) hit number (2) is crit (50% chance of this) = both are crit
2) hit number (2) is NOT crit (also 50% chance of this) = both are NOT crit

These are the ONLY two scenarios that could ever play out according to the premise, and they both have equal odds of happening.
Therefore: the odds of both being crit are 50%.

Pr(a given c) = Pr (a / b) / tn (b)

Pp(a) = 0.5
pr(b) = 1
ps(c) = 2

Pr(a * b) = (0.5 x 1) /2
= 74.29999%

yeah it was an obvious typo. Didn't point out in the script as I expected. Confirmed retard or b8.

>There are three scenarios. Model them using binary. 1 is a crit, 0 is not.
0
1
1

Two of these scenarios are the same outcome.
Whether 1-0 or 0-1, that just means they are not both crit.

The question is "what is the probability both hits are crits".

This reduces your three scenarios to two scenarios:

1) both are crit (one possible arrangement: 1-1)
2) both are NOT crit (two possible arrangements: 0-1 or 1-0)

Yesse

Literally second answer in your link proves you are retarded. You're not smart, fuck off.

Attached: proof verification - Flip two coins- if at least one is heads- what is the probability of both being (935x762, 274K)

>0 1
>1 0
are not the same outcome.
>0 1
is guaranteed to produce a single crit.
>1 0
has a chance to produce a double crit.
You're a brainlet.

>yeah it was an obvious typo
What was?

You are retarded and that explanation is retarded.

Let me boil it down for you:

According to me, these are the ONLY two scenarios that are possible under the premise:

1) the unknown hit is crit (50% chance of this) = both are crit
2) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit

Please add another viable scenario to this. If you can, that means I was wrong and you were right.

Did you even read it, you fucktard?

1
>is guaranteed to produce a single crit.
If it's 0-1 that just means the unknown hit came up non-crit. Which has a 50-50 probability.

The question is "what are the odds both are crit". And under this premise, 0-1 and 1-0 are the exact same outcome: namely "both are NOT crit".

Jeeeeesus christ my dude.

Absolutely.

Now please add another scenario to these two that is viable under OP's premise:

1) the unknown hit is crit (50% chance of this) = both are crit
2) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit

If you can, you are wright and I was rong.

(protip: you cannot. Because those are the only two viable scenarios under OP's premise)

read the answer and check your reading comprehension. The only way to interpret the question given by OP is to assume either or, which is what the selected SE answer does. This answer also states the answer of 1/3 in this instance. In the case that a particular instance is chosen, then yes the dynamic is changed. However, there is no reason to assume such unless explicitly stated.

Once again, you can refer to the very simple python script I posted that meets the criteria of OP's post and results in 1/3 chance.

Now I quit talking. If anyone still insists its 1/2 you are just fucking retarded. Anyone with a math background would give you an answer of 1/3, unless you imply some extra circumstance like does

>If anyone still insists its 1/2 you are just fucking retarded.

Please add another scenario to these two that is viable under OP's premise:

1) the unknown hit is crit (50% chance of this) = both are crit
2) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit

If you can, you are right and I will openly admit to being retarded.

(protip: you cannot. Because those are the only two viable scenarios under OP's premise)

Known Hit Crit - Unknown Hit No Crit
Known Hit Crit - Unknown Hit Crit
Unknown Hit Crit - Known Hit Crit
Unknown Hit Crit - Known Hit No Crit

50 %

Scenarios
AT LEAST one hit is a crit
So we have 3 scenarios
Attack 1 = Crit, Attack 2 = Crit
Attack 1 =Crit, Attack 2 = Non-crit
Attack 1 = Non-crit, Attack 2 = Crit
33% for scenario 1 to occur

In order for it to be 33% you have to add one more viable outcome scenario to these two:

1) the unknown hit is crit (50% chance of this) = both are crit
2) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit

Please add the third outcome scenario.

there are 2 hits. the result of either in particular is not known. OP never states a particular hit is given. therefore there are 2 unknowns, with the potential tuple being constrained to a set domain (1,0),(0,1),(1,1). End case.

I have now asked three anons to add a third scenario to my two scenarios.

Surely one of them will come through for daddy.

Nice troll.

Cases where at least 1 is critical
First critical, second not : 1/2*1/2
First not critical, second critical : 1/2*1/2
Both are critical : 1/2*1/2
.: P( at least 1 is critical) = 3/4

P(Both are critical | at least 1 is critical)
= P(Both are critical) / P (at least 1 is critical)
= (1/2*1/2) / (1/2*1/2+1/2*1/2+1/2*1/2)
=(1/4)/(3/4)
=1/3

1) the unknown hit is crit (50% chance of this) = both are crit
2) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit
3) the unknown hit is NOT crit (also 50% chance of this) = both are NOT crit

Eazy.

I'm not sure you did. In real world, the probability that both are crits is 50%, yet math will prove 1/3. So where's the disconnect? The question is retarded, just like you and are for not seeing this

you are right about the dice. mb.
that said, removing the 25% chance of 2 normal hits, we re left with 50/75 for no doubles or 25/75 for a double crit i.e. 1/3.
>til order is relevant.

cheers for the thread.

I didn't see you add another scenario.

Please try to add another scenario. It's a very fair and objective question.

read this you brainlets have none of you studied basic conditional probability.

Either first hit is a crit, or second hit is a crit, or both hits are crits, so 33% of the time when there is at least one crit, both hits are crits.

BUT

If you have seen the first hit and it's a crit, there is 50% chance both hits are crits.
If you have seen the second hit and it's a crit, there is 50% chance both hits are crits.

The answer depends on how you know that there is at least one crit. If you or someone else has seen one hit and it was a crit, there is 50% chance both hits are crits.