How smart is Sup Forums?

How smart is Sup Forums?

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courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf
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50%. so fucking easy.

test

%50.

The outcome of the second coin flip has nothing to do with the outcome of the first.

16% ??

It's 50 fuck off

1/3

900% probability that the one flipped first will be different than the second one flipped.

If they're flipped at the same time, that probability goes down to 842%

Math harder, plebs.

Retards detected. It's not 50%. Try again. I'll give you 1 more try before I explain why you're retarded.

God, not this bullshit again.

Answer is 1/3. Don't argue.

There's a difference between
>given that at least one of them landed heads
and "the first of the coins was heads"

Since there's no guarantee that the first one has landed heads, the answer is 33.333333333333333%

2 coins
25% chance both are heads
25% chance both are tails
50% chance of one head and one tail.

Therefore, the answer is 25%.

>"at least" one is heads
meaning 2/3 dumbass

Depends on who flips the coin.

Retard. At least 1 landed heads. They can't both be tails. 1/3

lol you fucking retards i'll explain it

4 outcomes:
HH
HT
TT
TH

>given at least one was heads
that eliminates TT. only 3 possibilities now. out of those, only 1 is both heads (HH), the other 2 are only one heads (HT, TH)

therefore 1/3

It was 25% before any coin was flipped. The fact that you flipped a coin does not affect the chances you had before you flipped it.

Would be 1/4 with two coins since there is a 1/2 chance of landing on heads for each coin.
If one coin is guaranteed to land on heads, then it would simply be a 1/2 chance of the other lands on heads.

Fucking brainlets I cant believe this
It's OBVIOUSLY 75%
OP's asking for the probability of both coins being heads, since we already have one we increase that probability by half. Fucking thirteen year old summerfags

>bout three fiddy

Seriously though, this question has no real life meaning as one cannot "guarantee" one will be heads. Therefore OP still a faggot

Not enough info. Are the coins fair? Are the outcomes independent of each other?

Yes it does, retard. It's called conditional probability.

1/3.

Nope.

OP states that two coins are flipped, one lands on heads.

Therefore, there are only two possible outcomes; HH, HT.

Which happens depends on the flip of one coin, which is a 50/50 odds of landing one way or the other.

>OP states that two coins are flipped, one lands on heads.
>Therefore, there are only two possible outcomes; HH, HT.

TH is also possible. OP never says teh FIRST coin landed heads, just AT LEAST ONE landed heads. there are 3 equally probable ways to get at least 1 heads coin for a 2 coin flip.

HH
HT
TH

1/3

OP states
>AT LEAST one of them landed heads

Threfore three possible outcomes; HH, HT, TH

Learn to read, brainlet

Theyre fucking coins. Dont over complicate things you know the answer to.

Have we looked at the coins and collapsed the wave function?

lol obviously 50% chance either it happens or it doesnt

HT and TH are the same outcome. Either way, the probability of the desired outcome, HH, hinges on the outcome of a single coin flip because you know that the question assumes one coin did land on heads.

Given that 1 landed heads and seeing as each coin flip is an independent event (the probability of landing heads is not affected by previous coin flip) your probability of landing heads is 50%. Why 50%? Well if the coin is equally balanced on both sides then you only have 2 possible outcomes, heads or tails. In each independent coin flip you have 50% chance of heads or 50% of tails. The law of averages says that given a large enough sample size you can expect the 50/50 results.

>HT and TH are the same outcome.

Now we know you're REALLY retarded.

Allow me to explain: Imagine using a penny and a quarter. You flip both. At least 1 landed heads. So you could have
>penny=heads, quarter=heads
or
>penny=heads, quarter=tails
or
>penny=tails, quarter=heads

Can you see how those are all different outcomes?
Can you see that they all contain at least 1 heads coin?
Can you count how many there are?

They are all equally probable.

3 possible and equally probable outcomes.
1 of those 3 is both heads
1 of 3
1/3

Stay in school, kid.

>What is the probability that both landed heads given that at least one of them landed heads.

Read the question again.

If one of them landed heads up then the question is really what is the probability that the other lands heads up. Which is %50 regardless of what order you flip them in. One of the two will always lands heads up and you can ignore it for all intents and purposes.

It's a stupidly worded question meant to start an argument.

The same amount of probability of jews running this country, and use shekel math schemes to distract us.

HT and TH are different, not the same at all.

The question asks
>"Out of the outcomes where at least one coin (i.e. one or two coins) lands heads, what is the probability both landed heads?"

There are 3 outcomes where at least one lands head.
(first coin H, second coin H)
(first coin H, second coin T)
(first coin T, second coin H)
Out of those, 1 outcome is where they both land heads. 1/3.

>Given that 1 landed heads and seeing as each coin flip is an independent event (the probability of landing heads is not affected by previous coin flip) your probability of landing heads is 50%.
>previous coin flip

you're assuming the "FIRST" coin flipped is the heads coin. this is a false assumption. First coin could be tails and second heads

3 equally probable outcomes.
Heads - Heads
Heads - Tails
Tails - Heads

1/3

See Bayes Theorem

thank you, at least one other person has the reading comprehension merit badge

People not understanding that HT = TH


kek.

Out of the outcomes where at least one of them landed heads. Not the 'first one landed heads'. You can flip T first, then H. That counts as at least one of them landing heads.

3 outcomes.
Out of those, 1 is double heads. The other 2 are only single heads. 1/3.

>If one of them landed heads up then the question is really what is the probability that the other lands heads up.

EEERRRRR incorrect, buddy. You lack both reading comprehension and math literacy.

Read this

This is called mutual exclusivity

Read the question.

>What is the probability that both landed heads given that at least one of them landed heads.

Of the three outcomes you've listed, two are functionally identical because you know that one of the two coins will always be heads up. Whether it's the first or second doesn't matter. Flip them at exactly the same time and it wouldn't matter. The order is irrelevant.

Flip one and it lands tails, you know the second is going to be heads. Flip heads and the second one might be tails or it might be heads.

The only thing that can vary is whether one lands heads or tails, and that's a 50/50 shot.

gr8 b8 m8

> getting this triggered

This.

%100

Proof: just tried it, it worked.

Checkmate, Christians.

This is the solution...

There is absolutely nothing which has a 100% probability.

Apart from OP being a faggot.

Future events cannot affect the past.
Prove me wrong.

>Do we count the answer after one of them has been determined it is heads?
50%
>If not.
1/3

>Of the three outcomes you've listed, two are functionally identical

No, they are not. they are separate and distinct outcomes and are both equally likely to occur.

How the FUCK can
penny=heads, quarter=tails
be functionally identical to
penny=tails, quarter=heads ???

ITT: Mathematically illiterate plebs

Here, I drew you retards a Venn diagram too.

1/3

Youre a special kind of stupid aren't You? Just smart enough to be dangerous

>explaining probability makes you triggered

1/3

you need to get laid

The question only asks what is the probability that both are heads up, not which order or which coin has what result. Just what is the probability of a specific outcome.

HT or TH doesn't satisfy the outcome, only HH does. The chance of getting either HT/TH or HH is 50/50.

>Do we count the answer after one of them has been determined it is heads?

Of course we do. It's stipulated in the prompt.

this is correct reasoning

>Future events cannot affect the past.
>Prove me wrong.

Look up delayed observation double slit experiment. Quantum mechanics says you're wrong.

Answer - 1/3

we're dealing with coins, not photons

>All of these kids forgetting that the coin can be balanced on its edge.
There are three terminal states, children.

flip both coins at the same time, we know one of them will be heads, what we are left with is 50/50 percent chance of getting another heads

it is 50%

>you need to get laid

I'm laying the smackdown on mathematically illterate retards, son.

Pic Related: 1/3

Given that at least 1 landed heads, then TT is not an outcome. Given that you weren't given a coin flip order HT = TH and your only possible outcomes are HT and HH. Bayes Theorem applies to things like cancer where your probability or having cancer is affected by family members having/had cancer. The coin flips are independent events.

HT = 1/3
TH = 1/3
HH = 1/3

Stay in school.

(everyone be nice to the autist, he can't do much so give him a moment)

>Two regular coins were flipped. What is the probability that both landed heads given that at least one of them landed heads?

That's what the image says. Don't trust me? Scroll up and check.

Two coins got flipped. One of them is heads, full stop. %100 guarantee that one of the coins is heads. There is no ambiguity there. You've got a heads coin, every single time.

Got it?

Sure?

Ok.

So then. If one of those coins is heads, what is the probability that the other coin is heads or tails?

1/3 obviously

The result of the coin isn't affecting the past probability. It affects the FUTURE probability.

This is basic conditional probability.

If we flip 2 coins, we get HH 25% of the time.

If we flip 2 coins and are told that AT LEAST ONE landed heads, we get HH ~33% of the time.

Why? Because the outcome of TT is no longer possible given AT LEAST ONE coin landed heads. So from thi spoint forward, the probability of HH has changed from 1/4 to 1/3

Answer to OP is 1/3?

>knowing basic conditional probability and gettin lulz from explaining it to retards means you have autism

What's it like being retarded, lad?

there are only two results
(HT = TH) and (HH)

you always have 50% chance of getting HH, because HT and TH are the same thing

>Given that at least 1 landed heads, then TT is not an outcome.
Correct.

>Given that you weren't given a coin flip order HT = TH
Incorrect. These are 2 separate and distinct outcomes, just as HH is distinct from TT.

>your only possible outcomes are HT and HH.
and TH

>Bayes Theorem applies to things like cancer where your probability or having cancer is affected by family members having/had cancer.
HAHAHHAHAHAHA Bayes' Theorem applies to conditional probability, lad, which is what the OP question is about.

>The coin flips are independent events.
Correct.

1/3

If you still don't understand why, answer pic related.

is correct.

P(heads = 2 | heads >= 1) = P(heads = 2) / P(heads >= 1)
= (1/4)/(1-1/4) = 1/3

There is no difference between TH and HT because one is always heads in this scenario. TT cannot happen because there is a head already present, and because we already have a head then HT = TH
We take H as a given. Therefore the remaining chance is H/T. 1/2 chance.
The easy way to do this is to flip two coins and remove any times that you don't get at least one head. You will only have scenarios wherein HH happens and HT happens, roughly 50% of the time each.
I just tried it myself and I had 11 HH and 9 HT
Which we can agree us closer to 50% than 33.3r%
Anybody who considers TH == HT needs to understand that because the only counted occurrences are ones in which one heads is present, if you distinguish the two coins it becomes
TH 1/4
HT 1/4
HH 1/2

it would be funnier if you were right

>So then. If one of those coins is heads, what is the probability that the other coin is heads or tails?
50%, but that's a different question to the OP question.

No its not.

>there are only two results
>(HT = TH) and (HH)

If you want to pretend that TH and HT are the same outcome, then you must add their probabilities together, in which case
HT/TH = 2/3
HH = 1/3

Answer is still 1/3

Exactly
Even if oyu want to take HT and TH as the same outcome, fine. But you must also realise that it is TWICE AS LIKELY as HH. This question is about probability. Therefore one H and one T, no matter what order it's in, is twice as likely as HH.

So, excluding TT:
Probability of HH: x
Probability of only one H: 2x

Answer is stil 1/3, dumbos.

ITT: negros struggling with basic probability

No it isn't, because OP question does not ask for the probability of TH or HT, only HH
If one H is always a given, then the probability of HH exclusively depends on the second coin toss. TH = HT not because they aren't different outcomes, but because they're both simply not HH. TH happens 1/4 of the time, HT happens 1/4 of the time, TT happens 1/4 of the time and HH happens 1/4 of the time.
This means the chance for two of the same faces is 2/4 and two different ones is 2/4
Since TT is not an option, it cannot occur, however it two of the same face still occurs 2/4. TT's quarter is not shared equally between the other three possibility's but rather given exclusively to HH
This mean its a 2/4 chance for two different faces (1/4 each) and a 2/4 for HH
50% chance

>There is no difference between TH and HT

You're dumb, son.

Flip a penny and a quarter. If they do not both land tails, then you are left with 3 equally likely outcomes
>penny=heads, quarter=heads
or
>penny=heads, quarter=tails
or
>penny=tails, quarter=heads

1/3 to get both heads

>The easy way to do this is to flip two coins and remove any times that you don't get at least one head.
Correct. I suggest you do this at least 100 times with a pen and paper and write down how many times you get both heads, You will soon notice that you get both heads about 1 out of every 3 cases where you get at least 1 heads. Because I'm not retarded, I wrote a program that simulates this exact coin flipping. Oh wow, look at that result. Pic related.

TH 1/3
HT 1/3
HH 1/3

Fixed that for you, user.

dear sir you can't have conditional probability alongside independent events. You yourself have admitted that the coin flips are independendent events. Stay in school lad and don't go gambling.

>The coin flips are independent events.
Correct.

It is different, because you're asking for the probability of a SINGLE coin.

OP is asking for the probabillity that BOTH coins landed heads given AT LEAST ONE landed heads.

the heads coin could be either coin, meaning you can't just ignore a coin and pretend its the heads coin because it could very well be tails.

In the OP question, the only possibilty that can no longer occur is tails, tails.

the other 3 outcomes are all possible and equally likely. i know this is a difficult concept for you lads to understand because probability is often counter-intuitive, but the answer to OP question is 1/3

If the OP stated that the FIRST coin landed heads, then th eprobability would be 50% or 1/2.

But since we don't know WHICH coin landed heads, either coin could be tails, just not both at the same time, and the probability of getting both heads is 1/3

Pic related

Source: Washington University Math Department (See slide 4)

courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

>he thinks he is not retarded because he can program

you sir have autism

I understand your point of view cause its the most immediate one. This riddle is just a joke made by a certain wording.

You believe basically only one coin is being tossed, because you already know one landed, and it lande heads. In this case the question about two coin tossed is a non question beucase only one coin is tossed at the end of the day, so 50% is correct.

But by the original question both coin are flipped together and only after they both have landed you are given to know that at least one of them is heads up. In this case the user answering up is right with his reasoning. You get to exclude tt. In you assumption instead you exclude tt and th. Yours is an entire different scenario. Basically there's no first coin. Only -t and -x

2/3 are NOT both heads, moron

50%
one is already flipped on heads. the chances that the other flips head too is 50%.
basically the question is "how are the chances one coin flips head" since the other will flip head anyways.

I actually have now reached 50
22 HH, 28 TH/HT
I welcome other people to try this too. Maths attempts to solve probability, probability is not dictated by maths. No matter how you look at it, if these tests keep coming out at in and around 50% then the answer is 50%
I'm not disputing that 3 outcomes exist, they just simply aren't all equally weighted, at least in practice. One is always H, meaning the only factor is the second coin. It doesnt matter if its the first or second coin you flip, because one of them always has to be heads, meaning the only variation is in the second coin.
Look at it this way;
In the scenarios remaining, we have:
TH HT HH
Or
HHHH TT as possible individual coin results.
4 heads and two tails
If we take the three given heads out we get
H TT
T and T are not two distinct entities, they are the same result as we still only have one coin.
H will appear 50% of the time. And the two possible occurances of T will appear 25% of the time each.
The flaw in your logic is that you think HT and TH are seperate, and yet you don't see the difrerence in HH and HH, when the given H is on either side.
HH in which the first H was the given H will appear 25% of the time
HH in which the second H was the given H will appear 25% of the time
TH will appear 25% of the time
HT will appear 25% of the time

it doesn't matter

>first coin go heads
>50% chance the other will flip heads too
>first coin go not heads
>the other will go heads

only one coin matters since the other goes heads anyways. it doesn't matter if the first or second is the magic heads coin, only the other one is important.

One of the coins HAS to be heads.
Therefore, only one coin can be tails or heads. 50/50.

>can't have independent events in conditional probability

Damn, you're a moron.

Buy a lotter ticket. you either win or you dont. One event

Buy another lottery ticket. you either win or you don't. One event.

What is the probability that you win twice?

What is the probability that you win twice when you win at least one time?

OMG CONDITIONAL PROBABILITY 2 SEPARATE EVENTS HOW IS RTHIS POSSIBLE????!!!?!?!?!?

Answer to OP is 1/3.

they are independent events, so the probability of one event happening doesn't effect the probability of the other. the answer is 50%. statistic major here

>can't program
>can't do basic probability
>you, sir

Go back to reddlt you fucking faggot.

1/3

*tips fedora*

HHAHAHHA this is rich a fucken mong calling me stupid when he doesn't know the meaning of independent event. Here is a hint retard "Idendent"

Incorrect. Read the explanations in this thread.

1/3

2 outcomes:

HH
HT

since the first coin has to be heads. hence 50%

jesus they are independent events, the conditional probability makes no difference

Justin Trudeau says, "100%"

No, the first coin does not have to be heads. The second coin can be heads if the first isn't.

3 outcomes:
HH
HT
TH

Out of those, 1/3 probability of both heads.