How smart is Sup Forums?

How smart is Sup Forums?

Other urls found in this thread:

math.stackexchange.com/questions/1805277/if-two-coins-are-flipped-and-one-gets-head-what-is-the-probability-that-both-ge
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf
twitter.com/SFWRedditImages

0.25

1/3

50%

1/3

Either it landed head or it didnt, so 50%
/thread

Lmao at 25% and 50%
At least one coin is always heads, so there's only 3 outcomes

>Coin A heads, coin B tails
>Coin A tails, coin B heads
>Coin A heads, coin B heads

1/3

HH
TT
HT/TH

HH is the only way.

420.69%

I see no head on those coins. You guys have been duped, OP is a prank artist

1/2 duh

Trick question, those arn't regular coins, that is europoor money.

Are you retarded? That doesnt even make sense

retards

>worth more than US niggerdollars

3/4
Now fuck off

Retard.

It is though, and calling everyone retard isn't fooling anyone

...

Firetruck.

It's right though.
It doesn't make sense if you're retarded.

>It is though

It isn't. Have fun explaining it.

LISTEN UP FAGGOTS

Here are all the possible options (excluding the side cause that's fucking stupid)

HH
HT
TT
TH

Bacause HT and TH are indistinguishable, they count as one. And because there are TWO ways to get TH, but only one for each of the other two, it is 50/50. YOUR WELCOME.

Just tried it...
,,,got two tails.

:(

There are 4 possible outcomes: HH, HT, TH, TT. However, we are also given the constraint that at least one must be heads.
Given this constraint, our sample space is now {HH, HT, TH}.
Now it is clear that the probability is 1/3 since HH is a double head flip.

Faggots.

>Bacause HT and TH are indistinguishable

Fucking retard.

do your own fucking homework this ain't Sylvan's learning center.

Haha! Hilarious!

TH=HT so they cancel out

roll

>TWO ways to get TH
>they count as one

Does not compute.

33.32%

>TH=HT so they cancel out

01 = 10

ab = ba

You're retarded is what I am getting at.

Retards think that TH is different than HT LMAO

Cant tell if trolling or actually retarded

You need better b8.

How are your first two listed outcomes different? They're essentially the same outcome. That should leave only two different results, 50%.

Sup Forums can't even count to 10.
This is mind blowin'.

Let A: p (both heads)=0.25
Let B: p (at least 1 head)=p (not both tails)=1-p (both tails)=1-0.25=0.75

P (A and B)= both heads and at least 1 head = both heads=0.25

P (A|B)=P (A and B)/p (B)= 0.25/0.75=0.33 (or 1/3)

If one is already given as heads then the question simply becomes a case of tossing one coin in which case it's 50/50

Allow me to explain, since it has become clear that you are a simpleton.

Let one of the coins be a penny
Let teh other coin be a quarter

Flip both. In the cases of there being one heads, and one tails, you could have:

penny=heads, quarter=tails
or
penny=tails, quarter=heads

2 distinct, separate and equally probable outcomes.

Answer to OP question is 1/3

Of course you could follow up with why they're retards instead of trailing off like a moronic pile of dead brain cells

1+2 = 2+1
Order shouldn't matter here.

If one is always heads you pretty much only have one coin to worry about, so it's 1/2 chance like a normal coin flip.

>How are your first two listed outcomes different?

Can you read/logic?

It depends. How pedantic do you want your answer?

If both are tossed simultaneously, it's 1/3.

if both are individual, it's 1/2

something tells me there's some bullshit way to get 0 but whatever.

Correct.

>explaining why they're retards

But it's already obvious to anyone who isn't.

Rain drop. Drop top.

We're not performing addition, you fucking simpleton.

You're assuming:

HT
TH
TT
HH

with TT not being allowed as a precondition. However, what if the first is always tossed to be heads? It never says how they are flipped.

>If one is always heads

False assumption

>Atleast one of them is head
So you basically flip one coin
So its 50%
Now kys 1/3 fags

You're an idiot. It does not matter if they're tossed simultaneously, sequentially, or if you toss a single coin twice.

In all cases where at least one coin (or one toss) landed heads, you will get 2 heads 1/3 of the time.

Prompt guarantees at least one lands heads.

What the hell even are "heads"

>given that at least one of them landed heads?

No, it's fair. Just depends on how you look at it.

1/3 and 1/2 are valid answers depending on how they are flipped. 1/3 for both at the same time and 1/2 for each individually.

TH and HT are the same thing, ya fuckin idiot. One is heads, the other is tails.
That's like me saying:
X+1=2
And
1+X=2
Are completely different equations, despite them both having the same answer.

> what if the first is always tossed to be heads?

then teh answer would be 1/2.

However, the question states AT LEAST one coin landed heads, so either coin could be tails, just not both at the same time.

HH
HT
TH

all equally probable and valid outcomes

1/3

Are you illiterate? Honest question.

>So you basically flip one coin
>2 coins were flipped

1/3

deciving

No, it DOES matter. If you have two coins flipped simultaneous with the precondition that AT LEAST one is heads, you have the Monty Hall problem all over again. This give you:
HT
TH
HH
with TT excluded.

However, if we flip one KNOWING it must be heads, then we get a simple 1/2. It's due to the poorly worded question.

4 outcomes:

HH
HT
TH
TT

TT is not allowed, so our sample is not HH, TH, and HT.

this isn't rigorous, but HH is one result out of three possible outcomes. 1/3

see this guy for the more rigorous mathematical justification. anyone not answering 1/3 needs to review their day 1 combinatorics.

>1/3 and 1/2 are valid answers

Not for OP question. Answer is 1/3.

>1/3 for both at the same time and 1/2 for each individually.

Incorrect. It's 1/3 for both, as at least one coin landed heads, not the FIRST coin

If the probability was 50% both coins would land balanced on their sides every time

But what if at least implies that you flip one until it is heads? The nature of how we flip the coins is important.

You better be baiting, faggot.

No, but clearly you lack basic reading comprehension.

If I tell you that I flipped a coin and it landed heads, does that imply that it ALWAYS lands heads, you fucking simpleton?

It's only asking the probability of them both being heads. We already know one is going to be heads, so it can be discarded. The answer is 50%.

33%

At least one of them ALWAYS lands heads, means T-T option is eliminated and never happens (magical lucky coins).

So the only options are TH HT and HH 1 out of 3 is 33.33333333%

if there was no sentence saying GIVEN ONE LANDS HEADS ALL THE TIME BECAUSE FUCK YOUR REAL WORLD LAWS, it would be 25%

Nope. If we flip both hoping both land on heads and given the precondition of at least one being heads, we get 1/3.

If we flip the coins until one is heads AND THEN the other, we get 1/2. It's still valid in OPs question which is why there is a discrepancy.

>if we flip one KNOWING it must be heads, then we get a simple 1/2

Incorrect.

Bayes' theorem, probability trees, punnet squares all confirm you are wrong.

Answer is 1/3.

Cry moar.

No no no no no. Remember, HT =/= TH. You want the probability that one or more is heads, which is the same as 1-P, where P is the probability that neither coin lands on heads. P = (1/2)^2 = 1/4. 1 - 1/4 = 3/4, so the probability of at least one heads is 3/4 or 75%.

>If we flip the coins until one is heads AND THEN the other

But that's not the OP question. Why are you making up different questions?

It's saying one coin will ALWAYS be heads, you stupid nigger. It's asking the what the chances are that they're BOTH going to be heads. Given that ONE is ALWAYS HEADS, it can be ignored. The only thing that matters is if the OTHER is heads or not.
Did me HIGHLIGHTING KEY WORDS help your middle school mind comprehend this simple line of reasoning?

Never said 1/3 isn't valid. Conditional probability is valid but the semantic matter.

I think this is one of the cases where logic/mathematics totally goes overboard on a trivial problem, and you (as well as your friend) are overthinking it. But it's fun anyway, so...

There are at least 3 possible answers which are equally correct, depending on how pedantically one tries to twist the wording one way or the other. But for practical purposes it does not make a lot of sense because there is only exactly one solution that is correct (and immediately obvious) in each situation.

The first obvious interpretation is that you toss two coins into the air at the same time (which is not what you describe!). There are 4 possible outcomes. One of these outcomes has two heads, and one has no heads at all. You have set up the precondition that one coin gets head, which rules out the "no heads" outcome, leaving 3 possible outcomes. Only one of the three has two heads in it, thus: 1/3. This is a dependent probability. It is also an example of a Monty Hall Problem.

The second obvious interpretation is you toss one coin, and it comes up head. That's the precondition. You could just as well not have tossed the first coin at all. You now toss the second coin. Alternatively, you can toss the two coins together, but ignore all cases where the precondition that the first one gets head isn't fulfilled.
Assuming the second coin is not weighted or a trick coin with two heads or such, the chance is, of course, 1/2. From the point of view of the second coin, the first coin doesn't exist at all. This is a single (independent) probability.

Anything worth doing is worth doing right. (Faggot)

51%

>You want the probability that one or more is heads

No, we want the probability that BOTH are heads given the condition that AT LEAST ONE landed heads.

HH
HT
TH

all valid. All equally likely.

1/3

Fuck the coins even, imagine a binary system where 1 number added to another.

Rule is : both numbers generated at random, if first is 0 , second is 1 if first is 1 , second is 0 or 1 , SUM of both is never 0 (equivalent of one coin always heads), what is the chance that sum of both is 2 ? yes, same shit 0+1 1+0 1+1 33%

Because it just says that two coins are flipped. Not "two coins are flipped at the same time thus giving you a perfectly valid reason for using only conditional probability".

It's poorly worded to cause infighting, not because it is hard.

math.stackexchange.com/questions/1805277/if-two-coins-are-flipped-and-one-gets-head-what-is-the-probability-that-both-ge

>The first obvious interpretation is that you toss two coins into the air at the same time
OK

>There are 4 possible outcomes
Correct

>One of these outcomes has two heads, and one has no heads at all.
Yes.

>You have set up the precondition that one coin gets head, which rules out the "no heads" outcome, leaving 3 possible outcomes.
Indeed

>Only one of the three has two heads in it, thus: 1/3.
Correct

>This is a dependent probability. It is also an example of a Monty Hall Problem.
It's a basic conditional probability question.

>The second obvious interpretation is you toss one coin, and it comes up head.
This is a false assumption given the information in the OP question. You are adding information to the question which invalidates your solution. You are essentially solving a question that was never asked, and therefore it is an incorrect answer to OP question. Get out.

Literally BTFO'd

They don't have to be flipped at teh same time you moron.

One of them could have been flipped last week.

The important piece of information is that 2 coins were flipped and at least 1 landed heads.

Nothing more, nothing less. With this information, the answer is 1/3.

50%

No this is important. It's in the same notion that choosing a door, one of the incorrect ones has been chosen for you.

Prove to me the question cannot imply this. It simply says two coins are flipped. One coin can be flipped to heads and then the other. That still satisfies OP's question.

If we know one (say the first) is heads then:

HT
HH

are the only two that can happen, correct? Thus it's 1/2. This is different than flipping both at once.

>Prove to me the question cannot imply this.

>2 coins were flipped. At least 1 landed heads
does not imply
>first coin was flipped and it landed heads

Do you understand?

Pic related

Source: Washington University (see slide 4)
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

Answer to OP = 1/3

3/4

No I don't. It doesn't describe the nature of how we are flipping the coins. It simply says that two coins are flipped and one must be heads. That isn't to say they cannot be flipped individually. The link you have provided does not address the poor semantics of the OP's scenario.

>2 coins were flipped. At least 1 landed heads does not imply first coin was flipped and it landed heads

It very well could. I can flip two coins with one hour apart from one another. The first coin was tails but the second was heads. This satisfies
>2 coins were flipped and one was heads

>If we know one (say the first) is heads then:
>HT
>HH
>are the only two that can happen, correct?

Yes, but we don't know the first is heads, you fucking pleb. That is information not given in teh question. We only know that at least 1 landed heads, and it doesn't matter how or when the coins were flipped, you imbecile.

Example:
I have 2 coins.
I flip the first one. I note the result.
I flip the second one. I note the result

I tell you that at least 1 landed heads. What is the probability that both landed heads?

What is your answer (they weren't flipped at the smae time)

Protip: Answer is still 1/3

>It simply says that two coins are flipped and one must be heads.
>must be

No, at least one coin landed heads.

>The link you have provided does not address the poor semantics of the OP's scenario.

kek. Sure thing, lad.

47/35

Well if one of them HAS to be heads, then only the other coin flip would really matter right? So it would just be 50/50 right? You guys can be so autistic sometimes

What the fuck

its fifty percent,if one is garunteed to land heads, the other still has a fidty fifty of getting heards

O.O who is that? I've been looking for her for ages