How smart is Sup Forums?

How smart is Sup Forums?

Other urls found in this thread:

en.m.wikipedia.org/wiki/Monty_Hall_problem
en.wikipedia.org/wiki/Boy_or_Girl_paradox
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf
math.stackexchange.com/questions/991060/flip-two-coins-if-at-least-one-is-heads-what-is-the-probability-of-both-being
twitter.com/SFWRedditImages

doesn't matter what anyone says. As soon as someone comes along with the right answer OP and others are going to sperg out and claim that they're wrong.

...%50? No idea

50%

Nice dubs, but both wrong.

I don't sperg out. i give the right answer and then call the other user retarded for not understanding it.

50%

1/3

you just claimed that both of those anons were wrong and proved my point

1/4

Out of the 4 possible states, one is removed, leaving 3 possible states, one of which is the condition for the answer.
1/3

That's because they're wrong. No sperging out necessary. Your point is invalid.

oow sorry thought how big is the chance the both got head

75%

the question is what is the probability of both coins landing heads, if it were a single coin it would be 50%

Answer the OP question

1/3

4 possibilities
TH
HT
HH
TT


Given that one landed H, the TT is eliminated

Would go for 1/3

Outcomes for coins:
heads heads
tails heads
tails tails

33%

TH and HT are the same

>TH and HT are the same

No, user. They are not.

This is not how reality works. The question is basically "what's the probability of getting heads on one coin" since the other has already happened and >>cannot affect the other coin

It's taking into account each coins as a separate entity. As either one of the two is guaranteed to land on heads you have to take that into account.

It's 1/3

this was a problem on my last stats test
lel
this nigga thinks learning a simple formula makes someone smart

25% perhaps? If both coins have a probability of 50% each then after two flips they would both have landed heads I guess.

TH and HT are not the same. IF you want to combine those two results into one, you have to combine their probabilities as well.

TH/HT - 2/3
HH - 1/3

At least 1 landed heads.

TT
TH
HT
HH

If the first coin is heads, the full outcome is either HH or HT, so 50%
If the first coin is tails, HH can't be achieved so 0%

Daft question that requires fiddling with probability in a way that makes an actual answer impossible

It doesnt matter because they are euro's and they are worthless.

this fucking reeks of verysmart

You're missing the point, faggot.
It's not that being able to give the correct answer means you're smart.
It's that being unable to give the correct answer means you're stupid.

If at least 1 landed heads then we only care about the other coin. And that coin has a ~50% chance.

So about 50%

So a two coins flipped independently will affect each other?

what is conditional probability?

It's a basic conditional probability question.

When you flip 2 coins and at least 1 coin lands heads, you could have

heads, tails
or
tails, heads
or
both heads

All 3 are equally probable.

1/3

Doesn't matter, the entire thing depends on the first coin and leads to two different answers, so there realistically and usefully is no answer

A & B happening: 1/4
A happening: 1/2

B happening given A: 1/2(A) X 1/2(B) = 1/4 , ans.
It is a simple branch

(heads/tails)*(heads/tails)=

1/2*1/2=1/4

you forgot both are tails xD

That's how reality work.
Should read about monty hall problem
en.m.wikipedia.org/wiki/Monty_Hall_problem

The question is not
"if I already launched one coin which landed head and then launch a second coin"
But "if I already launched 2 coins, and at least one landed head"

75%

A man flips a coin and its heads and bets me a dollar the next flip will be heads, what the chances this flip will be heads?

Yes, and since you don't know which coin will be heads there are two separate answers

>at least 1 coin lands heads

Try again.

50%, but that's a different question to the OP question.

Read the OP question again.

1/3

If at least one HAS to be heads then they're counterfeit coins and the whole puzzle is pointless

You didn't read about monty hall problem right ?
Or Just trolling me

50% for one coin flipping.
25% for 2 coin flip bet.
In probabilities it doesn't matter if the object are similar.
They will be always called A & B even for same values.
(A & !B) != (!A & B)

>The question is basically "what's the probability of getting heads on one coin" since the other has already happened

No, it isn't, retard.

the question is
>What is the probability that both coins landed heads given at least 1 coin landed heads?

4 possible outcomes from a 2 coin flip

HH
HT
TH
TT

3 of them contain at least 1 heads

1 of those 3 is both heads

1 of 3

1/3

Stay in school, kids.

The answer is ONE IN THREE.
There are four possible outcomes:
1.Heads/Heads
2.Heads/Tails
3.Tails/Heads
4.Tails/Tails
You already know that one of the coins came up Heads which means you can discount Tails/Tails. There are now only THREE possible outcomes.

I know about Monty Hall and it's another retardedly phrased question whereby math autists don't take into account the obvious human interaction in the events. For there to be 100% probability that at least one coin lands heads (as per OPs question) one of the coins has to be double headed, in which case it's just a 50% that rides on the other coin. Doesn't matter what the numbers in your little textbook say, when you put something like this into a physical example there are things that change how it's got to be worked out

The confusion comes from the way the problem was stated and typically visualized. Most people are imagining looking at one coin, seeing that it's heads and then being aware that the second coin is independent assigning a probability of 50% for that coin. The question is actually asking you to flip both coins, and then have a second person tell you, at least one of them is heads, after looking at both coins, which gives the probability of 1/3. This is only confusing because in normal interactions, we naturally try to conceive of how we got each piece of information, and the most natural way of getting the information that one coin was heads is to look at a coin.

tl;dr problem is counter-intuitive because most people imagine needing to look at a coin to know the given information.

>No, it isn't, retard
Yes it is, double retard. For there to be a "given" in a situation, that has to be 100%. If one HAS to be heads, that coin doesn't affect anything else in the sequence

If you're not trying to troll here, I'm hope you don't vote.

If you're trolling....well good job, you got me

>For there to be 100% probability that at least one coin lands heads

What the fuck are you talking about, you idiot.

The OP question never says that 1 coin is 100% guaranteed to land heads.

2 coins were flipped and at least 1 landed heads. This is a perfectly realistic scenario and a basic conditional probability question.

>Doesn't matter what the numbers in your little textbook say

So you're saying that math is incorrect and you're right, then?

kek retard

1/3

yall fucking retarded except for these two anons

You can get the feel for coins and get the right height and flip speed to land 80% of shots called. And when u do land on the other 20 u learned what height or flip speed is to little or much and can adjust. Only solid thing to a coin toss is it landing and bouncing oba a hard surface so now it is influenced by the floors probability of pure randomness. Unless ur some calculating robot and take the bounce, stop time of the bounce, new flip speed distance from there to possible landing spot and even then you would have to do it twice to get a range on the bounce flip speed just to tune it in..... In theory

The "given" is that AT LEAST 1 coin landed heads.

That ius not the same as saying "1 coin is 100% guaranteed to be heads."

This is why you are retarded, because not only are you mathematically illiterate, you also lack basic reading comprehension.

I hope English isn't your first language.

1/3

Pic Related

>The OP question never says that 1 coin is 100% guaranteed to land heads. 2 coins were flipped and at least 1 landed heads.

...

You not being able to translate problems into real life isn't me trolling dear. When things in the real world are affected by outside forces (ie: the CERTAINTY of one coin being heads) it changes how the question works. 1+1=2 but two halves of an orange don't necessarily make a whole one. The physical world matters

So....explain it to me cause I still believe it's 1/3

You still don't understand how those are 2 different things?

Ok, let me explain it to you like you're a 5 year old child.

I flip a coin and it landed heads
IS NOT THE SAME THING AS
This coin is 100% GUARANTEED to land heads.

Do you understand now?

Nigger

>The OP question never says that 1 coin is 100% guaranteed to land heads.
Yes it does. GIVEN THAT AT LEAST ONE IS HEADS means that one HAS TO BE HEADS. You not being able to read doesn't make you right. Be angrier about your autism

>I flip a coin and it landed heads
That's not what the question says fucknuts

Read this Realize you're retarded.

1/3

No, 100% chance that one of two isn't the same thing that 100% of one

Realise that you have no grasp on how the physical and theoretical are different

No shit, retard. Why do you think I was quoting the OP question?

If they're identical then it amounts to the same thing. You're still looking for theoretical answers to physical questions

>the physical and theoretical are different

Bayesian probability isn't theory, it's theorem.

A mathematical proof:

This is a conditional probability question, so use Bayes' theorem to solve.
P(A|B) = P(A∩B)/(P(B))

Explanation for faggots:

COnditional probability questions take the form:
>What is the probability of Event A given Event B?

OPs question is
>What is the probability that BOTH coins landed heads, given that AT LEAST ONE coin landed heads?

So here

A = "both coins are heads" = {(HH)} = 1/4
B = "at least one coin is heads" = {(HH), (HT), (TH)} = 3/4

P(A|B) = P(A∩B)/(P(B)) = (1/4)/(3/4) = 1/3

Kys that's 2 questions at once

ITT:
en.wikipedia.org/wiki/Boy_or_Girl_paradox

I don't deny that the probability of (on paper) getting HH out of a choice of HH, HT, and TH is 1/3 but what you seem to not understand is that the world isn't a piece of paper, and tampering with the real world causes it to have different results

conditional probability: P(A|B) = P(A&B)/P(B) (apologies for incorrect notation). reads as probability that A occurs given B occurring is the probability of both divided by the probability of b occurring independent.

Changing this to our situation, A=B=H (not same coin though). The joint probability of getting both heads is 0.25, the probability of getting heads on one coin is 0.50. So 0.25/0.5 = 0.5.

The above is operating on the fact that the coins are distinguishable. Let us assume we cannot distinguish the coins, this changes the situation so that we can no longer determine which of the coins flipped landed on heads, we just know that one did. In this instance we know there are three possible outcomes: HH, HT, TH. In this instance we arive at the 1/3 answer.

>I don't deny that the probability of (on paper) getting HH out of a choice of HH, HT, and TH is 1/3 but what you seem to not understand is that the world isn't a piece of paper,

You fucking retard.

Do you beleve that math does not apply to real life?

Take 2 coins and flip them 100 times. Count how many times you get at least 1 heads coin (~75 flips)
Count how many times you get 2 heads (~25 flips)

25(both heads) out of 75(at least 1 heads)

=

1/3

Try it, faggot.

Or if you're not retarded, write a simulation.

Pic Related: 1/3

the 1/2 comes from the assumption that a specific coin was heads, the 1/3 comes from the assumption that any coin landed heads.

I actually have degree in math. The problem here is that people are failing to distinguish between two separate problems.

Combinations represent equivalence classes of permutations. In this problem no additional information is specified regarding any criteria by which we could differentiate the coins, so it's safe to say that we're looking at combinations of possible outcomes rather than permutations. Thus we have: {{H, T}, {H, H}, {T, T}}
(Yielding a 50%)

However, if we were looking at permutations, we'd have: {(H, T), (H, H), (T, H), (T, T)}, in which case the probability is 33.3%.

Im more disappointed than usual for a thread like this. This is not rocket science.

>Do you beleve that math does not apply to real life?
Of course that's not what I think, but there are extra factors to take into consideration. If I have 3 oranges and cut them in half then glue them together after shuffling them around, I don't have the same three oranges even though 0.5x6=3 every time

holy fuck, user, how can you make such rudimentary mistakes?

>The joint probability of getting both heads is 0.25
Correct

>the probability of getting heads on one coin is 0.50.
Correct. The probability of getting AT LEAST 1 heads in a 2 coin flip is 0.75 or 3/4

>The above is operating on the fact that the coins are distinguishable. Let us assume we cannot distinguish the coins

That makes ABSOLUTELY no difference to the probability.

It does make a difference, see

P(heads|heads) = (1/2)(1/2) = 1/4 or .25

>Combinations represent equivalence classes of permutations. In this problem no additional information is specified regarding any criteria by which we could differentiate the coins, so it's safe to say that we're looking at combinations of possible outcomes rather than permutations. Thus we have: {{H, T}, {H, H}, {T, T}}
>(Yielding a 50%)

If you have a math degree, your university and/or professors are shit.

Do you honestly believe that the probability of getting 2 heads coins where at least 1 coin landed heads CHANGES depending on whether you choose to view them as permutaions or combinations?

Do you understand how retarded that is?

I really hope you're trolling about having a math degree.

THIS HAS TO BE BAIT, PLEEEEEAASE. PLEEASE I'M DYYYING

how is it different

It makes NO difference to the probability whether you can distinguish between the coins or not.

You are retarded if you think it does. Genuinely. Retarded.

Just think about it.

Nope. The physical world matters and if you don't take everything into account when you try to solve problems then you're forcing yourself to be stupid. Try to refute the orange thing, you can't

I'm starting to think everyone is trolling here, the answer is 50%.

en.wikipedia.org/wiki/Boy_or_Girl_paradox

You fucking morons

Because his question is asking
>what is the probability that the next coin flipped lands heads? Answer = 50%

OP question is
>What is the probability that both coins landed heads given at least 1 coin landed heads. Answer = 1/3

Why?

Because when you flip 2 coins, there are 3 equally probable ways to get AT LEAST 1 heads:

coin1=heads, coin2=heads
or
coin1=heads, coin2=tails
or
coin1=tails, coin2=heads

All 3 are equally likely. HH is 1 of them.

1/3

Meant this reply for you, user

The answer is 1/3, you fucking mongoloid. read your article.

>From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3

If FIRST coin landed heads, answer = 1/2

If AT LEAST 1 coin landed heads = 1/3

So 1/3 for OP question.

Source: Washington University
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

>claim the answer is 50% and that you are trolling if you answer 1/3
>post wiki that explain why the question is ambiguous and both 1/3 and 1/2 are correct with different assumption

Penis. Penis is always the answer.

Except OP question is not ambiguous. It's a clear basic conditional probability question that is easily solved.

The answer is 1/3 btw.

this question is purposefully ambiguous which leads to the two answers 1/2 or 1/3 probability. both are correct, depending on your interpretation of the question

ITT OP trolls everyone with this shit question posted at least once every week on Sup Forums

read this:
en.wikipedia.org/wiki/Boy_or_Girl_paradox

and read it IN FULL. not once does it say that there is a definitive answer to problems that are presented in the manner that OPs is

if you actually read the wiki you could understand why it is ambiguous.
i know the answer can be 1/3.
but you need assumptions to make that the correct answer. i know, they are basic and standard assumption we are use to make in probability problem, but they are NOT in the question.
our usual attitute in making these assumptions doesn't make the 1/2 answer wrong.

claiming 1/3 is the only correct answer is just as stupid as claiming 1/2 is

>but you need assumptions to make that the correct answer

no, you don't. All information in the OP question yields an answer of 1/3.

If you want an answer of 1/2, THEN you need additional information/assumptions.

It's a basic conditional probability question with one correct answer and the answer can be demonstrated with Bayes' theorem.

1/3

sorry, but you seem to lack reading comprehension skills. it isn't a "clear basic conditional probability question" because the answer is rooted in assumptions made by the reader

>this question is purposefully ambiguous

It's not ambiguous. You're just unfamiliar with conditional probability. Actually read the article, lad.

The only way that the answer could be 1/2 is if we ASSUME that we know which SPECIFIC coin is heads. This assumption cannot be applied to the OP question.

1/3

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

why do you faggots argue about things when the answer is freely available?