How smart is Sup Forums?

How smart is Sup Forums?

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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/NSFWRedditVideo

>regular coins
>posts monopoly money

>worth more than Amerifat burger money

75% ?

Depends:
If we flip both and both land tails is one magically flipped to heads or is one guaranteed heads? it will either be 25% or 50% depending

50% - If one of them is definitely heads, technically you're only flipping 1 coin, and that leaves you with 50/50 chance

50% you fucking retard

how is this even a question

This? Im not very smart but: 1 lands heads, thus 50 percent of the task completed. The 2nd coin has a 50% chance to land heads so it gives u a 75% chance that both of the coins will land heads, assuming one has landed heads. But if the one coin always lands heads then shouldnt it really be 50%?

Conditional probability only works when events are not independent. In that case P[B|A] = P[B]

>3
>what is the probability that at least one landed heads
>given one of them landed heads
>one of them is already heads

"what is the probability of a coin landing heads" is now the question

the answer to which is 50%

jesus

Two (I assume fair) coins were flipped, at least one was heads. This are the possible outcomes

HH
HT
TH
(not TT, since at least one head)

since we have three options, and only one options gives to heads, the probability is 1/3.

>Dependent

>one of them is already heads

there are 3 different ways for two coins to be "one is already heads". HT, TH, HH. of the 3 ways for two coins to be "one is already heads", one of those is HH.

>answer is 1/3

50%

2 of those are the same option dipshit

33.33333%

stop trying to sound smart idiot, TH and HT are the same because the coins are the same
50 fucking percent

7

If P(A|B) = P(A), then events A and B are said to be independent. In such a case, having learned about the event B does not change our knowledge about the event A. Also, in general, P(A|B) (the conditional probability of A given B) is not equal to P(B|A). For example, if you have cancer you might have a 90% chance of testing positive for cancer. In this case what is being measured is that the if event B "having cancer" has occurred, the probability A - test is positive given that B having cancer occurred is 90%, P(A|B) = 90%. Alternatively, you can test positive for cancer but you may have only a 10% chance of actually having cancer because cancer is very rare. In this case what is being measured is the probability of the event B - having cancer given that the event A - test is positive has occurred, P(B|A) = 10%. Falsely equating the two probabilities causes various errors of reasoning such as the base rate fallacy.

Prove it. You can't just say things.

Anyone saying 50%, please kill yourselves or at least never have kids.

The answer is 1/3.

For a 2 coin flip, there are 4 equally probable outcomes:

Heads - Heads
Heads - Tails
Tails - Heads
Tails - Tails

At least 1 coin landed heads, so Tails - Tails is no longer possible, leaving 3.

HH
HT
TH

HH is 1 of those 3

1/3

...

if you want to combine the two options into one, then you have to combine their probabilities as well!

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

combine into

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

and the answer is still 1/3

75%.
Heads/Heads - Can be.
Heads/Tails - Can be.
Tails/Heads - Can be.
Tails/Tails - Can't be.

Anyone who argues otherwise is a complete fucking retard and should go back to school.

we hAve coins a and b

being heads =h and tail =t we have the following possibilities

ah bh
ah bt
at bh
so it's a 2/3 or 66% of a and b not being heads at same time

Fuck, user, that's freaky.

you're wrong in so many levels

>Fuck, user, that's freaky.

Are you 12 years old?

Fucking spergs. Trick question.

The probability that two independent events BOTH happen is product rule. So P(A) and P(B) occuring is P(A)*P(B).

> P(A) = 0.5 ; P(B) = 0.5.
> P(A) * P(B) = 0.25. [two fair coins both heads]

The question is asking, what is the probability given that one of them is known to be heads?

Well, these are independent events (one does not influence the next), so it's exactly the same. P(B) is the same 50%, but both occuring is 25%.

It's 33,333...%, 1/3 possible outcomes, the outcomes being tails/tails, tails/heads, heads/heads.

50%

Christ. Thanks common core.

>Trick question.
No. It's a basic conditional probability math question, you underaged faggot.

Conditional Probability:
Probabilty of Event A given Event B

Use Bayes' theorem.

A = "both coins are heads"
B = "at least one coin is heads"

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

So many morons in this thread.

For all you fuckers saying it's 1/2, I'm really not sure if you're trolling or just being thick as shit.

The question doesn't say "assuming coin #1 has landed heads" (call this scenario 1), it says "assuming at least one coin has landed heads" (scenario 2). In this explanation, HT means "coin 1 lands heads and coin 2 lands tails".

Normally when tossing two coins you have four outcomes:
HH
HT
TH
TT

In the case of scenario 1, you have two possible outcomes:
HH
HT
We can ignore TH and TT because we KNOW for a fact that coin 1 landed heads. Therefore there are 2 outcomes, 1 of which is HH, so the answer is 1/2

In scenario 2, you have three outcomes:
HH
HT
TH
We can ignore TT, as we know at least 1 coin landed heads. So we have 3 outcomes, 1 of which is HH, so it's 1/3.

Mathematically 2 of those are the same, idiot.

Here

One is heads. It is removed from future of equation. As it is a definite H.

You are now left with H or T.

The answer is 1/2, .5, or 50%

You're all stupid.

C O I N S
O
I
N
S

>Mathematically 2 of those are the same

You're retarded.

Now think about it again until you figure out how you are wrong.

Protip: Imagine using a penny & quarter as the 2 coins.

Confirming 1/3

>One is heads. It is removed from future of equation.

No, try again.

There are four different combination that one flip with two coins can produce.
heads, heads
tails, tails
tails, heads
heads, tails
(yes you differentiate the last to combinations since both coins got a heads and a tails side)

therefore the chance of getting double heads is one out of four or 25%

Now given that one coin already landed on head the chances are changing because there are only two possible combinations left:
heads, heads
heads, tails

therefore the chance is one out of two or 50%

That doesn't change the probability you fucking ape.

> Conditional
> Bayes' Theroem

AHAHAH you're serious. No fundamentals.

Using Bayes, TH and HT are separate events. You might as well be crossing off TH because "the first one was heads". Then you know what you get? 50%.

Conditional table is the ruse, HT and TH are not separable events, its:

> BOTH H
> ONE H ONE T
> [xxxxBOTH Txxxx] crossed out.

This is an elementary ruse in combinatorics. Treating combined independents as conditionals. You fell for it.

Underage faggot? I have an M.S. in biochem you gook. Get off my bengalese carpentry board.

Idk, I'm bad at social studies

fucking faggots stop treating TH and HT as different events, goddamn retards

The question is not just the probability that both landed heads (25%), it's the probability that both landed heads _given_ that at least one of them landed heads.

YOU ARE SO FUCKING CLOSE YOU DUMB NIGGER

That mathage will come in handy while you're flippin' burgers, kiddo

It depends. Are these feminist coins we're talking about?
Those always land on their face.

correct me faggot

They ARE different events. The two coins are different.

But they are.

If coin 1 is a quarter and coin 2 is a penny, then HT and TH are different events - HT is a heads on the quarter and a tails on the penny, and TH is a tails on the quarter and a heads on the penny - DIFFERENT EVENTS

1/3

This thread makes me worry about the future of our species.

Anyone saying 1/3, please kys to save the future

What's it like being retarded, user?

Given that AT LEAST ONE COIN (note it doesn't specify WHICH coin, just A coin) landed on heads leaves you with HH, HT and TH

no because coin A is fixed on heads you dumb fuck.

Correct.

>Now given that one coin already landed on head the chances are changing because there are only two possible combinations left:

Nope. Only tails, tails is ruled out, leaving 3

heads, heads
heads, tails
tails, heads

1/3

Revise your solution and correct your mistake.

>That doesn't change the probability

I know. The answer is 1/3.

COIN A ISN'T FIXED, WHY ARE YOU SO FUCKING RETARDED

Learn to read you god damn idiot, are you too stupid to understand op's picture?

>"the first one was heads".

Question never says that, read it again.

If the FIRST coin landed heads, the probability that both coins are heads is 1/2

If AT LEAST ONE coin landed heads, probability that both are heads is 1/3

Pic Related

Source: Slide 4 Washington University Math Dept
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

I hope you didn't pay for your shit-tier education.

>the probability that both landed heads given that at least one landed heads.

It doesn't specify which coin landed heads. You only know that ONE has already been thrown and landed heads. Therefore, this leaves you with one coin to flip and a 50% chance of both landing heads.

There's really only one case here. You are currently evaluating for Y, knowing that X (order unspecified) is already heads. Y has a probablility of 50% each scenario. If you wanted to evaluate all possible cases you'd actually have:

xH
xT
Hx
Tx

Where x is definitely heads. So a 1/2 probability.

Nope. This is a run of the mill undergraduate combinatorics trick question. If it was worded "what is the probability that both landed heads given that the FIRST one landed heads", you would know to use Bayes / separate time-series. Because it's worded "what is P(HH) given that AT LEAST ONE landed heads", it combines the events. As if I was throwing two coins in one hand, TH and HT are the same event. This gives you 50% using conditional table, which is, not coincidentally, the probability of the second coin being heads.

Separating coins. Doesn't matter because of "AT LEAST ONE" trick in question. Nope.

I don't know, really, the real retards are those who can't ever accept that we can be wrong. What does that feel like, exactly?

You're retarded. No coin is fixed.

Answer = 1/3

It's 2/6 and if you think it's anything else you should die because their mother will die in her sleep tonight if you don't reply to this post

>at least one coin landed heads
means something sifferent to
>coin A is fixed as heads

You retarded nigger.

OP question allows for either coin to be tails, just not both simultaneously.

There are 3 equally probable outcomes that satisfy the condition of having at least one heads

heads - heads
heads - tails
tails - heads

1/3

Cry moar

I say it's 75 % due to Tree Diagram. (Worked on the goat problem too, so why not here.)

You still don't get it, lol. "the first one was heads" is in quotes because it shows the idea that these events are seperate time-series, I.E. conditional probability. "AT LEAST ONE" means HT and TH are the same event.

Still giving me conditional probability rules for an undergrad-tier question in combinatorics? Give me a break, fag. And no, I didn't pay a cent, full ride, thanks SAT. I hope you're paying for yours.

>You only know that ONE has already been thrown and landed heads. Therefore, this leaves you with one coin to flip and a 50% chance of both landing heads.

No. 2 coins were flipped ant AT LEAST ONE coin landed heads.

For a 2 coin flip, there are 4 equally probable outcomes:

Heads - Heads
Heads - Tails
Tails - Heads
Tails - Tails

At least 1 coin landed heads, so Tails - Tails is no longer possible, leaving 3.

HH
HT
TH

HH is 1 of those 3

1/3

> still falling for this thread
> literally takes longer to flip a coin than to just google the fucking answer

And if we now used this to give an answer to the actual question, the solution'd be 25%

>provides no math solution
>tells others they don't get it

top kek

You just got BTFO by University of Washington.

1/3

Because Monty Hall is an exercise in conditional probability. P(A) influences P(B) because the number of doors decrease. That's the only reason why it doesn't work.

For those who haven't got it yet

> Conditional(HH/HT/TH/TT) -> 0.75
> Independent(2H/1H1T) -> 50% or probability of flipping a coin.

Just FYI, my probability friendo.

lolwat

>conditional probability
>trick question

Never reproduce.

> Conditional(HH/HT/TH/TT) -> 0.75
> Independent(2H/1H1T) -> 50% or probability of flipping a coin.

Your post makes zero sense

I don't quite get the difference between conditional and independent yet. I mean, if you ask "How high's the probability to get two heads in a row", you also use tree diagrams to see 50%*50% = 25%. At least that's how we handled that in school.

you are fucking retarded.

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

>provide math solution
>get called retarded

Sure thing, m80.

33%?

Forced myself to answer within 10 seconds or so

What is the probability that a coin never lands on anything but heads ever ever ever

The question is a weirdly worded math circle jerk, by asking the same question in different way the answer can be 1/2, 1/3 or even fucking 1/4

In this case, its 1/3

>by asking the same question in different way the answer can be 1/2
>same question

No. It would not be the same question.

Here is the OP question worded differently.

What is the answer?

protip: 1/3

>this

Lets ask it this way.

You flip 2 coins and see that 1 landed head before looking at the second coin, what is the probability of both landing heads?

Or

You flip 2 coins, 1 of which is a double headed coin, what is the probability of both landing heads?

Your question assumes that the coins are flipped at the same time and doesnt specify which of the coins was heads, while most people would assume you look at them 1 at a time, its a stupid question made to troll people into saying 1/2 so you can laugh at them.

You're asking a totally different question you fucking retard.

>ITT: people who don't realize both coins have landed already.

The answe is 50%

yes, but im asking the question most think YOU are asking, still dont get it?

>being this dumb
Your so stupid it hurts

*your

>OP clearly states one coin will always land on heads
>it suddenly becomes a question of "if I have a coin what are the chances it lands on heads if I flip it?"
>clearly 50%

Niggas what?

>most
I would hope not. The OP's wording is EXTREMELY transparent. The people here read it, processed it, and spat out an incorrect answer with confidence. There was no tricky wording or cheap tricks, it's just that most people aren't very smart when it comes to probabilities.

this doesnt measure smartness. It just tests whether someone can think in a technical or logical way. Its like saying: if youre bad at math then youre not smart.

I'm disappointed in you, user. You can't just keep giving me examples of conditional probability for this independent event. That's the whole point of this question.

Yes.

> probability of two heads = 0.25
> probability of two heads given -AT LEAST- one head: 0.5 (probability of a coin flip)
> probability of two heads given -THE FIRST- is head: 0.333~

Did you notice the words "the first" in that slide titled 'Cond_Prob'? I don't know how to tell you this, but it's just academic laziness. It's from the conditional probability unit, so they are using Bayes. You can see this exact same treatment, they make HT and TH separate events, meaning there is a difference between the "first head second tails" and "first tails second head", where these are forced to be the same instance by use of "at least". An oversimplification for undergrads where the nuances don't exist.

Thank you, user.

Incorrect

You guys are Retarded

HH
HT
TH
TT

Each Side has a 50% Chance of landing on Heads, you determine probability my multipling the two chances of each coin individually landing on heads together, which equals 25%. So out of two coins its a 25% chance of both landing on heads.

If we dismiss TT

HH
HT
TH

Each coin has a 66% Chance to land on heads, which multiplied together gets us 43.5%

No shit, everybody is wrong.

What the fuck, you... you can't be serious. Someone disprove this shit right now.