How smart is Sup Forums?

How smart is Sup Forums?

Other urls found in this thread:

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

1/2.

0.5

1/3

fuckwits

Anyone who says 50%, please kill yourselves or at least never have children.

The answer is 1/3.

For a 2 coin flip, there are 4 equally likely outomes:

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

If AT LEAST ONE coin landed heads, only
Tails - Tails
is eliminated, leaving 3 equally likely outcomes, each containing AT LEAST ONE heads coin.

Heads - Heads
is 1 of those 3

1 of 3

1/3

Pic Related

1 in 3

im sorry for you and your education

haha wtf man, do you have this prepared?

Its 50/50, as it it is for everything on this planet.
It either happens or it does not.

What do you think?

>your education
>you are education

exactly

Nice post, man, concise but at the same time completely explains everything. Well done.

19.735% and 96.432%

You fail to recognize that in fact 2 probabilities are removed. Assume each column is 1 of the coins. If one coin is already declared heads then both tails scenarios from one column would be removed. Thanks for the jnfographic.

I'm sorry that your mum didn't abort you

Is bait?
Is bait right?
I can never tell anymore.

To everyone saying it's 1/3, it's actualy 50 : 50. Why you ask?

We flip two coins, but one of them is granted to land HEADS, than we actualy flip only one coin, second can lay on the table with HEADS on it.

So possible outcomes are:
H : T
H : H

Why did I substract T : H?
Because it does not matter which coin landed H.

If we know that once coin will always be heads, than our only question is what will the other land on, it can land heads or tails (50:50).

That's why it's 50%

ps. sorry im little high atm

>For a 2 coin flip, there are 4 equally likely outomes:
>Heads - Heads
>Heads - Tails
>Tails - Heads
>Tails - Tails

heads - tails
tails - heads

these are literally the same, therefore there are only 3 original outcomes. a guaranteed outcome of at least one heads removes "tails - tails"

therefore there are only two outcomes. 50%.

It actually depends on why you are saying "at least one of them is heads".

Are you saying that because you already saw both coins or are you saying that because you saw only the first coin and depending on that coin's value were going to say either:

"At least one of them is heads" or "At least one of them is tails".

Without knowing more about what caused the speaker to say the fact you can't know the probability.

I'm close to the right answer but I've fucked up somewhere and can't see where.

Using conditional probability:

P(two heads|at least one head)=P(two heads AND at least one head)/P(at least one head)

=(1/2)/(3/4)

=4/6

=2/3

it doesnt say "already" anywhere in the post ya fuckin iiiiiiiiddddiiiiiooootttttt (y)

Donevin?

>given that at least one of them landed heads

retard

50/50

It doesnt matter one landed heads or not

What matters it the chances that new coin has to land on heads....

Which is generally 50%

P(a|b) = p(a and b)/p(b)
= (1/4) / (1/2) = 1/2

P(two heads AND at least one head) = P(two heads) = 1/4

Two heads always have at least one head.

>heads - tails
>tails - heads

>these are literally the same, therefore there are only 3 original outcomes

But in statistics, heads on coin 1 and tails on coin 2 is a different result to tails on coin 1 and heads on coin 2. If you toss two coins 1000 times, I can estimate that about 500 will be one heads and one tails, about 250 will be two heads and about 250 will be two tails

>Being this stupid.
Somehow, trying to point out someone is stupid (Even though they aren't) made you look 2x as stupid.

P(2 heads and at least...) = p(2 heads) = 1/4

It's 1/3.

The four equally likely possibilities are:
heads-heads
heads-tails
tails-heads
tails-tails

we know it is not T-T, so it must be one of the three others, hence 1/3.

The mistake that people who think it is 1/2 make is they think that you observe one of the coins, let's say coin A. If you know coin A is H, then yes, the p of B being H is 1/2. However you don't know which of the coin is definitely H, so it is 1/3 as described above.

Ah, of course.

That makes it (1/4)/(3/4)=1/3.

Thanks

It's worrying that I have a first-year uni stats exam on Friday

2/3, AKA 1/2...

You forgot about the side

math is made up so chances are truely imesurable. 1/3 if youre a nerd.

...

Correct.

Simulation confirms 1/3

the chance will be zero - it wont happen.

all that proves is that your CS101 skills made a shitty program to output the outcome you wanted

there was no need to flip coin1 as it's given that it's heads. had you programmed it according to the OP's image, it would have output 50%

I was stoned through most of middle school but


2 variables right? Each has 1/2 chance of being heads. You need to do some maths and shit so its like 1/2 * 1/2 or something, right?


So pic related we have heads and not heads I forgot what the other thing is. Rgjt? So we take 1/2 and 1/2 then tjetez the possibility of 1 coin being heads and the other being bottoms so then its 1/2 plus 1/2 makes 1.

Right?

So we take 2 coins, add another coin for statistical value and recognition and proof, so when the third coin turns around it proves my theorem:

Coins have 1/2 * 1/2 * 1/3 chance of flipping to heads.

So we have a 1% chance of Hilary being president.

I'm not sure if you're trolling or just retarded

you just resort to name calling because you're backed into a corner lel

explain to me why you need to flip coin1 when OP says coin1 is given.

Nice

...

Does nobody really thnk of sides?? so it could be like this, the coins are 3D, the have head, tails, and the fucking side

head - head
head - tail
tail - head
tail - tail
side - head
head - side
tail - side
head - side

So is it 1/7 or what?

>there was no need to flip coin1 as it's given that it's heads

Incorrect. No coin is specified, therefore EITHER coin could still be tails, just not both.

This is a conditional probability question.

The condition is that "at least one coin landed heads."

I account for this condition in the program.

If you were good at math and were aware of Bayes' theorem, you'd understand that the answer is 1/3.

NOTE: No coin is FIXED as heads. EITHER coin could be tails. This is where you're making your logic/math error.

1/3

Pic related: Washington University Math department

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

I'll wait for you to apologize. Be quick though. I have an omelette in the pan.

1/2.

>at least one lands heads
>other is heads 50%
>coin that landed heads 100% doesn't matter anymore
>1/2 chance to land on heads

Check

Probability of either coin landing on its side is negligible, user.

Coin 1 isn't given. The stipulation isn't "coin 1 has come up heads" (call this scenario 1), it's "one of the coins has come up heads" (scenario 2), which is a very different situation in statistics.

In scenario 1, we can get the following results:
HH
HT

We see that there are two options, one of which is two heads, so the probability is 1/2, BUT only if our stipulation is "coin 1 has landed heads".

Scenario 2: we get the following possibilities:
HH
HT
TH
TT

Out of these four outcomes, one of them (TT) doesn't have either coin come up heads. Since we KNOW that AT LEAST one of the coins has to come up heads (remember our scenario 2 stipulation), we can ignore this result. This narrows down our outcomes to:
HH
HT
TH
We see that we have three possible outcomes, all of them have at least one coin come up heads, and they all have an equal probability of occurring. So our probability of getting two heads, given that at least one of them (not saying *which* coin) is heads, is 1/3

QED

But still possible, so there is a chance, we don´t know where are we flipping the coins, or gravity, or nothing at all, we are just talking about posibilities, and side is a possibility

>NOTE: No coin is FIXED as heads. EITHER coin could be tails.

ONE coin is FIXED as heads. It's literally specified.

>hurr i'm an intellectual m'lady

good explanation. Also see image here

...

Shit you are right, I forgot about side - side

>ONE coin is FIXED as heads. It's literally specified.

No. Your reading comprehension sucks, user.

the question states that
>at least one coin landed heads

not to be confused with
>one coin is fixed as heads

Those are two very different things.

If either coin was fixed as heads, the answer would be 1/2.

If at least 1 coin landed heads, and it could be EITHER coin, then the answer is 1/3.

See the pic here

Damn, you're a dumbass. Are you black or something?

>Those are two very different things.
There is no world where both coins land tails.

You are a fucking idiot

You cannot discount a T/T occurrence
Therefore the probability is 1/2

>muh intellectual
>muh niggers

ur dumb suk a dik

this here is the real answer

>i can't explain why the order matters but IT DOES YOU RETARD REEEE: the post

Who gives a shit. At least one coin is heads. Okay. The other coin flip, whether first or last, will be either heads or tails. Two possibilities. 50%.

>
Fuck off back to .reddit you cunt

If there are four possible combinations, wouldn't it be 1/4?

No, because there are only two possibilities.

H, T is the same as T, H, the order is irrelevant

if you dont believe me, you have to list H, H twice, leaving you with 4 possibilities with equal chance for Heads and Tails

its 50%

>There is no world where both coins land tails.

Yes, but if 1 coin is FIXED as heads, it can't be tails.

The OP question does not FIX a coin, therefore EITHER coin could be tails, just not both.

3 equally likely outcomes

HH
HT
TH

1/3

>Who gives a shit.

Mathematicians

> At least one coin is heads
Correct

>Two possibilities. 50%.

No. 3 possibilities. See above.

1/3


Also, Bayes' theorem, mutherfuckers.

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

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

>If there are four possible combinations, wouldn't it be 1/4?

Yes. But there is a condition specified. At least one coin landed heads, therefore 1 of the 4 is no longer possible (Tails - Tails) so that leaves 3 equally likely outcomes, 1 of which is both Heads.

1/3

You need code to remove any result that doesnt have at least one head, original problem says one head is a given. 50% master race.

>H, T is the same as T, H, the order is irrelevant

Don't be retarded, user.

Take two different coins. Let's say, a penny, and a quarter. Each coin has a 50% chance of landing either heads or tails. You flip both coins. What are the possible outcomes? Well, let's see:

penny = heads & quarter = heads
penny = heads & quarter = tails
penny = tails & quarter = heads
penny = tails & quarter = tails

4 possible outcomes. Each of them equally likely to occur (25% or 1/4)

Now, surely we can all see how the results,
penny = heads & quarter = tails
and
penny = tails & quarter = heads
are different, right? Surely you can see how these are two distinct, separate and equally probable outcomes, yes?

Great, let's continue. Now, it DOES NOT MATTER if you are using two identical coins. HT & TH are two distinct results, and the probability of at least one of them occuring is TWICE the probability of a HH or TT outcome. Remember HT(1/4) + TH(1/4) = 1/2.

So, since we know that TT can't be the result in the OP question, it is discarded, and we are left with the possible outcomes:

HH (1/4)
HT (1/4)
TH (1/4)

Three equally probable outcomes containing at least one heads, hence the answer = 1/3

yes, this is why I added the second part, according to your theory, you have to list H, H twice, because either coin could be heads

H, H
T, H
H, T
H, H

= 4 possible outcomes, 2 are Heads, still 50%

Oh yes, sorry misread the question.

HH is 1 result, you stupid fuck.

HT is a different result to TH

Here, I drew you a diagram.

1/3

HT and TH are redundant.

If you think otherwise you're literally room temperature IQ. Did you go to a University?

Should we also make the assumption that the coins can land on their side?

This code is bad and you should feel bad. Its some good trolling though, if thats what you were going for

>HT and TH are redundant.

This is what retards actually believe.

Explain to me why you think they are redundant, and I will tell you how you are wrong.

Sure think, m8.

I get 33% too. Let's see your code, faggot.

There are two coins. One outcome is that the first coin is head and the second is tails. The other outcome is that the first coin is tails and the other is heads.

1 in 3

>There are two coins. One outcome is that the first coin is head and the second is tails. The other outcome is that the first coin is tails and the other is heads.

Correct.

And the final outcome is that the first coin is heads and the second coin is heads.

3 outcomes.

All equally likely.

1 of them is both heads.

1/3

His math is solid, HT and TH are both different outcomes, better go back to statistics 101

known as "boy or girl paradox""
/thread

this guy knows what he's talking about. all you 1/3ers are fucking retarded.

coin 1 heads, coin 2 heads
coin 2 heads, coin 1 heads
coin 1 tails, coin 2 tails
coin 2 tails, coin 1 tails
coin 1 heads, coin 2 tails
coin 2 heads, coin 1 tails

6 possible outcomes, the two TT results are removed. Thus, you are left with two HH possibilities out of four total, which is 50%

This is b8. Don't respond to it.

You fucking retarded? There are only two available outcomes:

H - H
H - T

One coin is already on heads, so the probability now only remains that the second coin has a 50% chance of heads or tails.

>assuming a first coin
>assuming it's heads

first coin could be tails and second heads.

HH
HT
TH

1/3

Stay in school.

You're fucking retarded mate. There are more than 3 possible outcomes.

T + H
H + T
H + H
S + H
H + S

S being where the coin lands on its side.

You see, this is where your retarded minds go wrong. The order does not matter and is not stated as an objective of the problem.

Whether you get H-T or T-H, you have achieved literally the same outcome. One heads, one tails. You cannot and have not been able to explain why it matters. It's because it doesn't.

lets try to explain this with no maths for the less inteligent among us

the given is that atleast 1 coin landed on head, not knowing the result yet this means it could be either 1 of 3 things
- coin 1 head, coin 2 tails
- coin 1 tails, coin 2 head
- both coins head

all of the above options are equally likely, you just don't know yet

now given this info, what is the chance it will be both heads, well easy math from here, 1 out of 3, 1/3

This is b8. Don't respond to it.

>Whether you get H-T or T-H, you have achieved literally the same outcome.

2 coin flip

HT has a probability of 1/4
TH has a probability of 1/4
HH has a probability of 1/4
TT has a probability of 1/4

At least 1 coin landed heads.

TT no longer a possible outcome, leaving

HT
TH
HH

1/3

>Whether you get H-T or T-H, you have achieved literally the same outcome. One heads, one tails. You cannot and have not been able to explain why it matters.

Point 1:
They are not the same outcome, just because they look the same to retards. Imagine one coin is blue and the other is red.

Do you think
red = Heads & blue = Tails
is the same as
red = Tails & blue = Heads

If you think they are the same, congratulations, you're retarded.


Point 2:
It matters because they are 2 distinct results and are equally likely to occur. So you have

likelihood of HT = 1/3
likelihood of TH = 1/3
likelihood of HH = 1/3

Do you understand now?

Refer to

You keep saying the same thing over and over again and you STILL sound retarded. Was there even an attempt at explaining why the order matters? Get it together man lol.

there is no "first" or "second" coin, they are thrown at the same time.
All this question sais is that the result set contains at least one Heads

so we have (and pay close attention to the notation, because it is scientifically accurate):
{ {H, H}, {H, T} }

you cannot list {T, H} in there, because it is equivalent to {H, T}. If you disrespect that, you may as well list {H, H} multiple times, because you are not following a rule to eliminate duplicate items in the result set.

>coin 1 heads, coin 2 tails
>coin 1 tails, coin 2 heads

these two are the same thing. you can already assume that one has fallen heads, since it says it in the problem. so you are left with HH and HT

>This

H + H
H + H
H + T
T + H
T + T
T + T

T + Ts are eliminated, leaving a 50% chance it's H + H, 2/4

You are a fucking retards.

Well lets do the math.

There is a 50% chance for either outcome, so lets look at the possibility of 'tails'.

A 50% chance the first coin lands on tails, and a 50% chance the second coin also lands on tails.

Add the two 50%s together and we receive a 100% chance that OP is a faggot.