Alright fuckers. Let's settle this

Alright fuckers. Let's settle this.

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25%

It's obviously 1/2 because the one flipped on heads becomes irrelevant when you KNOW what it will be

>because the one flipped on heads

And which one would that be exactly?

this.

there are 3 possible outcomes

heads
tails
edge

first coin is already heads so the possibilities are
heads/heads
heads/tails
heads/edge

so probability of heads/heads is 1/3 or 33.3%

1/3

What's it like being a dumbass?

For a regular 2-coin flip, there are 4 equally likely outcomes. There are:

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

If at least one coin landed heads, that eliminates the Tails-Tails outcome, leaving 3 equally likely outcomes, all containing at least 1 heads coin.

1/3

Pic Related

We got 4 outcomes:
>HH
>HT
>TH
>TT

Second and third are equivalent in the scope of this question, and the fourth is excluded a priori. So the answer is 50%

1/3

1 definitely landed heads, so what is % that other did too?

So we have
HT
HH
TH

Or

P(A∩B)=P(A)P(B)
P(A∩B)=P(A)P(B)⇔P(A)=P(A∩B)P(B)⇔P(A)=P(A|B)

...

50/50, either you get heads or you don't

I hope you're trolling. If you believe that the chance of a coin landing on it's edge is equal to the chance that it does not then you are retarded.

The answer is 1/3 but your explanation is wrong.

YOU PEOPLE ARE FUCKING RETARDED, TH AND HT ARE THE SAME. IT'S A 1/2 IN CHANCE.

thank god!

I just flipped two coins until they both landed heads. it took 4 tries. therefore the probability is 25%

>TH AND HT ARE THE SAME

Nice b8, m8.

HT and TH are the same thing though

I stand corrected. I'm a convert.

>HT and TH are the same thing though

No, they are not.

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

It's 1/2. TH and HT would be considered the same outcome because order doesn't matter.

if one coin (C1) is already Heads, that means that TH (C1 tails, C2 heads) is also ruled out

therefore it's 1/2

>TH and HT would be considered the same outcome because order doesn't matter.

But you are TWICE as likely to get a HT or TH as you are to get HH.

Just because TH and HT look the same does not mean that they each have a 1/3 probability of occurring.

probability of HT or TH = 2/3
Probability of HH = 1/3

1/3

Wow you're dumb. They are the same thing.

This. One coin is already fucking settled. The other can be heads or it can be tails. That is the only unknown remaining so it's 50/50.

>if one coin (C1) is already Heads

Except you don't know which coin is heads, meaning EITHER coin could be tails, just not both.

3 equally likely outcomes possible

HH
HT
TH

HH is 1 of those 3

1 of 3

1/3

Pic Related

Retard detected

>One coin is already fucking settled

Which coin?

the only way you can know one coin is heads is if a coin already landed heads

flip the other coin and it's 50%

One hundred coins are flipped. 99 of them land on heads, because I say so. What is the probability that the 100th lands on heads too?
50%

Doesn't matter. It's the same either way.

The one that landed heads

Well...shit. I see your point actually.

That's a different question to the OP question though.

In the OP question there are 2 coins which can give 4 possible outcomes.

HH
HT
TH
TT

We are told that at least one coin landed heads (we don't know which) so that eliminates the TT outcome only.

Leaves 3

HH
HT
TH

Answer = 1/3

1/4
they were flipped at the same time, just because you know one landed heads doesn't change the probability of the initial trial.

mfw i just took a stats class and forgot everything

>Doesn't matter.

yes it does.

If either coin could be tails, then that leaves us with 3 equally likely outcomes:

HH
HT
TH

1/3

Pic Related: Bayes' Theorem

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Take a sharpie marker and write an A on one coin, and B on another. The A lands on heads, the probability of B landing on heads is 50%. You can swap A& B if you want to.

>Take a sharpie marker and write an A on one coin, and B on another.
Ok

>The A lands on heads,
Ok

>the probability of B landing on heads is 50%.
that's correct

>You can swap A& B if you want to.
Yep.

So we can have

coinA = Heads & coinB = Heads
or
coinA = Heads & coinB = Tails
or
coinA = Tails & coinB = Heads

3 outcomes possible.

All equally likely

1/3

they are, the order is irrelevant

they're not. Read this

Actually, I have a better question:

What are the odds that you won't post this question again?

Bro you're trying to explain stats to Sup Forums. Nobody is ever going to understand you.

I think we both know the answer to that.

You do realize this is a troll thread right... I agree with the 1/3 result, if the original question did not say that one coin always lands on heads.

I made the thread.

>I agree with the 1/3 result, if the original question did not say that one coin always lands on heads.

the question does not say that one coin always lands heads, if it did, then the answer would be 1/2.

the question says that 2 regular coins were flipped and at least one coin landed heads.

there are 3 equally likely ways to get at least 1 heads coin for a 2 coin flip

HH
HT
TH

This is really pretty simple and basic conditional probability.

1/3

Read the original question 100 times in a row. Document how many times it says that one coin lands on heads. I got all day.

>2 regular coins were flipped. What is the probability that BOTH landed heads GIVEN that AT LEAST ONE of them landed heads?
>at least one of them landed on heads

It's right there, user.

1/3.

Simulation of 100 million 2 coin flips gives 1/3
Bayes' theorem gives 1/3
Punnet Square gives 1/3
Venn Diagram gives 1/3

you admit Atleast one is heads every single time right?

Take two coins, Pen in a number 1 and 2 on the coins. Flip them 100 times. Every time you get double tails don't count it.

Once you have flipped it 100 times you will have roughly 66 instances with 1 heads 1 tails

and 33 instances with heads... it a 1/3 since we don't declare which coin is always heads.

>you admit Atleast one is heads every single time right?

No. That's retarded. That would be the same as fixing 1 coin to always be heads. in that case, the answer would be 1/2.

OP's question does NOT fix any coin as heads. It simply tells us that at least 1 coin landed heads. It could be either coin, meaning either coin could be tails, just not both.

That gives 3 equally probable outcomes which contain at least 1 heads coin

HH, HT, TH

I don't know why you're having so much trouble understanding this.

HH is 1 of those 3.

1/3

That's some beautiful code right there

It literally states in the question, that one coin is heads. It doesn't say, maybe there's a head result. Atleast, bare minimum, no matter what, unless you change the conditions of the original post, one coin will be heads.