Please discuss

please discuss

50%; one outcome is decided, and the desired result relies on one coin flip.

Let H represent heads, T - tails.

Possibile cases for described situation are:

H/T
T/H
H/H

Out of 3, only is satisfactory. Therefore, it's 1/3.

You failed to consider one inverse scenario.

phew, done, had some hard time disussing this issue with my wife, mainly because she didn't want to, but I went full alpha mode and stuff. You're welcome.

1/3
P(A|B) where B is the event a head landed (3/4 times) A is the event that both are head so (1/4)/(3/4) =1/3

Nothing to discuss.
If actions A and B are independent (effect of one flipped coin doesn't depend on other) then:
P(B|A) = P(B)
P(B) = .5 = 50%

wtf?

You failed to consider one inverse scenario. It's 1/3.

66%, it says AT LEAST ONE OF THEM is landed heads.

My bad you're right I misread

Repetition didn't help me realize I was wrong. Just sayin.

Looking at truth table above should have.

1/3

2/3

The anwser is

2=1+root1

I also initially thought 1/2, although after reading this I see 1/3.

In genetics problems you do basic probabilities with pedigrees but you're really only doing one "coin flip" at a time. In those scenarios you'd do probability of A being Heads (1/2) * B being Heads (1/2) = 1/4, then add 1/4 because you have to do it the other way, to ultimately get 1/2.

Meanwhile, this is more of like a punnet square that excludes 1/4 of possibilities right off the bat, alternatives being HT, TH, and HH, as established.

The issue was I misread the question.

If it was P(B|A) with A being determined then the truth table would have been incorrect because the two options would be H/T and H/H

So no, asshole, looking at the truth table shouldn't have. Looking at the question again should have.

what was her opinion?

Two coins can't both land in head, if you do not have that at least one lands heads.

It is redundant. The probability is the same probability it would be if you removed that last sentence.

To make this clearer:

What is the probability of two coins landing in heads, given that both of them landed in heads?

50%
either it's heads or it's not

always the right answer

Wow, you're dumb. Given that someone has HIV, what is the chance they have AIDS?

Hint: not all people with HIV have AIDS

Only brainlets can't see that the probability is 50%

It could have been the other coin that flipped heads though, so 1/3 you missed the inverse scenario

It's a meme kiddo.

50%, it either happens or it doesnt

What is the probability of two heads given that at least one coin landed on heads and is minted in 2005?

Ten percent of one euro coins are minted in 2005, according to my ass.

No, the additional information is not irrelevant.

lmao i c wat u did thar

>The issue was I misread the question.
Excuses-excuses.

50%. One coin's flip doesnt change the probability of another coin's flip.

Suppose that the event "coin 1 lands on heads" is denoted A and the event "coin 2 lands on heads" is denoted B. The problems asks for Pr(A and B | A or B). In other words, the probability of A and B given A or B.

We know:
Pr(A) = 0.5
Pr(B) = 0.5

Since A and B are independent events:
Pr(A and B) = Pr(A)*Pr(B) = 0.25
Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B) = 0.5 + 0.5 - 0.25 = 0.75

Thus:
Pr(A and B | A or B) = Pr((A and B) and (A or B)) / Pr(A or B)
= Pr((A and (A or B)) and (B and (A or B))) / Pr(A or B) Distribution
= Pr(A and B) / Pr(A or B)
= 0.25 / 0.75
= 1 / 3

For reference:
Pr(X and Y) = Pr(X)*Pr(Y) For independent events
Pr(X | Y) = Pr(X and Y) / Pr(Y)
Pr(X or Y) = Pr(X) + Pr(Y) - Pr(X and Y)

What you said is true, yet you had the wrong question.

1/infinity. Someone could catch the coin, the coin may land on its side, the laws of gravity may collapse, a wormhole may emerge on the table and send the coin to another dimension, all things are possible.

19/39

1/3
There! Thread over! Everyone go home!

P(two given at least one) = P(two and at least one)/P(at least one)
= P(two)/P(at least one)
= 1/2^2/(1/2 + 1/2^2)
= 1/(2+1)
= 1/3

oh my word, people. this is all theory. there's theory, and all too often, there is then, REALITY. please tryo to live life in reality, based on evidence that is observable and reproducable.
again, theory is silly compared to reality. for example, in this specific example, the whole thread is talking about magic coins that don't exist and using math to answer something that has not happened. its silly.
for example, lets take a moment and imagine something real. take a coin. if it is a coin from earth, it is most likely going to have more matter on the 'head's' side. thus, if you flip a real coin, odd's are it will land on heads more often (nevermind all the other artifacts of human or machine flipping).
so the real answer, talking about reality, is: take a real coin, examine and flip THAT specific coin that exists in reality, and you'll just see.
everything else is just theory propelled by arrogance. theory and arrogance can most easily make one blind.
rely on what is observable, and able to be reproduced. take care

The inverse scenario is irrelevant. One result is already fixed. One is already heads, which one it is doesn't matter. The entire result hinges on what the coin that does NOT flip heads does, which is 50/50.

>what the coin that does NOT flip heads does
It could be either coin

it's 50% because Coin A or B is fixed, it doesn't matter which one is fixed at the time. It's fixed at the point that they already landed one coin heads (A or B is irrelevant). So it is one coin flip, which is 50/50.The experiment takes place after 1 coin is already given.

Fuckin autists

Statistics on Veeky Forums is utter memery.

But that doesn't matter. Think of it this way, if you just place one coin heads side up does it matter which coin you pick? Does it affect the probability of the second coin in any way? If you pick coin A to be heads up it doesn't matter because your result is now entirely dependent on how coin B flips. If you pick coin B to be heads up it doesn't matter because the result is now dependent on how coin A flips. Meaning it doesn't matter which coin you pick to be heads the scenario is still entirely dependent on the coin that has a chance to flip tails, 50/50. You're conflating two separate events and trying to mush them together.

You are answering the wrong question
>given that coin A is heads, what is the probability that coin B was heads?
1/2.
>given that coin B is heads, what is the probability that coin A was heads?
Also 1/2.
That's not what's being asked in the OP though.

The question is phrased like this
>What is the probability that both landed heads given that at least one of them landed heads
Meaning we're calculating post hoc. One of them landed heads, we know that for a fact, meaning the question hinges on what the OTHER coin did. You're attempting to answer the question as if the coins have yet to be flipped, which is incorrect. One of them LANDED heads, you can see it, the flip has already been performed, you're calculating the probability based on the SECOND coin which is hidden from you.

You are assuming you know WHICH coin landed heads which is where you go wrong. At this point I just assume everyone saying 1/2 is trolling, not sure why I bother responding.

It literally doesn't matter which coin landed heads. Again think about it like this, the question is in past tense, meaning the flip has already occurred. What you need to imagine in your head is the flip has been done, you can see one of the coins is heads up and the other is covered by a cup. Which of the two coins flipped heads up is an irrelevant point that would only matter if you were talking about probabilities BEFORE the flip. The fact we're talking about AFTER the flip and the fact the coin has already landed (again, past tense) means that we're already in a H/? scenario. From there you can go H/T or H/T. There are no other options.

Oops, meant H/T or H/H obviously.

> the fact the coin has already landed (again, past tense) means that we're already in a H/? scenario.
Nope, we could be in a ?/H scenario, in which case the possibilities are then H/H or T/H - but we've already counted the H/H possibility in our first scenario.

what's the probability that a simple question intentionally badly worded makes anons argue?

>Nope, we could be in a ?/H scenario
It's the same scenario. It's still a 50/50 chance based on what the other coin flipped. You're making a distinction that doesn't exist. Again the fact that one coin flipped heads is a fixed result and it doesn't matter which one it was.

H/? gives H/H or H/T which is 50/50

?/H gives H/H or T/H which is again, 50/50

See? Which specific coin was the heads is irrelevant. Both scenarios give a 50/50 probability because the answer is based on the hidden variable, the coin we don't know, not the coin we do.

>H/? gives H/H or H/T which is 50/50
>
>?/H gives H/H or T/H which is again, 50/50

You counted H/H twice.

You're going back too far. We already know which coin flipped heads in the scenario. We can see it. The question is what the coin under the cup flipped. There are no other factors.

Again, this is super important, the flip has been made. We can see the coin that flipped heads on the ground. We know it is either coin A or B because that is the starting point of the question. Think of it like this, you write A and B on two coins, get a friend to flip them until one comes up heads, then he covers the second coin. You enter the room and see the coin marked either A or B and the covered coin, this is the point where you are asked to determine the probability of both coins being heads. The answer now should be obvious. The uncovered coin is irrelevant and the answer is 50%

You're assuming you're being asked to calculate the probability BEFORE you enter the room and see which coin flipped heads, which is wrong. You're being asked AFTER you already know which coin flipped heads.

I hope this was basic enough for you to understand where you made the mistaken assumption

You're dumb

Sorry, I explained it as simply as I could. You might need to do a bit of statistics before you can understand it.

fuck I remember my stat course, too fucking bad it was a summer class and i hadn't taken math in 2 years prior to that so i fucking failed with a 49%, i decided to study that shit on my own.

Open a high school book and educate yourself retard.

>You might need to do a bit of statistics before you can understand it.

event X = probability both land heads
event Y = probability of getting at least one heads

We want p(X|Y).

p(X) = 1/4
p(Y) = 3/4
p(Y|X) = 1

p(X|Y) = p(X) * p(Y|X) / p(Y) = (1/4 * 1) / (3/4) = 1/3

Tell me where I'm wrong here.

>Tell me where I'm wrong here.
The part where you responded to a troll

>Tell me where I'm wrong here.
You already know which coin flipped heads. You're calculating the probability as if we don't know which coin will flip heads. Again, this is all important, the question is past tense. The coins have already been flipped and we already know which one flipped heads. It's not a probability at that point, it is fact.

>we already know which one flipped heads
No we don't

>You already know which coin flipped heads.
We don't.

>No we don't
Yes we do. The coins have been flipped. You're going back in time and calculating the result based on probabilities before the flip. After the flip we know which one flipped heads, it's already happened. All that is left to do is guess what the other coin flipped.

We already know if we are in a H/? or ?/H scenario. That is the point where you start calculating. You can't go back in time and pretend the coins havn't been flipped, the result has already been determined, we know for a fact which coin flipped heads because it happened and you're looking at it.

>We already know if we are in a H/? or ?/H scenario
So you don't know which it is.

>were
>given
>at least one
50%
there's not even any ambiguity

It doesn't matter. They're both the starting point, not anything before and the answer to both is 50%. That is the reason the past tense phrasing of the question is so important. You're already in one of those scenarios from the perspective of the question, going back in time before the coins flipped makes zero sense.

>You're already in one of those scenarios from the perspective of the question
You don't know which one

>You don't know which one
But it doesn't matter because that is the starting point from which we calculate and both scenarios give 50%. Going back to before the coins were flipped and calculating the probability of which coin flipped heads is outside the scope of the question, the starting point is after that event, when which coin flipped heads has already been determined.

Back to /pol/ with you

You'd be right if the question read

>Two regular coins WILL BE flipped. What is the probability that both WILL land heads given that at least one of them WILL land heads.

Sadly it doesn't say that, and you're wrong. The question is a test of reading comprehension, not mathematics.

>It's the same scenario.
order makes it a different scenario. that's how perturbations work.

lol is this guy serious? what are you talking about as "starting point"? nigga you have two possible events to take into account. they're not equal in the slightest due to order.

>will
>he thinks it has to occur for us to take into account the probability of both h/x & x/h

that makes even less sense than what you started off with.

>you have two possible events to take into account
One of them has already occurred. Past tense. The coins have been flipped. You already know which one flipped heads. Why are you calculating the probability of a scenario that did not happen? Are you guys being deliberately obtuse or what?

>You already know which one flipped heads.
Which one was it? A or B?

Doesn't matter the answer is the same for both.

It matters that you know. If you know it's A it's 1/2, if you know it's B it's 1/2, if you don't know which it is it/s 1/3.

Agree, Completely Right your probability does not change at all due to any outcomes from previous flips. Just think logically you don't need to "Break it down" into an equation by any means.

>just trust your gut you don't need dem librul "equations"!

The problem here is that the flip is completely irrelevant to the question. If you phrased it "You have a coin showing heads and a second coin under a cup, what is the probability both coins are heads" then you immediately get the correct answer. What's important to realize is at the point you're asked for the probability the state of the coins has been set. You know there is a coin showing heads, and the other coin is the variable. There is no theoretical third state where the coins are switched, that horse has come and gone, the immediate scenario you are asked to answer is one where you know the state of one coin and to determine the probability the other coin has the same state. That's it. Again, the flip is irrelevant and outside the scope of the question. The states of the coins are already set. One is heads and whether that coin is A or B is completely irrelevant, but that is the point from which you work from, NOT before.

People are over complicating this.

Eliminate the other coin from your consideration because it IS ALREADY HEADS.

Then consider the question "What is the probability that a coin (the other coin) lands on heads?"

Then the answer is - 50%.

I know the answer is 1/3 and I can prove it using basic conditional probability.
However, it's still fucking unintuitive.
It would be easier to understand if we use a 6-face die instead.

Mods should ban everyone in here who posted an incorrect answer, I don't want to share Veeky Forums with these retards.

>People are over complicating this.
trolls

If the question was phrased "What is the probability that coin A landed heads or coin B landed heads" or "the left coin or the right coin" then that would make sense. It is not phrased that way. It is:

Two coins are flipped.
One lands heads.
What is the probability they will BOTH be heads?

It speaks nothing at all to which coin - that is irrelevant.

Agreed. Please ban everyone that posted 1/3 as an answer

If the mods don't know what the correct answer is they're not qualified to mod this board.

One is already heads guaranteed, meaning it doesn't matter what the fuck the one of coins outcome is.

The only relevant factor becomes one coin, which has a 50% probability. Anyone who is saying 1/3 is wrong.

>meaning it doesn't matter what the fuck the one of coins outcome is
It does when we're only considering a subset of all possible outcomes. You don't seem to understand how conditional probabilities work.

33.3% recurring

Fucking retards jesus christ its 1/3. This is what happens when american education is left to its own devices and not assimilated by europe.

2/3

Is this troll the same faggot who doesn't get Monty Hall?

A lazy user never read the follow-up...

OH MY WORD PRETEND TO BE BLIND WHEN UR NOT

33%

How much do you have to be trolling to claim it's 1/3 rofl

A = probability that both are heads
B = probability that one is heads

Sample space HH, HT, TH, TT

[math]
P(A) = \frac{1}{4} \\
P(B) = \frac{3}{4}
[/math]

[math]P(B|A) = 1 (\text{obviously if both are heads, one is heads})\\
P(A|B) = \frac{P(B|A) P(A)}{P(B)}=\frac{1\cdot \frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}[/math]

If any of you 50% niggers want to keep arguing, take it up with Bayes first, then get back to us.

...

Yes OP we all know that math is abstract that doesn't convey reality that well
But did you really need to create this thread?

>The inverse scenario is irrelevant
Truth table above proves this wrong.

-1/12