Right Veeky Forums so this aspergers kid thinks he's right- his answer is 50%

>Which is obviously wrong, yes?
It's certainly wrong. Whether it's obvious is something I'll leave to you.

>TT doesn't happen, which leads us to three events, and HT and TH are separate events, so we have 1/3.
Correct.

>no they are not. There is no distinction between the coins. They are identical.
Whether you want to model a difference or not doesn't matter at all. What matters is that HT/TH is twice as probable as HH. Whether you want to think of this as TH and HT outcomes each with 25% probability, or a single TH/HT outcome with 50% probability, doesn't matter at all.

Are the coins bosons or fermions?

This, my probability theory professor explicitly went over this problem in one of the first lectures.

Goddamn it, finally figure this out.

The correct answer is: indeterminable - not enough information given.

Both 1/2 and 1/3 can be successfully argued for. The sticking point is this phrase: "given that at least one of them landed heads." The method that this information is "given" changes the probability.

1/2 Probability Example:
1. I flip two coins, then place one under each hand, hidden from you, on the table.
2. I tell you "the coin under my left hand is showing heads." (This is a reasonable interpretation of the question. I tell you that at least one coin is showing heads by telling you where a heads is located.)
3. To you, the probability that both coins are heads is 50% - 1/2.

1/3 Probability Example:
1. Same as before, I flip two coins, then place one under each hand, hidden from you, on the table.
2. I tell you, "there is at least one heads showing under my hands." (This is also a reasonable interpretation of the question. I tell you that a heads is showing, but I do not tell you where.)
3. To you, the probability that both are heads is ~33% - 1/3.

This is pure bait, designed to get you to argue statistics when the actual argument should be over how the "given" information was given to you. Because the method is not described in the question, the actual probability can't be determined.

fuck you and your trips

it's still 1/2. Intuitively it is 1/2. Practically, it's still 1/2.

If you disagree, we'll play 's game and I'll happily take your fucking money, you sperglord nigger.

>The method that this information is "given" changes the probability.
No, WHAT information is given changes the probability.

>I tell you "the coin under my left hand is showing heads." (This is a reasonable interpretation of the question.
No, it is a completely wrong interpretation. You are telling me strictly more than the fact that at least one landed heads.

1/3 is correct, 1/2 is plain wrong, and you have successfully confused yourself into believing it's a phrasing problem. It isn't. It's a problem with your intuition as to what exact information is revealed by a certain action.

Revealing a coin as heads falls under the description of "given that at least one of them landed heads," whether you like it or not.

If the question wanted to specify how that information was transmitted, it could have. But it didn't. So it is bait designed to watch the 1/3's and the 1/2's bicker.

>Revealing a coin as heads falls under the description of "given that at least one of them landed heads,"
no
no it's not
I'm sorry for your brain.

Well, at least we're arguing that actual issue with this problem instead of comparing probability trees, or some shit.

>Revealing a coin as heads falls under the description of "given that at least one of them landed heads," whether you like it or not.
No, it doesn't. It transmits that information, yes; but it also transmits more information than that.

OP's question says "given that (this information), what is (some probability)". The scenario you describe reveals more information than that, and thus is not the scenario described in OP.

I would agree that is the stricter interpretation of the question, but I would also assert that it is needlessly ambiguous, given the long threads with filled with people trying to convince each other using bayesian statistics.

Nigga it's not "this one is heads" it's "one of them is heads".
You don't belong on this board.

OP, your friend is retarded, but it's a pretty easy mistake to make if you don't know anything about probability.

Most of this board doesn't belong on this board: 4chandata.org/sci/well--sci--a332034

It is 50%
It either happens, or it doesn't.

this
probability is retarded
de finetti is alive

>needlessly ambiguous
It's the clearest concise way to state it unambigously.
Ambiguous would be "one coin is heads"

Given that we have two events for two coins

Coin A: 100% heads
Coin B: 50% heads 50% tails

The designation of the coins does not matter.
Thus the probability of both being heads is 50 percent.

What if coin b is heads and a is 50/50
Then you got 2 50/50 and 1 100 which is 3 choices. It's 1/3.

But the problem IS bullshit because the way it's worded, and the way it works in real life is 50%

-1/12

If I flip six quadrillion coins. What is the probability that they create a better question than OP written in binary?

Pretty freakin good I would say

>Practically, it's still 1/2.

Haha, how about you hold an experiment and we'll see?

Whats the probability that none of them landed on heads?

Either it happens or it doesn't happen. 50/50 50% chance.

No, you should always switch doors.

You're wrong but in a way subtle enough that it's worth pointing out explicitly.

The problem is this line

>I tell you "the coin under my left hand is showing heads"...I tell you that a heads is showing but I do not tell you where a heads is located

Saying "The coin under my left hand is heads" is more informative than "At least one coin is heads". It's true that knowing the left coin is heads implies at least one of the coins is heads, but not conversely (i.e., knowing at least one the coins is heads does not imply that the left coin is heads). So the problem is now how the information is given to you, because the information "At least one coin is heads" is not the same as the information "The left coin is heads".

Either coin could have been in his left hand.

>Most of this board doesn't belong on this board
That's true, but irrelevant to the current discussion.

>I'll happily take your fucking money, you sperglord nigger
Even if you're right, you'd ought to end up with the same amount of money.

>But the problem IS bullshit because the way it's worded, and the way it works in real life is 50%
No, the way it works in real life is that "at least one coin is heads" is never the exact information you actually have; you need to strain to find yourself in a situation on which that is the exact information you have, which is why your intuition disagrees with the math. Nonetheless, when you DO have only the information given in the OP, 1/3 really IS the correct probability.

Doesn't matter, if that coin is heads, the other has a 50/50 of being heads.

That's why this problem is bs because it can't be modeled irl.

I agree.
But there's no situation where that could occur because the nature of the problem automatically gives the needed info irl.

underrated post, genuinely laughed.

>But there's no situation where that could occur because the nature of the problem automatically gives the needed info irl.
No, that's not true. Situations where the information described in OP occurs in real life are rare, but they DO exist.

I'm sitting at a restaurant with my friends; I'm the designated driver, so I order a non-alcoholic beer twice during the evening. Afterwards, I learn that the waiter is pretty poor of memory, and he may have given me alcoholic beer instead on either occasion, independently. (Let's assume my taste is sufficiently shit that I can't taste the difference.)

When it's time to drive, I check my breath with a $0.99 one-off breathalizer gadget, which can only detect whether I have drank any alcohol at all that night; it's not sensitive enough to measure degrees of intoxication, just "yes" or "no", because it's a cheap piece of crap. It says "yes", indicating that at least one of my beers contained alcohol.

I'm over the limit of I drank two alcoholic beers; I'm road-legal with just a single beer. What's the probability that I drank two alcoholic beers this night? It's 1/3. Not 1/2; the only information the shitty $0.99 breathalizer gives me is that drank at least one alcoholic beer. Thus the 1/3 result is correct, per the mathematical analysis described in this thread.

Uncommon and artificial? Absolutely. Which is why human intuition for problems like this is total crap. But these situations CAN occur, and when it does, the intuitive answer is wrong.

/thread

Fair enough.

You should probably not be the designated driver if two beers puts you over. And get checked for anorexia

Given that both coins were flipped it's 1/3rd, the situations are
>HH
>HT
>TH

If you say two regular coins are flipped, but coin 1 is heads, then
>HH
>HT

Explicitly stating which coin is heads is different from saying at least one of the coins is heads.

Consider it from the opposite side. If one of the coins was heads, you are more likely to get heads and tails than both heads, because either coin 1 or coin could be tails.

B-but there are TWO coins. So there are TWO ways of getting HT (or TH)
Think about it.

odds that you have come to this thread having learned how its 2/3 and sensing an opportunity to feel smart... 8/10
9nth guy hasn't learned yet
..10nth is hapless shitposter

>this
>trick question

one outcome is predetermined

the chance of a separate coin flip is 1/2

the probabilities are not cumulative; there is no probability value assigned to the first outcome.

>it is already heads
>it has always been heads
>it will always be heads

probability of getting a H or T next is a fucking independent event.

correct answer is 1/2

> at least one of them landed heads
"at least" is physically impossible, we can't predict something like that, therefore this problem doesn't exist in the real world and we shouldn't even talk about it because it can't be tested/reproduced (like god)

But if one of the coins landed heads, the other one has 50% of chance of landing heads as well.

ITT: weekly reminder that the autism spectrum is vastly over-represented in STEM

>its 1/2 you fucking spastics

who are you even meme arrowing, newfag

Nah its still 1/3
It isn't a separate coin flip, they are both flipped at the same time, you are just told the result of one of them, not necessarily the first one, which is the key.

>have 2 coins lebeled 1 and 2
>flip coins in magic coin reading scanner machine
>machine says at least one of them is heads
>possible coin on heads is 1, 2 or both

Another way of looking at it is like this
possible outcomes are
>1 lands on Tails, 2 lands on Heads
>2 lands on Tails, 1 lands on Heads
>1 lands on Heads, 2 lands on Heads
>2 lands on Heads, 1 lands on Heads

looks like 2/4 or 1/2 right? Not quite, the last 2 options are the same so you combine them, leaving you with 1/3.

If it said the 1 coin was heads then the probability would be 1/2 as the only options are
>1H 2T
>1H 2T

>1H 2H

my bad

hint... if you did the flips one after another, you would be right. Thats not how the problem is laid out though

i follow your logic; the point is does coin #1 have a probability value of 1 ? or does a pre-determined outcome have no assigned probability value?

if its the former the answer is indeed 1/3

however the wording of the question implies the latter is correct friend

***** does coin #1 = H have a probability value of 1

> the last 2 options are the same so you combine them
gr8 b8 m8

but what's the probability of both coins actually existing?

P(Second coin lands heads | first landed heads) = P(Second coin lands heads && first landed heads) / P(first landed heads) = 1/4 / 1/2 = 1/4 * 2 = 1/2

Keep in mind the original probability of the problem was 1/4, but we were given more information to narrow it down.

The question asks what the probability is given/IF at least one coin is heads. This does not mean that when you flip the coins one will always be heads. But of all outcomes WHEN there is at least one head what is the probability there are 2 heads.

Since you can't differentiate the coins the final outcome becomes its own "coin" if that makes any sense. We can only look at the final outcomes of this 3 sided "coin"

We can just recreate the Monty Hall problem with these coins. This is the easiest way to understand.

>4 Doors
>behind each door is a possible coin flip outcome
>HH HT TH TT

What are your chances of guessing the HH door? 1/4

They then close a door, and tell you that the remaining doors are ones where the outcome has at least one heads. (original problem)

>HH TH HT
You have a 1/3 chance

If they said, the remaining doors are ones where the first coin is heads you are left with.
>HH HT
1/2

>ask mom for 1 dollar
>ask dad for 2 dollars
>dad says no, mom says yes = 1 dollar
>mom says no, dad says yes = 2 dollars
>dad says yes, mom says yes = 3 dollars
>mom says yes, dad says yes = 3 dollars

Possible outcomes
>1 dollar
>2 dollars
>3 dollars

>If they said, the remaining doors are ones where the first coin is heads

b...but thats what theyre saying user

I really hope you are baiting my autism right now because no one deserves to be this retarded.

>coin A = H
is not the same as
>coin A or B = H

You have flipped 2 dices. What is the probability that sum equals 3? 2/36, not 1/36.

>coin A = H
>coin B = H or T

problem senpai?

You cant add coins together dumbass. It isn't the same. I guess I did it first though so here's a better example.

>ask mom for money
>ask dad for money
>dad says no, mom says yes = dN + mY
>mom says no, dad says yes = mN + dY
>dad says yes, mom says yes = dY + mY
>mom says yes, dad says yes = mY + dY
>dN + mY != mN + dY
>dY + mY = mY +dY

This is the last bait I will reply to. Good night.

ignorant bait!
>two regular coins were flipped. What is the probability that both landed heads given that at least one of them is heads?

since we know the outcome of one of the coins there are only two possible outcomes, either they are both heads or they aren't.]
You KNOW the outcome for one coin, and now you are asked the probability if the other coin is heads... 50 / 50.

I seriously do not see how you can get 3 possible outcomes from one coin.

Why not just do it and see?
>Flipped 1000000 times with at least one head and got both heads 332974 times.
>0.332974 chance of getting both heads given at least one is heads.

>Why not just do it and see?

>finite cases
>computers can't produce true randomness

Typical

>I am just a CS major lol. Elementary probability? Lol I failed that class twice, changed it for office feminism 101. I can just solve it with the python scripts I steal from github pages,

You know as well as I do that this is a good enough approximation and that I'm replying to poor bait.

>what's the probability of getting two heads, when the first coin lands on heads
vs
>what's the probability of a coin landing heads, now that the first coin is heads
1/3 or 1/2, it's not that hard, the whole argument is over semantics

On a related note, why is this board so retarded when it comes to probability? Is it just memeing?

If which coin is heads matters then sure it's 1/3, but wtf you autistic spastics? In reality, the problem says one is always guaranteed to be heads, so the other, whichever one it is, has a 1/2 chance of making the coins be both heads.

sci is so goddamn fun, more fun than /x/ or /pol/ when it comes to being retarded

Probability is hard ;(