How smart is Any Forums?

I concur

1/3 i think
4 possibilities
HH, TT, HT, TH
At least one lands heads so we dont count TT. That leaves three options and one is HH

Oh shit, I'm retarded. Thats the probability of both being tails, if at least one is tails

It's basic probability
There's two separate rolls with 2 possible outcomes - what's the chance of landing the same outcome twice?
It's literally 1/2^2 = 1/4

At least 1 landed heads, retard. Conditional Probability. Try again.

>At least 1 landed heads
lmao so what. Both flips are independent of each other

So TT can no longer happen, retard. That's the condition. At least 1 coin landed heads, so the probability cannot be 1/4 from that point forward. The answer has already been given in this thread.

This idea of it being 1/3 is so stupid.. Like yeah it might follow from conditional probability, but there's no way to ensure that the coins won't come out Tails-Tails while they are in the air so it's useless to consider that.

The only way to ensure at least one has heads is by checking one coin after flipping them which makes the probability of the other one being heads 50%

It's not a fucking condition, it's useless bit of information to filter retards like you.
Flip a coin. No matter how it lands it will not change the PROBABILITY OF LANDING TWO IDENTICAL FLIPS which is constant and bound to the fact that there's always BOTH (TWO) COINS (discs that output TWO possible flip outcomes)

>This idea of it being 1/3 is so stupid..
Not stupid. Counter-intuitive. Probability is often counter-intuitive and conditional probability even more so.

>but there's no way to ensure that the coins won't come out Tails-Tails
You're not ensuring that. It simply happened by chance.

>The only way to ensure at least one has heads is by checking one coin after flipping them which makes the probability of the other one being heads 50%

That's different though. The point is you do not know WHICH coin landed heads, so EITHER coin can still be tails, just not both. This isn't a difficult concept if you think it through logically. Let's say you flip 2 coins 1000 times. You will get, on average;

HH 250 times
HT 250 times
TH 250 times
TT 250 times

So there are 750 flips where at least 1 is heads
And there are 250 flips where both are heads
So the probability of both being heads when at least 1 is heads is 250/750 = 1/3

That's how conditional probability works. That's the math. The correct answer is 1/3

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>It's not a fucking condition
Stay in school, you brainlet. You don't know what conditional probability is.

Here's University of Washington Math Department saying the answer is 1/3, retard. Slide 4 related
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

Now eat my shit, you fucking brainlet.

Attached: Uni_Wash_Cond_Prob.jpg (921x631, 137.03K)

You can flip two coins as long as you want, you'll get average of 50% chance of landing two heads/two tails because each time you make two separate flips with 50% chance each. That was literally the question: "What is the probability that both coins landed heads?"

He schooled your dumbass. Just admit you're retarded and move onto the next trap thread, faggot.

HAHAHA look at this fucking moron thinking he's correct and a University Stats course is wrong. kek

>had one class on probability

Good for you that you learned that probability is often counter intuitive. Now ask some stats phd's about this, you'll get the same debate between 1/2 and 1/3 because of the setup of this question

I'm sorry you're too dumb to understand probability. At least we helped you do your homework right.

No you won't because the correct answer is 1/3 and is proving by Bayes' theorem, chud. There is no debate, only math. The 50% people are children, retards, trolls or some combination of those.

Without the condition specified, the sample space for the problem is
S = {(H & H), (H & T), (T & H), (T & T)}....(|S| = 4)
Since the event E = TT does not satisfy the condition that there is at least one head, the sample space is
S' = {(H & H), (H & T), (T & H)}....(|S'| = 3)
Thus, given that at least one head is tossed, the probability of two heads is
P(H & H) = 1/3

A more mathematical approach for problems of this sort is to apply Bayes' Theorem, where for two events A and B
P( A given B)
ie
P(A|B) = P(A & B) / P(B)
So in our problem we have
P(H|H) = P(H & H)/P(H)
= (1/4) / [1 - P(T & T)]
= (1/4) / [1 - 1/4]
= (1/4) / (3/4)
= 1/3

Attached: 1377124646016.png (412x86, 7.8K)

I get what you're saying, it just depends how you interpret the question.

Do we interpret it as after observing those 1000 coin flips we reflect and when this question is asked that means we ignore the 250 TTs as they do not adhere to the restrictions we set and so we get to 250/750.

Or we interpret it as well at least one needs to land on heads for the initial restriction to hold so we check one and if that isn't heads we can't ensure we adhere to the restrictions so we'll redo the experiment. That's how you get to 1/2.

Neither of the two answers are wrong or right it just depends on your interpretation of the experiment

All of these anons are stupid.
1/3 is right

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laws of logic dictate that the the former would require different wording, like "What is the probability that both coins landed heads if at least one of them landed heads?".