How smart is Sup Forums?

How smart is Sup Forums?

1/3

50%. you can ignore the first coin since it is given that one landed heads; so you only have to worry about the second coin. ez

Still 50% if you know one of them is heads but If you flipped both and didn't know one was going to be heads it would be 25% chance of both being heads

Probability Tree is as follows:

H | H
H | T
T | H
T | T

The Tails/Tails can be excluded because one landed heads. So the answer is 1/3.

This is only true if you knew which of the 2 coins landed heads.

The probability is both 0% and 100% until we actually observe the coin.

Given a coin A and B, assuming that coin A is going to land heads,
P(A)=1
Then, the probability of coin B landing heads is
P(B)=1/2

The probability of the coin A and B landing heads is P(A∩B) = P(A)*P(B) = 1*1/2 = 1/2

So, the answer to your question is 1/2

you do not know if coin A or B is the one stated

no we don't know if A or B landed head, you can't choose one. The answer is P(2 tails)/P(at least one tail)=0.25/0.75 = 1/3

guys you don't need to know which coin lands what, the probability is still 1/2, you know for sure that one of them lands heads, so it's like putting already one of the 2 coins on the table on the heads side and you just have to flip the other one

75%

no its not you fucking retard, take a probability course

what are the odds of any one coin landing heads?
50/50

what the fuck does it matter what some other coin did?

>regular coin
>posts fake money

Post real coins faggot.

that's basic probability mate, this post break it down for you:

Stop and read the entire problem. There's a condition in there that everyone misses.

It's asking what the probability of getting HH is when out of the coin flips where there is at least one H showing. The choices which fit that condition are HH, HT, TH. Of those choices, HH is 1/3, so 33% chance.

Once Sup Forums has tackled this problem, try this one:

Prisoners A, B, C are all sentenced to be executed the following morning.

The governor has selected one of them at random to be pardoned.

The warden knows the party who will be pardoned but is not allowed to tell the identity of the one to be pardoned.

Prisoner A pleads with the warden to tell him about the next day, the warden tells him he cannot tell him if who is going to be pardoned and cannot tell him his own fate, but tells him prisoner B will be executed tomorrow.

What are the odds prisoner A will be executed?

possibilities:
1.Heads Heads
2.Heads Tails
3.Tails Heads
4.Tails Tails

There must be at least one heads, so possibility 4 is invalid. Both need to land on heads, so the answer has to be possibility number 1. That means the answer is 1/3 chance

While this logic is sound, there's no guarantee that the first coin will be heads. The "at least one of them landed heads" condition can be met with the second coin landing heads and the first one landing tails.

And this is what makes this question good for trolling Sup Forums, but this is the right answer.

This issue here is some shit about knowing one of the coins is heads & knowing which one is heads.

If you take this faggots post he explains why it's 1/3. But if you know which one will land on heads 100% guaranteed, it becomes 50%.

>Current 1/3 chance
H | H
H | T
T | H
>First Coin Heads 50% chance
H | H
H | T
>Second Coin Heads 50% chance
H | H
T | H

It's a stupid question that uses word play to fuck with you rather than actual math.

I think the wording is clear, but it takes advantage of the fact that your average Sup Forums user will not read it all and just see 2 coins, something about probability, and something about 2 heads and run with it, calling anyone who disagrees with their answer a faggot along the way.

Appreciated!