Let's do this

Let's do this.

Other urls found in this thread:

braingle.com/brainteasers/teaser.php?op=2&id=29228&comm=0
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf
twitter.com/SFWRedditImages

50%

25%

Anyone who thinks the answer is 50%, please kill yourselves.

You are completely useless and will bring nothing of value to the human race.

The answer is 1/3.

1/3
bayes rule and such
give me something hard, user

If you've ever taken Statistics and have basic reading comprehension, it's 50%.
The phrase "given that at least one of them landed on heads" limits us to always have 1 head, the percentage is based on the second coin which still has a 50% chance to land on land no matter which of the two coins that isn't guaranteed to be heads

I eat dicks, listen to them

> given that at least one of them landed heads.

Thats irrelevant.

Or they both land heads or only one land head or the other land head or both land tails

answer is 25%

Pissers.

The answer's 1/3 and if you didn't write this you're a moronic fucking faggot

12.5%

Kill yourselves

1 in 3
Heads heads
Heads tails
Tails tails

Correct!! hense why i was first to say 50% good to know there's some intelligence left on Sup Forums

says the nigga who works in Walmart

50%.
Two coins, one is already heads, thus already out of the equation.
One coin, heads or tails.

>given that at least one of them landed heads
>tails tails

>landed heads
>tails tails


u dun goofd user

b8/3

It's now the probability of one coin being heads. 50%

No, not fucking bayes (go eat a dick)

Which coin?

Answer is 1/3

Actual fucking retards

Its a play on words you dumb cunts

It's obviously 50%

The one that landed heads. One coin always lands heads, and one coin has the choice of landing heads or tails.
1/2.

You're retarded. Congratularions.

Answer is 1/3, plebs.

It's a basic conditional probability question, you dumb cunt.

the people giving answers other than 50% will post in IQ threads claiming theirs is 138

oh Sup Forums.

Are you a fucking retard?

Look at Bayes' formula and you'll see it's actually fucking 50%

>One coin always lands heads

You're an idiot.

>[...] given that at least one of them landed heads?

are you all trolling or completely autistic?

you're basically saying a single coin as 1/3 of a chance to land on heads.

>guaranteed heads on the other coin

Wrong, this would only be true, if both coins would be flipped after each other, and if we were looking at the propability of 2 heads after the first toss has resulted in a head.

What the text actualy asks for is the propability for 2 heads, if atleast 1 head is present after BOTH have been tossed, and for some reason you know that one is head, but not what the other one is.

Thus we can caluculate it as

(Propability of 2 heads) over (Sum of Probabilty of all combinations that contain 1 head)

this is true because the alternative of "no heads" has been eliminated by us knowing that atleast one head has been seen.

now the combinations

heads tails
tails heads
tails tails
heads heads

are a-priori all equally as probable, with a chance of 1/4 each.

inserting this into bayes rule as mentioned before we get:

(1/4)/(1/4+1/4+1/4)=(1/4)/(3/4)=1/3

These are the combinations that can result from flipping two coins:

HH
HT
TH
TT

TT can be eliminated, because we know that at least one coin is heads:

HH
HT
TH

Out of those 3 possibilities, 1 has both coins landing heads.

The probability is thus 1 out of 3, or 1/3.

I think this is a question of interpretation of the question. You can look at it in two ways:

"One coin landed heads. What is the probability that the other one lands heads as well?" -> in that case the probability is 50%, because the first coin doesn't influence the second one.


The other way you could understand this question is akin to the monty hall problem.
Let's say that 2 coins were flipped. The possibilities are:
TT, TH, HT, HH

That's 1/4 per possibility, so normally there's a 25% chance that the coins will land HH. So if someone flipped two coins, you'd have a 25% chance that you'd be right if you guessed HH.

However if someone flips two coins and then tells you that one of them landed H, the possibilities are now:

HT, TH, HH

Obviously the TT possibility was eliminated because you know that at least one coin landed H. So now you're basically guessing which of the options HH, TH, HT the coins landed. That's 1/3 per option, so a ~33% possibility.


There. Now OP can go fuck himself with his bait

Wrong, faggot. It's like you can't read or something.

Two regular coins WERE FLIPPED. Past tense. It happened already.

>but it says that one.....

but-but-but-but-but-buh-buh-b-b-b-b-

NO!. LISTEN, you stuttering imbecilic fuck; When you flip TWO REGULAR FUCKING COINS, there are 4 (count them, 4) EQUALLY PROBABLE outcomes possible. These are as follows (probabilities in brackets):

HH (1/4)
HT (1/4)
TH (1/4)
TT (1/4)

Now, the question is a CONDITIONAL probability question. It specifies a condition. The condition is that AT LEAST ONE COIN LANDED HEADS. Note: NOT to be confused with "One coin WILL land heads" or "One coin ALWAYS lands heads" Or "One coin is GUARANTEED to land Heads." No. None of those are true based on the question. The question simply tells us that at last one coin landed heads. It doesn't say which coin it is or give us any other information about the flip.

So, we have to look back up there to our list of outcomes and count how many of them satisfy the condition of having AT LEAST ONE HEADS coin: Weel, let's look at what we get:

HH
HT
TH

WOW, look at that. 3 equally probable outcomes which ALL satisfy the condition. This means that the result could be ANY of these 3 outcomes.

Now, we next need to see how many of those 3 outcomes contains two heads coins. Well, look at that:

HH

So, 1 out of 3 equally probable outcomes contains both heads.

1/3

If you do not understand after this, just kill yourself, or at least never have children.

1 in 2. Same as anything in the universe. It will or it won't happen. No amount of known variables in all of existence will ever make that statement false.

this

Thank you

1/3
why?
lets list all cases
TT this flip is invalid since non landed heads
TH
HT this is different from TH because we have two different coins
HH this is what we are looking for
-> 1/3

>you're basically saying a single coin as 1/3 of a chance to land on heads.

No, we're not you stupid cunt. We're saying that the probability of both coins being heads when AT LEAST ONE coin landed heads is 1/3, which is correct.

its obviously 1/4
2 coins, 2 possibilities each
/thread

You are all missing something, read the fucking question you degenerate cunts.

>The condition is that AT LEAST ONE COIN LANDED HEADS. Note: NOT to be confused with "One coin WILL land heads"

are you actually retarded?

i can't deal with this autism

Yeah, I can deal with being wrong.
What're your stances on gold and silver balls in boxes?

Kek. If this not bait you are retarded

braingle.com/brainteasers/teaser.php?op=2&id=29228&comm=0

you're not tho. one coin is already guaranteed to land on heads, therefore the probability of the other coin landing on heads is 50%

kill yourself

But tails tails never happens

Hey retard one Lands heads 100%.
Ur possibilities are only heads heads, noheads heads, heads noheads.

exactly, because 0/(3/4) is still 0

0 obviously beeing the propabilty that the result will be tails tails after it is know that one toss resultet in heads.

this guy is the kind of nigga to sit at parties (when he gets invited) and try to chat up the landwhales using math

>only to still be rejected

>one coin is already guaranteed to land on heads
>gauranteed to
>in the future

Are you fucking retarded? The flip already happened, and at least 1 of teh 2 coins landed heads.

>ALREADY landed

it wasn't GUARANTEED, and we don't know which coin it is.

This means that EITHER coin could be tails, just not BOTH.

That gives us 3 equally likely outcomes, all containing at least 1 heads:

Heads - Heads
Heads - Tails
Tails - Heads

1/3

>inb4 HT is the same as TH

No, morons. They are 2 distinct outcomes.

1/3

if they were electrons you'd be right.

they're coins though.

1/3

HT is the same as TH

one is heads, the other is tails, no matter the order of the coins

it's 50%

I normally dont respond to these threads, but i saw this scrolling past, and its absolutely correct. The original question is framed imprecisely. The probability is different depending on whether the question is asked before the coins have flipped, or after one has flipped and landed heads.

Ok I was wrong

graduate highschool, faggot

You're either trolling or completely buttfucked retarded.

Either way, you're not worth explaining the solution to.

Everyone who isn't retarded can read the solution ITT already and see the answer is 1/3

Your b8 has failed.

Why are people so fucking retarded... It's obiously 50%. Fucking shit-tier education

/thread

ok faggots we already have 2 coins

There's 2

If flipped there's a 25% chance they'll both be heads

Now if 1 stays head all the time, you have 1 coin with a head/tails 50/50 chance being heads

100% probability coin with the 50% chance coin.

So it's 50/50

>thinks a single formula applies to every given situation

just end yourself, you mad virgin boi

No we can clearly see who the retard in this thread is

Saying it isn't 50 percent because the first flip being heads doesn't factor into it is the same as saying "Oh, flipping a normal coin and getting heads is less than a 50 percent chance, because you have to factor in the chance that coins were ever even invented!"

not the other user, but what the fuck are you talking about?

>Now if 1 stays head all the time

It doesn't.

1/3

...

the question isn't asking the probability of a given outcome

HH = 1/3
HT = 1/3
TH = 1/3

the question is asking whether a coin that has a 50% probability of landing heads or tails, lands heads, given one coin is guaranteed to land heads.

i don't see why you are all struggling with this so much

>It doesn't.
Read that last fucking sentence

I thought that to but both coins have an equal chance of being tails. Its not just one coin allways landing heads

>the question is asking whether a coin that has a 50% probability of landing heads or tails, lands heads, given one coin is guaranteed to land heads.
>one coin is guaranteed to land heads

CRITICAL ERROR

CRITICAL ERROR

ABORT

ABORT

No coin is guaranteed to land heads.

2 coins were already flipped, and at least 1 of them landed heads.

it could be coinA or coinB.

This gives us 3 equally likely outcomes

CoinA - BoinB
Heads - Tails
Heads - Heads
Tails - Heads

1/3

You said
>1 stays head all the time

OP's questions says
>at least one coin landed heads

Do you not see the difference between those 2 statements?

Your statement implies that 1 coin is FIXED as heads.

OP question doe not imply any fixed coin, simply that at least 1 coin landed heads after the flip. it could be EITHER coin, meaning EITHER coin could be tails, just not both simultaneously.

So no coin is fixed in the OP question. It is in yours.

Answer to your scenario is 1/2

Answer to OP question is 1/3

Pic Related

your autism has surpassed me

>one coin is guaranteed to land heads
>it could be coinA or coinB

so yes, one coin?

holy shit

the answer to this question is 50%, it's a play on words, and you nerds aren't acknowledging that. End of story.

Retard, there is a difference between
>one coin is guaranteed to land heads
and
>at least one coin landed heads

If you don't understand the difference, work on your reading comprehension.

Answer is 1/3

Source:
courses.cs.washington.edu/courses/cse312/11wi/slides/04cprob.pdf

> preassumed heads and tails are the only outcomes of a coin flip
50 percent. it doesn't matter WHICH coin landed heads (=: B). the other coin can land heads or tails, thus 50 percent possibility that it lands heads.
A := "other coin lands heads"
P(A | B) = P(A n B) / P(B) = //A and B are independent
= P(A) * P(B) / P(B) = P(A) = 0.5

The average is 37% of the time it will land opposite of what face you flipped from.

Anyone who thinks it's not 50% is fucking stupid

You're a fucking moron, reread the problem and come back when your autism isn't so heavy

If order matters then it's 1/3. If order doesn't matter then it's 1/2.

Incorrect.

A = "both coins are heads"
B = "at least one coin is heads"

P(A|B) = P(A∩B)/(P(B)) = (1/4)/(3/4) = 1/3

If you think it's 50%, you're retarded.

Congratulations.

You're a fucking idiot.

/thread

The phrase "at leas" implies that you have to use counter probability and have to assume that bot dont land on heads, so its

1/2*1/2= 0,25

Now you have to subtract this from 1 and you get

P(E) = 0,75% for at least one to land on heads.

You're literally autistic. If one is guaranteed to land on heads then it may as well be ruled out of the equation because the outcome is 100% heads for that one. That leaves one coin and if you think the probability of a coin landing on heads is 1/3 then you must be a fucking second grader

50%

retard confirmed

>If one is guaranteed to land on heads

It isn't. There's your mistake.

ITT: idiots who assume that the coins were only flipped once each

One coin is not guaranteed to land on heads. One of two lands on heads

Which one?

wrong.
A := the other coin is heads
B := a coin is heads
A | B := "the other coin is heads GIVEN that a coin is heads"
P(B) = 50 % /= P(A)
P(A | B) = ... = P(A) = 50 % as I already wrote in the other post.

Get ready for me to blow your mind.

In standard mathematics, you determine the probability of multiple events occurring by multiplying the probability of each event by one another. So in the case of a coin flip. Each coin has a 50% chance of landing heads, and therefore you multiply 0.50 by 0.50 to get the result of 25%. The chance of both occurring is 25%

In this situation. Your four results (Heads/tails, tails/heads, tails/tails, and heads/heads). Have one permanently disqualified (TT). Which leaves three situations. HT/TH/HH

If you look at the right coin individually, in two of those three scenarios it can be heads.
If you look at the left coin individually, in two of those three scenarios it can be heads.

So each coin individually has a 66% Chance to land on heads. 0.66*0.66=43.56%


The system has a 43.56% chance for both coins to land on heads if T/T is disqualified.

Fuck, now I see what you mean

does not matter. we didn't give them numbers or another thing to divide them.

>In standard mathematics, you determine the probability of multiple events occurring by multiplying the probability of each event by one another.
wrong, retard

then how do you know it landed heads?

Damn, you niggers are dumb.

This is a conditional probability question.

In simple words, it is asking what is the probability of event A given event B.

To answer a conditional probability question, we use Bayes' theorem.

Which in simple words looks liek this

(Probability of event A) DIVIDED BY (Probability of event B)

Event A = 'Both Heads' = 1/4
Event B = 'At least 1 Heads' = 3/4

So the probability of event A GIVEN event B is
1/4 DIVIDED BY 3/4
and that equals 1/3

1/3 is the answer.

Don't be absurd. Someone who is autistic would know it's 50%.

Step one: H or T
Step Two: Depending on the H or T, HT, HH, TT or TH. We want HH, since we want both to land head, and the first one did.

HH is one possibility out of four. To get HH, we have a 25% probability.

>(Probability of event A) DIVIDED BY (Probability of event B)
wrong, fucker
if you're going to shout HURR DURR BAYES, you'd better know what Bayes Theorem is.

Holy shit, you're retarded.

You've never heard of bayes' theorem before have you?

I keep rereading what he wrote but it makes complete sense. could someone elaborate on what's wrong?

Did anyone get the right answer yet?