How smart is Sup Forums?

How smart is Sup Forums?

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en.m.wikipedia.org/wiki/Boy_or_Girl_paradox
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50 percent chancr

Possible outcomes.
TT
HT
TH
HH
We can strike out double tail. Which leaves us with three outcomes and the chance for double head is 33%.

Retard detected.

This.

(1/4)/(1/2) = 2/4 = 1/2

Retard detected.

There is 100% probability that I don't give a fuck

...

Nigger.
There are 4 outcomes total. 3 of those have at least one of them landing on heads. Of those 3, only 1 has them both landing on heads.
1 out of 3 = 1/3

Nope
(1/4)/(3/4) = 1/3
Prob of conditioning event is 3/4

Morons. Stay in school faggots.

Given that at least one landed on heads, there's a 50 percent chance that they will both land on heads

How dumb are you faggot? The answer has already been given and explained.

1/3

There is a 100 percent probability that this will go on for way too long even though the answer was given and explained correctly at the second post.

Anyone that answered 1/3 is a fucking retard.

In this case you already know the state of one of the coins. This removes that coin's possible states (2) from probability calculations involving the second coin, so the answer can only be base on the chance of the second coin landing heads or tails, which is 1/2.

lrn2simplelogic

>assuming 1st coin is heads

No retard.

Learn2Probability

>If you play the slot machine enough you increase the odds of the next roll being all bars
Please don't breed.

You are incorrect sir. That is not how it works. This is a well known and well discussed paradox that even the smartest of mathematicians have fought over for years, so do not feel bad. It's all about that ambiguity son.

en.m.wikipedia.org/wiki/Boy_or_Girl_paradox

Dude is called conditional probability

>this
>given that at least one of them landed heads
So what's the probability that the other one landed heads ?

You don't know which one is head, it might be first one or second one, it's not like you flip one coin, see it's head, then ask yourself what's the probably to have another head.

since one of the two middle choices are at least guarenteed and the last is impossible its only 50%

Get out here with your b8. Discussion is over. Let it die.

No the experiment you're thinking at is different from the one in the image. You're not doing two flips one after the other, but simultaneous flips and some how are assured that one of the two coins will be head. That's a little weird, but that's what the problem states.

no

Negative. Each coins probability of landing head or tails are completely independant of eachother.

If you are calculating the probabilty based on not knowing the state of either coin then it can be calculated as 1/3, because there is a known state that is illegal (tails-tails), which leaves 3 possibilities. However, the state of one coin is known, which rules out both of the 2 possibilities of it's opposite state. (Tails-Tails, Tails-Heads).

The answer to this question is 1/2, because it isn't worded correctly for the answer you are looking for.

This is like that 'How many triangles are there?' problem that's also floating around all the time that people answer 80 to, where the correct answer is 64, because the question is never worded correctly.

Whats the difference between HT and TH?

These threads are always great because it is such a well known and debated paradox that messes with even the smartest of people. I could respond with a well worded and thoughtful rebuttal citing proper sources, but I really don't feel like doing that, so instead ... Your a dingus. The answer is 1/3 faggot.

>However, the state of one coin is known, which rules out both of the 2 possibilities of it's opposite state. (Tails-Tails, Tails-Heads).

At least 1 coin being heads DOES NOT rule out Tails-Heads you fucking retard. Tails-Heads satisfies the condition of having at least 1 heads coin and is JUST as likely as Heads-Heads or Heads-Tails

1/3. See Bayes Theorem

Stay in school.

Pretty sure it's not as you say. Still this conditional probably thing always bummed me. Like the 3 boxes game where one box contains money and the other one is empty. You pick one of the three then the announcer shows one of the other two is empty and asks if you want to change your mind. Conditional probably says you've better chances of you pick again between the two remaining.

>Whats the difference between HT and TH?

They are 2 separate and distinct outcomes. Imagine 1 coin is a penny and the other coin is a quarter

penyy=heads and quarter=tails
is different to
penny=tails and quarter=heads

They are both EQUALLY likely to occur however.

3 equally probable outcomes possible;
HH
HT
TH

HH is 1 of those 3
1/3

JESUS CHRIST YOU DUMB MOTHERFUCKERS.

The state of one coin is _KNOWN_, do you not understand that concept?

If the state is known it can NOT magically become the opposite state in order to satisfy every possible combination. ALL combinations involving the opposite state of the KNOWN coin are invalid!

Come back when the state of both coins is unknown.


To use the slot machine analogy that was brought up, it's like asking what the odds are of getting 3 BARS when one when one wheel is already a BAR. That one wheel is a KNOWN state, so the odds that there will be 3 bars lies SOLELY with the state of the remaining 2 wheels.

You aren't calculating what the odds are from an unknown state of all 3 wheels.

But what if one of the bars is actually a heart?

>it's like asking what the odds are of getting 3 BARS when one when one wheel is already a BAR.

Incorrect retard. You don't know which SPECIFIC coin is heads and so you would not know which SPECIFIC wheel is a bar. There are 7 EQUALLY PROBABLE WAYS to get AT LEASE 1 Bar for a 3 wheel spin.

There are 3 EQUALLY probable ways to get at least 1 heads coin for a 2 coin flip
coin1=heads and coin2=heads
OR
coin1=heads and coin2=tails
OR
coin1=tails and coin2=heads

All 3 of those satisfy the condition. All 3 are equally probable

HH is 1 of those 3
1 of 3

1/3

Since one is said to be heads 100% , there is 33% other also is heads.Coz it can also fall on side and tails.easy shit.

The second coin isn't affected by the first coins outcome. Why can't anyone understand this?

> When one wheel is already BAR
That's your mistake: you know one of the wheels is gonna a be a BAR, but don't know which one until you stop them all.
There's not an already stopped wheel yet.

not sure if thread is dead or not, and not sure if this proves anything, but a simple program to illustrate the point given by earlier anons.

>The second coin isn't affected by the first coins outcome. Why can't anyone understand this?

You're making the false assumption that the "FIRST" coin is the heads coin.

The condition dictates that either coin could be heads or both coins are heads so ;
HH or HT or TH

1/3

1/3

Simple probability

P(A|B)=P(A and B)/P(B)

In this case A is the probability of double heads and B is the probability of at least one of the coins being a head.

P(A and B)=0.25.

P(B)=0.75

P(A|B)=0.25/0.75=1/3

Kinda useless as the math once accepted the "double T is not possible" is quite simple and the program itself doesn't explain why the statement itself is right.

yea i know, but as i said, it's just to illustrate the point, for the people who still don't believe what the others are explaining

>You don't know which SPECIFIC coin is heads and so you would not know which SPECIFIC wheel is a bar.

It doesn't fucking matter which one is. A known state is STATIC.

If a wheel lands on BAR, and you are calculating the odds of the remaining 2 wheels landing on BAR, then the wheel that is a KNOWN BAR may as well be removed from the fucking machine and a Post-It note put in it's place with the word BAR on it, leaving only 2 wheels to calculate both landing on BAR.

A STATIC STATE CAN NOT CHANGE, PLEB.

If we were calculating from an unknown state of all 3 wheels (or 2 coins) then things would be different.

TH and HT are the same thing. There is no variation or difference. One is tails, one is heads. It doesnt matter which is left or right. This makes only two probabilities. It is 50%

>It doesn't fucking matter which one is. A known state is STATIC.
The only thing known is that AT LEAST 1 COIN landed heads.

There are 3 EQUALLY PROBABLE ways to get AT LEAST 1 heads coin for a 2 coin flip. This is what you don't understand you retarded fuck.

If you still don't understand answer pic related and explain how/why you think it's a different question to OP

Protip: Answer is 1/3

Dumb cunt detected.

Read this 1/3

I hate this meme.

But you don't know which one to remove, you cannot just go and take one wheel out because you don't know which one has the BAR state yet. Let's say a friend of yours flips two coins and doesn't show you the outcome, but takes a peek and says: one of the two coins landed head!
Now what's the probability that both are actually heads? The only thing you can deduce from what your friend said is that they cannot be both tails as one is for sure head, then you're left with three possible outcomes, one of which is both heads.

Nope.
It does not matter if one is tails first or one is heads first.
You have a 50% of getting heads on your first coin. The 2nd coin is automatically heads.

Or, rather, the first coin is heads. You have a 50% of getting heads again.

Stop thinking about it so hard. Placement doesn't actually matter

I'm not thinking too hard dumbfuck. It's basic conditional probability.

This is a conditional probability question, so use Bayes' theorem to solve.
P(A|B) = P(A∩B)/(P(B))

Explanation for faggots:

COnditional probability questions take the form:
>What is the probability of Event A given Event B?

OPs question is
>What is the probability that BOTH coins landed heads, given that AT LEAST ONE coin landed heads?

So here

A = "both coins are heads" = {(HH)} = 1/4
B = "at least one coin is heads" = {(HH), (HT), (TH)} = 3/4

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

Nope.

Eeehm, Yes?

If you still don’t get it, you never will

- friend flips a quarter and a cent and hides the outcome
- friend peeks the outcome and says one is head
- either pence is head, cent is head or both are
- one of the tree above is what you're looking for

>What is the probability that both coins landed heads given that at least one of them landed heads?
>given that at least one of them landed heads
>given

If one landing heads is given, the probability is 50% that the one remaining coin will land heads.

Herp

I do statistics, can confirm

Nope

The order of the coin orientation does not matter. H|T has the same weight as T|H so they can be grouped as a single outcome. It's 50%.

1+1 is 2 divide by 3, 2/3 quick mafs

No, because you calculate it after the given fact of one coin being heads. That's when the probability is being observed.

That's nonsense, permutations are not counted as one

It is whatever the lord's will

Read this dumbfuck

It all depends on the time each coin is flipped, obviously. If they are flipped at the same exact time, the chances are not dependant of each other, and both/either coin can land on heads. However, if one is flipped, you see the result, and then the next coin is flipped, then everything changes, since the chances of the second coin's flip are dependant on the first coin.

I'm a mathemetician major in one of the higest colleges in the world.

5girls and 8boys go in a classroom. When they go out, one by one and randomly, how many boys are followed after a girl in average???

Are they wearing clothes?

You're dumb as fuck. None of the scenarios you explained are the one we're facing here. Good luck with your autism major.

This, although your post sounds retarded, nobody gives a shit where you work at, that doesn't qualify your logic.

Anons are using time and sequence when they aren't factors. One is an absolute and didn't affect the outcome.

>One is an absolute

Both coins are variable as either could be tails just not both simultaneously.

HH or HT or TH all possible and equally likely

1/3

Actually , 2 of the girls have big udders and they are underage. Does that help in solving the problem user?

>chances are not dependant of each other
That's not where chance dependency kicks in user, read the fucking problem: you have an hint on the outcome, that's what conditions the probabilities.

> one is flipped, you see the result, and then the next coin is flipped
That's not the case either, you flip the fucking coins together at the same time, but you cannot see the actual outcome and only know at least one is head.

Almost. Could you also describe some of the boys' penises in great detail, not because I'm gay or anything, it just helps with my imagination?

They're distributed logarithmically from 5-9 inches.

Oh look it's this thread again.

"I dont undertsand whats written there and i dont believe in that yourre wrong you fukken idiot"

...

>probability of both being heads when one IS heads
>simulated
>works 1/3
Faggot.

>One coin
>Two possible values
>1/2
>50%

Now THIS makes sense

trips confirmed

Trips say truth

1 in 4 fgts jesus fucking christ

Let us assume that each coin has a 50% chance of landing on heads during its flip. This can be directly deduced from the fact that a coin has two opposing "faces" or sides, with such a small margin in between that we can discard the chance that landing on this "side" will occur.

Now, we first seek the probability that both land on heads with no initial conditions. Since each coin has a 50% chance and we are flipping 2 coins, we end up with 50% of a 50% chance that both coins land on heads, or a 25% chance. This is also because results of the flips will have 1 of 4 distinct results: TT, HT, TH, HH. Since each result must have an equal chance of occurring, each must have a 25% chance

Now we must account for the initial condition, "provided that at least one of them landed heads". This does not specify which coin fulfills the initial condition. This means that either coin can land on heads and that coin flip will be included in the probability.

So, we have 3 possible scenarios now. TH, HT, HH. Remember that we will always perform 2 rolls regardless of the result of the first roll, and as long as one of the two flips results in a heads, that flip will be counted in our "probability total".
Imagine for a minute a theoretical experiment to determine this probability. Experimenters would flip 2 coins then tally the flips. Then, any flip pair not including a heads would not be included in the final probability calculations. So, let's assume for a moment that they performed 200 individual coin flips for 100 total pairs, with the following distribution:

TT - 24
HT - 27
TH - 24
HH - 25

Let's now assume that the researchers throw out the TT results, as these results do not meet the "at least one flip results in heads" requirement. This leaves them with the following probability calculation:

27+24+25=76
25/76=0.3289

With this small sample size, the imprecision is obvious, so we assume this means a probability of 33% that both land on heads given one coin lands on heads.

>33%
Like, what's the third possibilty, a transcoin?

obviously 5/366, taking into account leap year

Noobs. Both coin flips already happened independently. All we know is that double tails is not an option. Therefore, the probability that it is double heads is 1/3.

However, if phrased differently: the first coin is heads, what is the probability that they are both heads? It becomes 1/2.

Some weird ass simulation aint no proof.

I want you to give me an asymptotic formula for that error term.

How can you have 3 out of only 2 possible outcomes? It's 1 coin either heads or tails. First coin doesn't factor in at all.

Read this faget

There is another outcome that all of you are overlooking, and that is if the coin lands on one of its edges. Although this is a very rare occurrence, it will actually lessen the chance of both coins landing on heads since it is also a possibility.

Not related to the question.
Clearly biased example.

That's exactly what the question is about dumbfuck

50% or 1/2

Suit yourself.

...

Just finished a discrete math course.
Lets call the event E one coin landing on heads and the event F the other coin landing on heads.
So, P(E|F) = P(E∩F)/P(F) = (1/4)/(1/2)=1/2=50%
So is correct.

I'll use this as an example. The chance 2 & 3 are effectively the same thing in reality. Really, it's a change between (1,0 / 0,1) and (1,1), one or the other. (1,0) and (0,1) are both the same thing when one of the two is always heads. No matter the case, you are effectively only flipping one coin, the chance one, the one that is really only heads or tails. The other is always heads, period. Only 1 coin has a chance to be tails or heads. It's 50/50 for all the non-autists.

I hope you didn't pay money for your course faggot.

.25 percent chance.
2 sides per each quarter and the chance of 1 landing on heads is 50 percent