How smart is Anon Babble?

# How smart is Anon Babble?

25%. The condition "if at least one of the coins" is irrelevant to the question. Either they both land on heads, which has a 25% probability, or they don't.

Mint mark on reverse

That's a 90% silver quarter.

Three possible outcomes

A coin heads, B coin tails

A coin tails, B coin heads

A coin heads, B coin heads

1/3

That was the chance before the flip with no prior knowledge of the result, but now that there is post-flip information, the odds are now 1 in 3

You forgot:

A coin tails, B coin tails

Making it 1/4 or 25%

A coin tails, B coin tails

At least 1 coin landed heads, retard.

Nope. It never says one coin is flipped and if it lands on heads, the other is flipped. It says both coins are flipped and give no other premises. So the chance is always and with no exception 25%.

It is a trick question. The outcomes of past coin flips has no bearing on the outcome of subsequent ones.

Ah, so you can't read? Indian or nigger?

basic conditional probability

trick question

Are you actually this retarded, anon?

The outcomes of past coin flips has no bearing on the outcome of subsequent ones

Who said they do, moron?

Quote where it says it's conditional.

It never says at what time the second coin is flipped in relation to the first coin. Are they flipped at the same time? Or is the second coin flipped after the first. If the latter, then 50% chance heads.

I say 50%

Have a good day

It never says at what time the second coin is flipped in relation to the first coin

That doesn't matter as long as both are always flipped. The only time it would matter is if it was conditional, which it is not.

Are you retarded or is English not your first language?

if at least one of the coins landed heads

That's the condition, brainlet.

first coin

second coin

LOL you're dumb. It doesn't matter if the coins are flipped at the same time or one after the other. All that matters is that both are flipped and at least 1 landed heads. That gives 3 possible outcomes which are all equally probable

HH

HT

TH

1/3

since we know one coin is heads surely we can just eliminate it and say there's 2 possible outcomes

i mean you either autism that the if heads condition is irrelevant and call it 25% or assume its relevant and call it 50% not split the baby for 33%

You're right. It's 50% all the time, regardless

Prior to that, it says "two coins are flipped". It does not say "one coin is flipped, and if it lands on heads, another is flipped". The purpose of these kinds of questions is to see if the subject is able to understand instructions, or if they insert their own interpretation outside of the information that was given. Since you clearly fail at the the former and gleefully do the latter, you should stick to skillsets more in line with your level of competence. Perhaps sorting paperclips according to color? Or chewing Lego's? Regardless, the answer is 25% and it's time for you to shut the fuck up.

PROBABILITY THAT BOTH COINS

LANDED HEADS?

post picture of two coins that landed on tails

I'm taking both quarters and you get..

Retard

two coins are flipped

at least 1 landed heads

what is the probability both landed heads?

Heads-Heads

or

Heads-Tails

or

Tails-Heads

1/3, brainlet.

IF AT LEAST ONE OF THE COINS LANDED HEADS

Condition #1: One of the coins has to land heads for the question to proceed

Condition #2: When and instance of Condition #1 happens then what is the probability that both of the coins was heads.

The answer is 1/3

You're right but your pic is wrong. Swap out HH for TT and we're good.

Learn to read, both coins are always flipped.

Both coins are always flipped. Says so in the first sentence. It does not say both coins are flipped on the condition that one of them lands on heads.

Hello stupid.

LOL are you genuinely retarded or trolling? We know 2 coins were flipped. We also know at least 1 coin landed heads. That's why the answer is 1/3 you simpleton.

1/3 is the midwit trap

the coinflip is a 0.5 probability

depending on problem formulation its multiplied by 0.5 or 1

"If at least one of the coins landed heads" is not negating that both coins are flipped. How come you don't understand this?

It's irrelevant what one of them landed as. Both coins are always flipped and the winning condition is that both land heads. There aren't three conditions possible, there's four.

Tails - Tails

Heads - Tails

Tails - Heads

Heads - Heads

Or, if you prefer, 0.5x0.5=0.25.

A coin landing head or not is irrelevant, it's a distraction from the question. It does not say there is a requirement of a coin to land on heads before the second one is flipped.

Test

Borderline retarded.

75%

The if statement has to be interpreted as:

One has already landed so 50% of the coins are already heads.

Now the status of the second one (is already flipping or has to be flipped) is technically irrelevant at this time but has the probability of 50% to land heads too so the total coins probability at this point is 75%

is that right or did I miss something?

The probability changes when moving in time. But the statement suggests that we’re looking at the probability after one has landed and the second hasn’t yet

is not negating that both coins are flipped.

Who said it did negate it, retard?

There aren't three conditions possible, there's four.

Tails - Tails

Heads - Tails

Tails - Heads

Heads - Heads

Yes there are 4 equally probable outcomes possible for a 2 coin flip. We know this, idiot. Your problem is that you have never seen a conditional probability question before. In the OP question, there is a condition, the condition is that at least 1 coin landed heads. GIVEN this CONDITION, the question THEN asks you to calculate the probability that both coins landed heads. As at least 1 landed heads, the tails-tails outcome is no longer possible, leaving 3. Seriously, faggot, just do a quick google for conditional probability to realize you are a fucking simpleton.

1/3

is that right or did I miss something?

Yeah, you're missing a brain, fag.

midwit on midwit violence

This is the correct answer.

Virtually all responses to this answer are not by people who actually don't understand it or actually don't agree with it, but instead by people who are trolling

Don't waste your time responding to trolls

They can't be made to understand, because they already understand

You can't be made to agree, because they already agree

They're just pretending to waste your time

midwit

You're the moron who thinks the answer is 0.5 LOL

Your problem is that you have never seen a conditional probability question before

I have, but this isn't it. It's not formulated as a conditional question, and you "filling in the blanks" making it a conditional question is why you believe it's 1/3. Your problem here is not the math, it's understanding what is being asked of you. If the question had been formulated as a coin toss where the condition of the first coin decided whether or not the second would be flipped, you would be right and the answer would be 1/3. But that is not the question. You "filled in the blanks" of the question and made it a conditional math question, when it is not.

I have, but this isn't it.

HURR DURRR HURRRR DURRRR I'M RETARDED NOW I'LL PRETEND THE PROBLEM IS THE QUESTION AND NOT ME

LOL no, retard.

Word problem

Step 1 = disregard irrelevant information

there are two coins

remaining steps = figure the actual problem

1 coin = 2 equal possibilities

EITHER heads or tails

1 desired outcome = 50% chance.

[GDGXR]

God damn you geniuses are retarded.

Thanks fag, now I explain it to you

If 2 coins get flipped the probabilities are 25% that they land heads

If one has landed it goes up to 75%

It cant be 1/3 since under normal consitions you cant make one of the coins not fall on tails. There is no way you can claim one will fall on head

If one has landed it goes up to 75%

God. I didn't think you could get dumber than the 50% anons, but there we go.

the wording is purposely confusing to start arguments

at a quick glace the question makes it sound like its asking ' if one quarter is glued to the table what are the odds the other coin lands heads?'

but whats its really asking is 'if we flip two coins and ignore the both landing tails outcome what are the odds the other coin lands heads?'

1 coin

There are 2 coins, retard.

at a quick glace the question makes it sound like its asking ' if one quarter is glued to the table what are the odds the other coin lands heads?'

Maybe if you are fucking retarded.

Kek

"If at least one of the coins landed heads..."

Is strictly equivalent to

"Ignoring outcomes where both land tails..."

Does this make the question easier to answer?

Maybe if you are fucking retarded.

they wont even be able to read the question.

Okay than answer this: two coins get flipped. One landed heads. what’s the probability that both land heads

Does this make the question easier to answer?

Not for the 50% people because they are dumb.

still want to...

1/3

First sentence is literally that both coins are being flipped without any condition. There is a period after that statement, so both coins are being flipped regardless of what the outcome of any of them are. If this had been conditional, it would have been worded:

One coin is flipped, and if that coin lands on heads, another coin is flipped. What is the probability that both coins landed heads?

But that is not how it's formulated. It's formulated so that regardless of outcome, both coins are always being flipped. Therefore there are four possible outcomes and only one of them is satisfactory. The probability is 25%

It's obviously 50 percent

0

It's 0.

According how smart you’re you have to have an astonishing career and life, now take a look inside the room you’re sitting and the building you’re in. I think those facts speak louder than your skills

Fag

One coin is flipped, and if that coin lands on heads, another coin is flipped. What is the probability that both coins landed heads?

That's a completely different question, you fucking mongoloid. You might be the dumbest cunt in this thread. Good job.

According how smart you’re you have

Answer is 0, pleb, but you probably don't understand how.

That is the question every tard in this thread is trying to answer, and it's the wrong question. I'm sorry you don't understand your own native language, but that is no one else's fault.

That is the question every tard in this thread is trying to answer

Why are you assuming that, moron?

50%

Give me my two quarters!

it's 0, moron.

that bit of information is disregarded as irrelevant, because the definition of coin in this ord problem indicates 2 possibilities per coin, but one of the coins is predetermined, thus negating it's uncertainty and it's contribution to possible outcomes; Retard. If you want a simple algorithm for finding coin flip proabilities, as in the case of adding more coins or even multiple-outcome mediums...

ab-cb=x

1/x to convert x to percentage

a= number of mediums

b=number of possibilities per medium

c=number of predetermined medium outcomes

3 six-sided dice

2 predetermined

(3·6) - (2·6) = 18-12 =

6 possible outcomes

1/6

durrhurr 1 six-sided dice

1 desired outcome

≈16%

If either of the pre-determined dice doesn't match the one desired outcome, the remaining probability for the third die to make it fit is 0%. Out of the original pool, given no predetermined dice:

(3·6)-(0·6) = 18-0 =

18 possible outcomes

1/18

≈5% chance of a single desired outcome of possible combinations.

one of the coins is predetermined

False

1/3

Reversing the percentages yields missing percentages, those are considered margins of error and uncertainty [in uncertainty.]

at least one coin lands heads

That's a GIVEN predetermined factor dumb fuck. That negates the uncertainty of one of the coin and renders its contributions to the determination to be moot. Only ONE remaining coin (out of the two) is uncertain and calls for mathematically expressed resolution.

Because they're arguing whether it's 1/3 or 1/2 and the only way you'd end up arguing between those two is if you missed that the first claim and the condition didn't have an "and" between them.

2 uncertain coins possiblilities=

Heads-Tails

Heads-Heads

-Tails-Tails

-Heads-Heads

At least one coin is heads

negates two possibilities

Tails-Tails

Tails-Heads

REAMINING possibiltities

Head-Tails

Heads-Heads

50%-50%

EVEN IF you designate the rpredetermined coin flip as the SECOND flip which is unintuitive as fuck, then your remaining possibilities are

Tails-Heads

Heads-Heads

50%-50%

What the serious fuck are you trying to inject into a simple problem? Pseuds always try to imply givens and factors not applicable.

2 uncertain coins possiblilities=

Heads-Tails

Heads-Heads

Tails-Tails

Tails-Heads

God dammit, you got me flustered.

predetermined

I don't think you know what that word means. Let me help you, it means 'set in advance' which isn't the case here. At least 1 coin landed heads by chance, it wasn't 'set in advance.' Your scenario is basically leaving 1 of the coins heads up and just flipping the other coin. This doesn't allow for the possibility of EITHER coin being tails, which is allowed in the OP question. Again, this is a case of you not having a basic understanding of language and probability. Your brain doesn't function correctly.

Look at the pic to see if it helps. In your scenario the red coin would always be heads, and there would only be two outcomes possible giving 50%.

In the OP question, EITHER coin could be tails, just not both simultaneously. That's the difference that you fail to understand, anon.

1/3

No, it's a core part of learning conditional statements in both logic and math.

50%ers on SUICIDE WATCH

two quarters

Your picture is misleading in that it represents two distinct designations. EITHER the "at least one lands heads" is the first coin OR it's the second coin. Your picture implies an amalgamation of separate possibilities. Even IF your try to convolute the wording, the fact that one lands heads eliminate TWO possible outcomes, not just one. Fucking mongs.

the fact that one lands heads eliminate TWO possible outcomes

Oh yeah? Which two outcomes does it eliminate?

Hey nigger, Here are the outcomes for a 2 coin flip

HH

HT

TH

TT

At least 1 landed heads. Which 2 outcomes do you eliminate lol?

From a basis standpoint, your stupid picture is missing the following combination:

Always 50/50. It either will or won't.

if one lands heads, TWO possibilities are erased, not one you fucking morons.

picrel*

At least 1 landed heads, you simpleton.

if one lands heads, TWO possibilities are erased, not one you fucking morons.

Oh yeah? Which 2 lol?

If the red coin lands heads, both of those outcomes are rendered certain. If the blue lands heads, two of the outcomes are rendered certain. One of the two for each didn't happen and one did. Hence certainty of that flip.

See, this entire breakdown is a good example of Bayesian logistics and its philosophical aspects in approach to more accurate math. The fact anyone feels based in arguing anything other than a 50%-50% is proof how classic pre-Bayesian mathematics were inherently less accurate.

So what you're saying is you could have

red=heads and blue=heads

or

red=heads and blue=tails

or

red=tails and blue=heads

So 3. And 1 of those 3 is both heads?

So 1/3

picrel

THESE are the possibilities you're advocating as viable...

but in the final outcome, the very bottom one CAN'T happen, if the final outcome stipulates at least one landed heads.

[KAXW2]

No, you're amalgamating two different possibilities. What' you're missing is that you're listing THREE of the possible SIX outcomes, not 3/8.

know who's two quarters that is, bitch?

the probability is 0, because nobody has 2 coins in this economy

vote trump

There are 4 possible outcomes, and you're only listing 3. Either we're talking 2/4 possible combinations OR 4/8 individual possibilities. What you're doing is convoluting the math by mixing factors to present a misleading representation of the reality.

possible SIX outcomes

LOL moron. List the 6 outcomes LMAO

The six outcome AS THEY RELATE to YOUR representation

1h

1h

1t

2h

2t

2h

In your misleading representation you're missing the additional 2/8:

1t

2t

[ARWXS]

LOL are you retarded, son? Show me the outcomes for a 2 coin flip that have at least 1 heads coin. This should be funny LOL

You're trying to amalgamate the beginning and the end. As a beginning you're one pair shy of the reality, and as an end representation, you have one pair too many.

After reading this thread and drawing out the possibilities on ms paint, I honestly went from believing 50% to 1/3.

Obviously the question is retardedly asked. At the end of the day, I personally interpret the question as involving TWO coins. It specifically states "IF AT LEAST ONE OF THE COINS LANDED HEADS", as such, any one OF TWO COINS could have landed on heads, and we have to answer the probability for two consecutive heads in that specific scenario.

To those who still believe in 50% or 25%, can you please explain why you disagree?

You're trying to convolute physical representations with Mathematical conception.

Picrel is the mathematical conception.

example 1

tbc...

Obviously the question is retardedly asked

It's not.You're just a retard. Conditional probability is often counter-intuitive and you have an illogical, female brain which makes you prone to being a fucking mongoloid and thinking intuitively that the answer is 50%. You should cut your penis off, you faggot.

At least you got the correct 1/3 answer in the end.

THIS is the physical representation. You're misrepresenting the reality by presenting anything different.

Two of those are the same, retard, leaving 3 equally probable outcomes possible.

HH, HT, TH

1/3

see:

EITHER the first coin is the predetermined OR the second coin is the predetermined. One of the possibilities SEEMINGLY overlaps conceptually. This si what I'm talking about idiot, you're attempting to misrepresent reality in one form or another, either the physical reality or the mathematical concept of reality.

[KGGN0]

EITHER the first coin is the predetermined OR the second coin is the predetermined.

There is no predetermined coin, fucktard. At least one coin landed heads. It could be either coin, so there again are 3 possible outcomes. you are literally braindead.

be a Jew with money

misrepresentation

convince plebs that 8 is only 6

or that 2 is actually 3

It's equal because I keep two and split the rest

it's fair because there are three and I rounded in my favor and kept two.

You're attempting to convolute the problem with word play in that case you fucking idiot. "Predetermined" substitutes as "Stipulated." One of the motherfucking coins was stipulated to have landed heads. WITH THAT GIVEN, TWO possibilities are removed from the potential 4, NOT ONE removed of 4 leaving 3.

WITH THAT GIVEN, TWO possibilities are removed

Only 1 possibility is removed, retard. TT is the only outcome that does not have at least 1 heads coin. You are retarded LOL

NO, there are FOUR probable outcomes and only TWO are viable when one coin is a given heads. As seemingly logical as your chart is, it's a misrepresentation of the probabilistic accuracies, HENCE why Bayesian logistics and philosophies are superior to stupid classical overly simplified, cold logic, dumb shit like you and your Newtonian Gravity.

You can only have two out of that chart, not three.

[DKGMT]

coin1 = heads, tails

coin2 = heads, tails

possible outcomes:

coin1: heads

coin2: heads

coin1: tails

coin2:heads

coin1: heads

coin2: tails

coin1: tails

coin2: tails

answer: 25% or 1/4.

explanation: there are four total possibilities; 1) you can have both coins be heads, 2) you can have coin1 be tails while coin2 is heads, 3) you can have coin1 be heads while coin2 is tails, )4 and finally you can have both coins be tails. only one of these four possibilities is what will satisfy the condition of BOTH coins being heads, so 1 outcome divided by 4 potentialities = 1/4 or 25%.

NO, there are FOUR probable outcomes and only TWO are viable when one coin is a given heads

LOL you can't count. 3 of the 4 have at least 1 heads coin, dumbass. LOL

Here's the Bayes' proof again. 1/3

Get a load of this retard. Try reading the question again, you mongoloid.

These are the two viable outcomes based on THAT chart.

[0NJND]

picrel*

HH or HT or TH

Gotcha. 1/3 Thanks

yeah that makes sense.

Anyone at this point not on my side of this idiotic math line is in fact a pseudointellectual who doesn't know what the fuck they're talking about or how reality fucking works; ESPECIALLY in regards to mathematics. Stupid fucking monkies.

What's your answer?

To all the 1/3 fags who are calling other fags retards because they don't get this correct:

Remember when you were a conditional probability virgin it took your mind some effort to get its head wrapped around the idea, too

Or maybe I'm just a slow fag

You're just a fag.

I got something for you!

Not just a fag, I'm a nigger too