I flip two coins

I flip two coins.
At least one of them lands heads.

What is the probability that both land heads?

Attached: flip.jpg (800x1013, 64K)

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25%

50%

HA HA YOURE WRONG YOU FELL INTO MY TRAP NOBODY UNDERSTANDS PROBABILITY I'M THE GENIUS YOURE STUPID HA HA LOL ROFLBBQ!

25% Right? fuck im too dumb for this

50 % brainlet

1/4

100% the coins are double headed.

Burgers detected

tree fiddy

OP here, i went afk and came back to realise, not ONE of you has given the correct answer yet.

Boy girl paradox

wikipedia org/wiki/Boy_or_Girl_paradox

If A is the event that the first coin lands heads, and B is the result that the second coin lands heads, then what we're looking for is P(A∩B|A∪B). The probability of both A and B is 12, so A∪B is 34. The probability of A∩B is P(A|B)P(B), which is 12(12)=14. Therefore, the probability is 1434=13

If A is the event that the first coin lands heads, and B is the result that the second coin lands heads, then what we're looking for is P(A∩B|A∪B). The probability of both A and B is 1/2, so A∪B is 3/4. The probability of A∩B is P(A|B)P(B), which is 1/2(1/2)=1/4. Therefore, the probability is 1/4 ÷ 3/4=1/3

if the first one is heads then it's 100% so if the second could be heads that lowers the thing to 75% so is the answer.

Possibilities of the throws:
T/T
T/H
H/T
H/H

But you said one of them landed head, which gives us those reduced choices:

H/T
T/H
H/H

Both of them landing heads thus represents a 1 out of 3 chance.

Finally someone that understands

if you're tossing the coins in order and the first lands heads, then its a 50% chance the second will land heads. But if you've already flipped both coins and you somehow know one is heads, then its a 1/3 chance they're both heads

He didn't say "what's the probability a coin lands head" but "both of them" with the information that one of them already landed head, you dumb cunts

H/t and T/H are the same thing

In the UK (or Wales at least) this kind of thing is taught at AS-Level, roughly age 17, but it's not too tough. Coin flips are mutually independent, meaning the outcome of one test doesn't affect that of the next. For mutually independent tests, the chance of consecutive identical outcomes is multiplied, such that for an initial outcome possibility of 50% (i.e. coin-flip), the same outcome happening again is 50% * 50%, or 25%. For three heads in a row, it's 12.5%, four is 6.25%, five 3.125%, and so on.
HOWEVER, when we know the first outcome, the possibility is only that of the second outcome itself. This is because there's only two outcomes possible - heads and tails, or heads and heads. When we don't know it, there are four - two tails, two heads, or tails first then heads, or heads first then tails. So in THIS case, the answer to OP's question is actually 50%.

For simple cases like this, is much easier to think about it in terms of the possible outcomes rather than the mathematical chances. You get the same answer, but you're less prone to slip-ups like the one being made by everyone saying 25%. Easy mistake to make.

That is why you disregard it. , in other terms , it's like asking what are the possibilities in one landing head , which is 50% , burger

No, it isn't. Coin 1 Heads, Coin 2 Tails vs. Coin 1 Tails, Coin 2 Heads.

1/2

30%

Attached: coins.jpg (2442x1998, 716K)

Those who answer 50% don't realize that 1 head 1 tails is twice as likely to happen than two heads. The correct answer is 33%. Other answers are just bait/single digit IQ.

No it isnt. There are 2 coins in play.
One of them is going to be heads but we dont know which coin..
So t/t gets removed
Leaving only
H/t
T/h
H/H

Which is 1/3 or 33% repeating probability

No. See this: You fail at probabilities.

just here to play devil's advocate and let all you faggots know that a coin toss isn't technically 50/50 due to the way coins are cast and tend to have a heavier side making it more like 51/49 in favor of the bottom side during the minting process
also the answer is zero because the other coin was flipped into my gaping asshole and will never be seen again

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Flipping two coins, flipping one coin, checking if it's head, flipping the second one, checking if it's heads and flipping coins by one until you encounter two are three entirely different scenarios with entirelyy different outcomes.

Its really effing basic. If one of them is always going to be heads then it is fully out of the equation. Given no special facts other then this you can assume the coin is not screwed with and is of equal weight and probability for either side. Therefore you have a 50 percent chance. Nothing from the previous quarter has any effect on the probability of the second quarter landing either way.

*until you encounter two heads in a row

That doesn't make sense. The way it is worded is as good as guarunteeing one is always heads. It doesnt matter which is flipped. The other is fully independent of that flip It isn't all of a sudden governed by new physics. It has an equal chance of landing heads or tails and be re added into the original equation. The only way there are three probabilities there is if both quarters have a chance of flipping heads or tails in the first place. Which means it would have four probabilities not three anyway. One is always going to land as heads in this problem which takes any influence it has on probability out of the equation. If you ADD to the problem that the one that WILL be heads can somehow be done in a purely random way then sure this problem makes sense and would be 33%. But its not so the problem itself is just illogical in the first place.

>The way it is worded is as good as guarunteeing one is always heads.
No it isn't. The wording is "I flip TWO coins." The case where both are tails is discarded.

Yes, one is always heads. It can be just the first one or just the second one - twice as likely as both landing heads.

Jesus fucking Christ.

This is worse than when someone posts the Monty Hall problem.

The issue is not that you aren't able to solve it correctly on the first try, the problem is that this paradox is old as balls. They show you this in every beginners statistics class.

Alright let's do a bet faggots.

14 throws, I disregard those which are both tails (since we know one of em is heads).
On each throw, I remove the one we know is heads as said in the OP, leaving out the 2nd coin.

Since you say 50%, you're gonna pick heads for that second coin since you don't care. I pick tails. $100 per win.

Looks like we got 7 tails and 4 heads. You owe me $300. And look at that, 4/11=0.36 which is fucking close to 1/3

Attached: brainlet.jpg (936x2408, 659K)

A variation of this paradox:

google.com/amp/s/whyevolutionistrue.wordpress.com/2018/02/20/the-answer-is-2-3/amp/

The question is invalid because you are manipulating probability with an outside influence, making it impossible to solve.

There's no outside influence. Only additional info about the throw (which is one of em fell Heads)

Have you never been to school ?

That is not the circumstance. The circumstance is an outside force makes sure one coin always lands on heads, not that a double tails flip is discarded. This is a big difference. Also with a sample size that shitty the number is going to be nowhere near true probability, on something like this you got to have a sample size in the thousands at least.

Its explicitly predicting the future in a vague and unrealistic manner, taking away the ability to use probability on the equation.

At 11 throws it's already pretty much spot on on the true probability, which is 33%. Double tails flips are not in the realm of possibilities and have to be discarded.

It's not influencing. It's textbook Bayes theorem task.

Not sure if trolling or retarded

33 %

2 sides two outcomes.
You will get 2 heads or you won't.

Probability is a construct.
The universe is about chaos not order.
Order is a human invention to try and make things understandable so we can feel better about the chaotic godless world we live in.

Stay woke