I flip two coins

I flip two coins.
Atleast one of them lands heads.
What is the probability that both coins land heads?

*Do not assume that one coin already landed heads*

First to answer correctly wins 10 interwebs and everyone getting it wrong is a filthy nigger.

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

Other urls found in this thread:

youtube.com/watch?v=4Lb-6rxZxx0

25% i guess. im not good at math but im not an idiot

wait no "at least one lands heads" implies one has heads on two sides so the answer has to be 50%

1/3

Yes

It's 50%

This person wins the interwebs, the rest of you are filthy niggers with IQ lower than your shoe size.

1 magma

i like this option, but im having troble explaining

so 1/2 for both coins two event happen together are multiplied and we remove one possiblity from one coin leaving 3 options and looking for one outcome the other coin. right? but shouldn't it still be a 1/2 because one coin is decide

p=1/3

but you dont know which coin it is. both TH and HT can happen, while TT cant.
making HH 1 out of 3.

im sure youre right and that in a more advanced calculation this is important but if you had two physical coins itd be 50%

Easy as shit, 50%.

50% it does.
50% it doesn't.

I use this same strategy when buying lottery tickets, either I win, or I don't, 50/50 chance. I also happen to be extremely unlucky :(

you can test this stuff, even with just 1 coin.

do 100 flips, record every result (ignore double tails) and see what % of them is double heads.
it is 1/3, i swear.

Y'all are stupid. By Bayes' thm, we have that P(two heads | at least one head) = P(at least one head | two heads)P(two heads)/P(at least one head) = P(two heads)/P(at least one head) = (1/4)/(3/4) = 1/3. Thus, the probability that we see two heads given that we have already examined one to be heads is 1/3, not 50%.

or as an alternate solution, consider the event space.

You can have:
HH, HT, TH, TT.

Out of the 3 (equally probable) options (HH, HT, TH), only one (HH) has 2 heads. So, then 1/3.

just did it, every single flip was HH

try harder, its Sup Forums, we are quite used to trolls here.

No, just the double tails you remove.

>One lands heads
>Dont assume one lands heads

what the fuck m80

Attached: simulate.png (879x504, 22K)

square the gamblers fallacy with laws of probability

Actually, I hear there's some truth to the gambler's fallacy. I mean, doesn't it make sense since if you live in an area with hundred year flooding, and it's been 99 years since the last flood, then it means you're due for a flood?

Like don't assume you have one down heads and then flip the other

Then you'll get HT and HH
instead of HT TH and HH

1200

if the flood is like clockwork thats not gamblers fallacy. gamblers fallacy is if you flip 49 tails on a coin you bet the next one will come up heads.

Read the code. I didn't assume that, i flipped, checked to see if one was heads, and went from there.

Ok, let's see...

2 coins, and each coin has 2 possible outcomes.

So, the answer would be Bulgaria

and you didnt assume because OP said not to, thanks for playing.

Well, hundred year floods have a probability of happening once every hundred years, so for any given year, there's a 1/100 chance of it happening, right?

>gamblers fallacy is if you flip 49 tails on a coin you bet the next one will come up heads.
thats just stupid, i would bet its tails, there is a clear pattern here, maybe its a trick coin or something.

I think there is a pattern actually.

See now this one is 1/3 for sure
But I still stand by the 50% chance on that golden and silver ball boxes shit

What's the golden and silver ball boxes one? I'm not familiar.

and you are wrong for almost the same reason.
probability is very counter intuitive, this coin proble, the monty hall door problem and the ball boxes all play on our inability to count probability.

This is essentially the same as the gold box one.
youtube.com/watch?v=4Lb-6rxZxx0

Something like
There are 3 boxes, each has 2 balls in them.
Box1 has 2xgold balls, box2 1 gold 1 silver and box3 2 silver
You pull out a ball and its gold. What chance is it that the other ball is gold as well. You cant see into the boxes

In it op clearly states that you pulled out gold already and doesnt say that you dont know that its gold

Yeah but in the ball one it clearly states or rather doesnt state that you dont know what the first ball is. I understand why it should be 2/3, but to me logically its 50%

thank you

and your logic is flawed, i told you, probability is very counter intuitive.
its the opposite of the monty hall problem.
you choose 1 of the 3 balls at random, there is a 2/3 chance you chose the a gold ball from the box with 2 gold balls because 2 of the 3 balls are in that box. meaning the other ball will be golden in 2/3 experiments.

Attached: balls.jpg (320x269, 30K)

>At least one of them lands heads.
>*Do not assume that one coin already landed heads*
Wat?

>One lands heads
>Dont assume one lands heads

Which one do i follow, fag?

they are not mutually exclusive, neither coin has landed yet, are you stupid?

>Atleast one of them lands heads.
Can you actually even have this situation outside of theory that does not require manipulation?
How can you guarantee one undetermined and randomly selected coin will land heads without manipulating statistic probability?

Because OP is being a faggot.

>*Do not assume that one coin already landed heads*
>already

Your reading comprehension is terrible

I honestly understand that 66% is correct mathematically, just the way its worded in the pic always gives me the impression that we go from step 2 of the process disregarding what came before.
Maybe its because English is not my first language, maybe I'm just retarded but my brain keeps saying 50% cause it can be either A or B since box3 is impossible and whatever box we didnt pull from is irrelevant

Nicee one Elon...

Attached: 1811573135152726.jpg (1410x1123, 91K)

OP worded the question poorly.

If you flip two coins, assuming at least one coin will land as heads, what is the probability of both landing heads? would read better.

Rolling

english is not my first language either and i have not studied probability ever.
and it CAN be either box A or B, but box A is twice as likely because if you chose the box B, half the time you choose the silver ball. Now that cant happen so you just start over.

not really any better, no matter how i word it i end up either giving away the answer or having people think its just a matter of flipping 1 coin, because people are stupid.

Basic statistics;

You have 2 coins, given 1 has already landed on heads, whats the probability both do? given these are 2 independent incidents, coin 1 being heads already has no impact on coin 2's probability

P=(C2 HEADS | C1 HEADS)
P=1/2

2 coins.
H-H
H-T
T-T
T-H
That's 1/4 or 25% chance right there, but since it's not a pure 50/50 on a single flip, it's probably more around 23-24%

Of course if you don't count the last possibility it's 33% and of course closer to 32ish%

Are you retarded? It's a sample theoretical question, meant to test if you know the "given" property of a probabilistic function. aka P=(something | something else) the "|" being 'given'. These are two independent incidents, so given one coin lands on heads, there is no change in the other coins probability.

did you not read the part that said

>*Do not assume that one coin already landed heads*

Are you retarded, it says one lands on heads. They are independent incidents. There is no change in probability from one landing on heads. Thus P=(C2 HEADS | C1 HEADS) and P(C2HEADS)=1/2

You flip two coins. Assuming the coins never land tails-tails, how often will they land heads-heads?

This picture literally doesn't apply to this situation? If you made 2 sets of 3 boxes, then yes. The probabilities of each coin landing on heads are identical. If you take a coin out of a box, it removes 1 possible option for the next draw, meaning your new probability is P=(chance | 1 gold removed)

basically its something like

You throw 2 coins at same time. 1 WiLL land heads but you don't know that yet, they are still in the air. Whats the chance no.2 will land heads in the case no.1 lands heads

You're right, I see it now

just saying, you guys dumb.


Also for all the other niggers who can wrap their heads around this.

Attached: coins.jpg (2442x1998, 721K)

1/3

If 1 will land on heads, you are just calculating the probability of 1 coin. It's like you are flipping 2 coins, one with only heads on both sides, and asking what's the chance of a heads on both? It's still 1/2.

Retarded

if getting a really simple math problem correct is retarded, then i guess im retarded.

Yes, but you are not calculating probability of just the second coin. You are also taking into the account the possibility of the first coin landing heads

Essentially, pretend you have 2 mystery coins that you don't observe until mid flight. You know every time you throw it, one of them will always land heads, and the other will land either heads or tails. Given that you know one of the 2 will land heads, regardless of which one, your only question remaining is what will the second coin do. Given that the second coin, regardless of which one of the 2 physical coins it is, has a 1/2 chance of landing on heads or tails, it is a simple 1/2 probability given the predetermined heads from one of the mystery coins.

Can't tell if baiting or just retarded.

No,
read

damn you beat me to it, was just gonna point out that you are trolling. well played man, real uno reverse card tier shit right there.

you dont know which coin it is, TH and HT are both options, leaving HH to be only 1 out of 3 options. you are stupid, get over it.

no you fucking idiot, you dont have a trick coin or some shit, its not guaranteed that 1 lands heads EVERY TIME, its just that in this case 1 did or does.
you can test this shit, just get a coin or 2 and start flipping, or are you too scared of finding out you were wrong all alone?

The situation is analagous to having 1 real coin, and 1 coin with heads on both sides, mixing them up so you don't know which is which, and then flipping them before you observe the coins. After flipping and seeing the coins spin quickly, you don't know which of the coins mid flight is the fake heads coin, however one of the 2 will be heads, so your only remaining question is what about that second, real coin, regardless of which one of the 2 coins it is mid flight.

Dude, it says 1 will land on heads. For this spin, it is a certainty, so you caclulate it as such.

You don't know if coin 1 lands heads tho

>You don't know if coin 1 lands heads tho
OP:
>Atleast one of them lands heads.

Bro....

Attached: 1544499246138.jpg (255x200, 10K)

Also OP:
>*Do not assume that one coin already landed heads*

Did you never solve math problems in school?

but you dont know which one...
you are actually just retarded, it has been explained to you many times and you have seen the solution be posted, still you insist its 50/50 when you can just test it and see the light.
get out of this thread, get out of your house and jump off a bridge before you spread your genes.

Coin 1 and coin 2 are just the names of 2 different coins. It doesnt matter if its coin 1 or coin 2. We just know one of them will be heads.

yea bro didnt you read Op
>you have dick in ass
>dont assume dick is in ass.

You don't know it though. In the actually experiment you are doing none of the coins landed yet

or we dont know, but we are talking about a situation where 1 is heads.

If that instruction is to be followed, OP is essentially contradicting one of his first statements. At least one of them will land heads. Its an absolute certainty regardless of any other factors. So we know from square one before any land that 1 coin, regardless of which one.

Thanks

You flip two coins. Assuming the coins never land tails-tails, how often will they land heads-heads?

>being this retarded
you don't understand that 01 doesn't equal to 10, those are completely different values in binary
00, 01, 10, 11 are seperare outcomes
what are the chances that the coins land 01? 25%; 10? 25%
what are the chances that one of the coins land on head (1)? (01+10+11)÷4=75%; tails (0)? (00+01+10)÷4=75%
what are the chances that both coins will fall on heads, if we take in consideration all the values where atleast one coin falls on heads [note: the first coin doesn't have to fall on heads]
if the first coin lands tails it's already a fail, 01
if the first coin lands heads, but the other lands tails it's also a bust, but a different one, 10
if both coins land on heads it's a win, 11
if both coins land on tails it's a loss, 00, but it's disregarded and doesnt fall into the equation
so 11÷3=33,33%, where 3=(01+10+11), unlike before when 4=(00+01+10+11)
also note that 00=01=10=11 have a same chance of happening, therefore they all have a value of 1 (1÷4=25%; 1÷3=33,33%)

no you idiot, he is talking about a situation where you dont flip double tails and wanted you to understand that its not just a matter of flipping 1 coin, he pretty much spoon fed you the biggest clue you can get regarding this problem and you just ignored it to say stupid shit like that instead.

No contradiction, it's "do not assume that one coin ALREADY landed heads", which is different from assuming at least one WILL be heads.

It says, 1 of the 2 coins will for certain be heads. It says specifically AT LEAST 1, not COIN 1.

You seem to think that you are even calculating the probability of both coins. One coin is already determined to be heads before it lands according to OP, it will happen, so your only remaining quesiton is what about the other coin. Whether its coin 1 or coin 2, it doesnt matter. The only uncertainty is whether or not the non-heads coin will be heads or tails. Its 1/2.

But you are calculating the probability of both coins in a scenario where TT is impossible

>One coin is already determined to be heads before it lands according to OP
i am OP, in this situation, you will get 1 heads yes, but you dont know which coin it is, you are misunderstading it even after i gave you a hint that i dont usually give. you are a nigger.

No, It's not about TT, it's about a guarantee of one coin being heads, so whats the chance of the other one being heads?

You seem to think probabilistically it matters which coin it is. It doesnt. The odds are the same regardless if you assume coin 1 is always heads, or coin 2 is, or if its random every time. The odds of both being heads are always 1/2. You are a bigger nigger faggot.

Three outcomes of flipping two coins in which at least one is heads: TH, HT, and HH. HH is one of three. 1/3. How is this confusing?

It matters which though

except that it does matter. if you took the time to actually do the test and flip the coins you would quickly realised that TH and HT are 2 separate situations and have the same chance of occurring.
leaving HH only 1/3 chance, do the test and sto....
wait, are you a nigger and therefor so poor you dont own a coin? damn im sorry.
you can do it with dice too, you got one of those?
1-3 is head and 4-6 is tails, you can also paint 1 side of a button.

1/3
ht
hh
th

The OP's question isn't "you have a coin, whats the chance of it landing head" for the same reason it isn't "you have 2 coins, whats the chance of HH"
Which coin it is matters, and OP justs asks a simple question
"Whats the chance of HH if TT is impossible"

Forgot tt you faggot.

tt isnt possible faggot.

>implying it doesn't matter
it most certainly matters
in an explosive two part mechanism (lets say a grenade) it's pretty fucking important
if you don't pull the ring, and the handle doesn't somehow break off you're not getting an explosion, 00
if you pull the ring but the handle is stuck and won't budge even if you throw it onto the ground you're not getting an explosion, 01
if the handle is so loose that the grenade will blow off in your hand if you pull the ring, but you don't, theres no explosion, 10
if both the ring and the handle get then it finally explodes, 11
in a scenario where you handle the grenade (atleast one (ring or handle) gets) what are the odds of grenade exploding? 1/3 (if other factors are excluded, like the coin dropping into the sewer or landing perfectly onto the side, those aren't in thevequation)
two coins aren't the same entity, but two individual entities, you don't get to take them as a singular entity just because one (which one) of the two has a desired outcome

TT is impossible in OP's question

75% chance AT LEAST ONE of the 2 quarters lands heads

4 total possibilities and 3/4 possibilities include a heads

1. Heads and Heads
2. Heads and Tails
3. Tails and Heads
4. Tails and Tails

You equally have a 75% chance that AT LEAST ONE will be tails

Your odds change if you require a specific outcome vs an "at least" outcome

You have a 50% chance of getting a heads and tails outcome while you have a 25% chance of getting a heads and heads outcome or a tails and tails outcome

>neither coin has landed yet
So the problem is "what is the probability of flipping two heads". You can't have it both ways; either both coins are subject to random variation (1/4) or one coin is predetermined (1/2). You cannot arbitrarily remove a random option just to make a bait thread.
>are you stupid?
No, but you clearly are

It's not that hard. You are too focused on determining which of the two coins are heads before it happens. All we know is one will be, so the remaining question is what is the other. The only possibilities are Heads heads, or heads tails. Which coin doesnt matter.

Attached: Untitled-1.png (2550x3300, 765K)

in this case both coins are subject to random variation (1/4)
but we are not counting in a double tails option, because in this case atleast 1 coin is heads. the question essentially is "what is the probability of HH if TT is excluded".