# Its amazing how many people get this wrong

Its amazing how many people get this wrong. Do you know the answer?

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50/50
its not like the other three flips influence the fourth one.

well it depends if it lands on heads the first, second, third, or fourh time.

t. retard

the probability of getting 4 heads in a row is 1/16

False. It can land on the edge. Also if it lands where you cant see it then you must assume it is simultaneously heads, tails, and edge.

Take that shit to
//boards.4channel.org/sci/
Math is for niggers and faggots

There are 16 possibilities.

O = Tails

XXXX (1)

OXXX (2)
XOXX (3)
XXOX (4)
XXXO (5)

XXOO (6)
XOXO (7)
XOOX (8)
OOXX (9)
OXOX (10)
OXXO (11)

XOOO (12)
OXOO (13)
OOXO (14)
OOOX (15)

OOOO (16)

There are 16 possibilities, and so the chances are 1/16.

but how is the last flip affected by the previous ones

But we're talking about the last flip only. So X = heads, O = tails
Possibilities are
XXXX
XXXO
50/50

You have no idea how horribly wrong you are.

Get a load of this retard. Never made it to junior year of high school did ya champ?

if you start from the beginning and flip 4 times, the probability is exactly 1/16

you already flipped it three times

Y'all need to learn that when a conditional statistics problem shows up on an imageboard it's always Bayesian.

P(A|B) = P(B|A)*P(A)/P(B) = 1 * .0625 / .125 = .5

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The correct answer is 1/2. No need to praise me

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This isn't math, it's one of those stupid questions like escaping a room with a table. The probability is 100% because it lands on heads each time.

for someone so educated you sure do have a shit taste in waifu

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not this again

5/7

wait this guy is onto something
explain room with a table too btw.

The statement either contradicts itself (3 times, no fourth time) or its 1/2 because its a fucking coin nigger, thats how they work

At least she can participate in a high five

Any other answer is factually wrong.

I hope this is bait, nobody can be that illiterate

Listen here, fucko. I dropped out of high school and don't even know what Bayesian is. I think its an asian from the San Francisco Bay Area.

This. According to previous observations the only educated conclusion possible is that it will land heads 100%.

You're in a room with no doors or windows. Just a mirror and a table.

cum on the table while looking at myself in the mirror.

Please tell me you're shitting me. You dont unironically believe the chance is 100% do you? I think its pretty safe to assume its a normal head/tails coin

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Do you have any proof that the coin even has a tails side?

Hmm, if I were in that situation, I'd look in the mirror, then at the wall and back at the mirror to see what I saw, then use the saw to cut the table in half and join the two halves to make a whole, and lastly put the "hole" on the wall and climb out.

Jesus fuck, you're all fucking retarded.

It's another conditional probability question.

A = all four flips are heads
B = first 3 flips are heads

P(A|B) = probability of all four heads given that the first 3 flips were heads <- what' we're trying to find
P(B|A) = probability of the first 3 flips being heads given that all four flips were heads <- obviously this is just equal to 1.
P(A) = probability that all four flips are heads = 1/16 = .0625 (since there are 16 possible head/tails combinations on four coin flips, and this occurs only once)
P(B) = probability that the first 3 flips are heads = 2/16 = .125 (since you can have HHHT but also HHHH)

Plug this all into Bayes and you get:
P(A|B) = P(B|A) * P(A) / P(B) = (1 * .0625) / .125 = 0.5

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The probability of heads in this circumstance is 100%. That doesn't mean if you flip the coin it will
be heads again, probability is just a means to guess the future.

If the question asks what the probability of ONE coin flip is, then the answer is 1/2.

However, the question is more like, what's the probability of flipping the same coin FOUR times, and getting it heads, all four times.

Always fucking hated that "riddle". It's just total nonsense that no one would ever actually guess without having heard it before.

You could have skipped the Bayesian method if you understood that prior coin flips have no impact on the final coin flip and applied that knowledge. Of course it's 50-50, you overly educated twip

the question is too vague. if it is asking the probability of it landing on heads 4 times in a row, the answer is 1/16. if asking for just the 4th toss, then 1/2 retard

or you could say just its 1/2 because one flip is not logically dependent on the others

post some vague ass probability question
have a bunch of autismos fight over who's right
truly evil

I dont understand how its vague. Its not asking for the chance of 4 in a row. Its asking for the chance of the last flip to be heads, which is 50%.

what is the probability of landing on heads a fourth time
its a stupid trick question that could be interpreted either way.

It really isn't user, you're just unfamiliar with the way statistics textbooks word things. It is very explicit and its interpretation is singular.

what are you doing here then mr. big brain

The OP question isnt vague. pic related is an example of a problem with no clear answer

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Other than statistics? Just getting drunk and hanging out in some feels threads.

Did you expect to find people who weren't interested in stats in a stats thread?

It's 9. You could get an answer of 1 if you put the 2(1+2) term in the denominator, but you didn't, so it's 9.

It's 2 -- PEMDAS dammit.

This.

The history of the coin does not impact future probabilities. It's always 50:50

The answer is 1 though. In PEMDAS multiplication/ division happen at the same time from right to left and subtraction/addition happen at the same time from right to left.

then because 3 in still in parentheses, you do 2 *3
then 6/6 = 1

right to left
wut? When do you ever parse an equation right to left? It doesn't matter anyway, because division is non-commutative, so if you want that result you have to format your equation appropriately, i.e. put the 2(1+2) in the denominator.

then because 3 in still in parentheses, you do 2 *3
That's where you're wrong. Once the 3 has been obtained the parentheses goes away, it's already been dealt with.

Your argument would mean that it equals 1, dumbass

You thought you did something here huh

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Six objects, divided by two groups, made up of one object plus two objects. That's three objects per group, and two groups is six objects.

Six divided by six is one.

Six objects, divided by two, multiplied by 1+2 equals 9. If you wanted to put your sentiment into an equation, you would need to structure it with you two groups in the denominator. Since that is not the case, you can not assume that it is one object divided by two groups.

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that would be (6/2)x(1+2)

Yeah, quick question op. who fucking cares?

literal autism jh

gamblers fallacy you nignogs

Coin flips are independent events. So, assuming that we are dealing with a 2 sided coin, the answer is 50%.

welI, you got me there.

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0%, you only flip it three times, so it can't land a fourth time.

Daily reminder that probability only works with large data sets.
It is a junk sciene to predict any one event.
Out of 10,000 flips patterns emerge, but you cant say /anything/ about any one flip

holy shit user what a fucking genius

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This question is (I assume on purpose) worded ambiguously so you can realistically come to both 1/2 and 1/16 as the conclusion.

Depends on what viewpoint we're talking about here. If its already happened 3 times in a row, and we're just talking about the 4th time, then yes, it would be 50/50. But if we haven't flipped any coins or anything yet it would be a 1/16 chance of that happening.

Dont go near a Casino

This is a trick question. The question is worded in a way that it can be interpreted as
a) what is the chance of getting heads when you flip a coin
b)what is the chance of getting heads 4 times in a row

Therefore 1/2 and 1/16 are both right