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kys

Except it's more narrowed down because we KNOW at least one is heads. That means the other coin is either going to be heads or tails. 50% chance

First sentence correct. Second sentence incorrect.

A nickel has a 1/6000 chance of landing on its edge. I'm sure a euro could be approximated but it would b e 50% - whatever that comes out to.

ui.adsabs.harvard.edu/abs/1993PhRvE..48.2547M/abstract#:~:text=Results of the experiments and,approximately 1 in 6000 tosses.

Checked
Math checks out
TT isn't an option
So if you don't understand the maths think of it like:
HH/(HH+HT+TH)

if you agree with the first sentence then you should agree with the second. i think I know what you're doing wrong. you're treating HT and TH as different things. They are the same thing. The options are either HT/TH, or HH. 50% chance. Not 33%

Knowing that one will land heads doesn't mean one of the coins will always be heads, the other coin can be aswell. Remember that both coins are flipped at the same time. You can have the 1° coin (H) and 2° (T), or you can have 1° (T) and 2° (H), or 12 (H) and 2° (H).

Knowing that at least one of them will always land Heads only removes the probability of both landing (T) (T).

The answer is 1/3

Are the coins distinct? E.g. one is explicitly coin A and the other is coin B?

It's 50% because you can eliminate one of the coins as a probability. You are essentially only flipping one coin. The other coin will always be heads.

if one coin is always going to land heads like the op suggests, then there is no difference between TH and HT. Because the heads is a given.

this. you could even have one coin marked A and one marked B. doesnt matter because one coin is always going to be heads.

I swear to god either this thread or the 3 boxes with 2 balls each is posted here 100% of the time. I think it's been well established the answer to both is 1/3. Now let's let this fade away quietly into the night.

>you're treating HT and TH as different things. They are the same thing.
This is how I know you're retarded. But go ahead, try to explain how they are the same when they're objectively not.

>you're treating HT and TH as different things
Because they are, the only coin tosses we throw out are TT
So possible outcomes are HH/HT/TH
1/3
Go get 2 coins
Flip them at the same time
Tally HH vs HT
Do this 100 times
Report back results

So you could have
coinA=heads, coinB=heads
or
coinA=heads, coinB=tails
or
coinA=tails, coinB=heads

1/3

if we're going to perform the experiment as closely to the rules in the op then you basically take one of those coins and leave it heads up on the table. because if you're flipping 2 coins there is no way to make sure one coin is always heads up, so in reality you just leave one coin heads up. so you flip the other coin, which is either going to be heads or tails. 50% chance of being HH

But there is a difference. If OP isn't specifying which coin flipped Heads, it's 1/3.

i.e. Coin A and Coin B are flipped and Coin A always lands on (H). What are the chances of flipping both coins at the same time and getting (H) (H)?

math.stackexchange.com/questions/1805277/if-two-coins-are-flipped-and-one-gets-head-what-is-the-probability-that-both-ge
math.stackexchange.com/questions/3100704/if-i-flip-two-fair-coins-and-then-tell-you-that-one-is-heads-what-is-the-proba

Sorry friendo, but I think you'll find the answer to Bertrand's Box Paradox is 2/3.

BTW I am the creator of both of these memes. Thank me later.

Attached: BertrandAnon.png (1024x812, 49.44K)

because one coin is always heads up. there is a major rule in this game that you're basically ignoring. one coin is always heads up. this is not a realistic game so I understand your confusion but dont go pretending im the retarded one

50/50
if two coins are flipped but the outcome of one is already determined then the total probability of the scenario is 50/50 since the remaining probability relies solely on a single coin toss

We don't know which coin has landed heads; we only know that at least one of them did. The description "at least one is heads" subsumes HT and HH; however, while there is only one way to obtain HH, there are two ways to obtain HT, namely HT and TH. Therefore, we have to include all four possible ordered pairs of coin flips in our sample space to ensure that we've accounted for every possibility. That's why it makes a difference.

>as closely to the rules in the op then you basically take one of those coins and leave it heads up on the table.
This is wrong, you retard. You're fixing a specific coin as heads doing that. The OP question doesn't fix either coin, it simply states at least one landed heads after being flipped. this is purely by chance, and could be either coin, which meant that for the OP question, EITHER coin could be tails. IN your Scenario, One of the coins definitely cannot be tails. This is why you are intellectually inferior to me. You lack basic logic and comprehension, as well as being mathematically illiterate.

The answer is 1/3.

Attached: 1384551078849.jpg (222x228, 14.97K)

Oh yeah you're right, 2/3. Misremembered the specific question but knew it was the answer with 3 in the denominator as opposed to the idiots who say it's 50-50.

I flip two coins
You see one but I have the other covered
The one you see is heads
What percent chance is the one I have covered heads?
That is OPs argument

...what? Are you retarded or something? Since when (H) (H) and (H) (H) are two different results?

Attached: aGd1X3K_700b.jpg (700x701, 50.28K)

>We don't know which coin has landed heads
it doesn't mater. one coin is always going to land heads. the probability is based on whichever coin isn't the coin that doesn't always land heads. so it's like flipping just one coin. and the probability of that coin landing heads is 50%

Kek

>because one coin is always heads up.
Incorrect, retardo. Stop being a simpleton and try again. I have faith in your ability to figure it out by yourself reading this thread.

OPs Question*
Fuck

50% you fucking spacker

Think like this. Get 2 coins and put one in each hand. You flip them both and you get (H) on your right hand and (T) on you left hand. That's one result. If you flip again and get the opposite, (T) on your right hand and (H) on your left hand, that's another result, and on both cases you always get (H) on one coin, which OP didn't specify which. Getting (H) (H) is the third result.

AB and BA are the same thing. Why does the order of the coins matter? Lmao. One is heads therefore you only haver to calculate the probability of 1 coin landing on heads.

>You're fixing a specific coin as heads doing that.
I'm only fixing a specific coin because you want to perform an experiment with real coins and it simply is not possible because it's not possible to have one coin flip heads every single time. but in this game there is, so you have to choose one coin to be heads every time.
it doesn't matter though because either way it's a 50% chance. one coin is going to land heads. the other coin is either going to land heads or tails. 50% chance it's also heads

This isn't summatory, where the otder of the number don't affect the result. This is probability, and the order does indeed affect the final result

Look at all these retards, holy shit.

Math below for those who aren't retarded. If you are retarded, the picture might help you understand.

Without the condition specified, the sample space for the problem is
S = {(H & H), (H & T), (T & H), (T & T)}....(|S| = 4)
Since the event E = TT does not satisfy the condition that there is at least one head, the sample space is
S' = {(H & H), (H & T), (T & H)}....(|S'| = 3)
Thus, given that at least one head is tossed, the probability of two heads is
P(H & H) = 1/3

A more mathematical approach for problems of this sort is to apply Bayes' Theorem, where for two events A and B
P( A given B)
ie
P(A|B) = P(A & B) / P(B)
So in our problem we have
P(H|H) = P(H & H)/P(H)
= (1/4) / [1 - P(T & T)]
= (1/4) / [1 - 1/4]
= (1/4) / (3/4)
= 1/3

Attached: coinsvisualexplanation.jpg (2442x1998, 716.15K)