3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

The reason Monty Hall fucks with people is because it's a stepped process that eliminates variables as it goes, and that fact is really important to the somewhat counter-intuitive "always change doors."

This frog problem is also a stepped process because you already know one of the coin toss results. It's just pretending it's not and trying to treat the entire distribution as two simultaneously observed events, when it's very much two sequentially observed events. It's bad math.

3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

The video asserts that it's asking "what is the probability that two coins tossed will contain at least one tails."

The video is actually asking "if you've flipped one coin and can see that it's heads, what is the probability the second coin will be tails." Trying to hide the result by going "oh, you can see that it happened but not which one it is" is mathematical sleight of hand and irrelevant to the fundamental statistical question. To belabor the point:

The video is saying "you've flipped two coins, you can't see them, I tell you one is heads" and wants the answer to be that you have probabilities:

[H H]
[H T]
[T H] [T T]

and a 67% chance of there being one tails

Except that what you actually have is a matrix of two possible states:

[H H] [H H]
[H T] [T H]

you're in one of these, and you just don't know which one yet. In either one your chance is 50/50 that there's a tails. There is no coin superposition where until you observe it and collapse the waveform either coin can be either outcome. The outcome's already happened.

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3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

One of those moments where I wonder about the stories of "engagement farming" and putting in errors to get comments and reaction. This feels like a bad variant of monty hall that doesn't quite work and doesn't really communicate anything new (in the ted version). We have the fuck you stats example. It's a goat.

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3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

One of those moments where I wonder about the stories of "engagement farming" and putting in errors to get comments and reaction. This feels like a bad variant of monty hall that doesn't quite work and doesn't really communicate anything new (in the ted version). We have the fuck you stats example. It's a goat.

Yeah, I landed on "deliberately incorrect as engagement bait" as well. It's a variant Monty Hall that arrives at an incorrect solution by deliberately doing the exact thing you're not supposed to do with a Monty Hall, it certainly nerd sniped me extremely effectively.

The video asserts that it's asking "what is the probability that two coins tossed will contain at least one tails."

The video is actually asking "if you've flipped one coin and can see that it's heads, what is the probability the second coin will be tails." Trying to hide the result by going "oh, you can see that it happened but not which one it is" is mathematical sleight of hand and irrelevant to the fundamental statistical question. To belabor the point:

The video is saying "you've flipped two coins, you can't see them, I tell you one is heads" and wants the answer to be that you have probabilities:

[H H]
[H T]
[T H] [T T]

and a 67% chance of there being one tails

Except that what you actually have is a matrix of two possible states:

[H H] [H H]
[H T] [T H]

you're in one of these, and you just don't know which one yet. In either one your chance is 50/50 that there's a tails. There is no coin superposition where until you observe it and collapse the waveform either coin can be either outcome. The outcome's already happened.

Nah.

I flip two coins sequentially.
The chance that I get at least one head is
50%*100% + 50%*50% = 3/4
(H then whatever on the second coin + T and then H)

If I tell you that one is H without showing you, the chance of getting two heads is now 1/4 / 3/4 = 1/3, and so there's a 67% chance of one of the coins being T.

(The full analysis with the probability of a frog croaking wants to get to your matrix by saying 'If frogs rarely croak, then having two male frogs will be twice as noisy as one male one female, so you'll be twice as likely to hear the frogs croaking in the first place'
But that's butchering the question)

JedocIn the scupperswith the staggers and jagsRegistered Userregular

The coins have no memory. If you know that one is heads, the other one is still 50/50.

Fuck, I got into the math part. This is a frog thread.

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3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

edited February 2022

My post explains how and why that way of modeling it is wrong. The model that results in 67% assumes either coin can be tails. We already know that one of the coins can not be tails (the fact that we haven't observed which is which is completely irrelevant here - this is what I mean by "mathematical sleight of hand," it sends them down the path of using the wrong model because they misunderstand how to model sequential events), therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer in Monty Hall :P

I'm not gonna keep arguing it though, there's not much more I can say to try and show that the HH/HT/TH model is wrong without repeating myself and it'll just become the two of us bickering, which is no fun for anyone.

My post explains how and why that way of modeling it is wrong. The model that results in 67% assumes either coin can be tails. We already know that one of the coins can not tails, therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer i. Monty Hall :P

I'm not gonna keep arguing it though, I said what I had to say on it and if you don't buy it then no amount of me trying to explain it will convince you.

Where is my math wrong then?

Unless me stating the result to you has somehow increased the chance of me flipping two Heads after I've flipped them (probability of croak), then the chance that I've flipped two heads is after I've told you I've flipped at least one is:
Original chance of HH (25%) / Chance of having flipped at least one H (100%-TT= 75%)

Whereas your matrix should look like
[H H] [T H]
[H T] [T T]

And I don't see where you justify doubling the chance of HH at all.

It looks like you've double counted it.
An extremely common probability math error.

The reason Monty Hall fucks with people is because it's a stepped process that eliminates variables as it goes, and that fact is really important to the somewhat counter-intuitive "always change doors."

This frog problem is also a stepped process because you already know one of the coin toss results. It's just pretending it's not and trying to treat the entire distribution as two simultaneously observed events, when it's very much two sequentially observed events. It's bad math.

Coin flips are independent events.
It doesn't matter if they're simultaneous or not, the probabilities of the results are the same.

Frog gender is not a good example of an independent event.

Key takeaway is,
If there are X frogs in the clearing and you hear X-1 different croaks, and X is a very large number, you lick all the frogs in the clearing.
Because there is almost certainly a female in the dogpile, and not a 50% chance.

My post explains how and why that way of modeling it is wrong. The model that results in 67% assumes either coin can be tails. We already know that one of the coins can not tails, therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer i. Monty Hall :P

I'm not gonna keep arguing it though, I said what I had to say on it and if you don't buy it then no amount of me trying to explain it will convince you.

Where is my math wrong then?

Unless me stating the result to you has somehow increased the chance of me flipping two Heads after I've flipped them (probability of croak), then the chance that I've flipped two heads is after I've told you I've flipped at least one is:
Original chance of HH (25%) / Chance of having flipped at least one H (100%-TT= 75%)

Whereas your matrix should look like
[H H] [T H]
[H T] [T T]

And I don't see where you justify doubling the chance of HH at all.

These two situations:

-I have flipped two coins, one is heads. What are the odds the second is tails?

and

-I have flipped a coin. It is heads. I will flip it again. What are the odds it will be tails?

are the same. The answer in both cases is 50% because the second flip is its own independent event. Sequential independent events have to be modeled differently from aggregate events.

Since we know the results of one flip already, we have to discard the aggregate model of the whole set and move to individual modeling of the next partial set (this is also the step that gets you to 50/50 in Monty Hall).

Whether the problem is "you flipped two coins, one is heads" or "you flipped a coin twice and got heads once" the math doesn't change. "You don't know which coin" doesn't matter because the coins are both already flipped, so you have to discard any solution in which the coin that came up heads could be tails, because it can't - the flip already happened and we know it was heads. The framing screws up the model.

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3cl1ps3I will build a labyrinth to house the cheeseRegistered Userregular

Okay now I'm done for really real because that's as good a job as I can do.

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JedocIn the scupperswith the staggers and jagsRegistered Userregular

You have been poisoned. To be cured, upload a video of yourself licking a frog to our site, FrogLickVids.Biz. Our sophisticated algorithm will tell you whether you have been cured or if you need to upload another video.

There is no coin superposition where until you observe it and collapse the waveform either coin can be either outcome. The outcome's already happened.

Yes there is.
Until you know the result, the coins are in a superposition for you.
It's only when you entangle yourself with the coins by observing them that you can know their state, and then now you are in the coin superposition as well.

My post explains how and why that way of modeling it is wrong. The model that results in 67% assumes either coin can be tails. We already know that one of the coins can not tails, therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer i. Monty Hall :P

I'm not gonna keep arguing it though, I said what I had to say on it and if you don't buy it then no amount of me trying to explain it will convince you.

Where is my math wrong then?

Unless me stating the result to you has somehow increased the chance of me flipping two Heads after I've flipped them (probability of croak), then the chance that I've flipped two heads is after I've told you I've flipped at least one is:
Original chance of HH (25%) / Chance of having flipped at least one H (100%-TT= 75%)

Whereas your matrix should look like
[H H] [T H]
[H T] [T T]

And I don't see where you justify doubling the chance of HH at all.

These two situations:

-I have flipped two coins, one is heads. What are the odds the second is tails?

and

-I have flipped a coin. It is heads. I will flip it again. What are the odds it will be tails?

are the same. The answer in both cases is 50% because the second flip is its own independent event. Sequential independent events have to be modeled differently from aggregate events.

Since we know the results of one flip already, we have to discard the aggregate model of the whole set and move to individual modeling of the next partial set (this is also the step that gets you to 50/50 in Monty Hall).

Whether the problem is "you flipped two coins, one is heads" or "you flipped a coin twice and got heads once" the math doesn't change. "You don't know which coin" doesn't matter because the coins are both already flipped, so you have to discard any solution in which the coin that came up heads could be tails, because it can't - the flip already happened and we know it was heads. The framing screws up the model.

You don't know which of the coins came up heads though.
So you can't discard all solutions in which the coin that came up heads came up tails.

So I flip a coin, it comes up heads, what is the chance that it comes up tails next?
Looks like:
[H H] [H T] [T H] [T T]

But I flip a coin twice, it came up heads at least once looks like
[H H] [H T] [T H] [T T]
Because we can't eliminate either T H or H T.

The framing of
-I have flipped two coins, one is heads. What are the odds the second is tails?
Tricks the reader into the first 50% scenario by putting an ordering on the results with the word 'second'.
Yes it you flip two coins and the first comes up H then the second has a 50% of tails, but if you can't distinguish between the two coins then it is twice as likely that you have one tail and one head than it is if you have two heads.

JedocIn the scupperswith the staggers and jagsRegistered Userregular

You're choosing between two sets! You know one of them has one coin that may be good or may be bad, and a second set where one is definitely bad and one that may be good or may be bad! They're the same!

Fuck!

Frog!

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MrMonroepassed outon the floor nowRegistered Userregular

The pair of frogs is 75% likely to save your ass

I didn't read your poll options properly and have shamed myself by picking the one that favors the clearing without reading the numbers.

But the video is correct, your odds are (confusingly) 2/3rds if you lick both clearing frogs.

You know that male/female are evenly distributed among any pair of frogs, so if you sampled a large number of pairs of frogs you would see roughly even splits between:

MF
MM
FM
FF

This means that you get 50% mixed gender samples, 25% all-male samples, and 25% all-female samples.

But that you can strike any solutions with no males because there's at least one:

MF
MM
FM FF

So of the remaining randomly selected pairs of frogs, 2/3rds will be a pair with (exactly) one female.

The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox as follows:

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[1] Its answer could be 1/2, depending on the procedure by which the information "at least one of them is a boy" was obtained. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,[3] and Raymond S. Nickerson.[4]...The paradox has stimulated a great deal of controversy.[4] The paradox stems from whether the problem setup is similar for the two questions.[2][7] The intuitive answer is 1/2.[2] This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sex of the second child (i.e., boy and girl),[2][8] and that the probability of these outcomes is absolute, not conditional.[9]

There's no way to embed this one without fucking the formatting, but the actual answer is we're both right depending on the framing and resultant model. Sweet.

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JedocIn the scupperswith the staggers and jagsRegistered Userregular

The way you get to 50% in the clearing is with an unreliable narrator:

I flip two coins, and then I either tell you at least one is Heads or at least one is Tails.
This time I tell you at least one is Heads.
What is the chance that they both are?

So my outcomes are normally
TT, HT, TH, HH
but now I also tell you extra info, at least one is Heads (h), or at least one is Tails (t).
So my possible results are
tTT, hHT, tHT, hTH, tTH, hHH.

So the possible results are hHT, hTH or hHH.
The probability of hHH is the same as HH.
But if I randomly pick which piece of information I tell you for HT or TH, then hHT = 50% * HT and hTH = 50% * TH.
So now these both only happen 1/8th of the time each.
So there's a 1/8 chance of hHT, 1/8 chance of hTH and a 1/4 chance of hHH.

Which makes HH as likely as HT or TH.

If I only say 'at least one coin is Tails' for TT and never TH or HT, then we're back to 1/3 chance of HH.

If I only say 'at least one coin is Heads' for HH, then it's 100% chance of two Heads.

Which means to say if I have ten coins and you see me counting out nine heads before telling you there are at least nine heads, then there's a 50% chance of the next coin being either heads or tails.

But if I just throw ten coins in the air and then say there are at least nine heads, then there's a 1:11 chance of a head on the last coin.

My take is that of the possible combinations (HH, HT, TH, TT), we can easily eliminate TT, because we have one heads. But we can also eliminate either HT or TH, because a specific coin landed heads (and so can’t be tails). I don’t know which one was eliminated, but that doesn’t matter; it’s still a specific coin. The only possible options here is that the left coin is the revealed one, or the right coin is. Either option eliminates the possibility of that coin being tails. Which leaves 2 options, with a 50% chance of either.

## Posts

This frog problem is

alsoa stepped process because you already know one of the coin toss results. It's just pretending it's not and trying to treat the entire distribution as two simultaneously observed events, when it's very much two sequentially observed events. It's bad math.https://youtu.be/4wGuqsNcW3s

The video is

actuallyasking "if you've flipped one coin and can see that it's heads, what is the probability the second coin will be tails." Trying to hide the result by going "oh, you can see that it happened but not which one it is" is mathematical sleight of hand and irrelevant to the fundamental statistical question. To belabor the point:The video is saying "you've flipped two coins, you can't see them, I tell you one is heads" and wants the answer to be that you have probabilities:

[H H]

[H T]

[T H]

[T T]

and a 67% chance of there being one tails

Except that what you actually have is a matrix of two possible states:

[H H] [H H]

[H T] [T H]

you're in one of these, and you just don't know which one yet. In either one your chance is 50/50 that there's a tails. There is no coin superposition where until you observe it and collapse the waveform either coin can be either outcome. The outcome's already happened.

Don’t eat random mushrooms you find in the rainforest.

This one even has angel in the name, so it can't be bad!

Yeah, I landed on "deliberately incorrect as engagement bait" as well. It's a variant Monty Hall that arrives at an incorrect solution by deliberately doing the exact thing you're not supposed to do with a Monty Hall, it certainly nerd sniped me extremely effectively.

Nah.

I flip two coins sequentially.

The chance that I get at least one head is

50%*100% + 50%*50% = 3/4

(H then whatever on the second coin + T and then H)

If I tell you that one is H without showing you, the chance of getting two heads is now 1/4 / 3/4 = 1/3, and so there's a 67% chance of one of the coins being T.

(The full analysis with the probability of a frog croaking wants to get to your matrix by saying 'If frogs rarely croak, then having two male frogs will be twice as noisy as one male one female, so you'll be twice as likely to hear the frogs croaking in the first place'

But that's butchering the question)

discrideronFuck, I got into the math part. This is a frog thread.

notbe tails (the fact that we haven't observed which is which is completely irrelevant here - this is what I mean by "mathematical sleight of hand," it sends them down the path of using the wrong model because they misunderstand how to model sequential events), therefore that assumption has to be discarded. You (and the video) are doing the exact same thing that gets people to the "don't switch" i.e. wrong answer in Monty Hall :PI'm not gonna keep arguing it though, there's not much more I can say to try and show that the HH/HT/TH model is wrong without repeating myself and it'll just become the two of us bickering, which is no fun for anyone.

3cl1ps3onWhere is my math wrong then?

Unless me stating the result to you has somehow increased the chance of me flipping two Heads after I've flipped them (probability of croak), then the chance that I've flipped two heads is after I've told you I've flipped at least one is:

Original chance of HH (25%) / Chance of having flipped at least one H (100%-TT= 75%)

Whereas your matrix should look like

[H H] [T H]

[H T] [T T]

And I don't see where you justify doubling the chance of HH at all.

It looks like you've double counted it.

An extremely common probability math error.

discrideronCoin flips are independent events.

It doesn't matter if they're simultaneous or not, the probabilities of the results are the same.

Frog gender is not a good example of an independent event.

If there are X frogs in the clearing and you hear X-1 different croaks, and X is a very large number, you lick all the frogs in the clearing.

Because there is almost certainly a female in the dogpile, and not a 50% chance.

These two situations:

-I have flipped two coins, one is heads. What are the odds the second is tails?

and

-I have flipped a coin. It is heads. I will flip it again. What are the odds it will be tails?

are the same. The answer in both cases is 50% because the second flip is its own independent event. Sequential independent events have to be modeled differently from aggregate events.

Since we know the results of one flip already, we have to discard the aggregate model of the whole set and move to individual modeling of the next partial set (this is also the step that gets you to 50/50 in Monty Hall).

Whether the problem is "you flipped two coins, one is heads" or "you flipped a coin twice and got heads once" the math doesn't change. "You don't know which coin" doesn't matter because the coins are both already flipped, so you have to discard any solution in which the coin that came up heads could be tails, because it can't - the flip already happened and we know it was heads. The framing screws up the model.

*This is definitely NOT a sex thing

Until you know the result, the coins are in a superposition for you.

It's only when you entangle yourself with the coins by observing them that you can know their state, and then now you are in the coin superposition as well.

My 100% sincere view of quantum mechanics.

discrideronYou don't know which of the coins came up heads though.

So you can't discard all solutions in which the coin that came up heads came up tails.

So I flip a coin, it comes up heads, what is the chance that it comes up tails next?

Looks like:

[H H] [H T] [T H] [T T]

But I flip a coin twice, it came up heads at least once looks like

[H H] [H T] [T H] [T T]

Because we can't eliminate either T H or H T.

The framing of

-I have flipped two coins, one is heads. What are the odds the second is tails?

Tricks the reader into the first 50% scenario by putting an ordering on the results with the word 'second'.

Yes it you flip two coins and the first comes up H then the second has a 50% of tails, but if you can't distinguish between the two coins then it is twice as likely that you have one tail and one head than it is if you have two heads.

definitelybad and one that may be good or may be bad! They're the same!Fuck!Frog!But the video is correct, your odds are (confusingly) 2/3rds if you lick both clearing frogs.

You know that male/female are evenly distributed among any pair of frogs, so if you sampled a large number of pairs of frogs you would see roughly even splits between:

MF

MM

FM

FF

This means that you get 50% mixed gender samples, 25% all-male samples, and 25% all-female samples.

But that you can strike any solutions with

nomales because there's at least one:MF

MM

FM

FF

So of the remaining randomly selected pairs of frogs, 2/3rds will be a pair with (exactly) one female.

anyway here's your goat, thanks for playing

Frog.

https://youtu.be/HBxn56l9WcU

Only frog

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Analysis_of_the_ambiguity

There's no way to embed this one without fucking the formatting, but the actual answer is we're both right depending on the framing and resultant model. Sweet.

The lesson here is mostly just "don't listen to statisticians giving TED talks"

I flip two coins, and then I either tell you at least one is Heads or at least one is Tails.

This time I tell you at least one is Heads.

What is the chance that they both are?

So my outcomes are normally

TT, HT, TH, HH

but now I also tell you extra info, at least one is Heads (h), or at least one is Tails (t).

So my possible results are

tTT, hHT, tHT, hTH, tTH, hHH.

So the possible results are hHT, hTH or hHH.

The probability of hHH is the same as HH.

But if I randomly pick which piece of information I tell you for HT or TH, then hHT = 50% * HT and hTH = 50% * TH.

So now these both only happen 1/8th of the time each.

So there's a 1/8 chance of hHT, 1/8 chance of hTH and a 1/4 chance of hHH.

Which makes HH as likely as HT or TH.

If I only say 'at least one coin is Tails' for TT and never TH or HT, then we're back to 1/3 chance of HH.

If I only say 'at least one coin is Heads' for HH, then it's 100% chance of two Heads.

Which means to say if I have ten coins and you see me counting out nine heads before telling you there are at least nine heads, then there's a 50% chance of the next coin being either heads or tails.

But if I just throw ten coins in the air and then say there are at least nine heads, then there's a 1:11 chance of a head on the last coin.

alsoeliminate either HT or TH, because a specific coin landed heads (and so can’t be tails).Idon’t know which one was eliminated, but that doesn’t matter; it’s still a specific coin. The only possible options here is that the left coin is the revealed one, or the right coin is. Either option eliminates the possibility of that coin being tails. Which leaves 2 options, with a 50% chance of either.https://youtu.be/tvFqMSCOsPU

ONLY

https://youtu.be/tqwjw7sPAhc

FROGS