I've got some time.

Sweet! Do you agree, Draug?

(And by the way, thank you abge).

I do not agree because it still allows for TT. The rules state in this reality that GG otr TT *cannot* exist in this instance.
So the only way I agree is if one coin is laid down on heads (can be first or last, I don't care and it can be mixed up sometimes laying one down first other times tossing one first) and the other coin is tossed. The result will be 100 instances of
at least one coin is a head" regardless of whether the coin was tossed first or set down on heads first.

I happen to have a 2-headed coin, if that helps : )

OK.

Let the record show that the following two facts are true:

a) Draug says that, if two fair coins are tossed and that afterward, you learn that heads came up on at least one, there is a 50% chance that they both came up heads;

but

b) He says that the only way to test this is by setting one of the coins down with heads up to begin with, not by tossing two fair coins.

Another way the great mod abge could help settle this is by using basic probability which he has to know as a physics machine to tell us the answer.

No, let the record state that in an experiment where one rule has been estalbished as inviolate ( that one coin must always be heads), allowing a result that violates the rule also violates the scientific method.

Wait, Draug. Which of the above two points do you disagree with?

I would like the record to show that I went through the trouble of finding one of my favorite coins and am now rather disappointed I won't be flipping it.

I disagree with A) because it violates the rules of the question asked. There aren't two random coins. there is one random coin and one fixed coin but the fixed coin randomly moves between two potential coins.

Now I could write that code and output the result of 100 instances.

I'm sorry. : ( That would have been cool. I was looking forward to it, and had estimated there was even a 1/9ish chance Draug would win due to small sample size.

I'm still waiting to hear which of my two points you disagree with, Draug, since you have asserted both this evening.

False, Draug, here is what I said: "You know only that (a) two coins were tossed and (b) at least one was heads." You have never objected to this formulation, and you clearly stated that you compute 50% in that situation.

So you see, you are wrong in contradicting A.

It appears to me that there is some serious disagreement about what you two are actually talking about.

Draug in another thread just now:

"""If you know only that (a) two coins were tossed and (b) at least one was heads, there is a 50% chance that both coins were heads." "

Yes! Yes! Yes! "

(I am collecting here these for easy reference for those who want to refer to this thread in the future).

I wish you would have kept this in one thread. Seriously. I spent 14 hours coding today and my brain is fried enough that trying to follow two is hurting.

Eh, I didn't think abge was following the other one, and we needed somebody we'd both trust.

I don't really plan to post more here though.

Good to the other thread then

I am, indeed, not following the conversation in any other thread. Should you need me, I'll be here.

"We will do the following experiment. 150 times, you will toss two coins. If at least one head came up, you will indicate accordingly in this thread. That is all you will say. "At least one head came up." Draug will then indicate that he believes that two heads came up."

I wrote a little routine for Excel VBA which demonstrates this experiment:

It toss two coins but only records the values if at least one of the coins is heads.
After 150 throws it displays the percentage of throws where both coins were heads.
Are those the correct parameters of the experiment?

Yes, spyman, but Draug will never admit that.

Sub CoinExperiment()
Application.ScreenUpdating = False
Range("A1").CurrentRegion.Clear

HeadsTails = Array("Heads", "Tails")
Range("A1:B1").Value = Array("First Toss", "Second Toss")
i = 1
Do Until i = 150
' 0 = Heads
' 1 = Tails
T1 = Int((2 * Rnd) + 1) - 1
T2 = Int((2 * Rnd) + 1) - 1
If T1 = 0 Or T2 = 0 Then
Cells(i + 1, 1).Value = HeadsTails(T1)
Cells(i + 1, 2).Value = HeadsTails(T2)
If T1 = 0 And T2 = 0 Then HCount = HCount + 1
i = i + 1
End If

Loop

DoubleHeads = Format(HCount / i, "0.0%")
Application.ScreenUpdating = True
MsgBox DoubleHeads

End Sub

"Yes, spyman, but Draug will never admit that."

Then the discussion sounds pointless, if you are essential talking about different problems. I'll have to read back and see what Draug is saying exactly.

Not quite, spyman, when you say HCount/i, you're still counting all the ones that were TT. You need another variable that counts just the ones with at least one head.

Spyman, we are talking about the same problem. He will just not admit how to simulate it. Here is my handy summary of the situation:

"
a) Draug says that, if two fair coins are tossed and that afterward, you learn that heads came up on at least one, there is a 50% chance that they both came up heads;

but

b) He says that the only way to test this is by setting one of the coins down with heads up to begin with, not by tossing two fair coins. "

HCount is when both values are Heads.

Yes I know. You need an additional variable for when only one of them is heads. i, at the end, is just 150, which counts all the times that were TT. That will give you about .25.

If T1 = 0 And T2 = 0 Then HCount = HCount + 1

HCount only increments when boht T1 and T2 = 0 (ie Heads)

I am confused now.

The percentage shows is HCount / i
i is the number of throws were either T1 or T2 where Heads.

Are you saying that I should modify it to show the total number of throws including where neither T1 or T2 where heads.

I can include that too.

Yes, I know. The problem isn't HCount. It's i. You shouldn't be using i in the final division.

Oh, my mistake. You're not incrementing i when there isn't a head. I was reading the i increment and the end if in opposite order in my mind. Sorry.

Exactly.

Take it back to the other thread, K? I just posted a program and it's result over there.

"b) He says that the only way to test this is by setting one of the coins down with heads up to begin with, not by tossing two fair coins. " "

THAT is a totally different question. I don't understand why Draug is saying this now. He was in agreement with the result of the boy girl question, which is identical to this question. Of course the answer to that question is 50%.