Wow this discussion is still going.

1. I have two children. The older one is a boy. What’s the probability they are both boys?
2. I have two children. One is a boy. What’s the probability they are both boys?
3. I have two children. One is a boy. What is the probability my oldest son is a boy?

Answer to question 1 is 1/2
Answer to question 2 is 1/3
Answer to question 3 is 2/3

Or are you saying that answer for each is 1/2 Socrates?

Uclabb, draug is right, completely! it is either the case it is on left so ht or hh, or it is on the right so th or hh, and the two scenarios, being exclusive, are ones that we have to pick between. And you and semck can look at my last post and respond.

And how can you say draug is definitely wrong? Not only can mathematical proofs be debated (e.g. cantor's infinities, monty hall), but you could be wrong! And philcore was enjoying this conversation too, so lets look at the logical road to get to the answer, because even if I knew you guys were right I would still think all these points should be examined

In answer to your explicit question, the odds arte 50/50 if you know one coin is head that both are heads. See my previous post as to why.

"In answer to your explicit question, the odds arte 50/50 if you know one coin is head that both are heads. See my previous post as to why."

I take you to be saying that if you know only that (a) two coins were tossed and (b) at least one was heads, then there is a 50% chance they are both heads.

This is a very easy process to simulate. Please write (but don't run!) a simple program doing so and paste the code on here. If I agree that you have correctly modeled the described process, I will re-enter our $100 bet (to be paid to the site).

Thank you.

semck you are misunderstanding the monty hall problem and i suggest draug's post for reading. but yes let's go with the option where there was no deal to tell if there was a head but you just did (like with the child example), it is a half! even though there is the option to label it ht th hh, which would indicate a third by your logic.

spyman, it goes 1/2, then really 0 unless you mean one is a boy and the other is ambiguous, in which case it is 1/2 for previously listed reasons by draug and myself, then the third case is 1/2; unless you are not commenting on the other child and the other child can be both, in which case the odds are actually 3/4 and not 2/3.

semck, your suggestion is wrong, what you ahve to do is throw a coin, and if it is heads then throw another coin and then count, you are invalidating the experiment by not ensuring that at least one coin is a head (i don't count not counting the results that don't suit you ensuring at least one coin is a head). with the op we were assured of having one boy before knowing the sex of the other boy!

SD, I understand the Monty Hall problem very well.

"it is a half! even though there is the option to label it ht th hh, which would indicate a third by your logic."

You could label it, but at that point you would be ignoring information that you have that actually differentiates the coins. The appropriate way to calculate at that point is to label, not by toss order, but by "seen/unseen." There are only two possibilities. HH and HT. So it's 50/50.

But this situation introduces asymmetry between the coins that is not introduced in the initial situation.

"semck, your suggestion is wrong, what you ahve to do is throw a coin, and if it is heads then throw another coin and then count, you are invalidating the experiment by not ensuring that at least one coin is a head (i don't count not counting the results that don't suit you ensuring at least one coin is a head). with the op we were assured of having one boy before knowing the sex of the other boy! "

SD, no we weren't. We were assured of their being "at least one boy" at a point in time when both children already existed and had determined sexes.

but you should order in our case by known/unknown! then there are only two possibilities, so it's 50/50!!!

i did not mean to cause offense with calling your understanding of the mhp into question, apologies.

There is no known boy in our case, SD.

i didn't say before the other child was born, i said before knowing the sex of the other boy, which is correct, we were assured of having one boy before knowing the sex of the other child which could be either.

Socrates wrote:", in which case it is 1/2 for previously listed reasons by draug and myself"

Draug. Really? Have you changed your mind about this problem?

In the children case: both are already determined, and we are then told whether there is AT LEAST one boy.

In the coin case: both are already tossed, and then we are told whether there is AT LEAST one heads.

You said: "what you ahve to do is throw a coin, and if it is heads then throw another coin and then count,"

Your case is disanalogous, because it EXPLICITLY disallows the TH (or GB) case. You have to find out AFTER both exist whether there is AT LEAST one heads. That is precisely the situation I described.

semck, we know we have a boy, there is a child that is known to be a boy. the other child is the only sex that is in question. so we can label it known/unknown like you would wish to label it seen/unseen. But i'm actually going to sleep now, think about this some more, hoping we come to an agreement but accepting that we probably wont :)

I already laid out an algorithm up in the thread. You won't agree it is valid any more than I agree yours is valid. The key difference is that coins and kids are concrete fully observable things. So either the one on the left or the one on the right is head. My algorithm says run it where the left is head, then run the same number where the right is heads. Add the set of results and you will see that the opposing coin is always 50/50.

But lets start with this basic principle. Even though we are not tolds which coin is heads we can deduce that it is either the left coin or the right coin (there can be no other alternative) so there fore we have to run it once with the left coin head and once with the right coin heads. Can you agree to that?

we don't have to find out after both exist that there is at least one, it is a precondition for the thought experiment!

There is a child that is known to be a boy, SD, but we know there are two children; so for all we know, it took two "flips" to get the boy.

That is why you cannot stop throwing after the first toss if you don't get a heads/boy. It took two to get a boy. We don't know the first is a boy.

I didn't see your algorithm, Draug. Please paste it or tell me the timestamp.

Or just write code. It's very trivial to do. Again, let me describe the situation:

We toss two coins. We find out at least one was heads. We check if both were heads. You say it's a 50% probability.

Note that this is a simulation, so no interpretation is needed to write the algorithm. Just, simulate two coin tosses.

But the children *do* exist, they aren't thought experiments. It is stated that he has two children. There is no "what if". So either the first one alphabetically is a boy or the second one alphabetically is.

SD, if we knew when there was only one that there was at least a boy, then we know the oldest is a boy. In this case, we don't.

Correct, Draug.

So please write the simulation already. Put up or shut up.

No, you don't simulate two coin tosses, you simulate one coin toss in two runs with a known entity for the second coin each time. It is either coin A or coin B that is heads, so run 1000 with A as heads and 1000 with B as heads. It's that simple.

@semck - I would but why waste my time? We have differing views on how to simulate it.

semck, you are living in a world where it is possible to get four solutions, one of which we then get rid of, but in the case we have we cannot have it being possible to get the four solutions as we must have a boy, so to stop this problem we give ourselves one boy.

or we can do it draugs way that the second one is a boy so we have the first as a girl/boy (fifty fifty), or that the first is a boy so the second is a girl/boy (fifty fifty again)

"No, you don't simulate two coin tosses, you simulate one coin toss in two runs with a known entity for the second coin each time."

Nope. YOU are the one claiming that that is equivalent to the situation that I described, so it is up to you to prove that they are the same by simulating what I said.

I said: "you know only that (a) two coins were tossed and (b) at least one was heads."

That is trivial to simulate, and it says nothing about the second one or the first one. If you insist on introducing your interpretation into the simple simulation of a physical situation, I think everybody will be able to see how bankrupt your argument is.

Again: it is up to you to SHOW that simulating the situation *I* described gives the results *you* claim.

So do it. Simulate two random coin tosses. You are the one who keeps talking about probabilistic independence, yet you refuse to write a simulation in which both tosses are actually random.

Unbelievable. Put up or shut up.

semck, we don't know the oldest one is a boy, this is your mistake of conflating the idea of assigning the known boy to be the first boy and saying the first child is the eldeset and a boy. but i can see that we are not going to come to any agreement....

SD, when you say that we have to toss a coin and only toss a second one if the first one is heads, that is saying that we're in the situation where we know that the first coin is heads.

We don't. We only know that at least one of them is.

How is this non-obvious? How is it non-obvious that if you stop throwing after the first one if you don't get a heads, you're talking only about the case where the FIRST one is heads?

Because coinjs are discrete objects and because you have stated that at least one coin is a heads, we can deduce that either coin a or coin b is heads. Therefore we find all the combination with coin A heads and all the combinations with coin b heads. This si four combinations because Hh and hH are two different combinations due to the assignment of coins a and b.

Cool, Draug, then you'll be fine with my simulation. It will simulate two discrete variables (or "objects,") randomly initializing each to 0 or 1 (independently). Then it will call a subprocedure if at least one of them is heads.

The subprocedure will be in the case you know about -- it can assume it can call the heads "a." So when it counts how many get two heads, it will get 50%. Right?

So, you'll write the simulation now, right? A simulation that includes two actual random coin tosses?