Comment author: William_Kasper 09 December 2012 06:37:19AM *  0 points [-]

Mr. Bonaccorsi:

Here are two links to classic posts by Eliezer Yudkowsky that you may find pertinent to the second dialog from your last comment. I hope you enjoy them.

How to Convince Me That 2 + 2 = 3

The Simple Truth

Comment author: JohnBonaccorsi 10 December 2012 07:16:03PM 1 point [-]

Thank you for those links, Mr. Kasper. In taking a quick first look at the two pieces, I've noticed passages with which I'm familiar, so I must have encountered those posts as I made my initial reconnaissance, so to speak, of this very-interesting website. Now that you've directed my attention to those posts in particular, I'll be able to read them with real attention.

Comment author: William_Kasper 08 December 2012 05:44:57AM 0 points [-]

Let's establish some notation first:

P(H): My prior probability that the coin came up heads. Because we're assuming that the coin is fair before you present any evidence, I assume a 50% chance that the coin came up heads.

P(H|E): My posterior probability that the coin came up heads, or the probability that the coin came up heads, given the evidence that you have provided.

P(E|H): The probability of observing what we have, given the coin in question coming up heads.

P(E&H): The probability of you observing the evidence and the coin in question coming up heads.

P(E&-H): The probability of you observing the evidence and the coin in question coming up tails.

P(E): The unconditional probability of you observing the evidence that you presented. Because the events (E&H) and (E&-H) are mutually exclusive (one cannot happen at the same time as the other) and the events (H) and (-H) are collectively exhaustive (the probability that at least one of these events occurs is 100%), we can calculate P(E):

P(E) = P(E&H) + P(E&-H)

P(E) = P(E|H) P(H) + P(E|-H) P(-H)

Using Bayes' Theorem, we can calculate P(H|E) after we determine P(E|H) and P(E|-H):

P(H|E) = [P(E|H) P(H)] / [P(E|H) P(H) + P(E|-H) P(-H)]

Let me try this. You come upon a man who, as you watch, flips a 50-50 coin. He catches and covers it; that is, the result of the flip is not known. I, who have been standing there, present you the following question: "What is the chance the coin is heads?"

In this case we can assume that our lack of knowledge is independent of the result of the coin toss; P(E|H) = P(E) = P(E|-H). So

P(H|E) = P(E) (50%) / [P(E) (50%) + P(E) (1 - 50%)] = [P(E) / P(E)] (50% /100%) = 50%.

The next day, you come upon a different man, who, as you watch, flips a 50-50 coin. Again, he catches it; again, the result is not revealed. I, who have been standing there, address you as follows: "Just before you arrived, that man flipped that same coin; it came up heads. What is the chance it is now heads?"

Again here, your probability of observing the first result is independent of the second result. So P(H|E) = 50%.

You come upon a man who is holding a 50-50 coin. I am with him. There is the following exchange:

I (to you, re the man with the coin): This man has just flipped this coin two times.

You: What were the results?

I: One of the results was heads. I don’t remember what the other was.

Here we can note that there are four mutually exclusive, collectively exhaustive, and equiprobable outcomes. Let's call them (HH), (HT), (TH), and (TT), where the first of the two symbols represents the result that you remember observing. Given that you remember observing a result of heads, our evidence is (HH or HT). The second coin is heads in the case of (HH), which is as probable as (HT). Given that P(HH) = P(HT) = 25%, P(HH or HT) = 50%

P(HH|HH or HT) = P(HH or HT|HH) P(HH) / P(HH or HT)

P(HH|HH or HT) = 1 (25% / 50%) = 50%

After I tell you that one of the results was heads but that I don't remember what the other was, you say: "Which do you remember, the first or the second?" I reply, "I don’t remember that either."

We can use the same method as in Question C. Since the ordinality of the missed observation is independent from the result of the missed observation, the probability is the same as in Question C, which is 50%.

Comment author: JohnBonaccorsi 08 December 2012 08:03:54AM *  3 points [-]

Thank you, Mr. Kasper, for your thorough reply. Because all of this is new to me, I feel rather as I did the time I spent an hour on a tennis court with a friend who had won a tennis scholarship to college. Having no real tennis ability myself, I felt I was wasting his time; I appreciated that he’d agreed to play with me for that hour.

As I began to grasp the reasoning, I decided that each time you state the chance that the coin is heads, you are stating a fact. I asked myself what that means. I imagined the following:

I encounter you after you’ve spent two months traveling the world. You address me as follows:

“During my first month, I happened upon one hundred men who told me—each of them—that he had just flipped a coin twice. In each case, I asked, ‘Was at least one of the results heads?’ Each man said yes, and I knew that, in each case, the probability was 1/3 that both flips had been heads.

“In my second month, I again happened upon one hundred men who told me—each of them—that he had just flipped a coin twice. Each added, ‘One of the results was heads; I don’t remember what the other was.’ I knew that, in each case, the probability was 1/2 that both flips had been heads.

“Just as I was about to return home, I was approached by a man who had video recordings of the coin flips that those two hundred men had mentioned. In watching the recordings, I learned that both flips had been heads in fifty of the first one hundred cases and that, likewise, both flips had been heads in fifty of the second one hundred cases.”

In considering that, Mr. Kasper, I imagined the following exchange, which you may imagine as taking place between you and me. I speak first.

“My dog is in that box.”

“Is that a fact?”

“Yes.”

“In saying it’s a fact, you mean what?”

“I mean I regard it as true.”

“Which means what?”

“Which means I can imagine events that culminate in my saying, ‘I seem to have been mistaken; my dog wasn’t in that box.’”

“For example.”

“You walk over to the box and remove its lid, and I see my dog is not in it.”

“Maybe the dog disappeared—vanished into thin air—while I was walking over to the box.”

“That’s a possibility I wouldn’t be able to rule out; but because it would seem to me unlikely, I would say, ‘I seem to have been mistaken; my dog wasn’t in the box.’”

“How much is 189 plus 76?”

“To tell you that, I would have to get a pencil and paper and add them.”

“Please get a pencil and paper and add them; then tell me the result.”

“I’ve just done as you requested. Using a pencil and paper, I’ve added those two numbers. The result is 265.”

“189 + 76 = 265.”

“Correct.”

“Is that a fact?”

“Yes.”

“Please add them again.”

“I’ve just done as you requested. Using my pencil and paper, I’ve added those numbers a second time. I seem to have been mistaken. The result is 255.”

“Are you sure?”

“Well—”

“Please add them again.”

“I’ve just done as you requested. With my pencil and paper, I’ve added the numbers a third time.”

“And?”

“I was right the first time. The sum is 265.”

“Is that a fact?”

“That the sum is 265?”

“Yes.”

“I would say yes. It’s a fact.”

“How much is two plus two?”

“Four.”

“Did you use your pencil and paper to determine that?”

“No.”

“You used your pencil and paper to add 189 and 76 but not to add two and two.”

“That’s right.”

“Is there any sequence of events that could culminate in your saying, ‘I seem to have been mistaken; two plus two is not four.’”

“No.”

“Is it a fact?”

“That two plus two is four?”

“Yes.”

"Yes. It's a fact."

“In saying that, you mean what?”

“—I don’t know.”

Thank you again.

Comment author: William_Kasper 06 December 2012 10:18:40AM *  1 point [-]

No, the chance that the kidnapped child is a boy is 1/2.

In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys.

In the correct version of the story, you do not gain access to any information that allows you to differentiate between the mathematician's two children and identify a specific child as a boy.

A woman says, "I have two children." You respond, "What are their sexes?" She says, "At least one of them is a boy. The other was kidnapped before I was informed of its sex."

In your story, you are able to partition the woman's children into "the kidnapped one" and "the other one", and the woman provides you with the information that "the other one" is a boy. The sex of "the kidnapped one" is independent of the sex of "the other one". That is,

P("the kidnapped one" is a boy | "the other one" is a boy") = P("the kidnapped one" is a boy)

Comment author: JohnBonaccorsi 06 December 2012 10:35:36PM *  0 points [-]

Thank you for the reply, Mr. Kasper.

Let me try this. You come upon a man who, as you watch, flips a 50-50 coin. He catches and covers it; that is, the result of the flip is not known. I, who have been standing there, present you the following question:

"What is the chance the coin is heads?"

That's Question A. What is your answer?

The next day, you come upon a different man, who, as you watch, flips a 50-50 coin. Again, he catches it; again, the result is not revealed. I, who have been standing there, address you as follows:

"Just before you arrived, that man flipped that same coin; it came up heads. What is the chance it is now heads?"

That's Question B. What is your answer?

If you and I were having this discussion in person, I would pause here, to allow you to answer Questions A and B. Because this is the internet, where I don't know how many opportunities you'll have to reply to me, I'll continue.

You come upon a man who is holding a 50-50 coin. I am with him. There is the following exchange:

I (to you, re the man with the coin): This man has just flipped this coin two times.

You: What were the results?

I: One of the results was heads. I don’t remember what the other was.

Question C: What is the chance the other was heads?

Let’s step over Question C (though I'll appreciate your answering it). After I tell you that one of the results was heads but that I don't remember what the other was, you say:

"Which do you remember, the first or the second?"

I reply, "I don’t remember that either."

Question D: What is the chance the other was heads?

Comment author: RobbBB 06 December 2012 10:35:46AM 2 points [-]

Initially, there are four possibilities, each with probability 1/4:

  • A) Boy, Boy
  • B) Boy, Girl
  • C) Girl, Boy
  • D) Girl, Girl

If you learn that one of them is a boy, then that eliminates option D, leaving the other three options (A, B, C) with 1/3 probability each. So the probability that both are boys given that at least one is a boy (ie., Pr[A] given A-or-B-or-C) is 1/3.

On the other hand, if you learn that the first child is a boy, that eliminates options D and C. You've ruled out more possibilities -- whereas before 'Girl, Boy' (C) was an option, now the only options are 'Boy, Boy' (A) and 'Boy, Girl' (B). So there's now a 1/2 chance that both are boys (i.e., Pr[A] given A-or-B). And the same calculation holds if you learned instead that the second child is a boy, only with B eliminated in place of C.

Comment author: JohnBonaccorsi 06 December 2012 10:15:54PM 0 points [-]

Thank you for the reply, RobbBB. As I mentioned in my reply to shinoteki (at 03 December 2012 01:48:47AM ), I followed my original post (to which you have just responded) with a post in which there is no reference to birth order. As I also said to shinoteki, that does not mean I see that birth order bears on this. It means simply that I was anticipating the response you, RobbBB, have just posted.

At 06 December 2012 10:18:40AM, as you may see, William Kasper posted a reply to my said second post (the one without reference to birth order). After I post the present comment, I will reply to Mr. Kasper. Thank you again.

Comment author: shinoteki 02 December 2012 08:50:51PM 1 point [-]

No. To get the 1/3 probability you have to assume that she would be just as likely to say what she says if she had 1 boy as if she had 2 (and that she wouldn't say it if she had none). In your scenario she's only half as likely to say what she says if she has one boy as if she has two boys, because if she only has one there's a 50% chance it's the one she's just given birth to.

Comment author: JohnBonaccorsi 03 December 2012 01:48:47AM 0 points [-]

Although I don't see what you're getting at, shinoteki, I appreciate your replying. Maybe you didn't notice; but about half an hour after I posted my comment to which you replied, I posted a comment with a different scenario, which involves no reference to birth order. (That is not to say I see that birth order bears on this.) I will certainly appreciate a reply, from you or from anyone else, to the said latter comment, whose time-stamp is 02 December 2012 06:51:25PM.

Comment author: JohnBonaccorsi 02 December 2012 06:51:25PM 1 point [-]

Let me try another scenario. A woman says, "I have two children." You respond, "What are their sexes?" She says, "At least one of them is a boy. The other was kidnapped before I was informed of its sex." You're saying that the chance that the kidnapped child is a boy is one out of three, not out of two? To repeat: That's what I gather from the present post, near the beginning of which is the following:

In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys.

Comment author: JohnBonaccorsi 02 December 2012 06:23:31PM 0 points [-]

Having no training in probability, and having come upon the present website less than a day ago, I'm hoping someone here will be able to explain to me something basic. Let's assume, as is apparently assumed in this post, a 50-50 boy-girl chance. In other words, the chance is one out of two that a child will be a boy -- or that it will be a girl. A woman says, "I have two children." You respond, "Boys or girls?" She says, "Well, at least one of them is a boy. I haven't yet been informed of the sex of the other, to whom I've just given birth." You're saying that the chance that the newborn is a boy is one out of three, not one out of two? That's what I gather from the present post, near the beginning of which is the following:

In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys.

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