Jeff Atwood’s exercise in Bayesian probability

It all started with this post by Jeff Atwood:

“…Let’s say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?…”

More comments on his next post as he answers:

“…Most people answer 50%.

Unfortunately, this isn’t correct.

This problem, although seemingly simple, is hard to understand. For cognitive reasons that are not fully understood, while our intuitions regarding a priori possibilities are fairly good, we are easily misled when we try to use probability to quantify our knowledge. This is a fancypants way of saying there were almost a thousand comments on that post, with not a lot of agreement to be found.

The key thing to bear in mind here is that we have been given additional information. If we don’t use that information, we arrive at 50% — the odds of a girl or boy being born to any given pregnant woman. That’s true insofar as it goes, but it’s the answer to a different, much simpler question, and certainly not the answer to the question we asked…”

okay. I guessed 50% also. But I’m not an expert of Bayesian probabality. Guess who disagreed with Jeff Atwood?

Give up?

Paul Bucheit, founder of Gmail begged to differ with Jeff Atwood. Nice. In his blog post THE QUESTION IS WRONG:

“…The problem with the question as originally posed was that it didn’t specify which of these algorithms was being used. Were we arbitrarily told about the girl, or was a selective process applied?

By the way, if we’re applying a selective process, then 100% is also a possibly correct answer, because at step two we could have eliminated all parents that don’t have both a boy and a girl. Likewise, all other probabilities are also potentially correct depending on the algorithm applied…”

hahaha. wow. you have to read the whole blog post to see the answers. But wait!

Rob Dickerson weighs in:

“…So, as far as I can tell, the real answer is that we don’t know enough about the mother’s behavior to give a definitive answer of how this scenario plays out on average.

Of course, this kind of reasoning is pretty tricky to pull off correctly, so I wouldn’t be surprised if a good counter arises. But I’ve thought about this problem for a while, and, for now at least, I’m pretty thoroughly convinced that this solution is correct….”

and in a later post, he even posts pictures! (I LOVE PICTURES!):

Rob Dickerson on Jeff Atwood

oh wow. there’s more pictures to illustrate Rob Dickerson’s answer. Feel free to view it here!

LESSON: Sometimes you can’t figure the answer becasue you DIDN’T KNOW what the question was! hahaha

Too much Bayesian Probability for a day? Maybe a very, very, very practical application (how to board a plane quickly!) of Markov chain Monte Carlo simulation? 🙂

UPDATE:

just saw a more “appropriate” version of this on HN:

…What do you think of this story? I get a frantic call from my friend at the airport. He’s been captured by the TSA for bringing a nose hair trimmer onto the plane. He was taking his brothers two kids (whom I know nothing of) back home, but now needs me to pick them up at the airport. The line goes dead. I show up at the airport and there are 4 pairs of children waiting behind the glass to be claimed, conveniently arranged in pairs of GG, GB, BG, and BB. Each pair is equally likely to be the one I am supposed to pick up.

If my friend had told me “their names are Sarah and–” before the line went dead, I would mentally eliminate BB. Sarah could be any of the four girls. The three remaining options are not equally likely to contain Sarah, because the first one has two girls, and either one could be Sarah. So, the probability of Sarah being the girl in BG is 1/4 and of being the girl in GB is 1/4. The probability of the other child being a girl is also 1/2. It’s not so much that BG or GB is eliminated in this case, but that the probability of her being in the GG group is better. But wait a second, I am not trying to find Sarah, I am trying to find the pair of kids that is his pair of kids. Is the probability that a group contains Sarah different from the probability that a group of kids is his? Maybe that’s my bad assumption…maybe “at least one is a girl” is some precise formulation I don’t understand that means the GG group isn’t twice as likely to contain that girl. I’ll get back to that. (Possibly her name is “at least once”?).

But what if he instead told me something over the phone that would eliminate the possibility of it being two boys only? What if when he had looked out over the groups of children waiting to be picked up before they blindfolded him and panicked and said “it’s not the two boys–“? Is knowing that one of the children is a girl different from knowing they both aren’t boys? I think it is, because the latter is a statement about the set of events, and the former is a statement about a single event. When we talk about the pairs of children, the set of events, we are not conveying direct information about the individual events. In this second case he is not conveying information about the individual events and the three remaining choices remain equally likely, 1/3 each.

Looking back to “at least one is a boy”- if we interpret that as a statement about the sets of probabilities, I would more precisely restate Eliezer’s question as “Does the set of two children contain at least one boy?”. This is why the important part of the story is the word “mathematician”. The mathematician is talking about eliminating sets of events when he says “Yes”. Ordinary people would just talk about the gender of a child…”

well? was that better?

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