This week it is a lovely problem, courtesy of my pal Mike…..

**Mrs Brown has two children. One of them is a boy called Albert. What is the probability of her other child also being a boy?**

The answer to this puzzle, and 100 others, can be found in a new kindle ebook called **PUZZLED**, and is available in the UK here and USA here.

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This is one of those annoying, subtly ambiguous probability puzzles, isn’t it. I look forward to the ensuing arguments.

This reminded me if another puzzle which helped me solve it after a minute of thinking.

I like the picture! Where’s it from?

An interesting question to consider against this one is:

“Mrs Brown has two children. The elder of them is a boy called Albert. What is the probability of her other child also being a boy?”

I don’t think anyone can actually give a distinct number…

You could give a probability density function. But I don’t think that’s the answer anyone is looking for.

I know what the “logic” answer is…and I also know about things not taken into account with the logic answer to know the percentage is lowered.

If you watch QI think of the time flipping the coin was asked.

Careful Shaded. Thin line. :o)

I’m just thinking of the “You’re a Good Man Albert Brown” song now. Thanks a load.

How often does Mrs. Brown eat bananas?

I hate this, especially after what happens if you ask if Albert is older or younger than his sibling (irrespective of Mrs Brown’s answer). My intuition tells me that introducing new information into the problem should adjust the probability – the question doesn’t introduce new information (inasmuch as the answer doesn’t matter) but has a dramatic effect on the odds.

I think I have the answer as have seen this before….only it wasn’t a Mrs Brown.

The answer seemed obvious, but at the same time i knew it couldn’t be that easy. Thought about it for a minute, and it turned out it wasn’t. Intution and common sense suck at probability calculation.

Very nice!

4 seconds to solve, by listing the combinations involved, and elimination the impossible.

But I did stats in Uni.

(And damn near passed!)

This one is breaking my head and I thought I was OK at statistics! I thought I had it solved but I think it might be more complicated than that. Compare it to this slight restatement of the same problem:

You meet Mrs Brown in the street, she has a boy with her that she tells you is her son. You ask if she has any other children and she tells you she has one other child.

So you know only that she has two kids, and one is a boy, same as before. But what is the chance the other is a boy?

Is knowing that his name is Albert meant to help?

umm

Its all random so could be either

I don’t know

I give up

Its a trick question isn’t it?

Oh, I think I’ve got it. It helps to think of the children as goats standing behind closed doors.

Got me stumped on this one. I can work out the probability but I think there is some lateral thinking to solving it that I can’t get my head round.

Damn you Mike/Wiseman!!!

Had some food, did some work, had some coffee. Came up with an answer.

Heres hopeing it’s right.

I’ve heard the question before so I know the correct answer, and the reasoning behind it, but I can never seem to fully agree with the reasoning.

Garrett, I don’t think there is just one correct answer.

There is one correct answer if you assume that there is a 50/50 chance of any child being a boy or a girl. I think it works better as a coin flip question. Biology is messy.

Conditional probability:

P(A|B) => Probability of event A, given the occurrence of event B

P(A|B) = (P(A) ∩ P(B)) / P(B)

conditional probability is only useful if the events are not independent. admittedly, the sex of the first child may reveal something about the underlying biology of the parents that may indicate a sex preference for the 2nd child, but that seems to be beyond the point of the question. The odds that any child would be a boy from a ‘fair’ uterus is .5 – it’s unaffected by the other outcome from the same woman.

Either you all are trying to read too much into this, and the answer is incredibly simple and obvious, or I’m missing something incredibly simple and obvious. I’m hoping it’s the former, but knowing me, it’s the latter.

I am in the exact same place as you. Damn that bugs me. Not that we are in the same place, but that I know I am missing something…

Probability is 0%. Mrs. Brown has a lovely daughter.

I have two possibilites. One seems logical enough, the other one is right. Or at least according to a tree diagram, a method that usually works. But there is a very good question, and there are different oppinions about this one.

I’m looking forward to this answere, to see which side you are on, Richard. However, I do susspect that you do not only have one, obvious right answer.

I have an answer which took a few seconds I suppose, but the long part is going to be waiting for next week sometime to see if it’s the right one.

Ahhh… Can I get the parents DNA and get back with you???? 😉

Why can’t I work it out!!

There’s obviously a fiendishly easy answer which eludes me…

I keep staring at the question hoping it’ll suddenly give me the answer, but I am truly stumped! Now I have to wait till Monday 😦

I thought I had it, then I ran a forced solution. I changed my answer but I still don’t quite get why (or if) it’s right. I had to sleep on it for the second answer.

Shamed,

W R

Is this a lion-type question, or genuine probability? Because I know the answer if it is straightforward.

Easy! Solved in 5.3 seconds.

why do you to praise? You are very vain person. 😀

I was completely stumped but my husband has explained it to me now and it’s complicated, well it seems complicated to me! 🙂

My husband says it’s not ambiguous because of the particular way the question is phrased. 🙂

Спасибо, хорошая статья. Подписался.

Wondering if Charlie Brown had a brother called Albert, if so then its two son’s unless its more scientific than that in which case I haven’t a clue

Can we have “Friday chocolates” instead of Friday puzzles please

Очень полезная вещь, спасибо!!

Think that I have an answer to this.