That's the title of an interesting essay by author of *Innumeracy* John Allen Paulos. Some highlights:

"...there is a tension between stories and statistics, and one under-appreciated contrast between them is simply the mindset with which we approach them. In listening to stories we tend to suspend disbelief in order to be entertained, whereas in evaluating statistics we generally have an opposite inclination to suspend belief in order not to be beguiled."

-and-

"Of course, the contrasts between stories and statistics don’t end here. Another example is the role of coincidences, which loom large in narratives, where they too frequently are invested with a significance that they don’t warrant probabilistically. The birthday paradox, small world links between people, psychics’ vaguely correct pronouncements, the sports pundit Paul the Octopus, and the various bible codes are all examples. In fact, if one considers any sufficiently large data set, such meaningless coincidences will naturally arise..."

I still can't wrap my head around the two-son problem.

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## Stories vs. Statistics

By
Brian Burke

published on 11/05/2010

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Is the two son problem just a restatement of "the price is right paradox?"

Maybe the Monty Hall paradox?

Yes, it seems similar to Monty Hall. The way I finally accepted it was to think Bayes.

For all two-child families, having a boy-girl combination (in either sequence) is twice as likely as a boy-boy combination. You have to start with that prior distribution.

Alternatively, the 4 possible combinations (with probabilities) are:

BB (1/4) BG (1/4) GB (1/4) and GG (1/4)

Now remove the possibility of GG. We are left with:

BB (1/3) BG (1/3) and GB (1/3)

Therefore the probability of having the other child be boy is 1/3. The difference is in the sequence. If we were told the

youngerchild is a boy, then the answer would be different. We would have:BB (1/2) and BG (1/2)

There can be no GB combination.

Yeah, the selection process is key. There is a great discussion of the two-son problem (and a related variant) here:

http://www.sciencenews.org/view/generic/id/60598/title/Math_Trek__When_intuition_and_math_probably_look_wrong

See I wonder if I'm not understanding the two-son problem properly, because it makes sense to me. The important thing is understanding exactly how much information you are being given.

Normally when you're told 'at least ...', that's when you need to think about it. Maybe I'm just more wary than I should be.

Brian, I have a project to do in my Probability Theory class that is a variation on the two son problem. It's some pretty interesting stuff...and yeah, Bayes to the rescue. I'll send you a copy of the power point when I'm done if you want.

John

Thought everyone here would appreciate this quote. How many times have you heard this before?

Igor Olshanksy, Cowboys DE: "I mean, there's a lot of things that can go wrong when you pass the ball. You can get sacked, the ball can get tipped, you can get an interception. If you run the ball, you can get tackled and get a forced fumble. See, there's a lot more things [that can go wrong on pass plays]. ... You've got to run the ball."

Paulos' book Irreligion is also quite good and not filled with the type of inflammatory rhetoric many such books include.

Brian, you are completely wrong on the 2 sons problem. No matter what way you calculate it, you get 50%. The only way not to get 50% is to use statistical sophistry that makes 0 logic.

Not wrong. It's in the wording of the question.

If you are told that at least

either oneof the two children is a boy, and are asked for the probability the other is also a boy, then the answer is 1/3.If you are told that

this particular childis a boy, and are asked for the probability that the otherparticular childis boy, then the answer is 1/2.Nice try. But I do admire your defiant certainty.

Ian hit it on the head above.

You are still wrong Brian. Birth order is irrelevant because the question didn't ask that and even if it did, it is still 50%

If you are not asked for an order then it is:

A: Boy

B: Girl

That is 50%

If you are asked for an order then it is:

A: Boy A: Boy B

B: Boy A: Girl

C: Girl: Boy A

D: Boy B: Boy A

Either way it is 50%.

Wrong again, my stubborn friend. You'll figure it out eventually.

If you remove the GG combination from the 4 possible combinations, you get:

GB

BG

BB

each with 1/3 probability.

But hey, I, together with thousands of mathematicians, could be wrong. Could you be?

Yes they are wrong. Mathematicians can be wrong. It's not like peer reviewed work isn't false around half the time. Smart guys are just as wrong at times.

Here are all the possible combinations:

1) Boy(known), Boy

2) Boy(known), Girl

3) Boy, Boy(known)

4) Girl, Boy(known)

That is still 50%

Anonymous: Your #1 and #3 *are the same thing* in the "either" question. You need to stop thinking about sequence, and then you'll get it.

Brian, the order is not relevant at all. The question didn't ask just like the Tuesday question is wrong.

Think of it like this. Say you have a 4 baskets with red balls (girls) and blue balls (boys) and you are blindfolded. 1 has 2 red balls. 1 has 2 blue balls. 2 has 1 red ball and 1 blue. If you randomly pick a blue ball, there is a 50% chance the other ball is blue.

Boy, Boy

Boy, Girl

Girl, Boy

There are 4 boys here. If you pick the first boy, he has another boy with him. If you pick the second boy, he has another boy with him. That is 2 with 2 boys. If you pick the 3rd boy, he has a girl paired with him. If you pick the fourth boy, he has a girl paired with. That means 50% of the boys are in 2 boy families.

To the other Anonymous - In your basket example, if you randomly pick a blue ball, it means you picked from baskets #2, #3, or #4.

1/3 chance of basket #2 x 100% chance the other ball is blue.

+

1/3 chance of basket #3 x 0% chance the other ball is blue.

+

1/3 chance of basket #4 x 0% chance the other ball is blue.

= 1/3 chance the other ball is blue.

Anonymous #2 again. Actually, please ignore this last comment - I think the chance that you've picked basket #2 is actually higher than 1/3.

There is an important distinction between the basket problem and the 2-sons problem. They're not quite the same thing. Knowing that you have at least 1 boy does NOT increase the probability of being BB vs. BG or GB. It's still 1/3 chance of being BB. In the basket puzzle, having more blue balls in a basket DOES increase the chance that you've picked that basket. It's a bad analogy.

This is awesome. I could read the back-and-forth of the two sons problem all day.

(FYI: I'm in the 33% camp)

The basket problem isn't analogous to the two sons problem, but in the two sons problem, you aren't choosing between 3 choices. They say the parents have a boy. Here are the possibilities:

Boy, Boy

Boy, Boy

Girl, Boy

Boy, Girl

Half of them are BB

Another way to look at it:

Boy, Boy

Boy, Girl

Girl, Boy

There are 4 boys here. If you randomly pick one, there is a 50% chance he is in a 2 two son family. They aren't asking you to pick a certain pairing. They certainly didn't ask for order. They are asking if the boy is in a two son family. 50% of the boys are in a two son family. It's as simple as that. If you don't understand, you are using bad math. You know that there are 4 possible boys and 2 of them are in a 2 two son family.

A set of parents have 2 children. At least one is a boy. What is the probability both children are boys?

GG (eliminated)

BB

BG

GB

Answer = 1/3

It does depend on how you ask the question.

Of course Brian is exactly right. Mr. Smith has two children. There is 1/4 chance to have two sons. You see him with a son. You trim the sample space, eliminating two girls. You are left with boy/boy girl/boy or boy/girl....1 in 3.

Now, there is a 90% chance that if Mr. Smith goes to the theater, it will be with his daughter (if he has one). You see him at the theater with a son. What are is the probability he has two sons?

JudoJohn

90%? Here's my logic, let me know if I'm doing this right. I think its a matter of thinking backwards. So, if he has a son and a daughter. There is only a 10% chance that his son happened to go with him that time. If it wasn't that 1/10 chance that is happening in this scenario, then it must be a boy/boy family. Hence 90%?

We use Bayes. What is the probability of two boys, given there is one boy?

The numerator: (1)(1/4) which = (1/4).

The denominator: (1)(1/4)+(1/10)(1/4)+(1/10)(1/4)+0 which =(12/40).

So, (1/4)/(12/40)=5/6, which is our answer.

This makes sense. After all, if we see Smith with a son, it's very likely he doesn't have a daughter.

Anyway, Chris, you are exactly right to think backward, that's the whole point of Bayes.

JudoJohn

For the Anon poster who is having trouble understanding why the correct answer is 1/3, perhaps you should try running a simulation. If you assign equal probability of having a boy or girl, you will find that 25% of two-child families are BB, 25% are GG, and 50% are a mix (25% with the youngest being a boy, 25% with the youngest being a girl). So, let's say you were lucky enough to get exactly 25% with only 100 samples, or in other words, 100 2-child families.

You would have:

GG - 25 Families

BB - 25 Families

GB - 25 Families

BG - 25 Families

Now let's restate the question: A set of parents have 2 children. At least one is a boy. What is the probability both children are boys?

The "At least one is a boy clause" eliminates GG as a possibility. That leaves us with three types of families left:

BB - 25 Families

GB - 25 Families

BG - 25 Families

Now when we consider the OTHER child (i.e. the one that is not definitely a boy), we find that only 25/75 (or, simplified, 1/3) families have a boy as the other child - and that is the BB family.