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Think you understand Monty Hall? Try the Tuesday boy problem.

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Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#41

Post by monorailkitty »

user24 wrote: post #2

Doesn't this rest on the simple ambiguity in the phrasing? > I have two children and one is a son born on a Tuesday. If by that is meant: > I have two children. Here is some information about one of them: son, born on Tuesday. Then the probability of the other child being a son is 1/2. If on the other hand we mean: > I have two children. One or more is a son. Exactly one of them was born on a Tuesday. Then we get the…

> I have two children. One or more is a son. Exactly one of them was born on a Tuesday.

I'm not sure that's what you are supposed to infer.

Looking at your earlier statement:

> I have two children. Here is some information about one of them: son, born on Tuesday.

There are two ways to interpret this.

(1) I am a man pulled at random from the set of [families with two children of indeterminate gender]. Here is some information about one of them: son, born on Tuesday.

(2) I am a man pulled at random from the set of [families with two children of indeterminate gender, one of whom was born on a Tuesday]. Here is some information about one of them: son, born on Tuesday.

We're not selecting from the same initial set in each case - set (2) is more restrictive. A difference in probability is maybe not surprising.

Still, I totally agree with you that it's a bit of a jump to conclude that the man is referring to scenario (2) in which the birthday information is used to narrow the initial set while the gender information is used to determine the probability. Just seems like a trick question to me.

To expand on the numbers a bit more... In scenario (2) the possible combinations are:

(Combo A) G G (0/49 at least one boy TB)

(Combo B) B G (7/49 at least one boy TB)

(Combo C) G B (7/49 at least one boy TB)

(Combo D) B B (13/49 at least one boy TB)

If the man has one boy, then combo A does not apply and the probabilty he has two boys must be 13 / (7 + 7 + 13) == 13/27.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#42

Post by ajanuary »

rubergly wrote: post #8

Let's try a simpler problem. Suppose we know that a certain man has two children and we also know that the older one is a boy. In this case we would say that the probability that the other child is a boy is 1/2. After all, the sex of one child is independent of the sex of the other child. That the older child is a boy has no bearing on the sex of the younger child. Now suppose we know simply that a man has two childr…

Using age as the ordering is not important. What matters is that when enumerating possibilities you count the probability of the first and then the second, and the second and then the first.

If manlier child = boy one+ child = boy then surely in your final equation one of the sides boils down to Pr(two boys | impossible event) as the probability of either manlier child = boy or one+ child = boy must be 0?

The problem comes from the fact you've asked an impossible question, not from applying an ordering. If the ordering is used consistently, then it should all work fine.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#43

Post by ars »

nadam wrote: post #38
ars wrote: post #27

Earlier quoted context omitted.

If both can be born on Tuesday then telling me one was born on Tuesday does not uniquely identify him! It's not unique if both can do it. And since it's not unique the rest of your analysis is based on a faulty assumption. (BTW I did not downmod you, in case you were wondering.)

It is about specifying a child, and then saying it is a boy. The one extreme is: one of my children is a boy. in this case the other is a boy with 1/3 probability. The other extreme is: One of my children has a national unique id=... He is a boy. The other is a boy with 1/2 probability. And other cases are in between. The more information you provide on the first child (the less chance there is that the other can hav…

"the less chance there is that the other can have the same property"

No! That is not true! The two events are independent, specifying information on one has zero impact on the other.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#44

Post by incremental »

rubergly wrote: post #18
incremental wrote: post #14

Earlier quoted context omitted.

For a randomly selected family with two children, there are four possible boy/girl combinations: B B, B G, G B, G G In the first case we are told that the older child is a boy. This leaves only two cases: B B, B G Therefore, there is a 50% chance the second child is a boy. In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities: B B, B G, G B Therefore, the probability…

But you are assuming that each of those three possibilities has equal probability; can you explain the rationale for that? (it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case, but less clear why BB, BG, and GB have equal probability in this restricted case) Besides, this is just a rephrasing of the original article's argument, and doesn't counter mine at all. I am open to the possibili…

So you say: "it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case"

The restricted case is just the unrestricted case + one additional bit of information, that is, you're told that GG is not an option. This eliminates GG from the unrestricted case, but says nothing more about the probabilities of the other options. So the probabilities stay equal, although they now equal 1/3 each (if you eliminate options, the remaining options all become more likely).

What you're missing is this: The statement "the older child is a boy" has more information than the statement "one of the children is a boy". The first statement allows you to eliminate two options (GB and GG), while the second statement only allows you to eliminate one option (GG).

The "older" part is not fundamental to the problem. Equally, the statement "the taller child is a boy" has more information than "one of the children is a boy". The problem with this is that probabilities for height are not so friendly like the 50/50 probabilities for birth order (e.g., boys are likely to be taller than girls, older children are taller than younger, etc), which introduces unnecessary complexities to a logic problem. So that's why birth order is used for these types of puzzles.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#45

Post by copper »

user24 wrote: post #40
dmd wrote: post #37

Earlier quoted context omitted.

I think it's fairly clear that nadam means 1/2, not 2.

it is now he's edited it. Wasn't clear before, to me at least. But then I "don't get it" so what do I know ;) edit: can't reply to you as HN isn't giving me the option. But: No worries, I didn't take it personally, just a little dig :)

Personally, I find Tanya Khovanova's explanations (also linked to by the article) much easier to read: she's discussed this and related problems quite a few times.

http://blog.tanyakhovanova.com/?p=221

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#46

Post by user24 »

ColinWright wrote: post #9
user24 wrote: post #2

Doesn't this rest on the simple ambiguity in the phrasing? > I have two children and one is a son born on a Tuesday. If by that is meant: > I have two children. Here is some information about one of them: son, born on Tuesday. Then the probability of the other child being a son is 1/2. If on the other hand we mean: > I have two children. One or more is a son. Exactly one of them was born on a Tuesday. Then we get the…

No, you shouldn't "also discount every other symmetrical pair", for exactly the same reason as there's a 1/36 chance of rolling double 6, but 2/36 chance of rolling a six and a one. It's all to do with labellings, and it's the most common source of error[1] in statistics. [1] By "error" I mean calculations that then don't agree with the experimental results.

What about this:

I have two teenagers. One is a boy of 13.

Do we encounter a similar situation with regard to the odds of the second teenager being a boy?

edit: I'm thinking the odds are exactly the same, 13/27, by coincidence, as there are seven possible teen ages.

So then, what about this: One is a boy named George. Or One wears a black shirt. Or One likes chocolate.

Doesn't this mean that the more information we gain about the boy, the less likely it makes it that his sibling is a brother?

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#48

Post by ajanuary »

ars wrote: post #29
tallanvor wrote: post #28

Earlier quoted context omitted.

No, you are not supposed to assume that only one of them is a son born on a Tuesday. The 13/27 probability comes from the knowledge that one is a son born on a Tuesday, but you don't know which child this refers to - the other child being a son born on a Tuesday is covered in the 13/27 probability.

No, the 13 was arrived at by excluding the possibility that the second child was a boy on Tuesday. (Read the article again and see.) Once you include that possibility it's back to 14/28.

When considering the case that the older son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on Tuesday.

When considering the case that the younger son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on a Tuesday.

But woops, we've already covered the case that both were born on a Tuesday in our first count. Removing the duplication gives the 13/27.

You then combine those two probabilities.

The 13 wasn't arrived at by excluding the possibility that the second child was a boy on Tuesday, it was arrived at by discounting the case that they were both born on Tuesday as it had already been covered in the previous sub-calculation.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#49

Post by nadam »

user24 wrote: post #40
dmd wrote: post #37

Earlier quoted context omitted.

I think it's fairly clear that nadam means 1/2, not 2.

it is now he's edited it. Wasn't clear before, to me at least. But then I "don't get it" so what do I know ;) edit: can't reply to you as HN isn't giving me the option. But: No worries, I didn't take it personally, just a little dig :)

I did not mean 'not getting it' in a negative way. (English is not my mother language.) Getting a paradox or not is a state. First I did not get it also. And I only wrote my comment, because at the time it was almost a consensus in this thread that the paradox is an uninteresting language ambiguity. I wanted to say that it is more interesting than that.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#50

Post by gcp »

extension wrote: post #24

The more sons you have, the more likely that one of them is born on a tuesday. Thus, if you take all the two-child families with at least one son, and eliminate the families without a son born on tuesday, the two-boy families are more likely to remain than the one-boy families, and you will end up with a higher proportion of two-boy families than before. Simple. (edited)

Simple.

And wrong. Note that the correct answer is 13/27 probability of a boy, which is less, not more, than 50%.

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