Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

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

scienceblogs.com

Post reply on HN
Print view View original

21–30 of 152 posts

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

#21

Post by nadam »

As the currently top voted comment does not get it, I try to intuitively explain the paradox.

No, it is not ambiguity of language. It says formally:

I have two children. There exists a child of mine who is (boy and born on tuesday).

And yes, the probability of the other child being a boy is 13/27.

To understand this, try a more extreme case:

When a child is born we generate a random number: rnd(1billion)

Now the man says:

'I have two children. There exists one of them which is boy with generated id=456765234'

As we can see, the probability of the other child being a boy approaches 1/2 as we increase the bound of our random number!

Why is this intuitively?

Because by using a very specific id we are closer and closer to specifying one of the boys, as the probability of the id not being unique decreases. We are closer and closer to saying this:

'Randomly pick one of my children. He is a boy. What is the probability of the other being a boy?'

Yes, that would be 1/2.

Edit.: Seriously downvoting this??? For most commenters the correct result was not intuitive. (It was not intuitive for me at first also). I tried to make this intuitive to them. (a bit deeper than usual paradxes though) Or do you still think the 13/27 result is not correct? In that case you are fighting mathematical truth with the downvote button.

Edit: Thanks, corrected probability from 2 to 1/2

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

#22

Post by ars »

nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

Why can't the other boy also be born on Tuesday?

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

#23

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.

That sounds convincing.

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

#24

Post by extension »

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)

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

#25

Post by nadam »

ars wrote: post #22
nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

Why can't the other boy also be born on Tuesday?

The other can also be born on Tuesday. (generated id = 1 in rnd(7)) And the other can also have a generated id = 456783123 in rnd(1billion). But as the bound of the rnd is bigger and bigger, the probability of that approaches to zero. So as we know more and more information of one of the children, we are closer and closer to simply uniquely identifying him, so the probability of the other being a boy approaches 2.

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

#26

Post by NickPollard »

twelvechairs wrote: post #19
incremental wrote: post #16

Earlier quoted context omitted.

Since the constraint simply says that at least one child is a boy, we don't need to distinguish between two types of BB. This is a curious misconception, by the way. The typical failure mode I see on these questions is people having difficulty accepting BG and GB as different possibilities.

If we are determined to look at order of birth (ie. have a BG and GB) then we should consider all cases by order of birth, so where 'B' represents the boy we know and 'b' or 'g' represents a child we dont, and the first character represents the first child, and the second the second childe - we have: Bb, bB, Bg, gB 50%.

You need to look at it in a Bayesian sense.

Consider all the possibilities and then look at the outcomes they produce.

BB BG GB GG

These are all equally likely, right? They each have P = 1/4. If we took 1000 fathers who had two children each, and lined them up, we would expect 1/4 of them to have children matching each of the above pairings. That means 250 of each. With me so far?

Now, we know that we are dealing with a father who has at least one son. If we went along the line and asked each father 'Do you have at least one son: Yes/No?', then 750 would answer yes, and 250 (the GG fathers) would answer no.

If a random father comes up to us and says 'I have at least one son', we know that he is from those 750 - we have selected a subset to deal with. Of those 750, only 250 have a second son, so the probably is 250/750 or 1/3.

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

#27

Post by ars »

nadam wrote: post #25
ars wrote: post #22

Earlier quoted context omitted.

Why can't the other boy also be born on Tuesday?

The other can also be born on Tuesday. (generated id = 1 in rnd(7)) And the other can also have a generated id = 456783123 in rnd(1billion). But as the bound of the rnd is bigger and bigger, the probability of that approaches to zero. So as we know more and more information of one of the children, we are closer and closer to simply uniquely identifying him, so the probability of the other being a boy approaches 2.

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.)

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

#28

Post by tallanvor »

ars wrote: post #3

I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.

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.

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

#29

Post by ars »

tallanvor wrote: post #28
ars wrote: post #3

I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.

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.

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

#30

Post by reitzensteinm »

[deleted] wrote: post #17
ars wrote: post #3

I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.

[deleted]

OK, after rereading the article, I'm wrong about this - that's not what the author is doing. Post deleted to not spread confusion myself...
Post reply on HN