Monty Hall Problem (statistics/choice game)

[Mise]

If he randomly selects a child then he is thinking of a distinct daughter (eg Sandra), thus the gender of the other child is independant.

I'll have to think about that... I'm not convinced, because we don't know whether it's his first or second child, so we have to calculate the probability that he picked his first child vs his second child.

Re: Birthday problem.

n=23. IIRC that is.

I got 18.6... (or 19 in whole people). How is it calculated? Do you remember?

 

[Brighteye]

n=24, 36 if p>0.9

I just calculated it by multiplying 365/365*364/365*363/365... until I got less than 0.5. I got to 343, which is 23. I've been told it's 24, so perhaps leap years make a difference.

 

[Truronian]

http://en.wikipedia.org/wiki/Birthday_problem#Calculating_the_probability

 

[brennan]

Let's just agree it's an ambiguous question.

Here's my favorite bit of counterintuitive probability:

Say you have a room with n people in it. How big must n for the probability that two people share a birthday to be greater than a half?

It's 21 IIRC.

 

[Defiant47]

Guys, consider it this way.

You have 200 people in a room, all with two children. 50 of them fall into each of the following situations:
A) s/s
B) s/d
C) d/s
D) d/d

You pick one person at random. You ask him to pick a child at random (i.e. first or second child) and for him to tell you the gender of that child. He says daughter.

Exactly! 50 people will have option B. Of those 50 people, 25 will say daughter. 50 people will have option C. Of those 50 people, 25 will say daughter. 50 people will have option D. Of those 50 people, 50 will say daughter. 25+25=50 people will have option B or C, meaning they have a son. 50 people will have option D, meaning they have another daughter.

The trick is that although twice as many people have a son (when ruling out the s/s option A), only half as many of them will first say that they have a daughter.

 

[brennan]

Exactly! 50 people will have option B. Of those 50 people, 25 will say daughter. 50 people will have option C. Of those 50 people, 25 will say daughter. 50 people will have option D. Of those 50 people, 50 will say daughter. 25+25=50 people will have option B or C, meaning they have a son. 50 people will have option D, meaning they have another daughter.

The trick is that although twice as many people have a son (when ruling out the s/s option A), only half as many of them will first say that they have a daughter.

...right. But if one of these people at random tells you he has a daughter, then you know he has to be one of the 150 people with at least one daughter, 100 of whom (2/3) also have a son.

 

[Defiant47]

...right. But if one of these people at random tells you he has a daughter, then you know he has to be one of the 150 people with at least one daughter, 100 of whom (2/3) also have a son.

This is true. However, those 100 are half as likely to tell you that they have a daughter. So you will only hear 50 of them tell you they have a daughter.

 

[brennan]

This is true. However, those 100 are half as likely to tell you that they have a daughter. So you will only hear 50 of them tell you they have a daughter.

I'm fairly sure that is already taken into account...

 

[Defiant47]

I'm fairly sure that is already taken into account...

Excellent. So on average, 50 people with one son and one daughter will tell you that they have a daughter, and 50 people with two daughters will tell you that they have a daughter. Thus, if one person tells you that they have a daughter, the probability that they have a son is ~50/100=1/2.

 

[Mise]

I think Defiant's right...

If there are 2 people with 1 daughter and 999 sons, and 1 person with 1000 daughters, and you randomly selected one of them, who chose a kid at random and said it was a girl, chances are you've picked the one with 1000 daughters, and not one of the other two dads who happened to have chosen the 1 daughter out of 1000 children.

 

[brennan]

No.

On average 100 people will have told you they have daughters. But this represents 3 groups of 50 people (the other 50 told you they had sons), which gives you 100/150.

Edit: Nonono, I think I see where the problem is, gimme some time to figure this out. I suspect we're not talking about the original situation anymore.

Edit 2: this is giving me a headache.

Edit 3: Okay, I think i've nailed it: the thing is that we're not considering the random chance that 50 people with 1 son and 1 daughter tell you they have a son. We are interested in the odds of the second child being either a son or a daughter based on the information we are given on the first child (not using first and second to mean chronological order here). That removes the random aspect Mise introduced and makes it:

We want to know about sons, so
50 people can't play the game (no daughters).
100 people with 1 son and 1 daughter all tell you they have a daughter.
50 people with 2 daughters tell you they have one daughter.

It would work the other way with sons. You just have to make the choice of whether to discuss sons or daughters before you ask everyone. If you let each one randomly decide to give you the information (s/d) it won't work.

 

[Masquerouge]

excellent! It went exactly as expected.

I've stopped trying to argue about this with people convinced it's 1/2, and not 2/3.

It either goes down to semantics or stubbornness (on both sides), while it's really a simple probability issue.

I'm wondering if you would get different results if you were to ask "I flipped a coin two times, one of the flip was tails. What are the odds the other flip was heads?"

 

[Defiant47]

No.

On average 100 people will have told you they have daughters. But this represents 3 groups of 50 people (the other 50 told you they had sons), which gives you 100/150.

This represents 1 group of 50 people with d/d that have ALL told you they have a daughter, 1 group of 50 people with s/d that have told you that they have a daugther, and 1 group of 50 people with s/d that have told you that they have a son (but they aren't counted since they did not say they have a daughter - they are not part of the equation). So that's 100 people total that are relevant. 50 out of those relevant people have a son.

Edit: Nonono, I think I see where the problem is, gimme some time to figure this out. I suspect we're not talking about the original situation anymore.

Original situation.

Problem 1: The person has two children. They say they have a daughter. What's the chance of them having another daughter (or son)?

Problem 2: You have two chips, one all black, one white-black. A chip's black side is revealed. What's the chance of it being the all-black chip?

Edit 2: this is giving me a headache.

Edit 3: Okay, I think i've nailed it: the thing is that we're not considering the random chance that 50 people with 1 son and 1 daughter tell you they have a son. We are interested in the odds of the second child being either a son or a daughter based on the information we are given on the first child (not using first and second to mean chronological order here). That removes the random aspect Mise introduced and makes it:

We want to know about sons, so
50 people can't play the game (no daughters).
100 people with 1 son and 1 daughter all tell you they have a daughter.
50 people with 2 daughters tell you they have one daughter.

That's not the original situation. If all 100 people make sure that they tell you they have a daughter first, then you have 100 people telling you they have a daughter and they have a son and 50 people telling you they have a daughter and they have another daughter. (which means 50/150)

The point is that you ask the person for the gender of one of their children. You must then assume that the person picked the child randomly. Thus, not all 100 of those people will say daughter first. 50 on average will say daughter first. (which means 50/100)

It would work the other way with sons. You just have to make the choice of whether to discuss sons or daughters before you ask everyone. If you let each one randomly decide to give you the information (s/d) it won't work.

You aren't asking the person "Tell me the gender of one of your two children, but if you have a daughter, claim daughter first". You're just asking them the gender of one of their children, and thus you have to assume that they choose randomly (i.e. do not assume that ALL people will always choose daughter first if they have a daughter).

If that is so, then only 50 out of 100 people with s/d will tell you they have a daughter. The other 50 will say they have a son, and thus not be included in our computations (since our computations include people and chances when someone says they have a daughter).

 

[brennan]

8 random people walk up to Masquerouge:

2 of them have 2 sons, they both tell Masquerouge they have one son. masq guesses the other child is a daughter. He is wrong both times.

4 of them have both a son and a daughter, two tell Masq they have one son, two that they have one daughter, in all four instances Masq correctly guesses that the other child is the opposite sex.

2 of the mysterious strangers have two daughters and tell Masq they have one daughter. In both instances Masq wrongly guesses that the other child is a son.

50/50 win rate. Same as a random guess.

 

[Evie]

They sort of have a point - if you formulate the problem in such a way that anyone with at least one daughter, will tell you daughter, then the problem works. However, if you formulate it in such a way that people with one child of each gender get a choice of which gender's child they tell you about, then the problem introduces a whole host of new factors (cultural bias toward one gender or another, for a start).

(Note that meeting a girl in the street falls on the "choice" side - that is, they get to chose which child they go out with)

In the end, though, arguing over it is sort of futile. Posting this problem online, outside dedicated statistics forums, is a case of borderline trolling to begin with - you don't post it to get any serious discussion started, you post it (and Masque's various "will not argue" posts makes that fairly clear) to get an argument started (and mock people on the "non-statistician" side of the argument).

Which means they'll never back down - they'll just consider you a moron who won't listen to reason and keep arguing semantics rather than bowing before their Almighty Logic.

 

[Brighteye]

That removes the random aspect Mise introduced and makes it:

We want to know about sons, so
50 people can't play the game (no daughters).
100 people with 1 son and 1 daughter all tell you they have a daughter.
50 people with 2 daughters tell you they have one daughter.

Mise introduced? 'Twas me!
Here you assume that everyone with one of each will tell you about their daughter.
If it's a stranger on the street, a far better assumption is that half tell you about a daughter and half a son. You happen to have encountered one of the 50 daughter people.
After all, with no other information about the person, and with no specific question from you, why would a person specifically want to tell you if he has one or more daughters?

 

[Mise]

To be fair, the problem is quite different to Monty Hall. In the Monty Hall problem, it doesn't matter if he deliberately picked the door with the goat, or if he picked it by random and it just so happened to be the goat -- you're still better off switching. That's what was confusing me...

I accept now that how/why he picked his daughter makes a difference. If out of the 200 people, we asked, "do you have a daughter," then it'd be 2/3 for a son. If we just asked, "what is the gender of one of your children," then it'd be 1/2.

But if someone came up to me on the street and said, "I have two kids, one of them is a daughter, what's the other one?" I'd probably say daughter, cos I reckon he's trying to trick me.

 

[Evie]

But if someone came up to me on the street and said, "I have two kids, one of them is a daughter, what's the other one?" I'd probably say daughter, cos I reckon he's trying to trick me.

If some stranger came in the street to tell me that, I would look at him, then depending on what he looks like, either run away real quick, or else tell them "go bother someone else, you freak,".

I mean, a stranger who tells me that is either a madman and possibly dangerous, or a trollish uber-geek. Either way, my geekiness has limits, and they're on the wrong side thereof.

 

[Mise]

For me, that's exactly the sort of question I can't resist answering

 

[brennan]

Me too, but i'm happy to be an uber-geek.