A woman has two children, at least one of which is a boy...

[Solly]

Since the boring smartarses have started delving into the pointless "but more than half of all births are male" argument, I might as well ask what the intentions of the game show host are...

[Chowlett]

Originally posted by Victor Galis
Nonsense. Once door #3 is revealed both door 1 and 2 offer the same odds. (1/2)

Well, that would be true if the host opened door#3 before you chose door number 1. However, you chose first.

[KrazyHorse]

Victor and Natan

Didn't you two pay attention a few months ago when we "did" this and scads of other similar problems?

This is all about information/knowledge flow. The seemingly paradoxical result that your chances increase if you switch doors results from the fact that the host has not eliminated any of the possible scenarios involving Door1 losing by opening Door3; all the gain comes to Door2. This result has been accepted as fact by the math community since the 18th century, so I doubt you'll change anybody's mind on this.

[Urban Ranger]

Suppose the host has knowledge of what's behind the doors.

You have 1/3 chance of choosing the right door. In this case, you're wrong to switch (1/3).

You have 2/3 chance of choosing one of the wrong doors. After the host reveals the other wrong door, the remaining door hides the prize. If you switch now, you are right (2/3).

[Victor Galis]

"Well, that would be true if the host opened door#3 before you chose door number 1. However, you chose first."

-Nonsense. The fact that there are two doors now increases your chance. Your choice can not increase the chance.

This is all about information/knowledge flow. The seemingly paradoxical result that your chances increase if you switch doors results from the fact that the host has not eliminated any of the possible scenarios involving Door1 losing by opening Door3; all the gain comes to Door2. This result has been accepted as fact by the math community since the 18th century, so I doubt you'll change anybody's mind on this

-They're on crack. #1 and #2 are arbitrary designnations. You could switch them around all you want, the odds for both are the same.

[Solly]

Nah, Victor, you are wrong.

[Chowlett]

Ok, the following situations are all equally likely:

Door 1	Door 2	Door 3
Car	Goat	Goat
Goat	Car	Goat
Goat	Goat	Car

As you pointed out, the numbering is arbitrary, so assume the contestant always picks door 1. The host, who knows where the goats are, opens either #2 or #3 to reveal a goat.

Now, in the first row, door one is right, and if you swap you'll lose
In the second row, door one is wrong. The host must reveal the goat in door 3, and if you swap you'll win
In the third row, door one is wrong. The host must reveal the goat in door 2, and if you swap you'll win

So in 2 out of 3 cases, if you swap you'll win.

[KrazyHorse]

Jesus Christ. The relevant theorem here is "Bayes' Theorem". The problem in question is the "Monty Hall" problem.

The problem, solved by the math department of Imperial College in London.

http://stats.ma.ic.ac.uk/~atw/Bayes.pdf

[Provost Harrison]

Re: A woman has two children, at least one of which is a boy...

Originally posted by Boddington's
The probability that the other child also is a boy is thus 1/3.

Is this correct or incorrect?

I'm with loinburger on this one myself. At least one child is a boy. You've already stated, indirectly, that one of the children is male. Therefore the gender of the other child is totally independent to the statement, and is thus 50% chance of either (assuming equal ratio of male:female)

[KrazyHorse]

Re: Re: A woman has two children, at least one of which is a boy...

Originally posted by Provost Harrison


I'm with loinburger on this one myself. At least one child is a boy. You've already stated, indirectly, that one of the children is male. Therefore the gender of the other child is totally independent to the statement, and is thus 50% chance of either (assuming equal ratio of male:female)

For the last frigging time:

1st: The woman has two children. Therefore there are 4 distinct possibilities

1) eldest: boy, youngest: boy
2) eldest: boy, youngest: girl
3) eldest: girl, youngest: boy
4) eldest: girl, youngest: girl

2nd: At least one of the children is a boy. Therefore possibility 4 is eliminated. There are 3 distinct possibilities, each of equal magnitude. 2 of these involve her having one boy and one girl. One involves her having 2 boys.

Note that this is a different problem than: "A woman has 2 children. Her oldest child was a boy. What is the probability that the other child is also a boy?"

In the original problem you can't claim that the gender of the "other" child is an independent piece of information, since you have been told something about the set of two children, not about any individual child.

[Dauphin]

Imperial College in London.

[KrazyHorse]

Your Alma Mater?

[devilmunchkin]

the fact that the woman has another boy has no bearing on whether or not her other child will be a boy unless they are fraternal twins.
otherwise, the chances of the other child being a boy are the ratio of X chromosome sperm to Y chromosome
other staticians will argue that it's got a 50/50 chance of being a boy..either you are or you're not.

[KrazyHorse]

Would all non-mathematicians please vacate the premises?

Thank you.

Your error is born of ignorance of probability theory, not of genetics, dm.

[Jon Miller]

for what iti s worth I agree with Krazy

take a probability course or study on your own if you think otherwise

Jon Miller