06-02-2002 15:20									#31
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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...

	06-02-2002 17:13									#32
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quote: 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.

	06-02-2002 19:23									#33
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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.

	06-02-2002 21:37									#34
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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).

	06-02-2002 23:39									#35
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"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.

quote: 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.

	06-02-2002 23:44									#36
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Nah, Victor, you are wrong.

	06-02-2002 23:54									#37
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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.

	07-02-2002 00:48									#38
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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

	07-02-2002 00:53									#39
	Re: A woman has two children, at least one of which is a boy... Support Apolyton

quote: 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)

	07-02-2002 01:06									#40
	Re: Re: A woman has two children, at least one of which is a boy... Support Apolyton, pre-order Civilization IV

quote: 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.

	07-02-2002 01:10									#41
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quote: Imperial College in London.

	07-02-2002 01:36									#42
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Your Alma Mater?

	07-02-2002 02:03									#43
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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.

	07-02-2002 02:14									#44
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Would all non-mathematicians please vacate the premises?

Thank you.

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

	07-02-2002 03:50									#45
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for what iti s worth I agree with Krazy

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

Jon Miller

	07-02-2002 05:03									#46
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To me, it looks completely different...I understand where you are coming from but surely you can reach it like this as well...

quote: A woman has two children, at least one of which is a boy... The probability that the other child also is a boy is thus 1/3. Is this correct or incorrect?

A dissection: We know that one of the children at least is a boy, yes?

Look at the middle statement and the language, stating the probability that the other child also is a boy which indicates that the first child is a boy. Therefore the first criterion that at least one child is a boy is fulfilled. Therefore the gender of the second child is undetermined and unrelated to the gender of the first child. Therefore the gender of the second child is independent of the first child. That is the way I reason it.

	07-02-2002 05:09									#47
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quote: Originally posted by Provost Harrison To me, it looks completely different...I understand where you are coming from but surely you can reach it like this as well... A dissection: We know that one of the children at least is a boy, yes? Look at the middle statement and the language, stating the probability that the other child also is a boy which indicates that the first child is a boy. Therefore the first criterion that at least one child is a boy is fulfilled. Therefore the gender of the second child is undetermined and unrelated to the gender of the first child. Therefore the gender of the second child is independent of the first child. That is the way I reason it.

It says other child

not second child

there for the other is refering to at least one child is aboy

Krazy is still right

Jon Miller

	07-02-2002 05:36									#48
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To those who still don't believe Krazy and Jon, I propose a wager. We'll survey 100 women who have exactly two children, including at least one boy. (Shouldn't be hard to set that up in an online chat room or something.) If at least 51 of them have a girl in addition to the boy, you pay me $1000. If not--if at least 50 of them have two boys--I'll pay you $1500. If Harrison is right and I'm wrong, I'm offering you 3/2 odds on a 50-50 bet. Any takers?

	07-02-2002 05:40									#49
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Let's make that monopoly money, and you can do the donkey work

	07-02-2002 11:28									#50
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krazyhorse: why don't you educate me rather than just telling me i'm stupid.

i still don't see how i could be wrong since the father is someone who determine's the sex of the child. Fraternal twins will be of the same sex because they are fertilized by one sperm..so that would help if that info was given.
otherwise, it really does depend of the amoung of sperm that man has in that one ejaculation containing sperm with X chromosones and those with Y chromosomes. THe number will vary dending on how many times in a single..umm...tryst that he's ejaculated into her (it changes the number of sperm inside her) and also it depends on how many are able to get part her natural sperm barriers (sperm enter into an acidic environment which most don't get past). You have to go further in the comlpexity of the numbers because the fertility of each man is different...he might be shooting some weak ones or he might be a loaded elephant gun.
it all depends....and no...you're right..i can't attach numbers to it.

	07-02-2002 13:38									#51
	Re: A woman has two children, at least one of which is a boy... Put an end to popups!

quote: Originally posted by Boddington's The probability that the other child also is a boy is thus 1/3. Is this correct or incorrect?

This is absolutely correct:

If a woman has 2 children the possibilities are

[B,G],[G,B][B,B][G,G]
(always order from young to old=> that's why the first two are different)
==>If there is at least one boy there are 3 possibiliyies left
[B,G],[G,B],[B,B]
-->Only 1 in 3 has 2 boys so chance is 1/3

Shade

	07-02-2002 13:44									#52
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quote: Originally posted by devilmunchkin krazyhorse: why don't you educate me rather than just telling me i'm stupid. i still don't see how i could be wrong since the father is someone who determine's the sex of the child. Fraternal twins will be of the same sex because they are fertilized by one sperm..so that would help if that info was given. otherwise, it really does depend of the amoung of sperm that man has in that one ejaculation containing sperm with X chromosones and those with Y chromosomes. THe number will vary dending on how many times in a single..umm...tryst that he's ejaculated into her (it changes the number of sperm inside her) and also it depends on how many are able to get part her natural sperm barriers (sperm enter into an acidic environment which most don't get past). You have to go further in the comlpexity of the numbers because the fertility of each man is different...he might be shooting some weak ones or he might be a loaded elephant gun. it all depends....and no...you're right..i can't attach numbers to it.

It is not biology, it is probability

if the biology is messing you up, take it out

instead imagine two coin flips

now if at least one is heads, the probability of the other being heads is 1/3

since the possibilities are
1 h, h
2 h, t
3 t, h
4 t, t

notice that number 4 has no heads result, so we can throw it out

that leaves 1, 2, and 3

of the three remaining, (that have at least 1 heads) how many have another heads

since that is one the probability in the question is 1/3

now replace two coin flips with two children
and heads with boy

better?

Jon Miller

	07-02-2002 13:55									#53
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To add on (for dm)...

Say the probability of having a boy is 50%. The first child is a boy. Therefore the odds of the 2nd child being a boy should be less because it eventually should end up at approximatly 50% .

	07-02-2002 14:19									#54
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no

that is the poiny

a child is a boy

not the first one

Jon Miller

	07-02-2002 17:19									#55
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And Imran's assumption is false anyway. It would imply that a coin that has been tossed 10 times and come up heads each time (roughly 1 in every 1024 trials) would be very much more likely to come up tails next time than a freshly minted one - which is ludicrous. In both cases, the second "happening" is independent of the first "happening".

	07-02-2002 17:34									#56
	Re: Re: Re: A woman has two children, at least one of which is a boy... Support Apolyton, buy Galactic Civilizations

quote: Originally posted by KrazyHorse 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.

That's wrong because you are distinguishing case 2 and 3.

For the question, these two cases are exactly the same. It doesn't matter if the oldest or the youngest is the boy.

There are 3 cases:

1. 2 boys
2. 1 boy and 1 girl
3. 2 girls

Having a boy eliminates situation 3, so the probability of the other child is also a boy is 50%.

	07-02-2002 19:08									#57
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UR there is no degeneracy. Both of the "One boy, one girl" cases have to be counted seperately.

	07-02-2002 19:18									#58
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quote: Originally posted by Sagacious Dolphin UR there is no degeneracy. Both of the "One boy, one girl" cases have to be counted seperately.

Since they are equally likely, and exactly as likely as boy-boy or girl-girl. If you poll a set of families with 2 children, you will find twice as many with boy-girl as you will with boy-boy.

	07-02-2002 19:20									#59
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This is amusing.

The difference is when you make the prediction. Before or after the births. Before the births it is approx 50% on any given birth. After two births, it's 1/3. As nicely shown by many posters.

Same thing with coin flips. If after 7 heads, whats the odds of the next flip being heads. 50%. But if you ask what the odds of 8 heads before any are flipped, then considerably higher

RAH
So much for making it clearer.

	07-02-2002 19:27									#60
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Is it just me or is all this bloody obvious? (Clearly not - since there is so much debate....)