A Rejection of Sklansky

[PairTheBoard]

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The answer is 1/3. Once we know that one child in the family is a girl, the chance the family has two girls is 1/3 or about 33.3%

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Are you saying that after the first girl is born in anticipation of the second child, you will give me 2/1 odds on my choice of the second child as a girl?


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No. Notice I said, The family has two children. One of them is a girl. I did not say the First Born child is a girl which is what you are saying when you assume the condition, "after the first girl is born in anticipation of the second child".

If all we know is that One Child is a girl the family can be one of three things, (g,b),(b,g),(g,g). Those three things Are equally likely. But only the (g,g) satisfies "the other child is a girl too".

So the Condition I gave you is that "one of the children is a girl". That is what must be conditioned on to answer the question posed in the problem.

You are now presenting a different Condition. "after the first girl is born in anticipation of the second child". That Condition cuts the possible outcomes down to two rather than the three of the problem's Condition. Your Condition cuts the outcomes down to (g,b),(g,g). You are right to say that Given Your Condition the probability the other child (which your condition defines to be the second child) is a girl, is 50%. But your Condition is different than the Problem's Condition. No fair complaining that they give different answers.

I agree there is something psychologically disturbing going on here. I've had years of training in mathematical probability, yet the first time I saw this problem I too experienced an unsettling feeling that something didn't seem right. I think it has something to do with our strong sense that girls happen with 50% probability.

Well, they do and they don't. Girls don't "happen" with 50% probability if you define Events in such a way that they don't, as in this problem. You can define lots of Events where girls don't happen with 50% probability. For example, go into the Bellagio and pick a random poker player. What is the probability your random pick will be a girl?

There's also some kind of psychology going on with the logical difference between the statements, "One child is a girl" and "The first child is a girl". I think we psychologically identify those two statements as being equivalent because for many purposes they serve about equally well for making inferences. In this case they carry very different implications relating to the question the problem asks us to answer.

Edit: The Psychology may also relate to our use of Symmetry when making inferences. And also our frequent ineptness when intuitively breaking things down into cases. We hear, "One child is a girl". We think, well if the first child is a girl it's 50-50 the other one is. And if the second child is a girl it's 50-50 the other one is. That covers all cases and by symmetry 50-50 must hold even if all we are told is that "one child is a girl". This is the kind of thinking that you will learn to unlearn when you study and practice more mathematics.

PairTheBoard

[carlo]

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Case in point.
Assume the chance a random child in random family is a boy is 50% for our population. Now,

A family has two children. There are 3 Events. E1=both boys, E2=both girls, E3=one of each. We are told that one of the children in this family is a girl. What is the probablity the other one is also a girl?

Students think it should obviously be 50%. But then they figure they should look at the three Events that were given. Maybe the three events apply and they should use the conditional probability they've been learning. So they think Sklansky to themselves. They see 3 equally likely events. One of them has no girls in it so they can focus on the remaining two that do. They see that E2 and E3 are therefore Equally likely. One is the two girl event and one is the boy-girl event. So even applying conditional probability on the three events they conclude the probability it is a two girl family given that one child in the family is a girl must be 50%. This agrees with their previous intuition and they are done thinking. Sklansky has led them astray again.

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Your reasoning only applies if TIME is eliminated. If one assumes that the children were born at the same time, out of the fount of whatever, then the probabilities hold. Your analysis also implies that Boy1 possibility is the same as Boy2 and likewise Girl1 equals Girl2 possibility. But if they were not the same then Boy1/Girl2 is different than Boy2/Girl1.There are not 3 events but 4. Therefore the combining of Boy/Girl possibilities is not allowed.

In the problem as stated, TIME is not eliminated and the practical and theoretical answers equal=.5. The idea of 11 flushes in a row as noted previously states that the probability of a flush on the next hand is indeed the same as before the first flush.

I really do get the sense that mathematicians who write word problems should get a sense for what words mean and not approach words as a perfunctory route to their reality. It seems that the abstruse level of these word problems are more related to misunderstandings of the spoken /written word than the mathematics involved.

This is a semi-rant of mine for off and on through these posts I see the Kantian parallel in which one states that the probability stuff is "in my mind"(and it is) and the other stating that what he sees does not relate to THE reality yet attempts to use this very "in my mind stuff" to prove the other wrong.

A better question concerning probability is "what is its justification in the perceptual world?".Taken as a method where does it fit and where are its sources? One can see geometry as a product of the human mind and one would not dare argue with the truths of the triangle,etc. Does probability contain this same certainty?

[carlo]

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No. Notice I said, The family has two children. One of them is a girl. I did not say the First Born child is a girl which is what you are saying when you assume the condition, "after the first girl is born in anticipation of the second child".

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Tricky [img]/images/graemlins/smile.gif[/img],so you are working backwards in time to the possibility at the beginning of conceptions. But the thing is that in your answer you have ASSUMED that the child is the second child and not the first. It can only be one or the other. In that case you still have the original premise that the earlier child was either male or female. Still 50/50. Because the female is less likely according to the original "out of time" question a reversal of time does not change the original probability of male or female only the probability of the combination which is moot.

[David Sklansky]

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PTB writes,

"Let's see. There are two alternatives. Either you are getting lucky with Brandi or you are not. Therefore, The probability David is getting lucky with Brandi is 50%."

I really need Persi Diaconis to chime in on this.

DY

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El Diablo already pegged it at 25%. Thus it is no longer a question regarding two alternatives without information.

[PairTheBoard]

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Your reasoning only applies if TIME is eliminated. If one assumes that the children were born at the same time, out of the fount of whatever, then the probabilities hold.

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I really do get the sense that mathematicians who write word problems should get a sense for what words mean and not approach words as a perfunctory route to their reality.

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No carlo. The phrase "one of the two children is a girl" is simple and the words are clear. The information the phrase carries is clear. The information your words carry is not clear.

PairTheBoard

[carlo]

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Quote:
I really do get the sense that mathematicians who write word problems should get a sense for what words mean and not approach words as a perfunctory route to their reality.



No carlo. The phrase "one of the two children is a girl" is simple and the words are clear. The information the phrase carries is clear. The information your words carry is not clear.

PairTheBoard

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Wasn't trying to be personal. I shouldn't have generalized about mathematical word problems or mathematicians who I greatly admire. Carry on.

[NotReady]

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I agree there is something psychologically disturbing going on here.


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3 situations

Couple has a girl and is expecting a child: 50% it's a girl?

Couple has 2 kids and says the 1st was a girl: 50% 2nd is a girl?

Couple has 2 kids and says 1 is a girl, but doesn't give order: 33% other is a girl?

3 more situations

Couple has 999 girls and is expecting a child: 50% it's a girl?

Couple has 1000 kids and says the first 999 were girls: 50% #1000 is a girl?

Couple has 1000 kids and says 999 are girls but doesn't give order: 1/1000 the last is a girl?

If the above is true then it would appear what matters is your knowledge of the order. I can accept this but I can't see why it's true. Gotta stop, getting dizzy again.

[PLOlover]

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Let's see. There are two alternatives. Either you are getting lucky with Brandi or you are not. Therefore, The probability David is getting lucky with Brandi is 50%. Maybe I should have posted this in NVG.

PairTheBoard

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He's dutch booking her.
He's taking both sides.
He's obviously taken pokher to a whole new level.

[TomCowley]

They have 2 kids in order. They either had boy-boy, boy-girl, girl-boy, girl-girl (with equal probability since the question specified as much).

Now we know 1 child is a girl, which rules out boy-boy, leaving girl-boy, boy-girl, or girl-girl, all with equal probability.

So out of these three possibilities, only girl-girl satisfies the question, while girl-boy and boy-girl do not (the other child is a boy), so one configuration works and two do not, and the probability is 1/3.

[PairTheBoard]

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PTB writes,

"Let's see. There are two alternatives. Either you are getting lucky with Brandi or you are not. Therefore, The probability David is getting lucky with Brandi is 50%."

I really need Persi Diaconis to chime in on this.

DY

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El Diablo already pegged it at 25%. Thus it is no longer a question regarding two alternatives without information.

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I will have to adjust my Prior to 33%

PairTheBoard