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  #31  
Old 06-10-2007, 12:47 AM
PairTheBoard PairTheBoard is offline
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Default Re: A Rejection of Sklansky

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Not thinking past your Sklanskyish assumptions.


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I probably think that way but my mistake here was I misread the problem - I always did good in math calculatons, etc., but lost points when we did word problems. I thought the couple had one child, a girl, and the question was the probability the second would be a girl. Am I right that that would be 50%? In other words, the order matters, which would also answer my question about the poker situation?

I've only thought about this briefly but it reminds me of the Monty Hall Problem. As there it seems the new information changes our knowledge of the probablity - I can see this conceptually but it makes me dizzy. Very non-intuitive.

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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?


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PairTheBoard
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  #32  
Old 06-10-2007, 01:01 AM
PairTheBoard PairTheBoard is offline
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Default Re: A Rejection of Sklansky

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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.



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.

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Actually, the more I think about it, the phrase "one of the children is a girl" might be a little fuzzy. We've been told the Family has two children. Now we are told "one of the children is a girl". Does that mean "exactly one of the children is a girl" or does it mean "at least one of the children is a girl". Logically, I believe the information it conveys is limited to "at least one". But I'm afraid common use would take it to mean "exactly one". So I can see somebody arguing it means "exactly one". But if you argue that then when asked what the probability is that the other child is a boy you should say 1, not 1/2.

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  #33  
Old 06-10-2007, 01:42 AM
PairTheBoard PairTheBoard is offline
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Default Re: A Rejection of Sklansky

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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.

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All your statements are correct except for the last one in bold. It's almost right though. It's probably better to say the 1000 child family has "at least" 999 girls. That probably would also have been a better way for me to state the 2 child problem. Notice in the two child case where at least one is a girl, there are two ways it can have a boy. As the first or second child. Only one way it can have two girls. So chance of two girls 1/3. Now in the 1000 child family with at least 999 girls there are 1000 ways it can have a boy. As the first, second,...,1000th child. And again 1 way it can have 1000 girls. So chance of 1000 girls 1/1001. The reason this works is because in this case, all these "ways" I'm counting really are equally likely.

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  #34  
Old 06-10-2007, 03:56 AM
ALawPoker ALawPoker is offline
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Default Re: A Rejection of 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.

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But David, you're a pretty sexy man. I think that alone pushes it back close to 50/50. El D just doesn't see your beauty.
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  #35  
Old 06-10-2007, 04:28 AM
PairTheBoard PairTheBoard is offline
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Default Re: A Rejection of Sklansky

5. Speaking of beauty, here is another example of misapplying the Sklanksy Principle. It is the Sleeping Beauty Paradox. A fair coin is flipped. If it comes up heads, Beauty goes to sleep on Monday and is awoken later that evening. She then goes back to sleep for a week. If the coin is tails Beauty also goes to sleep on Monday and is awakened and then goes back to sleep. But after she has gone back to sleep a magic wand is waved that gives her Amnesia. She will not remember her Monday Awakening. She then sleeps until Tueday evening when she is awakened once again, than goes back to sleep for the rest of the week.

Beauty is perfectly rational. She is told how all this is going to work. Now she has an awakening. She doesn't know if it's monday or tuesday. What is her Credence or Belief about how the coin landed? The Paradox argument claims she now believes there is a 1/3 chance the coin landed on heads. Why does she have that belief?

The argument goes that she knows she can have three kinds of awakenings. One due to heads. And two due to tails. She applies Sklanskyish thinking to this to conclude that since she knows nothing about which of these kinds of awakenings she might be having, she concludes they are equally likely. After all, she should experience on average 1.5 awakenings. .5 due to heads .5 due to tails and another .5 due to tails. They are equally likely so there must be 1/3 chance the coin landed on heads.

Here is a link to my treatment for resolving the parodox.
Resolution of the Sleeping Beauty Paradox

PairTheBoard
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  #36  
Old 06-10-2007, 09:25 AM
Piers Piers is offline
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Default Re: A Rejection of Sklansky

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The problem is that people in general, after being told what the two choices are, have a false intuition that the two Choices are equally likely.

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It might be false intuition in some technical sense, but nether the less I believe it is very useful.

Most decisions people have to make tend to between close to equally likely events. This is because of our inbuilt polarizing mechanism that tends to discard unlikely events as imposable or unbelievable.

So when faced with four options (A=5%, B=10%, C=40% and D=55%) our automatic processing simplifies to (A=0%, B=0%, C=50%, D=50%). Making our decision making much easier.

Your Sklannsky principle might sometimes be irritating for people who have an understanding of probability, but for virtually all the life forms on this plant that can grasp the Sklansky principle at some intuitive level, it’s a huge aid to decision making.

The Sklansky principle as applied by ones opponents is also a key mechanism for making money at poker.
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