A Rejection of Sklansky

[PairTheBoard]

David made the following statement on his "What Question?" Thread. That thread is about the meaning of his Probability Statements. If you don't know what he means by Probability you should read that thread. His is not a Frequentist Definition. This Thread is about this particular statement of his. I believe it needs to be Greatly Discounted.

Link to "So What's This Problem I Won't Answer?" (http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=10678680&an=0&page=0#Pos t10678680)
[ QUOTE ]

Sklansky -
If the only information you have is the number of choices, your personal probability should be equally divided among them. Perhaps because you have data that shows that when the only information is the number of choices the choices have come up equally. But in this particular case even if you don't have the data it is silly to say that you are falling back on subjective judgement. Rather you are falling back on not being a moron.


[/ QUOTE ]

I'm not Rejecting his statement. Rather, I am Rejecting his contention that people are not aware of this principle and he must preach it to them. That is exactly NOT the problem. Not only are people well aware of this principle. They are Too aware of it. In fact they are so aware of it that they apply it habitually without even thinking. The way they apply it in practice amounts to the very illogical principle,

"Whenever there are two Choices they must be Equally likely."

This causes people much much more problems in their thinking than not being aware of Sklansky's principle. And after all, Sklansky's principle in its pure form is totally Theoretical. You never have two Choices where the only thing you know about them is that there are two of them. If that's all you knew, you wouldn't be able to distingish which was which. That never actually happens. Only in a theoretical state of imagination can we imagine two choices without being told what they are. Once we are told what they are we know something More than that they are just two Choices.

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. If they have any difficulties seeing any implications for their relative liklihood from the description of the choices they automatically assume they are equally likely. They need to break that habit. They hardly need Sklansky telling them to apply it more freely.

Some examples from my exerience on these Forums and with students I've taught.

1. The Monty Hall Problem. There is a prize behind one of three doors A,B,C. After you pick a door, Monty Always opens one of the remaining doors and shows that it is empty. You know this. He then gives you the option of switching to the other unopened door. What is the probability the other unopened door contains the prize?

Many Many Many people automatically say 50%. Why? Because they are Not aware of Sklansky's Principle? No. They are Too aware of his principle. They are so aware of it they assume it applies here because they can't see any reason why it shouldn't apply. They are so convinced of this, that even after they are shown the logic of Conditional Probability (Bayes) as it applies to this problem, they remain convinced Sklansky's principle applies and they stick to their 50% assertion. Their conviction about using Sklansky's principle is so strong it Prevents us from teaching them to use Bayes' Theorem. They hardly need to be preached to by Sklansky that they are not using his principle enough.

2. The Two Envelope Problem. There are Two Envelopes. Dollar Amounts have been chosen in some way we know nothing about whereby one Envelope contains twice as much money as the other. You pick one at random, open it and see $100. You can either keep the $100 or switch to the other envelope. What is the probability the other envelope contains $200 and what is your EV for switching?

A Huge number of people will without blinking apply Sklanky's Principle immediately and say the probability is 50%. Even after consderabale thought they will say 50%. Even after they have been shown the logical contradictions from the resulting EV calculations they will still say 50%. Even after extensive Baysian analysis is shown to them they still think it should be 50%. Why? Because they are not aware of Sklansky's Principle? Sklansky's Principle is so engrained in their psyches they can't prevent themselves from applying it despite the abundance of evidence showing them they shouldn't.

3. Almost any situation where a Baysian analysis would be beneficial. Here Sklansky is trying to get people to take his Baysian approach to probability while at the same time preaching a Principle which only holds for sure in an imaginary Theoretical Setting and which is exactly the thing that Prevents people from taking advantage of his Baysian approach to Probability.

Case in point:

Disease testing. A TB test has 90% accuracy both for Positive and Negative results. Those who have TB test positive 90% of the time. Those who don't have it test Negative 90% of the time.

A teaching applicant must be tested. So he had the test done and it is Positive. He concludes he almost certainly has TB. Why? Because intuitively he is applying Sklansky's Principle. He figures he has two choices. Either he has TB or he doen't. So he applies Sklansky when looking at the TB results. He might just say, he has half a 10% chance of having gotten a False Positive so 5% chance he doesn't have TB. Or he could apply Bayes' Theorem,

(.1)(.5)/[(.1)(.5)+(.9)(.5)] = 5%

to see a 5% chance he doesn't have TB.

Sklanky's Principle is so engrained in people's intuition that it affects their judgement on many many things.

The Extension of Sklansky's Principle to multiple disjoint events where nothing is known except the number of the events.
==================

4. Elementary Probability Lessons
When teaching classes in Elementary Probability it is very easy for students to grasp models where there are atomic outcomes, all of which are equally likely. But when you pass to situations where the atomic outcomes are in the background and it's not clear where the atomic outcomes are in the Events being desribed by the model, the students begin to have difficulties. The reason is that they want to continue applying Sklansky's Principle to the Model of Events which contain hidden atomic outcomes. If they see there are N disjoint Events of this type which they don't have a feel for, they want to think of them as equally likely.

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.



For anything remotely tricky like this where it's not easy for people to see why events wouldn't be equally likely they will quite stubornly hold on to the Sklanksy Principle telling them they are equally likely. Even after being shown why the Events are not equally likely. Sklansky's Principle is hardly something people need to be made aware of. What they need to be aware of is the fact that in Reality it is skewing their judgement on many many things.

PairTheBoard

[TomCowley]

Saying that the situation is "only theoretical" is your standard nihilist irrelevant nonsense. "An event will either happen or it won't. What is the probability that it will happen?" You can draw absolutely no bias from the description, any more than you could if you were told there were two outcomes, frumplesnort and dingleberry, described in a language incompehensible to you, and you were asked for a probability. ANSWER the question as it DOES exist, don't say it can't exist.

Monty Hall is a MISapplication of sklansky's principle. If you couldn't assumethat all 3 doors were equally likely to begin with (skalnsky's principle), then you couldn't solve the problem at all.

Two Envelopes is not a mathematically valid problem. When converted into a mathematically valid problem using limits, guess what the odds asymptotically approach? That's right, 50%.

The third and fourth cases are simple math ignorance about how to calculate conditional probability (if shown this problem from a frequency standpoint, even relative idiots can understand the process, although they'll be shaking their head at the answer). It has nothing to do with Sklansky's principle, and is certainly no sort of counterexample to it. If you list the FOUR indexed outcomes, boy-boy, boy-girl, girl-boy, girl-girl, even idiots should get this right.

Some people are godawful at math. DS has nothing to do with this. Please tell me where you teach, so if I ever have kids, I can make sure they don't end up in one of your classes. The last thing I'd want is somebody who can't analyze trivial situations and instead has to resort to "we just have to know something about the outcomes by what they are" gobbledooygook gibberish.

[PLOlover]

[ QUOTE ]
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. If they have any difficulties seeing any implications for their relative liklihood from the description of the choices they automatically assume they are equally likely. They need to break that habit. They hardly need Sklansky telling them to apply it more freely.

[/ QUOTE ]

so you're saying the bent coin problem is a specific case of an erroneous general principle, but that in this particular case it happens to (accidentally) be correct?

Or are you saying (more correctly in my opinion), that the bent coin problem appears on the surface to be an application of an erroneous general principle, but upon further analysis has actually nothing to do with the e.g.p., but still you want to warn people about the dangers of the e.g.p.

[PLOlover]

just to clarify, the sklansky bent coin example specifically states that the two choices (heads, tails) are not equally likely. I mean, it's a bent coin. The bent coin example works no matter what the degree of bentness is, or no matter what the ratio of heads/tails is when you flip bent coin.

Is that in dispute?

[NotReady]

[ QUOTE ]

If you list the FOUR indexed outcomes, boy-boy, boy-girl, girl-boy, girl-girl, even idiots should get this right.


[/ QUOTE ]

OK, I'm an idiot, but this is driving me nuts if the chance isn't 50%. The set-up stated 50% probability of a boy, which means 50% probability of a girl. Unless the fact the first child was girl indicates a bias towards girls, I don't see how it can't be 50%. But then, I missed the Monty Hall Problem the first time, though I do get the explanation.

If it isn't 50% how is that different from say a poker situation. I miss 10 flush draws in a row. Isn't my chance of making the 11th flush try the same as the first?

If someone answers this and I'm wrong, please explain the difference in the two situations.

[David Sklansky]

Surely you realize that I'm not even reading the posts on this subject. So I would appreciate it if you don't label them in such a way that I think it is a post about Brandi which I AM reading.

[PLOlover]

[ QUOTE ]
Surely you realize that I'm not even reading the posts on this subject. So I would appreciate it if you don't label them in such a way that I think it is a post about Brandi which I AM reading.

[/ QUOTE ]

now thats funny

[yukoncpa]

Hi PTB,

It seems like there are now three different threads regarding your discussion, and I didn’t follow along on the first two and they got huge, so pardon me if this has been answered.

I can figure out the Monty Hall problem, but I can’t figure out the envelope problem. Before I look in the envelope, the chance of picking the one of two envelopes that has the most money, must be .5. Now it looks like, once I peak into the envelope, my peeking somehow changes the probability of the distribution of money in envelope number two into something other than 50% $50 and 50% $200, because, I don’t accept that I’m going to make money by doing a switch. If that were the case, I could make money from nothing. In other words, suppose you and I both chose an envelope. Now we both come to the conclusion that switching is correct, so we switch with each other, and we both make money. This would be nice, I’d be happy to switch envelopes with you all day long. But I don’t see this as working any more than I see cockamamie crap systems as working.

Can you explain how before the fact, there is a 50% chance of choosing the envelope with the most money. But after peeking, evidently there can not be a 50% chance that the remaining envelope either contains half of your amount or twice your amount, else you would create black magic. What’s the answer? If it’s already been answered on one of the two other threads, just tell me so and I’ll search some more. Somehow I never found the answer and decided a lengthy letter would get me a faster response.

Thanks (also, I’m being a bit humorous, I’m sure, peeking has nothing to do with it, but rather some sort of mathematical trick that I’m unaware of )

P.S. I see TomCowley has said that the envelope problem is not mathematically valid. I accept this using common sense, but where is the invalidity?

[TomCowley]

The invalidity is that there is no such thing as a uniform distribution over an infinite number of items. If you structure the problem in a way that probability theory can handle, taking limits as the range goes to infinity (as I did for a variant in the Probability forum- check it out), there's always a chance you're opening the "biggest" envelope, and switching it gets very very bad (enough to make switching 0 EV overall, despite the gains from every other switch). The odds of not gaining by switching approach 0, the odds of the other envelope being double approach 50%, but the EV stays 0. I confused myself for awhile working through it until I realized what I'd done wrong.

[Siegmund]

I agree with PTB's contention that this "no information, assume equally likely" principle is overused, not underused. Come to think of it, not sure I've ever met anyone who DIDNT subscribe to that notion in some form.

Not sure how that is a rejection of Sklansky's point - just an observation that "no information" is a situation that's a lot less common than a lot of people believe.

Bit disturbing that he doesn't read the threads he starts, though.

[PairTheBoard]

This thread has nothing directly to do with the Bent Coin. Click on the Link I gave to the Thread where the quote came from. It's in Sklanksy's OP there which says absolutely nothing about the Bent Coin. His OP is a general description of what he means by probability. At the end of his OP he makes the statement I quoted as a General Principle which he wants more people to become aware of. This thread intends to treat it as that General Principle.

As such, you may apply it to the Bent Coin example if you want to. But I don't want this thread sidetracked by another debate about the Bent Coin. There are already a gazillion threads about that. If you want to post on one of them with a link to this one go ahead. Howver, I'm not debating with you about the Bent Coin here.

PairTheBoard

[Max Raker]

I think I agree with PTB. Given two options people are too inclined to assign them as 50 50. Their is even the old poker joke, I am on a flush draw, either I hit or I don't, it must be 50 50! The reasons why all possibilities are not equally likely often has to do with calculating degenercies (atleast in physics) and this is somewhat abstract math wise. (or it requires clever grouping of the terms in a sum) If I missed the whole point, PTB can correct me.

[PairTheBoard]

[ QUOTE ]
OK, I'm an idiot, but this is driving me nuts if the chance isn't 50%. The set-up stated 50% probability of a boy, which means 50% probability of a girl. Unless the fact the first child was girl indicates a bias towards girls, I don't see how it can't be 50%. But then, I missed the Monty Hall Problem the first time, though I do get the explanation.


[/ QUOTE ]

I pointed out the three Events,

E1 = Two boys
E2 = One Boy One Girl
E3 = Two Girls

People assume these 3 events are equally likely. They are not. You have to think past your assumption to see that.

Notice I said that for the Family with two children, One of the children is a girl. I did not say their Oldest Child is a girl. That would have made a big difference. So let's depict possible two child families with ordered pairs,

(k1,k2) where k1 is the oldest first born kid, and k2 the youngest.

Now look how two child families come to be. First k1 is born with 50-50 chance of being a boy or girl. Then k2 is independently born with the same 50-50 chance of being a boy or a girl. So,

P(k1=boy, k2=boy) = (.5)(.5) = 25%
P(k1=boy, k2=girl) = (.5)(.5) = 25%
P(k1=girl, k2=boy) = (.5)(.5) = 25%
P(k1=girl, k2=girl) = (.5)(.5) = 25%

So the event E1 = both boys = {(boy,boy)}
and P(E1) = 25%

E2 = one boy one girl = {(boy,girl),(girl,boy)}
and P(E2) = P(boy,girl)+P(girl,boy)= 50%

Similiarly, P(E3)=P(girl,girl)=25%

So your assumption that E1,E2,E3 were equally likely was not true. You should have thought past your intuitive Sklankyish assumption and looked closer at what the descriptions of E1,E2,E3 implied. As you can see now, saying that one of the children in the family is a girl just means that the family is either of the type E2 or the type E3. That's all it tells us. The family could be either (boy,girl),(girl,boy),or (girl,girl) but it's definitely Not (boy,boy).

So it's E2 or E3 but not E1. We are conditioning on Not-E1. You should have no trouble just "seeing" that once we know it's not E1, the chance it is E3 is the relative probability of E3 to whats left of the probabilities after we've taken away E1's 25%.

It's the proportion of times we generally see E3 in those cases where we don't see (boy,boy). ie. In those cases where we don't see (boy,boy) we see E2 twice as often as E3.

So the proportion is 25/75 = 1/3

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%

Or you can continue being Sklanskyish in your judgement on these kinds of things and continue to think the answer incorrectly is 1/2. How many other things like this do you base your Sklansyish Judgement on? Not thinking past your Sklanskyish assumptions.

(speaking generally. not about you personally NR)

PairTheBoard

[PairTheBoard]

[ QUOTE ]
Surely you realize that I'm not even reading the posts on this subject. So I would appreciate it if you don't label them in such a way that I think it is a post about Brandi which I AM reading.

[/ QUOTE ]

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

[PairTheBoard]

[ QUOTE ]
I can’t figure out the envelope problem. Before I look in the envelope, the chance of picking the one of two envelopes that has the most money, must be .5. Now it looks like, once I peak into the envelope, my peeking somehow changes the probability of the distribution of money in envelope number two into something other than 50% $50 and 50% $200

[/ QUOTE ]

A lot of people don't think that far into it. They just look at two alternatives, $200 and $50. They know the envelope amounts were chosen by some method for which we have no information. So with Sklanskyish thinking they conclude $200 and $50 are equally likely. What they don't realize is that assumption implies they are thinking that the unknown method of picking envelope amounts makes all envelope amounts equally likely. This is similiar to a Baysian assumption of a Uniform Prior.

But it is impossible for an envelope amounts method to make all envelope amounts equally likely. That would require a Uniform probability distribution for all envelope amounts, ie. all positive real numbers. That is impossible. Whatever Prior disribution might have been used as a method to pick envelope amounts it must satisfy the condition that at some point large envelope amounts become less and less likely. So it is not logical for them to assume $200 and $50 are equally likely.

Also, seeing the $100 Conditions the prior distribution. You are not really looking at P($200) for the prior distibution. You are looking at P($200|sees $100), a conditional probablity on the Prior. There's no reason to think that equates to P($50|sees $100). Not when we know the Prior has got to make Extra Large Amounts less likely. We don't know if $100 is Extra Large for the unknown Prior. But it certainly might be. That should make us suspect the $200 may very well be less likely than the $50.

There happens to be a method by which you can improve your decision to switch, sometimes switching sometimes not. It's a method that requires you to pick a number p and switch if the amount you see in the envelope is less than p, otherwise don't switch. The thing is, you have to pick your p before opening the envelope.

On computing the probability you might just take the approach whereby you conclude you just don't have enough information to compute P($200|$100) as a conditional probability on a prior you know nothing about. With that you can also say nothing about the EV of switching. Not very Sklanskyish but better than saying something that's not true.

Your observation is often even more difficult for people to reconcile themselves to. It involves other things than Sklansky's Principle. You have to forget about Prior Distribtions for the Envelope amounts for a while and just consider the cureent Envelope amounts as Fixed. Once you realize they are fixed you must realize that it's not possible for there to be positive probabilities for both $50 and $200. Not for these two fixed envelope amounts. Once you see $100 you must conclude that either P($200)=1 or P($200)=0, you just don't know which. That statement makes people very uncomfortable. But if you carefully examinine it with respect to these two Fixed Envelope amounts you see it logically follows. Sometimes we just don't know certain things about probabilities.

The way you were thinking about what would happen with the switching before you opened the Envelopes means that you Were considering the Envelope amounts Fixed. You can't compare that Fixed Envelope amounts model to the Baysian Model people are using when they talk about the unknown Prior Distribution for Envelope Amounts. You have to go with one or the other if you're going to compare them. They are modeling the situation differently and will say different things. The Baysian makes additional Assumptions. You are only assuming the two fixed envelope amounts that you have in front of you.

The Problem has been well analyzed mathematically and there are several good treatments of it on threads in the Probability Forum which ran within the past 6 months or so. Just search on the Two Envelope Problem.

PairTheBoard

[yukoncpa]

Hi PTB, thank you and TomCowley for your responses. I’m a layman, but you both explain things so well that I was able to understand. I was thrilled to read your whole response, but let me comment on one sentence:

[ QUOTE ]
Once you see $100 you must conclude that either P($200)=1 or P($200)=0, you just don't know which


[/ QUOTE ]

This is interesting, I actually wrote this out in my original question, but then deleted it, because I’m mathematically inept, and didn’t want to come across as stupid.

[Divad Yksnal]

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

[carlo]

[ QUOTE ]
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%

[/ QUOTE ]

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?

Are you saying that if I flip 2 fair,ideal coins, one after the other that if I get tails on the first flip you will again give me 2/1 odds on my choosing the second flip as tails?

[NotReady]

[ QUOTE ]

Not thinking past your Sklanskyish assumptions.


[/ QUOTE ]

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.

[luckyme]

[ QUOTE ]
[ QUOTE ]

Not thinking past your Sklanskyish assumptions.


[/ QUOTE ]

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.

[/ QUOTE ]

Interesting. I've tended to look at these type of situations as there not being any new information, therefore the probabilities don't change.

We knew that of families that were not both boys, 1/3 are both girls. Finding out this is one of those 'some girls' families doesn't change that.

Monte Hall is similar in that regard. We knew we had a 1/3 chance of being right and that at least one of the other doors had a goat. Being shown a goat door 'Deliberately' added no new information, so we're still at 1/3.

Others find other ways to look at these and get a conceptual grip on them, checking to see if we have new information helps me along,

luckyme

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
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%

[/ QUOTE ]

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?


[/ QUOTE ]

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]

[ QUOTE ]
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.

[/ QUOTE ]

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]

[ QUOTE ]
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".

[/ QUOTE ]

Tricky /images/graemlins/smile.gif,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]

[ QUOTE ]
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

[/ QUOTE ]

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

[PairTheBoard]

[ QUOTE ]
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.

[/ QUOTE ]

[ 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.

[/ QUOTE ]

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]

[ QUOTE ]

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

[/ QUOTE ]

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

[NotReady]

[ QUOTE ]

I agree there is something psychologically disturbing going on here.


[/ QUOTE ]

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]

[ QUOTE ]
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

[/ QUOTE ]

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]

[ QUOTE ]
[ QUOTE ]
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

[/ QUOTE ]

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

[/ QUOTE ]

I will have to adjust my Prior to 33%

PairTheBoard

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]

Not thinking past your Sklanskyish assumptions.


[/ QUOTE ]

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.

[/ QUOTE ]



[ QUOTE ]
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?


[/ QUOTE ]

PairTheBoard

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]

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

[/ QUOTE ]

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

[/ QUOTE ]

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.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]

I agree there is something psychologically disturbing going on here.


[/ QUOTE ]

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.

[/ QUOTE ]

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.

PairTheBoard

[ALawPoker]

[ QUOTE ]
[ QUOTE ]
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

[/ QUOTE ]

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

[/ QUOTE ]

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.

[PairTheBoard]

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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&Number=10731898&page=0&vc=1 )

PairTheBoard

[Piers]

[ QUOTE ]
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.

[/ QUOTE ]

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.