A Rejection of Sklansky

[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