Any probability experts?

[Sauron]

I'm taking stats and probability. Can anyone explain this one to me?

A man and a woman both have two children. At least one of the woman's kids is a boy. The man's oldest child is a boy.

Is the probability of the woman's other child being a boy the same as the probability of the man's other child being a boy?

The answer is "no". :?

[ceptimus]

Look at it this way, using the first letter of man, woman, girl, boy for the diagrams. First just consider all the (equally likely) possibile ways of having two children (oldest child first):

GG
GB
BG
BB

Now for the woman, you have to cross out the top line, as you are told that at least one of her children is a boy, so you are left with:

W
GB
BG
BB

And for the man, you have to cross out the top two lines, as in both of those cases, the oldest child was a girl:

M
BG
BB

We are left with three equally likely cases for the woman, and only two for the man.

You can see that there is a probability of 1/2 of the man's other child being a boy, but only a probability of 1/3 that the woman's other child is male.

[BDS]

It's simple logic (one need know nothing about probability).

The man's oldest child is a boy, the chances are 50/50 that the other child will be a boy.

The woman has two children, which can be (B,B) (B,G) (G,B) or (G,G), arranged by order of birth. We eliminate (G,G) based on the premise of the puzzle. We then look at the other three possibilities, and see that the chances of the other child being a girl are 2/3.

ETA cross post.

[Ari]

"A man and a woman both have two children."

Obviously they have the same two children, the oldest of which is a boy, thus 1/2 probability all around (and prejudice against the multi and androgenous gendered)
:D

(Ok so I buy the other answers more)

[Sauron]

I understand the concept of listing out the possible options and eliminating the invalid ones.

But this seems to tell me that this is deceptive. It shows that the man has fewer permutations of [boy vs. girl birth order] available to him than the woman does. But I don't see how fewer available permutations of [boy vs. girl birth order] equates to a lower probability of the second child being a boy.

Another way of thinking about this bothers me. We were told last night in class that it doesn't matter if the man's son is an older or a younger child; all that matters to the puzzle is that he is identified as being one or the other. However, even though it is unstated in the problem, the woman's son must also be either an older or a younger child (even twins are not born at precisely the same time). So in the woman's case, there is a 50% chance that the boy is the older child, and a 50% chance that the boy is the younger child.

The only difference is that in the man's case, the birth order of the boy is identified. In the woman's case, it is unspoken. But we know it must be younger or older. There seems to be some kind of magic going on here, where identifying the birth order suddenly impacts the probability of the second child's gender.

[ceptimus]

While we're on probability paradoxes, try this one out.

A man gives his two children (which happen to be a boy and a girl) sealed envelopes. He tells them that each envelope contains only money, and that one envelope contains twice as much money as the other. He tells them that they can each look inside their own envelope (if they wish) but they must not reveal what is inside. If they both agree, they may swap envelopes, whether or not they have opened them.

Now the girl looks inside her envelope and finds one hundred dollars. She reasons like this: 'If I swap, then I will either lose fifty dollars, or gain one hundred dollars - both of these outcomes being equally likely.' This seems like a good bet to her - better than she could expect at any casino, so she decides she will swap.

Now obviously the boy, whether he found fifty or two hundred dollars in his envelope, reasons exactly the same - by swapping he is wagering half his money but stands an evens chance of winning double the amount wagered.

What is paradoxical is that such apparent cast-iron reasoning leads to such a nonsensical result. Using the above reasoning, the children could decide to swap without even opening their envelopes, and then having swapped, they could reason that it would be beneficial to swap again!

To resolve the paradox, you have to find the flaw in their reasoning.

[Ari]

I Think:
Basically the women gets one more flip of the coin.
Ignoring the question.
By looking at a single child we get either G or B
by looking at two children we get a second chance, so it's not just G or B but GG, GB, BG, BB. Thus the probability of at least one of the women' children being a girl is 3/4 and the probability of at least one of the women's children being a boy is 3/4.

Let's say there are 4 children, while there is only one chance the women doesn't have a girl BBBB, there are multiple chances that she does (BBBG BBGB BGBB GBBB for example).

I think.

[cappuccino]

The only difference is that in the man's case, the birth order of the boy is identified. In the woman's case, it is unspoken. But we know it must be younger or older. There seems to be some kind of magic going on here, where identifying the birth order suddenly impacts the probability of the second child's gender.

There isn't anything magical going on. The probabilities are reflecting the degree of information we have about the situation at hand.

In the man's case, we have one extra piece of information, the ordering of the boy (coming in first, being the oldest one) which allows us to narrow down the number of possible outcomes, BB or BG. In the woman's case, our knowledge is more fuzzy. We only know she has a boy, not where he comes in the ordering of the two children. In that case, we can't narrow down the outcomes as much as we did with the man; we're left with three outcomes instead of two outcomes, BB, BG, or GB. Hence that translates into different probabilities.

The key issue is how much information we have available to translate into probabilities which is our measure of our knowledge or conversely ignorance of the situation.

ceptimus,

Hmm, at first glance, I didn't really see the paradox. Why would the children decide to swap twice? Or thrice, or ad infinitum? I can see how blind reasoning would lead to infinite regression.

Need to think about it some more to spot the flaw.

[Clutch Munny]

While we're on probability paradoxes, try this one out.

A man gives his two children (which happen to be a boy and a girl) sealed envelopes. He tells them that each envelope contains only money, and that one envelope contains twice as much money as the other. He tells them that they can each look inside their own envelope (if they wish) but they must not reveal what is inside. If they both agree, they may swap envelopes, whether or not they have opened them.

Now the girl looks inside her envelope and finds one hundred dollars. She reasons like this: 'If I swap, then I will either lose fifty dollars, or gain one hundred dollars - both of these outcomes being equally likely.' This seems like a good bet to her - better than she could expect at any casino, so she decides she will swap.

Now obviously the boy, whether he found fifty or two hundred dollars in his envelope, reasons exactly the same - by swapping he is wagering half his money but stands an evens chance of winning double the amount wagered.

What is paradoxical is that such apparent cast-iron reasoning leads to such a nonsensical result. Using the above reasoning, the children could decide to swap without even opening their envelopes, and then having swapped, they could reason that it would be beneficial to swap again!

To resolve the paradox, you have to find the flaw in their reasoning.

Tricky, tricky ceptimus!

I think this is more like Russell's barber than a strict probability paradox. I.e., just as there's no such barber, the lesson here is that the father could not (truly) utter that promise to his children. He'd need to put aleph nought dollars in both envelopes in order to be able to guarantee what he seems to guarantee. So the flaw in their reasoning is that they trust their old man.

I could very well be missing something, but that's my story for now.

ETA: No, wait: that's stupid. What I should say is that the kids should recognize that their father would have to have put infinite amounts of money into each envelope in order to make their reasoning sound. But that's not a diagnosis as much as a symptom of their error! Back to the drawing board.

[JoeP]

I haven't worked out how the children's reasoning is false, but I've got some better reasoning.

Child: I know the envelopes contain $X and $2X. I also know that have a 50/50 chance of having either.

If I keep the envelope I've been given, I expect to receive $1.5X (assuming I am old enough to have been taught the probability definition of expectation).

If I swap envelopes, either I had $X and now have $2X - gain $X; or I had $2X and now have $X - loss $X; net expected change zero.

...

The flaw is something to do with a change of variables ... I open my envelope and see I have $Y, so X is either Y or Y/2.

[Clutch Munny]

Crap. Aren't they missing important information? That's got to be it. But then, I'm not even sure the puzzle is well-formed.

ETA: I think JoeP's onto it. The amounts in the envelopes really do make a difference.

[BDS]

I'm not sure on Ceptimus paradox either, but I think its similar to Sauron's. The money is already in the envelopes. Each player will get either 2/3 of the money, or 1/3 of the money. The amount of total money will not change. Since you are already guaranteed to get 1/3 of the money, your gain if you win is 1/3X, and your loss if you lose is 1/3X. So it's an even chance.

[Uthgar the Brazen]

How many consecutive lucky guesses does it take to earn the title of probability expert?

[BDS]

I'll bet Matt Holliday gets a hit in the first game of the World Series.

[Sauron]

Let me approach this from another angle.

It seems that we're looking at each case in isolation first, to make it easier. So first we estimate probability for the man's situation, then the woman's situation. And then we compare the two probabilities against each other to see if they are equal.

So forget about the man for a moment. If we just look at the woman's situation, we know:

1. she has two children
2. at least one of them is a boy

Now we're asked to calculate the probablity that the 2nd child is also a boy. Given only #1 and #2, doesn't that mean the probability of the 2nd child being a boy is 50%?

Thanks again for your help.

[BDS]

No, the probability is not 50%. That’s because of the stipulation that “one of them is a boy” – but we don’t know which one. Another way of looking at the odds: “One of the children will be a boy 75% of the time (the other 25% of the time, both will be girls). Half of the time, there will be one boy and one girl. This will qualify for the “one of them is a boy” stipulation – but the other child will be a girl. ¼ the time there will be two girls. ¼ the time there will be two boys. We eliminate the ¼ that are two girls (it doesn’t qualify). Of the remaining ¾, 2/3 will be one boy and one girl (the 2/4 of the original), and 1/3 will be two boys.

Suppose you flipped a coin two times. It can come up HH, HT, TH, or TT, all equally likely. If we say, what are the odds that IN CASES WHERE ONE TOSS CAME UP HEADS the other toss came up tails. Again, it’s 2/3. When heads came up first, it’s 50/50. However of the 2/4 of the time that TAILS came up first, the OTHER toss will ALWAYS be heads when the tosses qualify (i.e. when heads comes up on the second toss, as it will 50% of the time).

[Loren Pechtel]

Let me approach this from another angle.

It seems that we're looking at each case in isolation first, to make it easier. So first we estimate probability for the man's situation, then the woman's situation. And then we compare the two probabilities against each other to see if they are equal.

So forget about the man for a moment. If we just look at the woman's situation, we know:

1. she has two children
2. at least one of them is a boy

Now we're asked to calculate the probablity that the 2nd child is also a boy. Given only #1 and #2, doesn't that mean the probability of the 2nd child being a boy is 50%?

Thanks again for your help.

No. You have to count up all the cases and see what results. (You always have to do this in stat problems, it's just there are often formulas to do it for you and in big problems the formulas are the only realistic answer.)

With the man we know the gender of the first child there's no random information. The first child doesn't affect the second and can be completely removed from the picture. Thus it reduces to a simple case: A man has a child. What is the probability that it's a boy? Genetics will tell you the answer, it's a hair over 50%.

The woman is a different matter. While we know the gender of *A* child we don't know which child. Remember that the person who is telling us this knows all the data and thus this is *NOT* random--he can inspect both children and tell you "one is a boy" if either is--he gets to pick out the boys. As the sample has been non-randomly reduced it will now favor girls.

Lets try a different approach that might make it clearer:

I take a box of 1000 quarters. I dump them on the floor and then go through and pick out 490 of them that landed heads. I now ask you what the probability of a given remaining coin being heads is.

[Ari]

The probability that these problems will eventually mind-fuck you and make you go on a shooting rampage with a water pistol = 1

[cappuccino]

Let me approach this from another angle.

It seems that we're looking at each case in isolation first, to make it easier. So first we estimate probability for the man's situation, then the woman's situation. And then we compare the two probabilities against each other to see if they are equal.

So forget about the man for a moment. If we just look at the woman's situation, we know:

1. she has two children
2. at least one of them is a boy

Now we're asked to calculate the probablity that the 2nd child is also a boy. Given only #1 and #2, doesn't that mean the probability of the 2nd child being a boy is 50%?

Thanks again for your help.


Yes the probability is 1/2 when you're considering the woman's case in isolation and disregarding the ordering of the woman's children as irrelevant. Instead of considering BB, BG, GB, and GG where it's important to account for the oldest and the youngest gender, you're considering BG and GB to be equivalent since all we're caring about is the gender itself. Thus, we only have two possibilities, BB and (BG/GB). Then the probability of the other child being a boy would be 1/2.

With the man's case thrown in then then we're forced to worry about the ordering of the children and the probability of the woman's case becomes 1/3.

Probability is often very subtle that way. It's easy to overlook some critical piece of information and get your probabilities all wrong.

[Furby]

Let me approach this from another angle.

It seems that we're looking at each case in isolation first, to make it easier. So first we estimate probability for the man's situation, then the woman's situation. And then we compare the two probabilities against each other to see if they are equal.

So forget about the man for a moment. If we just look at the woman's situation, we know:

1. she has two children
2. at least one of them is a boy

Now we're asked to calculate the probablity that the 2nd child is also a boy. Given only #1 and #2, doesn't that mean the probability of the 2nd child being a boy is 50%?

Thanks again for your help.

Yes, the 2nd child has a 50% probability of being a boy since the information that at least one of the two children are boys gives us zero information on the gender of the 2nd child. (In this world, there is at least one male human, given that, what is the probability that I am male?)

However, in the original question, you are were not asked to estimate the probability of the 2nd child or the 1st child being male but to estimate the probability of the other child being male given that one of them is a boy.

[Furby]

ETA: No, wait: that's stupid. What I should say is that the kids should recognize that their father would have to have put infinite amounts of money into each envelope in order to make their reasoning sound. But that's not a diagnosis as much as a symptom of their error! Back to the drawing board.

I think the answer lies somewhere along the lines of your reasoning. If the father had simply stated in advance that he had put $100 dollar in one envelope and $200 in another, then there would be no reason to switch, since the gain/loss would be exactly $100. (Same for 50-100 or 200-400). It is only when your expected gain can always be twice as much as your expected loss is when you run into the above paradox.

I would guess that the probability distribution of the stated scenario would result in an infinite mean and since infinity is just weird, we don't actually have a paradox.

[Sauron]

To be clear here is how the problem phrases it:

A woman and a man (unrelated) each have two children. At least one of the woman's children is a boy, and the man's oldest child is a boy. Do the chances that the woman has two boys equal the chances that the man has two boys? Provided relevant probabilities to support your claim.

So why does specifying birth order on the part of the MAN, change the woman's odds of having two boys?

[Furby]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

[Furby]

I understand the concept of listing out the possible options and eliminating the invalid ones.

But this seems to tell me that this is deceptive.

I think that the reason intuition does not match with the actual probability calculations is that the way the problem is phrased does not reflect how we obtain information in the real world.

Let's compare the two different pieces of information in the problem:

1. "oldest child is a boy"
2. "at least one of the children is a boy".

In the first case, we can easily imagine a scenario on how this information can be obtained. You are walking in park and a friend you have not met in a long time greets you. You catch up a bit and the friend tells you that she is married and has two children. She also tells you that the elder one is a boy and asks you to guess the gender of the younger child.

Intuition here tells you that it is 50/50 and intuition is right and agrees with the probability calculations, so there is no problem here.

Now let's look at the second case. Again, you are walking in the park and you meet your friend but this time she tells you "one of my children is a boy" and then asks you to guess the gender of the other child.

Again, your intuition tells you that it is 50/50 and again your intuition would be right. Why?

The problem here is that P(other child is a boy | at least one child is a boy) does not equal P(other child is a boy | friend tells you at least one child is a boy). This is because your friend would be equally likely to tell you that one of her children is a girl (if she does have a girl), so telling you that one of her children is a boy does not actually give you any information about her other child. Hence, your a priori assumption of 50% remains unchanged.

The "at least one child" scenario can only be achieved if we lived in a world where people will only volunteer information if they had a boy but would remain silent if they had girls. Alternatively, you live in a country where somehow every family has at least one boy. These situations are simply not realistic, which is, I think, the reason why your intuition does not match up with the probability calculations.

[Furby]

<Edited because of double posting>

[JoeP]

ETA: No, wait: that's stupid. What I should say is that the kids should recognize that their father would have to have put infinite amounts of money into each envelope in order to make their reasoning sound. But that's not a diagnosis as much as a symptom of their error! Back to the drawing board.

Well, my dad would have put infinite amounts of money in each envelope. :hmph:



How many consecutive lucky guesses does it take to earn the title of probability expert?

Earn? You just Freethought Forum (http://www.freethought-forum.com/forum/profile.php?do=editprofile)



The probability that these problems will eventually mind-fuck you and make you go on a shooting rampage with a water pistol = 1

:sadyup:

None of the above responses relate to Sauron's original problem, hence the spoiler

[mickthinks]

There are really two problems for the price of one in conditional probability puzzles - getting the right answer, and explaining why it is the right answer. That's why they can be sources of seemingly endless discussion, because those who have the mathematical skills to solve the problem correctly often lack the communication skills required to explain it. Which is presumably where Sauron's tutors have failed him and why he brought the problem here in the first place.

Seeing as I have had no success at :ff: explaining the simplest issues such as why prejudice is a bad thing and why you'd have to be excessivley belligerant to take a "thank-you" as a sign of aggression; why I should feel I can explain probability is a mystery but anyway, here goes!...

(Incidentally, all this has been said in different ways by the others. I don't mean to imply that any of them have not said what I am saying. Rather I am repeating stuff in my own way, in the hope that it helps)

---:professor:---

Simple probability is a matter of some event about which we know nothing except the likelihood of it happening. It's often a coin about to be flipped or a die to be rolled, but maybe it helps to think of it as a piece of information hidden in a closed box. The probability of X is just a measure of the likelihood that the box will reveal X when it is opened. It gets more intricate if we have boxes for X and Y, because we can then talk about the probabilites of combinations of X and Y, such as "X and Y" or "X and not Y", but it's still all pretty simple.

On the other hand, conditional probabilty is like a closed box inside which someone, but not you, has looked and then written a message on the lid. If you know how what is written on the lid is related to what is in the box, then you know more about the contents beyond their simple probability. If you are very careful you can calculate a new set of probabilities for the box. But it may help to spend a few seconds imagining all the ways in which knowledge about how the message on the lid might relate to the contents; you'll begin to see just how difficult it is to intuit conditional probabilities.

So, a conditional probability problem has two aspects; a relatively simple random event (the closed box) and a scenario in which some other information about the event is given you (the message on the lid).

---:boxedin:---

In Sauron's problem, there are two boxes each containing a pair of children of unknown sex. On the lid of each is a message about a boy being inside. At first sight they appear to be equivalent messages, but that is because intuitively we think that the information about comparitive ages is irrelevant, and that intuition isn't quite right.

The age info is irrelevant in the sense that it could just as well have been height information ("the tallest is a boy") or weight ("the fat-butt") or intelligence ("the retard") or ... you get the idea. But the important point is that this information identifies the two children. We have the "oldest" and we have the other one, and the probability of there being two boys in the box is simply (and intuitively) the probabilty that the "other one" is a boy.

Whereas the message on the lid of the other box actually gives us much less information because it doesn't identify the children. It is a mistake to think about "the other one" because that could be either of them.

Of course I'm tempted to construct a range of possible scenarios in my head in which the second box is like a version of the first box; "Either the boy in the message is the oldest or he is the youngest. Suppose he is the oldest, then..." However, there isn't a boy in the message, there's one of maybe two boys in it, and so all of these imaginary scenarios would implicitly misread the message. The message could be rewritten "There are not two girls in the box". Try thinking "Either the two girls in the message are the oldest or they are the youngest" and see if that helps to explain where the mistake here lies.

Alternatively, consider these imaginary scenarios - either the oldest is a boy*, or the youngest is a boy*, or they are both boys. It so happens that these three possibilities are exhausive and equally likely, and so they give us the answer; the probability of two boys is 1/3. But this still isn't intuitive enough for me, and I'm guessing it won't appeal to Sauron's intuition either. So instead I'll return to the beginning and construct a new scenario.

---:think:---

Imagine billions of boxes representing all the two-children families in human history **. Imagine that whoever is responsible for writing cryptic messages on the lids of these boxes has written "No boys" on the lids of all the girl-pairs, and "One boy" on the lids of all the others. I choose a box at random. Theres a 1/4 probability of it being a boy-pair, because I chose without considering the message on the lid. But there's a 3/4 chance of the message being "One boy". Suppose that if the message is "One boy" I don't open the box but give it to you, and you read the message.

What is the probability that the box contains two boys if you are "given" "one boy"? I hope it is now intuitively obvious that it is twice as likely that your "One boy" box will contain a girl than that it contains two boys.

I hope that wasn't too passive aggressive!
:wink:

Mick
* and the other is a girl
** I'm ignoring the possibility of children of ambiguous sex here. I hope that doesn't offend anyone.

[slimshady2357]

Imagine billions of boxes representing all the two-children families in human history **. Imagine that whoever is responsible for writing cryptic messages on the lids of these boxes has written "No boys" on the lids of all the girl-pairs, and "One boy" on the lids of all the others. I choose a box at random. Theres a 1/4 probability of it being a boy-pair, because I chose without considering the message on the lid. But there's a 3/4 chance of the message being "One boy". Suppose that if the message is "One boy" I don't open the box but give it to you, and you read the message.

What is the probability that the box contains two boys if you are "given" "one boy"? I hope it is now intuitively obvious that it is twice as likely that your "One boy" box will contain a girl than that it contains two boys.


I think that is an excellent way of explaining the problem. In fact I think that 'enlarging' these problems in some way often gets the idea across. One quibble, that 3 above should be a 2. There is also a 1/4 chance of it saying 'No boys'.

Edit: My mistake, I was reading it wrong. No quibbles.

I also liked Furby's 'enlarging' of the problem to dropping 1000 quarters. Once you're told that 490 of them ended up heads, would you think the chances of a randonly chosen quater from the other 510 being tails would be 50/50? I hope not.

[Furby]

I also liked Furby's 'enlarging' of the problem to dropping 1000 quarters. Once you're told that 490 of them ended up heads, would you think the chances of a randonly chosen quater from the other 510 being tails would be 50/50? I hope not.

That wasn't me, you are referring to Loren.

[slimshady2357]

I also liked Furby's 'enlarging' of the problem to dropping 1000 quarters. Once you're told that 490 of them ended up heads, would you think the chances of a randonly chosen quater from the other 510 being tails would be 50/50? I hope not.

That wasn't me, you are referring to Loren.

Thanks for the correction, I'm starting to wonder why I even bothered posting! :D Got two things wrong and only really made two comments!

[viscousmemories]

I hope that wasn't too passive aggressive! :wink:
:ironymeter:

[Sauron]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.

[slimshady2357]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.

Why are you ignoring the GB possiblity?

[Furby]

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.


Why are you ignoring the GB possiblity?

What he said.

BG and GB are separate and you cannot lump them together. It's the same when calculating the heads/tails probability distribution of flipping two coins.

Also, "at least one is a boy" can be translated to "either the first child is a boy, or the second child is a boy, or both children are boys".

If you want to take it to mean "either a mixed gender pair or a boy pair" then I think you would agree that a mixed gender pair is twice as likely as a boy pair. So your probability distribution should be

BB - 1/3
BG - 2/3

instead of

BB - 1/2
BG - 1/2

[Loren Pechtel]

To be clear here is how the problem phrases it:

A woman and a man (unrelated) each have two children. At least one of the woman's children is a boy, and the man's oldest child is a boy. Do the chances that the woman has two boys equal the chances that the man has two boys? Provided relevant probabilities to support your claim.

So why does specifying birth order on the part of the MAN, change the woman's odds of having two boys?

No, the man exists only to mislead you. The probability her other child is a boy is 1/3 even if the man doesn't exist.

The key point of the question is that the information we know about them is subtly different.

[Loren Pechtel]

The "at least one child" scenario can only be achieved if we lived in a world where people will only volunteer information if they had a boy but would remain silent if they had girls. Alternatively, you live in a country where somehow every family has at least one boy. These situations are simply not realistic, which is, I think, the reason why your intuition does not match up with the probability calculations.

Or we are in a sex-selective environment of some kind.

It's not just the park, it's the boys soccer team. You find two old friends there. One tells you "my eldest son is playing". The other says "my son is playing".

[Loren Pechtel]

I also liked Furby's 'enlarging' of the problem to dropping 1000 quarters. Once you're told that 490 of them ended up heads, would you think the chances of a randonly chosen quater from the other 510 being tails would be 50/50? I hope not.

Expanding a problem like this often helps make things clearer. It's a good tactic to try when you're stuck on figuring out the effects of some factor on a situation.

[Loren Pechtel]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.

No. You are implicitly making this into the man's case by assuming it's the first child that's a boy.

The possibilities are actually BB, BG and GB.

[seebs]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.


No. BB, BG, GB.

1/3.

[Sauron]

But it does not. The probability of the woman having two boys depends only on the information you have been given about her. You can ignore the information given to you about the man and still get 1/3.

No. You can't. Because the woman's scenario makes no mention of older vs. younger. That only shows up in the man's scenario.

Without that distinction to worry about, the woman's situation simplifies to:

* she has two kids
* at least one is a boy

Q: What are the chances she has two boys?
BB
BG

50%.

Why are you ignoring the GB possiblity?

Because it only exists as a result of the older vs. younger child distinction. Which is part of the man's scenario, not the woman's. The woman's scenario should yield the same probability, whether (a) considered in isolation by itself, or (b) in a side-by-side comparison.

You're forgetting the steps here.

1. IGNORE THE MAN.
2. Focus on just the woman's case. You know:

* she has two kids
* at least one of which is a boy

Now what is the probability that the woman has two boys?

[Furby]

Okay, so let's do the math.

Pr(has two boys | at least one is a boy) = Pr(has two boys ^ at least one is a boy) / Pr(at least one boy)

Pr(has two boys ^ at least one is a boy) = Pr(has two boys) = 1/4
(since if you have two boys you must have at least one boy)

Pr(at least one boy) = 1 - Pr(has two girls) = 1 - 1/4 = 3/4

Therefore,

Pr(has two boys | at least one is a boy) = 1/4 / 3/4 = 1/3

That is, if you agree that Pr(has two boys) = Pr(has two girls) = 1/4

[cappuccino]

Look at this way, in an idealized abstract world we could add or drop assumptions. Assuming for the sake of the discussion we're ignoring the older and younger distinction utterly (or the two children were born at the exact same time) The probability works out to 1/2. BUT, in real life, we have to take the ordering into account. Then the probability works out to 1/3. The way probability is defined is that it's event we're considering divided by the set of ALL possible distinct outcomes. It simply tells us how often a particular event occurs out of a set of all possible events. We have to count up every single possible ways something can occur, that's part of probability theory definition.

In the woman's case we can't forget about the outcomes of girl being born first followed by a boy OR boy born first then a girl born. Because children being born at the exact same time doesn't happen. Unless possibly a twin is taken out of an uterus at the same time via cesarean section, but that's just extreme nit-picking. In normal sense, two children would be born in two distinct outcomes.

[Furby]

Look at this way, in an idealized abstract world we could add or drop assumptions. Assuming for the sake of the discussion we're ignoring the older and younger distinction utterly (or the two children were born at the exact same time) The probability works out to 1/2.

I don't see how you got that or why does the time difference between the births matter at all. You could take the scenario of flipping two coins together, assume they land at the same time and ask the question - given that at least one of the coins lands tails, what is the probability that they both land tails - and you would still end up with 1/3.

The birth order of the children only provides a convenient way of identifying them (younger vs older) but you can arbitarily name them child A child B (ignoring who was born first), left child right child, child 1 child 2 and it wouldn't matter since their births are assumed to be distinct and independent events.

[mickthinks]

Furby's right. cappuccino's post is a good example of mathematical skill unsuccessfully communicating understanding.

It occurs to me that it might help to shed light on the issue if we change the scenario slightly:

You are in the garden of a friend who is entertaining several families whom you don't know and have never met. The only thing you know about them is that they are all two-children families. You're sitting drinking iced tea with a man and his sister when two young boys run up, one to his dad, and the other to his mum. "Hello Son" their parents say, and each is sent off again with a kiss.

The man tells you "That was my eldest", but the woman merely says "little scamps!" Now, is the chance that the woman has two boys the same as the chance that the man has two boys?

Yes, in this case, even though the information you have been given for the two families appears to be the same as in Sauron's original problem. The question is, what exactly is the difference here?
Mick

[cappuccino]

I don't see how you got that or why does the time difference between the births matter at all. You could take the scenario of flipping two coins together, assume they land at the same time and ask the question - given that at least one of the coins lands tails, what is the probability that they both land tails - and you would still end up with 1/3.

The birth order of the children only provides a convenient way of identifying them (younger vs older) but you can arbitarily name them child A child B (ignoring who was born first), left child right child, child 1 child 2 and it wouldn't matter since their births are assumed to be distinct and independent events.

Yes, talking about time difference is irrelevant, my thinking is that you're assuming ordering to be important, otherwise how do we distinguish between BG and GB? or tail then head versus head then tail? Yet probability considers permutation of outcomes to be distinct and unique.

It seems to me that Sauron considers BG and GB to be equivalent. At first glance, it feels intuitive that it's not the case. Both BG and GB are two distinct outcomes that must be accounted for but as I thought about it, I can't explain why I think that's so.

[cappuccino]

Furby's right. cappuccino's post is a good example of mathematical skill unsuccessfully communicating understanding.Mick

Yeah, it's apparent that I'm not communicating very well. But now I want to fully understand the probability problem. Like I said earlier, at first glance at Sauron's problem, I could see that the answer was 1/3 right off, but as I read the other posts and thought about the problem, it's no longer apparent to me why it *should* be 1/3. I can easily see the paths leading to both 1/2 and 1/3 answers but now it appears a little arbitrary to choose 1/3 as the correct answer.

[Furby]

It seems to me that Sauron considers BG and GB to be equivalent. At first glance, it feels intuitive that it's not the case. Both BG and GB are two distinct outcomes that must be accounted for but as I thought about it, I can't explain why I think that's so.

Suppose I labelled one of the two children brown-eyed child and the other blue-eyed child. Then, "blue-eyed child is a boy and brown-eyed child is a girl" is clearly distinct from "blue-eyed child is a girl and brown-eyed child is a boy".

[mickthinks]

Thanks cappuccino!

I can easily see the paths leading to both 1/2 and 1/3 answers but now it appears a little arbitrary to choose 1/3 as the correct answer.

Yes, exactly! This is the meta problem and in many ways it is a more interesting, and perhaps even a more important problem to solve, yet some able mathematicians in my experience seem unable to discuss it.

It is obvious that "blue-eyed boy and brown-eyed girl" and "brown-eyed boy and blue-eyed girl" are distinctly different pairs. What is troubling is that for mathematical purposes we also seem to be required to distinguish between unlabelled cases. It seems we have to see the difference between "unknown-eye-colour boy and unknown-eye-colour girl" and "unknown-eye-colour girl and unknown-eye-colour boy".

Anyone who doesn't see a problem with the labelling trick hasn't given it enough thought, I reckon.

Mick

[Sauron]

Okay, so let's do the math.

Let's not.

If understanding the math were the issue, I wouldn't have needed to post the question in the first place.

The issue is understanding why the woman -- who knows nothing about the man, and his scenario -- why does this woman have three possible outcomes?

Erase the entire man/woman problem from your mind. Start over with a new problem:

The King of England has two children, one of which is a boy (prince). What are the chances that the king has two boys (princes)?

GG - 0%
BG - 50%
BB - 50%

[Furby]

Thanks cappuccino!

I can easily see the paths leading to both 1/2 and 1/3 answers but now it appears a little arbitrary to choose 1/3 as the correct answer.

Yes, exactly! This is the meta problem and in many ways it is a more interesting, and perhaps even a more important problem to solve, yet some able mathematicians in my experience seem unable to discuss it.


I still fail to see why a problem exists (but I am not a mathematician). The path that leads to 1/2 is erroneous as it fails to recognize that there are two distinct birth events that lead to the gender determination of the children.


It is obvious that "blue-eyed boy and brown-eyed girl" and "brown-eyed boy and blue-eyed girl" are distinctly different pairs. What is troubling is that for mathematical purposes we also seem to be required to distinguish between unlabelled cases.


Again, the labels only serve to distinguish between two separate events which are independent. We assume that any birth event has a 50/50 chance of producing a male child or a female child. Any 2 such events which do not affect each other will have 4 distinct combinations (which are equally likely), it would not matter if the events happened at both ends of the world and were caused by two different women.


It seems we have to see the difference between "unknown-eye-colour boy and unknown-eye-colour girl" and "unknown-eye-colour girl and unknown-eye-colour boy".


The phrasing above is misleading because it would seem that the event that led to "unknown-eye-colour boy" in the first statement is the same as the event that led to "unknown-eye-colour boy" in the second (similarly for "unknown-eye-colour girl "). If we think that, then since the events are independent and the ordering does not matter, we will mistakenly believe that the two statements are equivalent. This is the reason why we want to have labels in the first place.


Anyone who doesn't see a problem with the labelling trick hasn't given it enough thought, I reckon.

Mick


Why is that? The labels only help us define the problem, we use them in the same way we use variables in algebra problems. The naming of the labels isn't important, just that they must be present. For example, I could have

Birth Event XQ12320a-233 Birth Event AZUI199923-134
unknown-eye-colour boy unknown-eye-colour boy
unknown-eye-colour boy unknown-eye-colour girl
unknown-eye-colour girl unknown-eye-colour boy
unknown-eye-colour girl unknown-eye-colour girl

[Loren Pechtel]

I don't see how you got that or why does the time difference between the births matter at all. You could take the scenario of flipping two coins together, assume they land at the same time and ask the question - given that at least one of the coins lands tails, what is the probability that they both land tails - and you would still end up with 1/3.

The birth order of the children only provides a convenient way of identifying them (younger vs older) but you can arbitarily name them child A child B (ignoring who was born first), left child right child, child 1 child 2 and it wouldn't matter since their births are assumed to be distinct and independent events.

Yes, talking about time difference is irrelevant, my thinking is that you're assuming ordering to be important, otherwise how do we distinguish between BG and GB? or tail then head versus head then tail? Yet probability considers permutation of outcomes to be distinct and unique.

It seems to me that Sauron considers BG and GB to be equivalent. At first glance, it feels intuitive that it's not the case. Both BG and GB are two distinct outcomes that must be accounted for but as I thought about it, I can't explain why I think that's so.

The ordering is unimportant. The key thing is that in the case of the man the child has been uniquely identified but in the case of the woman there is no such unique identification.

The GB case is eliminated in the case of the man because we are looking at the first one and seeing a B. This leaves only the BB and BG cases.

With the woman, however, we don't know the ordering and thus both BG and GB are possible cases.

[Loren Pechtel]

Okay, so let's do the math.

Let's not.

If understanding the math were the issue, I wouldn't have needed to post the question in the first place.

The issue is understanding why the woman -- who knows nothing about the man, and his scenario -- why does this woman have three possible outcomes?

Erase the entire man/woman problem from your mind. Start over with a new problem:

The King of England has two children, one of which is a boy (prince). What are the chances that the king has two boys (princes)?

GG - 0%
BG - 50%
BB - 50%

No. You're still falsely assuming the known boy to be child #1, something we don't know. Thus it's BG (33%), GB (33%) and BB (33%).

You still haven't addressed my illustration involving picking out the heads from a box of quarters.

[Crumb]

Okay, so let's do the math.

Let's not.

If understanding the math were the issue, I wouldn't have needed to post the question in the first place.

The issue is understanding why the woman -- who knows nothing about the man, and his scenario -- why does this woman have three possible outcomes?

Erase the entire man/woman problem from your mind. Start over with a new problem:

The King of England has two children, one of which is a boy (prince). What are the chances that the king has two boys (princes)?

GG - 0%
BG - 50%
BB - 50%
Wow that's convincing Sauron. I was not leaning this way before this post. But if one event is completely independent of the other than the odds of the other child being a boy should be 50/50. It seems identical to flipping a coin and getting "tails" and then asking "what are the odd of getting tails again?" The answer should be 50%.

[mickthinks]

Hello Furby

The path that leads to 1/2 is erroneous as it fails to recognize that there are two distinct birth events that lead to the gender determination of the children.
It fails to recognise a whole load of events, such as whether one or both the children are adopted, or whether either of them has had gender reassignment surgery. I'm not clear why that makes it wrong.

Any 2 such events which do not affect each other will have 4 distinct combinations.
I think I know what you are getting at but this is a confusing thing to say. Two children will produce only one combination.

The phrasing above is misleading...
I would describe it as surprising and uninformative, which is kind of the point. Whether you are misled by it depends on whether you understand it properly.

...because it would seem that the event that led to "unknown-eye-colour boy" in the first statement is the same as the event that led to "unknown-eye-colour boy" in the second (similarly for "unknown-eye-colour girl ").
I disagree.

If we think that, then since the events are independent and the ordering does not matter, we will mistakenly believe that the two statements are equivalent.
Sorry, I have no idea what you mean by this.

This is the reason why we want to have labels in the first place.
The point I am getting at is that the mathematics should be clear with or without labels. If you can only explain with labels, you make it look as if the labels affect the probabilities. As I understand it, that is precisely what Sauron is worried about and I think he's right to ask for a better, labelless account of the matter.

Mick

[Furby]

Wow that's convincing Sauron. I was not leaning this way before this post. But if one event is completely independent of the other than the odds of the other child being a boy should be 50/50. It seems identical to flipping a coin and getting "tails" and then asking "what are the odd of getting tails again?" The answer should be 50%.

:cry:

[Furby]

Hello Furby

The path that leads to 1/2 is erroneous as it fails to recognize that there are two distinct birth events that lead to the gender determination of the children.
It fails to recognise a whole load of events, such as whether one or both the children are adopted, or whether either of them has had gender reassignment surgery. I'm not clear why that makes it wrong.


If we have no knowledge of such events, then Bayesian theory ignores them. If an event has X possible outcomes, and we do not know anything else, then we simply distribute the probabilities evenly between the X possible outcomes. In the above problem, we are first made aware there are 2 events where each has 2 outcomes. We can then create a uniform combined probability distribution for the 4 combinations. If other evidence comes to light (such as eldest child is boy), we redistribute the probabilities accordingly. If we don't have knowledge about events such as gender reassignment etc, then we simply don't care.



Any 2 such events which do not affect each other will have 4 distinct combinations.
I think I know what you are getting at but this is a confusing thing to say. Two children will produce only one combination.


Well, if you wish to take the sample space of all births that result in twins, and have a single event that leads to 3 equally possible outcomes (pair of boys, mixed gender, pair of girls) rather than 2 events with 4 possible outcomes, in that case, I agree that the probability is 1/3 for each combination and you will end up with 1/2 for the above problem.

If that is what you mean, then I apologize for misreading your previous post.

However, I would argue that the most intuitive interpretation would be that of 2 separate birthing events.


If we think that, then since the events are independent and the ordering does not matter, we will mistakenly believe that the two statements are equivalent.
Sorry, I have no idea what you mean by this.

Simply that if we have "E1: unknown-eye-colour boy E2: unknown-eye-colour girl" as well as "E1: unknown-eye-colour girl E2: unknown-eye-colour boy" and you remove the E1, E2 labels, then you will end up thinking that they are equivalent since the ordering of independent events do no matter.


This is the reason why we want to have labels in the first place.
The point I am getting at is that the mathematics should be clear with or without labels. If you can only explain with labels, you make it look as if the labels affect the probabilities. As I understand it, that is precisely what Sauron is worried about and I think he's right to ask for a better, labelless account of the matter.

Mick

But, surely you don't oppose the use of variables in algebra such as let x = .... . It certainly helps to define the problem more clearly.

[Crumb]

Maybe I'll do better in the morning... :(

[But]

Wow that's convincing Sauron. I was not leaning this way before this post. But if one event is completely independent of the other than the odds of the other child being a boy should be 50/50. It seems identical to flipping a coin and getting "tails" and then asking "what are the odd of getting tails again?" The answer should be 50%.

Your coin analogy would be identical to the man's case. You have flipped a coin and got tails. The first child is a boy. Now you have a 50/50 chance that the second one will be a boy.

Compare that to saying you have flipped a coin twice and got tails at least once (the woman's case).

Maybe it helps to think of a large sample, which is probably what probability is all about.

Let's take a large sample of women who all have two kids. 1/4 will be both boys, 1/4 both girls, and 1/2 a girl and a boy. Now you throw all those out who have two girls. Then 1/3 is both boys and 2/3 mixed.

Do the same in the man's case. It's identical (1/3 both boys, 2/3 mixed) until you specify that the older one is a boy. You could as well say that the one with the lower IQ or better hairstyle is a boy, doesn't matter. But now you have to throw out half of the sample where one is a girl, and the "both boys" part remains the same size, because if both are boys, the older one will always be a boy. That makes the two groups (both boys vs. mixed) the same size, ergo 50/50.

I hope you can see what I'm getting at, I'm not sure my language is all that clear.

[Sauron]

No. You're still falsely assuming the known boy to be child #1,

No I'm not. Go back and re-read what I wrote; I said nothing about "the first child". I'm merely assuming that at least one child is a boy. My example doesn't care if it's child #1 or child #2.


Thus it's BG (33%), GB (33%) and BB (33%).

However, BG is the same as GB. Remember, the new puzzle about the King of England doesn't mention (or care) about birth order. I'm forcing you to restrict yourself to the question of the gender of the children, and ignore issues of birth order.

You still haven't addressed my illustration involving picking out the heads from a box of quarters.

Because it was convoluted and too far afield. Care to address my example?

[Sauron]

Note:

I appreciate everyone's patience.
I'm not trying to pick an argument here just to piss people off. I'm looking for an explanation that puts this nagging question to rest.
:worry:

[mickthinks]

LOL! Sorry Furby, but if this was intended as a simplified approach to the problem, to make the solution intuitively obvious and place it within the grasp of an interested and intelligent non-mathematician, I'd say you have failed, spectacularly.


Hello Furby

The path that leads to 1/2 is erroneous as it fails to recognize that there are two distinct birth events that lead to the gender determination of the children.
It fails to recognise a whole load of events, such as whether one or both the children are adopted, or whether either of them has had gender reassignment surgery. I'm not clear why that makes it wrong.


If we have no knowledge of such events, then Bayesian theory ignores them. If an event has X possible outcomes, and we do not know anything else, then we simply distribute the probabilities evenly between the X possible outcomes. In the above problem, we are first made aware there are 2 events where each has 2 outcomes. We can then create a uniform combined probability distribution for the 4 combinations. If other evidence comes to light (such as eldest child is boy), we redistribute the probabilities accordingly. If we don't have knowledge about events such as gender reassignment etc, then we simply don't care.



Any 2 such events which do not affect each other will have 4 distinct combinations.
I think I know what you are getting at but this is a confusing thing to say. Two children will produce only one combination.


Well, if you wish to take the sample space of all births that result in twins, and have a single event that leads to 3 equally possible outcomes (pair of boys, mixed gender, pair of girls) rather than 2 events with 4 possible outcomes, in that case, I agree that the probability is 1/3 for each combination and you will end up with 1/2 for the above problem.

If that is what you mean, then I apologize for misreading your previous post.

However, I would argue that the most intuitive interpretation would be that of 2 separate birthing events.


If we think that, then since the events are independent and the ordering does not matter, we will mistakenly believe that the two statements are equivalent.
Sorry, I have no idea what you mean by this.

Simply that if we have "E1: unknown-eye-colour boy E2: unknown-eye-colour girl" as well as "E1: unknown-eye-colour girl E2: unknown-eye-colour boy" and you remove the E1, E2 labels, then you will end up thinking that they are equivalent since the ordering of independent events do no matter.


This is the reason why we want to have labels in the first place.
The point I am getting at is that the mathematics should be clear with or without labels. If you can only explain with labels, you make it look as if the labels affect the probabilities. As I understand it, that is precisely what Sauron is worried about and I think he's right to ask for a better, labelless account of the matter.

Mick

But, surely you don't oppose the use of variables in algebra such as let x = .... . It certainly helps to define the problem more clearly.

[mickthinks]

Hello Sauron

I'm not trying to pick an argument here just to piss people off. I'm looking for an explanation that puts this nagging question to rest.
I'm not sure why you are confining your questioning to those who seem to have most difficulty explaining it in simple terms. It's an approach seemingly guaranteed to produce maximum annoyance and frustration on both sides.

Mick

[Furby]

Okay, after re-reading the earlier posts, I think maybe we need to illustrate more clearly the difference between a combined probability distribution of two events with two possible outcomes and the probability distribution of a single event with three possible outcomes.

Case 1:

There is a machine that generates red and green cards. Everything you push a button, a pair of these cards shoots out. Given that you know nothing else about the machine, what do you estimate the probability of getting a mixed colour pair to be?

Case 2:

This time, you have a machine that generates cards with either red on both sides, green on both sides, or red on one side and green on the other. Everytime you push a button, one of the above three cards will shoot out.

Again, given you know nothing else about the machine, what do you estimate the probability of getting a red/green card to be?

Question 1:
Which case (1 or 2) relates more closely to the scenario in the OP?

Question 2:
Suppose you were warped to a dimension where every woman gives birth to conjoined twins (and they only give birth once).

Given that one of these women has at least one boy, and having no other information, what is the probability that her other child is a boy?

[But]

LOL! Sorry Furby, but if this was intended as a simplified approach to the problem, to make the solution intuitively obvious and place it within the grasp of an interested and intelligent non-mathematician, I'd say you have failed, spectacularly.


Just because you're an idiot doesn't mean everyone else is.

Please note that my comment is an example of bias stemming from reading lots of your stupid posts.

[mickthinks]

LOL! Sorry Furby, but if this was intended as a simplified approach to the problem, to make the solution intuitively obvious and place it within the grasp of an interested and intelligent non-mathematician, I'd say you have failed, spectacularly.


Just because you're an idiot doesn't mean everyone else is.

Please note that my comment is an example of bias stemming from reading lots of your stupid posts.

That illustrates your particular problem so clearly a child could understand it, But!
:yup:

[Sauron]

Hello Sauron

I'm not trying to pick an argument here just to piss people off. I'm looking for an explanation that puts this nagging question to rest.
I'm not sure why you are confining your questioning to those who seem to have most difficulty explaining it in simple terms.

Where did I ever confine my questioning to any particular person? This is a free-for-all; anyone can respond.

Oh, I see - well; the thread title should be taken tongue-in-cheek, given that I'm so bad at probability, that almost anyone is an expert by comparison.

[But]

Erase the entire man/woman problem from your mind. Start over with a new problem:

The King of England has two children, one of which is a boy (prince). What are the chances that the king has two boys (princes)?

GG - 0%
BG - 50%
BB - 50%

But case 2 (BG) is twice as likely as case 3 (BB).

[Loren Pechtel]

No. You're still falsely assuming the known boy to be child #1,

No I'm not. Go back and re-read what I wrote; I said nothing about "the first child". I'm merely assuming that at least one child is a boy. My example doesn't care if it's child #1 or child #2.


Thus it's BG (33%), GB (33%) and BB (33%).

However, BG is the same as GB. Remember, the new puzzle about the King of England doesn't mention (or care) about birth order. I'm forcing you to restrict yourself to the question of the gender of the children, and ignore issues of birth order.

Except you are saying that the order is significant when you chuck the GB option. That is the difference between BG and GB--order!

In the man's case we know the order and can rule out GB. In the woman's case we can't.

You still haven't addressed my illustration involving picking out the heads from a box of quarters.

Because it was convoluted and too far afield. Care to address my example?

All I did was increase the number of remove cases and replace kids with quarters. The principle remains the same.

Lets do it with kids. Lets take a million 2-child families and see what we get.

The first kid: 500,000 boys, 500,000 girls.
The second kid: 500,000 boys, 500,000 girls.

Combine these and we get:

250,000 BB, 250,000 BG, 250,000 GB, 250,000 GG.

Since it has been specified that there is a boy we know it's not the GG case. Remove them.

250,000 BB, 250,000 BG, 250,000 GB.

Now, look at the man's case. We know the first child is a boy. He can't be one of the ones that had a girl first. Thus for him we have:

250,000 BB, 250,000 BG. It's 50:50.

Now, look at the woman's case. We know she has a boy. All of the remaining options have a boy, we can't exclude any of them.

250,000 BB, 250,000 BG, 250,000 GB.

Now, remove the boy we know about:

250,000 B, 250,000 G, 250,000 G.

Note that we are left with 67:33 in favor of a girl.

[Dragar]

Edited out becasue I'm daft.

[seebs]

2. Focus on just the woman's case. You know:

* she has two kids
* at least one of which is a boy

Now what is the probability that the woman has two boys?

One in three.

Okay, let's start with "she has two kids":

1B, 2B
1B, 2G
1G, 2B
1G, 2G

You will note that she is twice as likely to have a boy and a girl as she is to have two boys, right? The chances are 25% BB, 50% BG, and 25% GG.

Now. We say "at least one of which is a boy". What's left?

1B, 2B
1B, 2G
1G, 2B
1G, 2G

Of the four possible sets of kids, three remain. Only one is "BB".

You have to count BG separately from GB for the same reason that 11 is more common on two dice than 12; 56 and 65 are two DIFFERENT cases, even though we count them as having the same outcome.

I think the problem here is that, since they are the same for your purposes, you're refusing to count the different ways that same result could happen as being separate possible outcomes for purposes of calculating probability.

Now, you may ask why it'd be different if it were "the oldest". Here we go. Two kids, and #1 is a B:


1B, 2B
1B, 2G
1G, 2B
1G, 2G

The qualifier removes one of the two "one boy, one girl" cases.

[Dragar]

Children (and dice) are distinguishable. This means that we can talk about Child 1 and Child 2, even before we know anything about the two children. We could label them by height, or age, or anything.

Because of this, (B, G) is a different state to (G, B). They are distinguishable outcomes. We can tell them apart, and we must count them when we consider possible outcomes of two birth events.

The two scenarios at the beginning of this thread provide different information. Ordering children number by age:

One provides information that Child 1 or Child 2 is a boy.

The other provides information that Child 1 is a boy.

Note that if instead of birth-order information were provided, and some other piece of information were provided, I could simply change my ordering of the children and preserve the above expressions.

Dropping this assumption of objects being distinguishable does produce different results, and is actually important when calculating statistics of identical particles.

[Joshua Adams]

The pertinent fact is that the two children are distinct somehow. Calling one the older and one the younger is simply that way we've chosen to label them. You could translate the problem into flipping quarters simultaneously, and it'd still be the same. Order doesn't matter, but you still have two different quarters, perhaps called the left quarter and right quarter. Two quarters times two faces per quarter is 4 different possibilities. It's simply wrong to conflate the mixed cases into one case.

Remembering that the problem corresponds to some hypothetical physical situation, it is easier to see this. Imagine you flip a quarter in your right hand and it comes out heads. The quarter in your left hand comes out tails. Is this the same thing as if the quarter in your right hand was tails, and the quarter in your left hand was heads? No! It's the exact opposite of what happened in the first case, actually.

If you do enough trials with this you will see that you get "one of each" about half of the time. This is because there are two ways to construct this case, whereas there is only one way to get two tails, and only one way to get two heads.

Considering the gender permutations of two indistinct children is a perfectly valid mathematical problem, I guess, but it's a nonsensical scenario. In real life, children are distinct from one another, and formulating the problem that way models reality better.

Edit: God damn Dragar!

[Dragar]

I've been sat here for about half an hour trying to express what I meant!

[Dragar]

Considering the gender permutations of two indistinct children is a perfectly valid mathematical problem, I guess, but it's a nonsensical scenario. In real life, children are distinct from one another, and formulating the problem that way models reality better.

It's interesting to note that the problem posed in the opening post has no analogue with indistinguishable children. The information ('my eldest child is a boy') no longer makes sense. That children are distinguishable is explicitly brought to our attention right at the start.

[Joshua Adams]

Considering the gender permutations of two indistinct children is a perfectly valid mathematical problem, I guess, but it's a nonsensical scenario. In real life, children are distinct from one another, and formulating the problem that way models reality better.

It's interesting to note that the problem posed in the opening post has no analogue with indistinguishable children. The information ('my eldest child is a boy') no longer makes sense. That children are distinguishable is explicitly brought to our attention right at the start.
I think Sauron was indicating that, while the men's children are distinguished, the woman's are not (because the problem only explicitly mentions birth order in the man's case). I'd argue that the metaphysical baggage of "distinguishability" has to be consistent across the problem. We don't need to be told that they are different children because it's just absurd to think of them as not.

Even if we don't know anything about the age of her children, or indeed anything else, we can still call them Child 1 and Child 2, as you said.

[Sauron]

Erase the entire man/woman problem from your mind. Start over with a new problem:

The King of England has two children, one of which is a boy (prince). What are the chances that the king has two boys (princes)?

GG - 0%
BG - 50%
BB - 50%

But case 2 (BG) is twice as likely as case 3 (BB).

Since the order of birth is irrelevant to the question of gender, why do you think so?

[Sauron]

No. You're still falsely assuming the known boy to be child #1,

No I'm not. Go back and re-read what I wrote; I said nothing about "the first child". I'm merely assuming that at least one child is a boy. My example doesn't care if it's child #1 or child #2.


Thus it's BG (33%), GB (33%) and BB (33%).

However, BG is the same as GB. Remember, the new puzzle about the King of England doesn't mention (or care) about birth order. I'm forcing you to restrict yourself to the question of the gender of the children, and ignore issues of birth order.

Except you are saying that the order is significant when you chuck the GB option.

I never chucked it.

That is the difference between BG and GB--order!

Order is irrelevant to the question of gender.
Gender is the only question I am asking.

[Sauron]

Wait.

Is this the same thing as a coin toss? Is this problem equivalent?

"You friend just tossed two coins. They were tossed simultaneously, so there is no toss order. One coin on the left, one coin on the right.

Your friend is an asshole, and won't tell you the results. All he will say is that there is at least one heads. There might be more, though - but he's an asshole and is going to make you guess. What are the chances that they are BOTH heads?"

HH
TT
HT
TH

One in three?

[But]

Exactly!

[Sauron]

Exactly!

And is it the same as:

"There are two bottles of milk in the refrigerator. Mom says that one of them is past its freshness date and is spoiled. What are the chances that both bottles are spoiled?"


'You got dressed at 4am. It was dark in the room. Your wife gave you two socks. She thinks at least one is black. You're GQ, so you only buy two colors of socks - blue and black. What are the chances that you are well-dressed and have two black socks?"

[seebs]

Wait.

Is this the same thing as a coin toss? Is this problem equivalent?

"You friend just tossed two coins. They were tossed simultaneously, so there is no toss order. One coin on the left, one coin on the right.

Your friend is an asshole, and won't tell you the results. All he will say is that there is at least one heads. There might be more, though - but he's an asshole and is going to make you guess. What are the chances that they are BOTH heads?"

HH
TT
HT
TH

One in three?

Right.

But, if he says the first coin was heads, then the chances of both heads are 1 in 2, because you also strike out TH.

[But]

And is it the same as:

"There are two bottles of milk in the refrigerator. Mom says that one of them is past its freshness date and is spoiled. What are the chances that both bottles are spoiled?"


'You got dressed at 4am. It was dark in the room. Your wife gave you two socks. She thinks at least one is black. You're GQ, so you only buy two colors of socks - blue and black. What are the chances that you are well-dressed and have two black socks?"


Yes, if it's reasonable to assume that the probabilities of one sock/milk bottle being in a particular state are independent of one another.

[Crumb]

Ok you guys win. The 1 million kids made it perfectly understandable.

[Sauron]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

[ceptimus]

Are we to assume that the ratio of ripe to non-ripe apples is 1:1 ?

The standard way to pose the follow-on question now is:

You encounter two bears in the woods, a brown one and a black one. You learn that one of them is male. What are the chances that the other one is male also?

[Loren Pechtel]

You have to count BG separately from GB for the same reason that 11 is more common on two dice than 12; 56 and 65 are two DIFFERENT cases, even though we count them as having the same outcome.

Brain check!

I'm sure you meant to say 12 is more common than 11.

[Loren Pechtel]

Wait.

Is this the same thing as a coin toss? Is this problem equivalent?

"You friend just tossed two coins. They were tossed simultaneously, so there is no toss order. One coin on the left, one coin on the right.

Your friend is an asshole, and won't tell you the results. All he will say is that there is at least one heads. There might be more, though - but he's an asshole and is going to make you guess. What are the chances that they are BOTH heads?"

HH
TT
HT
TH

One in three?

Yes. This is logically equivalent to the woman case.

[Furby]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Sauron,

Bayesian probability reflects the degree of belief you have in a certain statement or proposition and given whatever available evidence you have against or in support of that proposition. You need to take into account information that you have been given as well as the way you have recieved the information before you make a judgement.

Given a real life situation as you have described above, it is likely that Mom chose to tell you of the ripe apple so that if you were hungry, you could have a quick bite. There is no reason why she would not tell you both apples were ripe if she observed it.

Given that, the probability that the other apple is unripe is likely close to 100% hence, the probability that both apples are ripe is close to 0%.

If the apples were put into separate boxes and hidden from view, and Mom opened one of them and told you, "hey, this apple is ripe". Then, the chances of the other apple being unripe is 50%, since she is just as likely to open the box with an unripe apple (provided there is one) as she is to open the box with the ripe one. This assumes, you do not know the number of days that has passed since the apples were bought.

If this question was asked of you in an exam, then the questioner simply wants you to calculate P(both apples are ripe | one is ripe)
and you should answer 1/3..

You could probably argue that you have a different interpretation of the question and can come up with a different value, but that is unlikely to score you any points..

[Loren Pechtel]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Unknown, as we don't know the ratio of ripe & non-ripe apples.

[Sauron]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Sauron,

Bayesian probability reflects the degree of belief you have in a certain statement or proposition and given whatever available evidence you have against or in support of that proposition. You need to take into account information that you have been given as well as the way you have recieved the information before you make a judgement.

Given a real life situation as you have described above, it is likely that Mom chose to tell you of the ripe apple so that if you were hungry, you could have a quick bite. There is no reason why she would not tell you both apples were ripe if she observed it.

Given that, the probability that the other apple is unripe is likely close to 100% hence, the probability that both apples are ripe is close to 0%.

If the apples were put into separate boxes and hidden from view, and Mom opened one of them and told you, "hey, this apple is ripe". Then, the chances of the other apple being unripe is 50%, since she is just as likely to open the box with an unripe apple (provided there is one) as she is to open the box with the ripe one. This assumes, you do not know the number of days that has passed since the apples were bought.

If this question was asked of you in an exam, then the questioner simply wants you to calculate P(both apples are ripe | one is ripe)
and you should answer 1/3..

You could probably argue that you have a different interpretation of the question and can come up with a different value, but that is unlikely to score you any points..

What if there are three states for the apple?

1. green - not ripe yet, not fit to eat
2. ripe - ready to eat
3. rotten - black with flies buzzing; not fit to eat.

Is it still 1 in 3 chance? Or does the number of possible states matter?

During the coin toss question the 1 in 3 chance is predicated upon there being only two mutually exclusive states (heads or tails) - correct?

[Sauron]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Unknown, as we don't know the ratio of ripe & non-ripe apples.

That is a good point. Two possible states (ripe vs. not ripe) does not equate to each state being equally likely. A box of 99 white ping-pong balls and one black ping-pong ball gives two possible states. But they are not equally likely.

[Furby]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Unknown, as we don't know the ratio of ripe & non-ripe apples.

That is a good point. Two possible states (ripe vs. not ripe) does not equate to each state being equally likely. A box of 99 white ping-pong balls and one black ping-pong ball gives two possible states. But they are not equally likely.

This is only true of the frequentist (http://en.wikipedia.org/wiki/Classical_definition_of_probability) or classical interpretation of probability, where the probability of an event is interpreted to be the frequency it occurs in a large number of trials.

Bayesian probability allows us to make an assessment based on our belief of a proposition, even if it is impossible to test the validity of the proposition in the real world. We change our assessment according to new evidence. Even if we cannot make an accurate one, it does not mean we can't make one at all

[Furby]

What if there are three states for the apple?

1. green - not ripe yet, not fit to eat
2. ripe - ready to eat
3. rotten - black with flies buzzing; not fit to eat.

Is it still 1 in 3 chance? Or does the number of possible states matter?

During the coin toss question the 1 in 3 chance is predicated upon there being only two mutually exclusive states (heads or tails) - correct?

If you truely have no evidence either way, then yes, you assign a 1/3 probabilty to each state. An apple-seller who has a large sample of apples in his store would likely come up with a different probability distribution, and his estimate would likely be much more accurate than yours.

If one day, you happen to meet the apple seller, and acquire information about his apples from him, you reassess your probabilities accordingly.

According to Bayesian intepretation, if you say, "there is a 33% chance that this apple is ripe", it reflects the degree of belief you have in the proposition that "this apple is ripe". As mentioned earlier, this may or may not be accurate depending on how much evidence you have.

[slimshady2357]

You have to count BG separately from GB for the same reason that 11 is more common on two dice than 12; 56 and 65 are two DIFFERENT cases, even though we count them as having the same outcome.

Brain check!

I'm sure you meant to say 12 is more common than 11.

I think you are reading it wrong, which I did as well the first read through.

He is saying eleven is more common than twelve, not one-one is more common that one-two. That is why he mentions 65 and 56, they are the eleven.

[Loren Pechtel]

OK, so new question. This may be tedious, but trust me - I have a point here.

There are two apples on the dining room table. Mom says one of them is ripe. What is the probability that *both* apples are ripe?

Unknown, as we don't know the ratio of ripe & non-ripe apples.

That is a good point. Two possible states (ripe vs. not ripe) does not equate to each state being equally likely. A box of 99 white ping-pong balls and one black ping-pong ball gives two possible states. But they are not equally likely.

Exactly. With the child problem we have simply been counting states as all the states are equally likely (other than the ones that we can exclude entirely.) This is actually just a simplification, you really need to count the probability of each state.

You have two boxes that each contain 9 white and 1 black ball. Given that you have at least one white ball what are the odds you have two?

9 white/1 black * 9 white/1 black =

81 WW, 9 WB, 9 BW, 1 BB.

We can exclude the BB case and the WB is equivalent to the BW =

81 WW, 18 BW.

Thus the odds are 81/99 that we have two white balls.

[Loren Pechtel]

You have to count BG separately from GB for the same reason that 11 is more common on two dice than 12; 56 and 65 are two DIFFERENT cases, even though we count them as having the same outcome.

Brain check!

I'm sure you meant to say 12 is more common than 11.

I think you are reading it wrong, which I did as well the first read through.

He is saying eleven is more common than twelve, not one-one is more common that one-two. That is why he mentions 65 and 56, they are the eleven.

You're right. It's unclear at first glance and I fell for it.

[seebs]

You have to count BG separately from GB for the same reason that 11 is more common on two dice than 12; 56 and 65 are two DIFFERENT cases, even though we count them as having the same outcome.

Brain check!

I'm sure you meant to say 12 is more common than 11.

I don't THINK so.

12 = 1/36
11 = 2/36

Edit: Whoops. I get it, and I see what you mean.

Yes, "1,1" is less common than "1,2" -- because "2,1" and "1,2" get collapsed together.

Okay, another way of looking at it:

Much of probability comes down to trying to decide whether or not two things are actually the same thing.

If we articulate all the possible events, and don't coalesce them into higher-level categories (such as "sets of dice with the same sum"), they are all equally probable.

Now, when we start coalescing them, the question is which cases we're considering. When we specify a specific position, we're reducing the amount of aggregation that happens.

"I rolled two dice, getting a total of 7. What is the probability that one of them is a five?" -> 1/3
"I rolled two dice, getting a total of 7. What is the probability that the first one is a five?" -> 1/6

BTW, anyone who's interested in this REALLY ought to buy a copy of Lady Luck by Warren Weaver. I think Dover's got it still in print.

[Perry]

A man and a woman both have two children.

It's unclear whether they share the same children.

[mickthinks]

Thanks Dragar

Children (and dice) are distinguishable. This means that we can talk about Child 1 and Child 2, even before we know anything about the two children.

However convenient it may be for mathematical purposes to claim that we can distinguish the children and label them before we know anything about them, it seems to me to be a highly dubious contention. Especially when our subsequent explanation rules out treating the one who is a boy separately from the one who might be a girl on the grounds that we don't know which is which.

Children about whom we know nothing are indistinguishable.

Mick

[Dragar]

What about dice we know nothing about?

If so, would you care to play a game of chance with me? :)

[mickthinks]

Hello again Furby

It is only when your expected gain can always be twice as much as your expected loss is when you run into the above paradox. [from here] (http://www.freethought-forum.com/forum/showthread.php?p=433387#post433387)
I think some aspects of the paradox appear in cases which are finite. If we imagine envelopes to contain sums matching any two consecutive terms in the series 1,2,4,8, ... 1048576, chosen at random with equal probability, the chance that both children receive an envelope that might be doubled by switching is nearly 95%. The girl's reasoning in Ceptimus's OP (http://www.freethought-forum.com/forum/showthread.php?p=433139#post433139) is sound. She has $100 dollars and she's better off swapping. If she had $200 she'd still be better off swapping. If her brother is looking at $200, he, like his sister, can argue that he'd be better off swapping. Both children seem to be better off swapping, so what's going on?

I think the answer is connected to the fallacy that says because the girl has an equal chance of being given the meaner envelope and the more generous envelope means that the contents of her envelope, when examined, are equally likely to be the meaner amount and the more generous amount available. But this is still just an observation; I haven't been able to build it into a satisfactory explanation.

Mick

[Dragar]

If you prefer a less flippant answer, the issue is distinguishable in principle, not merely in practice. And in principle we can distinguish children (fortunately for many parents, in practice too!). That we know nothing about them is not the issue; what matters is that with sufficient information we could, in principle, distinguish them. If you prefer - the universe can tell them apart.

Contrast this with, say, two identical particles who occupy the same region of spacetime (I am being deliberately vague as to what that constitutes in order to avoid de-railing the thread). In such scenarios, you cannot, even in principle, distinguish one particle from another. They have no differing heights, ages, hair colour, DNA, clothes, or anything we might use with children - and with their positions smeared out in a probability distribution, even their location cannot be used to distinguish them. Even the universe cannot tell which is which! It is almost as if the question ceases to make sense. In such a scenario we have to adopt the alternative premise when calculating probabilities and, perhaps most crucially, it works.

[mickthinks]

What about dice we know nothing about? If so, would you care to play a game of chance with me? :)
Ha ha! I think I can see the way you are thinking, Dragar, but it would be a mistake to assume that someone who thinks an explanation is faulty must also disagree with its conclusions.

Mick

[fargazmo]

I think that the problem with the money one is in their relative expectation of gains. If the boy agrees to switch, the girl knows that he either expects her to have exactly what she has, or to have 4 times what she has.

[mickthinks]

Thanks Dragar

... the issue is distinguishable in principle, not merely in practice.
I don't understand "distinguishable in principle" as opposed to "distinguishable in practice". What does this mean in practice?

what matters is that with sufficient information we could, in principle, distinguish them.
But in that case the difference in the information we have in practice should make no difference in the probabilities we calculate. We should calculate probabilities on the basis of any and all information that we could in principle have even when we don't have it in practice. But not having the information in practice is how we arrive at the probability than the woman's chances of having two boys is one third.

Mick

[Dragar]

In principle, there are facts about the children that are different facts (other than (possibly) gender, which is the issue under consideration.

In practice, I may never have the capability to learn those facts (I may, for instance, be in another country, or on another planet).

[mickthinks]

In other words; in practice, there are facts about the children that are different facts including gender, which is the issue under consideration, and, in principle, I may never have the capability to learn those facts.

Are we calculating probabilites in principle or in practice? is there a difference? Does any of this help to explain what's going on?

I don't think so.

Mick

[fargazmo]

We're calculating probabilities in principle. Otherwise known as theoretical probability. "In practice" would be experimental probability, which on a long enough time frame should come to approximate the theoretical probability.

[Dragar]

Are we calculating probabilites in principle or in practice? is there a difference? Does any of this help to explain what's going on?

I don't think so

Fair enough. That one worked for Sauron; perhaps a different explanation would help illuminate matters for you.

[mickthinks]

perhaps a different explanation would help illuminate matters for you.
Thanks Dragar, but there's no need. I'm content if it is now clear that "Children (and dice) are distinguishable" leads nowhere, and that generating satisfactory explanations requires a deeper level of understanding than generating correct answers.

Mick

[Dragar]

I don't think that's the case, but I don't seem to be able to explain it very well. A book on probability theory might be more enlightening than I seem to be.

[mickthinks]

Okay, having given Ceptimus's problem (http://www.freethought-forum.com/forum/showthread.php?p=433139#post433139) some more thought, I think I have got most of the way there.

The 'blind' scenario:
There are two different scenarios and they are worth considering separately. In one, the children swap envelopes without looking inside. The paradox arises because it seems to be possible to argue that "whatever I have in my envelope, call it $x, if I swap I have an equal chance of doubling as of halving it, and my expectation if I swap is $x + $x/4. That's 25% more than my expectation if I don't swap, so the swap seems worth it.", and since both children can make this argument, they are both better off with the other's envelope.

The error here is to discount the possibility that one of the envelopes holds the largest amount possible, call it $M. Although this possibility is a small one, nevertheless the sum is considerable. The loss of $M/2 is not just greater than every possible gain, it is greater than the sum of all the other gains. The expected contents of the swapped envelope is approx $x + $x/4 - $M/N where M and N are both large. This term cancels the apparent benefit of the $x/4 term.

As others have pointed out, this $M/N term seems to disappear if N were infinite, but two things to note;
If all the possibilities are equal then N = (number of possible envelopes), M = 2 ^ (N/2), and (2^(N/2))/N does not tend to zero as N tends to infinity, so actually this term never disappears.
Contrary to opinion popular with children, no parent has infinitely deep pockets.

The 'revealed' scenario:
In another scenario, the children open their envelopes and look inside before considering the swap. If they don't know how much their father was willing and able to pay into the game, then they are in a very similar situation to the blind scenario above, and must include a negative term of $M/N in their calculations, and there's nothing to be gained by swapping.

But suppose they know what M is and therefore know that they haven't got it. This isn't as easy as it may seem but, for the sake of argument, just suppose that Dad tells them M = $512, Sis is looking at £128, and Bro at £256. I still see a problem here.

It seems to me that, although we 'know' that Sis should swap and Bro should stick, they don't know that. As far as they know, Sis has an expectation of $128 + 25% = $160 if she swaps and Bro has an expectation of $256 + 25% = $320 if he swaps, and it still looks as if they are both better off with the other's envelope. Something is still wrong here and I can't see quite what it is.

As I say, we know the whole truth and can easily reason that Bro will lose, but unless we argue that what has gone wrong for them is lack of perfect information (!), we still need to find the fault in their reasoning.

:knock: Still pondering...

Mick

[1Samuel8]

Permit me to barge in here.
The fault lies in the assumptions made in trying to solve the dilemma: unless the two agents involved are exactly the same, the whole concept of statistical probability in predicting their behavior is invalid.

There is something that is also unclear to me: we do not know if the decisions are made:
- simultaneously
- in sequence
- who gets to go first
- who has control (in other words, if one wants to trade but the other does not, what happens?)


Let me try to illustrate the principle/practice dichotomy in a different way. Consider the following possible practical scenario:
- the boy cares only about buying chewing gum and comic books (cheap stuff)
- the girl wants a deluxe doll-house that is bigger than all of her friends' doll-houses combined and nothing else makes her happier than having the newest doll-house available in the toy store (expensive stuff)

After they open the enveloppes, there is no rational reason to believe they should both follow the same probability of trading. If the boy's envelope contains enough to supply him with his cheap expenses, we might not want to risk losing it. If the girl's envelope does not contain enough to buy a new doll-house -- the one and only thing in the whole world that will make her happier than she already is now -- it would make sense to risk losing it to gain more.

The boy may just want to know how much money the girl is given, thus, the only way he can learn that is by trading.

Conversely, knowing that the girl wants more money, the boy may find excitement in refusing to trade.

[seebs]

Yes, but that's all irrelevant to the interesting question: Why is the calculation of "expected value" producing garbage? Just assume for the sake of argument that we hypothesize two clones of Scrooge McDuck, both of whom have happiness EXACTLY proportional to their material wealth.

[1Samuel8]

I think the reason can be summarized thusly: garbage in, garbage out.

Why is the calculation of "expected value" producing garbage? The question is garbage because it contains a logical flaw.

The logical flaw is the following assumption:
Thinking that risking losing "half of your money" to gain "twice your money" is always the same independent of the nominal amount of money.

Look at this statement again: "Now obviously the boy, whether he found fifty or two hundred dollars in his envelope, reasons exactly the same - by swapping he is wagering half his money but stands an evens chance of winning double the amount wagered."
That quoted statement is wrong. The boy is NOT facing the same decision in both cases. Thus, there is no valid reason to believe that he will do the same thing as the girl.

Wagering "half your money" is not the same as wagering a specific amount of money -- in this case, either fifty dollars or two hundred dollars. This is the logical flaw of the scenario.

Even though both people have the same preferences, they are still risking different amounts of money.

We make decisions on how to invest 50 dollars based on how much 50 dollars is worth to us not because it is a ratio of an arbitrary number.


---


Eventually, a person should be expected to stop gambling regardless of arbitrary ratios. There is no way of telling whether that threshold is above or below any of the amounts ($50 or $100 or $200) in question. However, we do know that the threshold is the same for both the boy and the girl. Therefore, one of the two may be at that threshold.

[ceptimus]

Thanks for the posts guys. I've never entirely resolved this paradox to my own satisfaction.

I agree that when the amount of money discovered in the envelope is very small (one cent) or very large (one million dollars) then the child has extra information on which to base its decision: 'There aren't any half-cent coins, so I know I will gain if I swap'; 'I know that dad doesn't own three million dollars, so I must already have the larger of the two amounts.'

There is also the fact that the value of money is not linear. To most individuals, $2 million is not worth twice as much as $1 million - once you have more money then you are likely to ever spend, then additional cash becomes, in some sense, worthless.

But when the amount found in an envelope is mundane - say somewhere in the $2 to $500 range, then these objections seem to evaporate and we are left with both children able to make the convincing argument: 'I hold $X, and my expectation of gain if I swap, is $X/4'.

I can show mathematically that if the value of money decreases geometrically as the amount rises, then the children's reasoning is false, but for mundane everyday amounts it doesn't seem to me that the value of money does have this inverse geometric form. :dunno:

[1Samuel8]

But when the amount found in an envelope is mundane - say somewhere in the $2 to $500 range, then these objections seem to evaporate and we are left with both children able to make the convincing argument: 'I hold , and my expectation of gain if I swap, is /4'. In essence, what you are assuming is that the person is indifferent to all of the amounts of money in that range. For instance, $3 is equvalent to $250 and $499 in all cases. That is an unreal assumption.

I can show mathematically that if the value of money decreases geometrically as the amount rises, then the children's reasoning is false, but for mundane everyday amounts it doesn't seem to me that the value of money does have this inverse geometric form. :dunno: If so, you can not assume your probability function would stay the same over the "mundane" range either.

[Dragar]

I don't see why not. That £50 is worth twice as much as £25 is probably pretty accurate until you start reaching (as Ceptimus said) moneys of order £250.

But besides which, these considerations are irrelevent. We are not trying to predict behaviour; we are trying to determine what each person will concude about which action will produce the most money. How much they value that amount of money justified or not is not something particularly interesting.

[Sorrel]

That depends entirely upon your utility function.

If someone has no or very little money, and *needs* £25 to, say, be able to stay in a hostel for a week, or to be able to eat for a month, a certain £25 is likely to be worth more to that person than an uncertain 0 or £100 with an expected value of £50. If they gamble, and get 0, that person has nothing. They are risk averse.

For a gambler, a bet with an expected value of £50 is worth a whole lot more to them than £50 in the pocket. Certianly more than twice £25. Such people love risk.

It's all about individual preferences.