A simple question in probability theory. The answer is not so simple. Take a look at ::
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox to see how mathematics can sometimes be bewildering to many of us.

Probability is bewildering because it can seem both simple and complex. When viewed as simple, we tend to trust intuitive answers over deduced ones. When viewed as complex, we rationalize why such complexity must be ignored. I'll generalize the first illustration of this controversy, from 1889.

Say you have N boxes, N gold coins, and N silver coins. You put two coins into each box, with exactly P*N boxes having one of each. We can see that (1-P)*N have two of the same kind of coin, (1-P)*N/2 have two gold coins (or two silver), and (1+P)*N/2 have at least one gold coin (or at least one silver). So, the chances a random box has two of the same will be (1-P).

What if I select a box at random, and tell you it has a gold coin? What now are the chances it has two of the same kind? It is tempting to say it should be the ratio of "two gold coin" boxes to "at least one gold coin" boxes, or (1-P)/(1+P). But you would have to give the same answer if I told you it had a silver coin. And if the answer is (1-P)/(1+P) regardless of what I tell you about the contents, it has to be (1-P)/(1+P) if I don’t tell you anything. Yet we know that answer is (1-P).

This apparent paradox is known as Bertrand's Box Paradox. He used N=3 and P=1/3. Bertrand's resolution of it was to point out that the set with at least one gold coin is not the same as the ser where I'd tell you about a gold coin, since I could tell you about a silver coin in some boxes with a gold coin. Yet, the only way to get (1-P)/(1+P) is to assume that was impossible. You can only assume I'd choose randomly in such situations, and the answer becomes [(1-P)/2]/[(1-P)/2+P/2]=(1-P).

Bertrand's version is an isomorphism of the similarly-famous http://en.wikipedia.org/wiki/Monty_Hall_problem, although it takes an obscure twist of logic to make the comparison exact. Suffice it to say that the intuitive answer, that the prize is behind the remaining door with probability (1-P)/(1+P)=(1-1/3)/(1+1/3)=1/2, is incorrect. The answer is (1-P)=(1-1/3)=2/3. Most mathematicians today will agree - usually after having it explained to them after they intuitively say 1/2 - that 2/3 is correct under the assumption the only assumption that can be made.

In the Boy or Girl paradox, two intuitive blunders get in the way of seeing that it, too, is an example of Bertrand's Box Paradox, with N=4 and P=1/2. The first is to count only three cases, not four; eliminating one makes the answer 1/2. The second is to say "the chances the OTHER child is a boy can't be other than 1/2." These blunderers end up counting two-boy families twice, and also answer 1/2.

Once both blunders are avoided, one often stops looking for others. But just like with Joseph Bertrand's box, and Monty Hall's doors, the answer of (1-P)/(1+P)=(1-1/2)/(1+1/2)=1/3 is wrong. You would have to give that as the probability that there are two children of the same gender no matter what gender you are told exists. That has to be the same as the unconditional probability that there are matched genders, and that probability has to be 1/2. This answer does not - and I'll repeat "not" because few who answer 1/3 want to believe it - make either of the two blunders I mentioned above. It counts only half of the N/2 families with a boy and a girl, because only half of the time that you know one gender will that gender be "boy."

The point is, that it really isn't that complicated. All you have to do is recognize all of the possibilities, either implicit or explicit, in the random process. What creates the illusion of complexity, is the depths to which people will go to discredit this procedure. Which requires, in turn, those with the correct answer to go into great detail, as I did, to close any imagined holes in the argument. I had to show not only why my answer was right, but why the other must be wrong, with indisputable arguments.

The conditional probability of an event E, given a condition C, is P(E&C)/P(C). In this problem, E is "the family has two boys." But C is confusing to some: they think, for naive reasons, it is "the family has at least one boy." Since 3/4 of all two-child families is in this C, and all of the 1/4 who are in E are also in C, this ration is (1/4)/(3/4)=1/3. But that is the incorrect C; it really is "You know the family has at least one boy."

One fourth of all two-child families have two boys. One half have exactly one boy, but in half of those (whom you know only one gender) you will know about a girl. That makes P(C)=1/2, not 3/4, so the answer is (1/4)/(1/2)=1/2. That took only about 40 words to explain.

But to force people to see which C is correct, it is necessary to resort to Bertrand's Box Paradox. A single family can't be in the C for each of companion problems that are based on whether you know about a boy or a girl.

Probability is bewildering because it can seem both simple and complex. When viewed as simple, we tend to trust intuitive answers over deduced ones. When viewed as complex, we rationalize why such complexity must be ignored. I'll generalize the first illustration of this controversy, from 1889.

ReplyDeleteSay you have N boxes, N gold coins, and N silver coins. You put two coins into each box, with exactly P*N boxes having one of each. We can see that (1-P)*N have two of the same kind of coin, (1-P)*N/2 have two gold coins (or two silver), and (1+P)*N/2 have at least one gold coin (or at least one silver). So, the chances a random box has two of the same will be (1-P).

What if I select a box at random, and tell you it has a gold coin? What now are the chances it has two of the same kind? It is tempting to say it should be the ratio of "two gold coin" boxes to "at least one gold coin" boxes, or (1-P)/(1+P). But you would have to give the same answer if I told you it had a silver coin. And if the answer is (1-P)/(1+P) regardless of what I tell you about the contents, it has to be (1-P)/(1+P) if I don’t tell you anything. Yet we know that answer is (1-P).

This apparent paradox is known as Bertrand's Box Paradox. He used N=3 and P=1/3. Bertrand's resolution of it was to point out that the set with at least one gold coin is not the same as the ser where I'd tell you about a gold coin, since I could tell you about a silver coin in some boxes with a gold coin. Yet, the only way to get (1-P)/(1+P) is to assume that was impossible. You can only assume I'd choose randomly in such situations, and the answer becomes [(1-P)/2]/[(1-P)/2+P/2]=(1-P).

Bertrand's version is an isomorphism of the similarly-famous http://en.wikipedia.org/wiki/Monty_Hall_problem, although it takes an obscure twist of logic to make the comparison exact. Suffice it to say that the intuitive answer, that the prize is behind the remaining door with probability (1-P)/(1+P)=(1-1/3)/(1+1/3)=1/2, is incorrect. The answer is (1-P)=(1-1/3)=2/3. Most mathematicians today will agree - usually after having it explained to them after they intuitively say 1/2 - that 2/3 is correct under the assumption the only assumption that can be made.

In the Boy or Girl paradox, two intuitive blunders get in the way of seeing that it, too, is an example of Bertrand's Box Paradox, with N=4 and P=1/2. The first is to count only three cases, not four; eliminating one makes the answer 1/2. The second is to say "the chances the OTHER child is a boy can't be other than 1/2." These blunderers end up counting two-boy families twice, and also answer 1/2.

Once both blunders are avoided, one often stops looking for others. But just like with Joseph Bertrand's box, and Monty Hall's doors, the answer of (1-P)/(1+P)=(1-1/2)/(1+1/2)=1/3 is wrong. You would have to give that as the probability that there are two children of the same gender no matter what gender you are told exists. That has to be the same as the unconditional probability that there are matched genders, and that probability has to be 1/2. This answer does not - and I'll repeat "not" because few who answer 1/3 want to believe it - make either of the two blunders I mentioned above. It counts only half of the N/2 families with a boy and a girl, because only half of the time that you know one gender will that gender be "boy."

WOW ! this is getting more complicated than I thought. I was a happy man until I started learning maths.

ReplyDeleteAny way, thanks for your contribution.

partha

The point is, that it really isn't that complicated. All you have to do is recognize all of the possibilities, either implicit or explicit, in the random process. What creates the illusion of complexity, is the depths to which people will go to discredit this procedure. Which requires, in turn, those with the correct answer to go into great detail, as I did, to close any imagined holes in the argument. I had to show not only why my answer was right, but why the other must be wrong, with indisputable arguments.

ReplyDeleteThe conditional probability of an event E, given a condition C, is P(E&C)/P(C). In this problem, E is "the family has two boys." But C is confusing to some: they think, for naive reasons, it is "the family has at least one boy." Since 3/4 of all two-child families is in this C, and all of the 1/4 who are in E are also in C, this ration is (1/4)/(3/4)=1/3. But that is the incorrect C; it really is "You know the family has at least one boy."

One fourth of all two-child families have two boys. One half have exactly one boy, but in half of those (whom you know only one gender) you will know about a girl. That makes P(C)=1/2, not 3/4, so the answer is (1/4)/(1/2)=1/2. That took only about 40 words to explain.

But to force people to see which C is correct, it is necessary to resort to Bertrand's Box Paradox. A single family can't be in the C for each of companion problems that are based on whether you know about a boy or a girl.

This comment has been removed by the author.

ReplyDeleteI bet you must be right. But it will take me some time to understand why you are right. Let me try.

ReplyDeleteThanks a lot for your attention to detail.

partha

BTW:: Who are you ? Balaji ? Jeffjo ? Can you mail me your mail ID ? I would like to get in touch with you directly. Thanks.

ReplyDelete