Game Show Problem

[Elmo]

Niles said:
My question to you, is why don't you just state that the question is ambiguous or ill posed? Instead you seem to be insisting that the interpretation which gives 1/3 is not a valid interpretation.

Niles,
Because, the interpretation which gives 1/3 is wrong. To get 1/3, boys must be selected prior to inspection. Our statement,"The woman has two children and at least one is a boy" cannot be stated, as a true statement, prior to inspection. When it's made after inspection, then the answer is 1/2.

You outsmarted yourself with the widow's peak demonstration. If that's analogous to our question, then "The woman has at least one girl" would have to have two girls with probability one.

I used the lights demo simply to show the meaning of "at least one is". "At least one is red" means there is a red light and another light of which we have no knowledge. Suppose that there is a red light and an orange light. "At least one is red" gives no knowledge of orange. For the 1/3 answer to be correct, "at least one is red" must convey something of orange.
The four ways are equally likely, up to and until our statement was made. After the statement, then there is a boy in evidence, then bb and bg are no longer equally likely.

There are two children, at least one is a boy. That is not an ambiguous statement. With boys and girls equally likely, the probability for two boys is 1/2. Unambiguously!!

Elmo

Eldon

[davar55]

This is getting out of hand. P == 1/3 is correct, not 1/2.
All of these puzzle/problems have the same root.
Marilyn was right on the MHP, 2/3 not 1/2, and same here.

I'll try to briefly summarize the essential error that leads to 1/2.

THE SAMPLE SPACE IS DISCRETE!

The possibilities that can't occur don't occur
and must be summarily excluded from consideration.

The ones that remain, when equally likely, then divide the
unitary total probability of one into equal parts.

Three parts, three into one yields P == 1/3 here.

The same illogic that produces P == 1/2 here
produced P == 1/2 instead of P == 2/3 in the MHP.

Marilyn is correct again. Read her book!

[niles]

davar55,

In an effort to better define your dice rolling experiment, I asked you a simple question, "What would you have done if ODD-ODD had occurred?"

You didn't answer the question. You said, "I created the problem. I am the host." I fail to see how that sheds any light on the situation. You reminded us how Marilyn reigned superior in the Monty Hall controversy. Again, that does not answer my question. You state, "Controversy be damned." Not only is that eminently unhelpful, it suggests a loss of temper.

You will not prove any points by constantly repeating, "Marilyn is correct, Read her book." There is no such thing as proof by fiat. Moreover, I don't need a lecture from you on discrete sample spaces. I may be dozens of IQ points shy of Marilyn's record, but I am quite capable of doing elementary probability.

What I hope to do is to formulate some guidelines on how to translate the often confusing words, nuances, assumptions, etc. of a verbal problem into the rather precise discrete sample spaces which we can manipulate using the laws of probability. I am taking advantage of the opportunity to ask insightful questions to people who have strong opinions regarding their interpretations.

Discuss the topic if you wish, tune out if you must, but please don't tell me that this is getting out of hand. You and I and others have had agreements and disagreements on this board, and that is fine. Never have I told you or anybody else what they should and should not write about. I don't like a lot of the stuff that Fred Snider or bill write, or reading about people's diets, so I skip it. I write about the stuff I like to write about. That's mostly probability, and the philosophy behind it.

Finally - a little bit of nitpicking on your post with the ODD-ODD dice example. I glossed over an error you made. Let's go through it briefly.

We roll the dice. You tell me honestly that ODD-ODD did not occur so we can exclude that case. (To make everyone happy, let's add that it was decided to report on the occurrence or non-occurrence of ODD-ODD before inspection of the dice.) We now have {OE, EO, EE} each outcome with probability 1/3. We conclude that the probabilty of EE (both even) is 1/3. So far so good.

Now you ask me to (presumably roll the dice again and then) pick one die randomly, and you will tell me whether it is even or odd. It is EVEN. In this case, the sample space is either {EO, EE} or {OE, EE}. The probabilty of getting EE (both even) from this experiment is thus 1/2.

But you said the probabilty is 1/3 for both cases. Please clarify.

[niles]

[Elmo]

Niles,
Our question: A woman has two children and at least one is a boy. What are her chances for two boys?

There is a woman. A statement has been made about her. It says, "A woman has two children and at least one is a boy." That's all we know.

For the 1/3 answer there must be prior prejudice toward boys. Martin Gardner said, in his book, "Aha Gotcha!" that to get one third the flipper must agree, in advance, to say heads every time possible and reflip on two tails.

Our question is a written question. Our advance agreement must be written into our question. IT AIN'T There.

You said. "Your only objection seems to be that we are selecting boys prior to inspection in the second school of thought. Please explain why it is wrong to select boys prior to inspection. Philosophically, I might just say that we are selecting boys prior to inspection because we can-there is no rule against it. Practically, I can think of several examples where boys are significant. For example, if you were a mother of two children during one of those many times in history where all firstborn sons were killed, then "at least one boy" is equivalent to "in mourning" where "at least one girl" tells you nothing about whether or not your family is intact. "

You can do anything you wish to do, but put it in the problem statement. "At least one" does not convey prior selection. Our problem statement is all we have. We have nothing about prior selection.

It is wrong because it adds to the information.

Our statement was said about our woman.

She has bb and the statement was said, OR
She has bg and the statement was said, OR
She has gb and the statement was said.

That's all we know.

Martin Gardner alluded to ambiguous questions. I've read everything he wrote about this question which I can find. So far I haven't found where he said exactly which question was unanswerable. I was in touch with Martin by fax. He would not discuss this question, said he'd spent too much time on it allready.

I defer to Martin's overall knowledge and Marilyn's intelligence. On this one question I have found no one who knows it better than me. (so far)
Eldon
_________________
Challenged Marilyn in 1997 on the sex of the other sibling question. Told her I would put $1000 into her favorite charity if my argument was wrong. She said, "You're wrong, send the money to the American Heart Association."

[Elmo]

Niles said:

[davar55]

[niles]

[niles]

Elmo,

I want to respond to your eleven points. I can't do it now. I need to give them more thought. Moreover, I have to do some work to earn a living. I have to raise my two children, at least one of which is a boy (really ).

Tonight, I just want to offer a few of my thoughts on why this problem is so tough and controversial.

I'll start with a brief summary of what I understand to be basic probability so that we're all on the same page. I've said many times that the mathematics - the probability - is not hard. Probability is all about experiments which don't always turn out the same way. The different ways that the experiments can turn out are called outcomes, and the set of all possible outcomes is called the sample space. Often times, we can assign probabilities to these outcomes and proceed from there. We're quite lucky in the problem at hand, since all of these are easily determined! The experiment is to determine the gender of each of a woman's two children. The set of outcomes for this experiment is {BB, BG, GB, and GG}; these four outcomes define our sample space. The probability of each is 1/4.

The next step in elementary probability is to define events. Technically speaking, events are subsets of the sample space. For example, BOYS = {BB} is an event. This is the event that the children are both boys. SAME = {BB, GG} is another event. This is the event that the children are both of the same gender. Since the outcomes are always mutually exclusive (i.e. one and only one outcome can happen), the probability of any event is given by adding the probabilities of the individual outcomes. For example, P(SAME) = P(BB) + P(GG) = 1/4 + 1/4 = 1/2. This says simply that the probability of both children being of the same gender is 1/2. How about the event that's been causing us so much trouble: AT LEAST ONE BOY = {BB, BG, GB}. It's easy to see that P(AT LEAST ONE BOY) = 3/4.

Next, we go on to define conditional probability. This is where things begin to get a little tough, but much more interesting, too. Conditional probability lets us talk about the probability of one event happening given that another event occurred. For instance, what is the probability of two boys given that both children are of the same gender? We defined both of these events (BOYS) and (SAME) above. In words, we want to know the probability of BOYS given SAME, or in symbols, P(BOYS | SAME). Although there are formulas, we often "talk" our way through the solution by means of the reduced sample space. For instance, we note that the probability of two boys given that both children are the same gender is 1/2 because {BB} is one of two equally likely outcomes making up the "both children are the same gender" event { BB, GG }.

Now, lets consider two very similar problems:

(1) A woman has two children, and both of these children are the same gender. What is the probability that she has two boys?

(2) A woman has two children, and at least one is a boy. What is the probability that she has two boys?

Problem 1 is easy. Nobody will have any trouble interpreting this as a simple conditional probability question. P(BOYS | SAME) = 1/2. Done.

Problem 2 is, well, problematic. If we follow exactly the same paradigm as established in problem 1, we interpret this to be P(BOYS | AT LEAST ONE BOY). Since BOYS = {BB} and AT LEAST ONE BOY = {BB, GB, BG}, we can see that BB is one of three equally likely outcomes in the reduced sample space, and therefore the answer is 1/3. Done? Not so fast....

There is, at least from Elmo and others like him, severe resistance to solving problem 2 in the same way that problem 1 is solved. What is the difference in these two? That is what I am trying to shed some light on.

Lets look at the event AT LEAST ONE BOY. What is it about this event that makes the problem so difficult?

Well, it's hard to observe this event directly. Recently we were discussing two lights, at least one of which is red. Elmo said that this means that one light is red and the other is unknown. Not exactly true. What it means is that one light is red and the other is unknown, or one light is unknown and the other is red. Think about that. It means that observing one light as red, with the other unknown is NOT equivalent to observing at least one red. This is because it is possible that I could observe a not-red light, and still have at least one red if the unknown one happens to be red. In other words, having at least one red is not "if and only if" equivalent to observing one red with the other unknown.

Herein lies a major difficulty with problem 2. If I observe one of the woman's children, and the other is still unknown, haven't I observed, "at least one boy?" Technically, yes, but technically I have also observed the "everything" event of my entire sample space {BB,BG,GB,GG}. And certainly, I have observed one specific child. I may not know which one I have observed, but I can say for sure that I have either observed {BB, BG} or {BB, GB}. So I can make three statements at this point.

{BB,BG,GB,GG} occurred. Therefore the probability of two boys is P( {BB} | {BB,BG,GB,GG}) = 1/4.

AT LEAST ONE BOY occurred. Therefore the probability of two boys is P( {BB} | {BB,BG,GB}) = 1/3.

{BB,GB} or {BB,BG} occurred. Therefore the probability of two boys is P( {BB} | {BB,BG}) or P({BB} | {BB,GB}) = 1/2.

It seems like a paradox, that we get three different probabilities from the same information. Obviously, it behooves us to use the most restrictive sample space possible given the information. Using the full sample space {BB,BG,GB,GG} to derive probability 1/4 in the presence of more restricted sample spaces seems silly - we're not using the information at hand. Similarly, it's silly to use the AT LEAST ONE BOY sample space when we have the more restrictive {BB,BG} or {BB,GB} at our disposal.

So how do we make the problem meaningful in practical terms? We can take what I might call the davar55 approach: "I told you about the event AT LEAST ONE BOY and only that event. It is therefore the best information you have. This is obviously a conditional probability problem, so shut up and do it."

That approach is unsatisfying, because it leaves us wondering how, in practical terms, we got the event AT LEAST ONE BOY as the best restriction of the sample space possible. And we have seen that it is not easy to do this, because the obvious way to find out that there was AT LEAST ONE BOY is to observe the boy, and in so doing, we have observed a specific child, and can further restrict the restricted sample space from three events to two, increasing the probability of two boys from 1/3 to 1/2.

This is why, in my recent discussions, I have always had to contrive something extravagant to explain how we found out the "at least one" information. For example, when we discussed two lights, at least one of which is red, we supposed the non-red light was black and the event was manifested by a red glow on a distant wall. When we discussed genetics, we observed the widow's peak, which occurs when a person has at least one of two genes for it. When we discussed at least one boy, we discussed whether the parents might be in mourning if some mighty power struck dead all firstborn sons overnight. These are all "if and only if" equivalent to the "at least one" event. Logically, it means that not only does the phenomenon imply the "at least one" event; but the absence of the phenomenon means that the "at least one" event did not occur.

This strongly suggests that you cannot observe "at least one" directly. You have to observe a side effect of it. If you do not describe that side effect in the problem, it leaves the reader with a confused and empty feeling, and sometimes prompts him to concoct extravagant schemes of his own, such as, "maybe they are telling me at least one boy as opposed to at least one girl." With that in mind, the problem is further analyzed using what I'll call meta-events: the event in which you told me a specific event occurred! In other words, what is the probability that there are two boys, given that you told me there was at least one boy. And we've seen where that leads.

------------------------------

That was quite a bit of writing, and I think the point comes across, but I am not sure. One thing I have noticed about people like Marilyn, who are in the one-in-a-million IQ club, is that they often have deep insights on what we might call philosophical issues, or paradoxes, and have clear and concise ways of resolving paradoxes and getting points across. It is said that Marilyn reads these posts; but what happens from there is anyone's guess. If you are reading, and you are amenable to re-opening the "two-child" issue, perhaps you could offer some insights as to why problem 1 above is universally seen as a simple conditional probability plug-in, but problem 2 causes many people to reject the notion of setting up a simple conditional probability situation, and proceed with a different set of assumptions. Respectfully, Mr. Niles.

[davar55]

[Elmo]

Martin Gardner, in his book “Aha Gotcha” said

[niles]

[niles]

Elmo,

We're getting into the realm of interpretation here. You are essentially arguing that "at least one boy" should be taken as opposed to "at least one girl." Given that the word "girl" does not even appear in the problem statement, don't you think it is a bit overreaching?

Sometimes, its best to show a counterexample to illustrate the absurdity. Consider this.

An integer lies between -10 and +10. It is equal to its cube. What are the chances that it is negative?

Answer 1: There are three integers in the given range which are equal to their cubes: -1, 0, and +1. Of these -1 is the only negative number. Since we have no reason to believe any integer in that range is more likely than another, the answer is 1/3.

Answer 2:

Our statement was made about our integer. All we know about our integer, we learned from the statement. For the one third answer to be correct, there must have been prior selection of cubes. "Equal to its cube" does not communicate prior selection. "Equal to its cube" could not have been stated, as a true statement, prior to investigation. When the choosing was done after investigation, then the probability for -1 is no longer equal to the probability for 0 or 1, because 0 and 1 are also equal to their squares, and -1 is not. Our integer is -1 and the statement was made, or 0 and the statement was made, or 1 and the statement was made. If the integer was 0 or 1 and the statement was made, then "at least one is a square" would also have been a true statement, so it is equally likely in case of 0 or 1 that cubes or squares are chosen.

Since cubes are twice as likely to be chosen for -1 than for either 0 or 1, and cubes were in fact chosen, then we can let the chances for cubes be x/2 for 0 and 1, and x for -1. Since all of these chances must add up to 1, we find that x + x/2 + x/2 =1, or x = 1/2.

Are you starting to see the problem? Here are two solutions. Each is logically correct, depending on the assumptions of the reader.

The first reader interpreted it as a simple conditional probabilty problem and got 1/3.

The second reader interpreted the outcome as a meta-event (what is the probability that we were told about cubes as opposed to being told about squares), and proceeded to evaluate the chances for -1 given that we were told cubes as opposed to squares, and got 1/2.

Most people would think the second solution was a little outrageous, because it introduced the concept of squares, and meta-events, when clearly only cubes were mentioned in the problem. Much in the same way you introduced the possiblity of "at least one girl", when only "at least one boy" is mentioned in the problem.

Given that we're arguing over interpretations here, don't you think we're relying on opinions and common sense, rather than a logical proof that the answer is 1/2 as opposed to 1/3? How on earth your eleven statements are going to constitute a logical proof of the invalidity of a particular interpretation is beyond me at this point.

[Elmo]

Niles said:

[niles]

Elmo,

Quick post this time. I'm getting ready to take my "at least one boy" someplace fun this weekend.

You wrote quite a bit. It's clear that you are under the fixed and immutable belief that "at least one boy" is to be interpreted as opposed to "at least one girl."

Of course I know that my "cubes and squares" example is silly. All you had to do is point out that heads and tails, boys and girls, odd and even, share a special relationship that cubes and squares do not. Of course heads implies not tails, boys implies not girls, odd implies not even, but what do cubes and squares imply about each other? Nothing. I was simply trying to show how your argument, with a simple verbatim substitution of a few words in a different problem, leads to an absurdity. Clever, was it not? So there must be something missing or false in your argument, something that would prevent me from using it to prove such an obviously flawed result.

You are interested in your $1000. Or maybe you have other motives. That is fine. I don't ever expect you to change your fixed and immutable mind. And you may never have to, because, after all, I'm with Martin Gardner on this one. I think that there are two equally valid interpretations. There are good arguments both ways. And as we all know, with human beings as arbiters, an argument is only as good as the eloquence of its proponent.

I, on the other hand, am interested in the hypothetical book of guidelines and rules for interpreting probability questions. Such a book of guidelines does not exist as far as I can tell, and I wish that it did, so we could simply refer to it and decide this problem once and for all. Even if it did, there would probably be one rule enforcing the 1/2 interpretation, and another enforcing the 1/3, and then who knows, maybe a judge would have to decide. Maybe I'll write the hypothetical book if I get enough insight from this issue. Most likely, I won't because I am very lazy.

Until then, we must rely on simple common sense in all cases to determine what context phrases such as "at least one XXX" or "equal to its cube" are to be interpreted in. Some of us have different sensibilities than others. Of course, you can always clear up any ambiguity by putting sufficient detail in the problem statement. So why not do so?

Marilyn should have restated the problem as follows. "A man and a woman (unrelated) each have two children. The woman is asked if she has any boys. The man is asked if his oldest child is a boy. They each answer yes. Do they have equal chances for two boys?"

If that had been the question, would Elmo have gotten his fifteen minutes of fame? Would we be discussing this now?

Have a nice weekend, all.