Monty Hall Question

Cecil's Columns/Staff Reports
61

Thank you for your random numbers, Frylock, it helps me explain this better. We are in total agreement on the 79 families that said they had a “girl.” Now the next question is… what about remaining the 21 families that said they have boys? Chosing the opposite for them is an automatic lose. As this community in our example prides itself on daughters not sons, you will always lose if a son is mentioned. So if a daughter is given, you have in this survey around a 62% chance of being right. If a son is given you have 0% chance of being right. I think if you average them together you will get 49%. So at the end of 100 families, you have 30 girl/girl losses and 21 boy/boy losess. 51 out of 100 sounds closer to 50% than 66%, wouldn’t you say.

To clarify in this example:
When Girl is given and the family really has GG…30 out of 79 families
When Boy is given and the family really has BB…21 out of 70 families
When either boy or girl is given and the family has both…49 out of 100 families

62

Nitpick: there are 79 families that have a girl. There’s nothing to be said about families “which said they have a girl.” Its just a fact that, out of the 100 families, 79 have a girl.

I just have no idea what you mean by “choosing the opposite” “for them” and “is an automatic loss.” Whoever said anything about choosing anything? Who’s the “them” you refer to? “Automatic loss” of what? Who said anything about winning or losing anything?

What are you talking about? No one has said anything like this.

Lose what? Why? What are you talking about?! :dubious:

Right about what? After saying what?

About what? After saying what?

Losses of what? And how is counting “losses” relevant to answering the problem?

[quote]
51 out of 100 sounds closer to 50% than 66%, wouldn’t you say.

Sure but what in the holy heck are you talking about in this post?

Why don’t you look at the number the problem asks for–the number where a girl is given and the family really has both a boy and a girl? (49 out of 79.)

I think you’re trynig to rephrase the problem in terms more directly parallel to the Monty Hall problem. That’s valid, but you haven’t made it clear how you would carry out the rephrasing. Since the two problems are not exactly the same problem, it is necessary that you make explicit how you concieve of the accurate rephrasing of the boy-girl problem in terms of the Monty Hall problem.

-FrL

63

Please explain something to me. You say that of the 1024 combinations, the fact that there are at least 9 girls out of 10 children eliminates all but 11 sequences.

If there are 10 children, and at least 9 are girls, the only sequences I can come up with are a)10 girls, and b) 9 girls with one boy. What are the other sequences that you speak of?

If I am correct, then the probabality would, in fact, be 50-50 would it not? Am I missing something? Thanks.

64

You’ve identified the two relevant categories of family, but that doesn’t mean those categories are distributed evenly, 50/50, in the population at large. The all-girl (and all-boy) families are much rarer.

A couple has 10 children. There are a grand total of 2^10 = 1024 possible patterns of boys and girls that might result. All these patterns are equally likely, given the same assumptions as in the original two-child problem.

Of these 1024 equally-likely patterns, exactly one of them is 10 girls and no boys. And exactly 10 of the patterns are 9 girls and 1 boy. So families of the second category outnumber those in the first by 10 to 1.

Therefore, if you are told that a particular person has 10 children, and that 9 (or more) of them are girls — and if you’re told nothing else, so that you might as well assume this person is “randomly” chosen from the population — then it’s ten times more likely the person has 9 girls and 1 boy, rather than 10 girls and no boys.

Or to put it like the original problem… A person has 10 children. You are told that 9 of the children (at least) are girls. What’s the probability he or she also has a son? The answer is 10/11.

65

I’m pretty sure you meant to say, “the other child is a boy”.

66

You’re right…

Thanks for clarifying!

-FrL-

67

I see what you’re trying to say, but this isn’t correct. Look at the 100 families from your post 60. Suppose the first column represents, say, the older child, and the second the younger, the person only knows the values of the first column, and he doesn’t know which child is taller. He must be talking about one of the 53 cases where the first column is 1, and it’s 50/50 that the other child is a boy.

Alternatively, suppose the first column represents the taller child, and the second represents the shorter, and the person knows both values. It depends on how he answers when I ask whether the girl is the taller or shorter one in the case where both children are girls. Is he going to answer “both”, or will he just pick one? If he just picks either “The taller one,” or “The shorter one”, it’s still a 2/3rds chance the other child is a boy.

What’s relevent is whether the original 50/50 sample is restricted to only cases where at least one child is a girl based on a single child (no matter how that child is selected) or whether it’s based on both children together.

68

It looks to me like you’re bringing information into the problem which is not given in the problem–namely information as to what the person giving you info on the family “knows” as opposed to simply the information given as to what he says.

But I’m not sure you’re bringing that information into the problem in your analysis. Do you think you are?

Do agree that the following two problems have different answers?

A. You are told, “One child in a family of two children is a girl.” What is the probability the other is a boy?

B. You are told, “The taller child in a family of two children is a girl.” What is the probability the other is a boy?

And do you agree that the following problem is substantially the same problem, with the same answer, as problem B?

C. You are told, “The older child in a family of two children is a girl.” What is the probability the other is a boy?

-FrL-

69

No. I’m pointing out that just because (for the first response in the part I quoted in my post above) the person doesn’t know which child is taller, that doesn’t tell you whether he is speaking of a specific child (like your B and C), or both of them together (like A).

For the second response, where he says “taller” or “shorter”, you don’t specify how the person would answer if he knows both children are girls. If he would answer “both” when they’re both girls, the probability is 1 (since either “taller” or “shorter” must mean there’s only one girl). If he just arbitrarily says either “taller” or “shorter” when both children are girls, the answer is still 2/3. To get the probability to be 1/2, he’d have to know the relative heights of the two children, but not know the sex of the other child. This last possibility is the case that’s equivalent to your case B.

Anyway, the answers for B and C are clearly 1/2, and A is 2/3.

(and before someone pipes up, we’re assuming girl/boy, older/younger, and taller/shorter are statistically independent)

70

I think I see what you’re saying.

It is my understanding that we do know whether the person is speaking of a specific child or both of them together–namely he is speaking of both of them together–exactly because he doesn’t specify one or the other child. Of course, there is a sense in which “one of the children” is ambiguous in general, in much the way you are talking about. But that ambiguity does not come into play here–my understanding is that it is simply a convention of the English language, and most especially, of puzzles like this one, that the person hearing the puzzle has been given all relevant information. The only reading which makes this condition satisfied is the one which reads “one of the children” as proffering information about the pair, rather than about one specific child.

He can’t answer “both.” The problem is put in terms of “one of” the children. If an answer of “both” were an allowable option, then the option’s allowability would be made explicit by information drawn from the problem’s wording. Since it’s not made explicit, either:

A. The problem is ambiguous, or
B. The problem does not intend “both” to be a possible answer. (Not that the problem doesn’t intend it to be possible both are girls–just that whatever situation the problem is describing, it is a situation where the man, while remaining honest, couldn’t or wouldn’t answer “both.”)

We can eliminate possibility A on the following grounds. When a problem like this seems ambiguous, if one and only one of the readings the problem could be given constitutes a problem which is interesting and determinately solveable, then that reading is the intended reading. That principle holds because of the convention I mentioned, that these kinds of problems are intended to be such that they do in fact give you all the necessary information, even if they make you think a little bit to figure out how they do so.

Possibility B is therefore actual.

Then as with the early Wissdok (the later Wissdok seems to have gone insane (just kidding Wissdok!)) it looks like any disagreement between you and I is over how to interpret the wording of the problem, not over how the problem should be solved given this or that interpretation. In this post I’ve explained my reasons for believing the problem should be interpreted as I have interpreted it.

Yeah, I said that in one earlier post, but to be safe I should be reiterating that point!

-FrL-

71

Should be “between you and me.”

If you didn’t notice what I’m talking about, then ignore this post.

:slight_smile:

-FrL-

72

If I can address this post, I don’t think others have addressed it in this way.

I want to suggest a similar experiment.

The original card is either a King or an Ace with equal probability.

You toss in your own card, which is either an Ace or a King with equal probability. But first you bend the corner back.

You draw out a card with no bent corners. It’s an Ace.

What is the probability that the original card in the hat was an Ace?

Clearly, the probability is 100%: because of the experiment you conducted, you gained additional information. Even though you originally thought the probability was 50% of the card’s being an Ace, new data leads you to revise your estimate.

If you can revise your estimate with this experiment (or similar experiments, in which you throw a card in and immediately take it back out, or throw two cards in and then look at both of them), why can’t you revise your estimate based on other experiments?

Same thing with the kids. I know a woman who has has two children, and one of them is a girl. Her other child is also a girl. What is the chance that both of her children are girls?

Obviously, it’s 100%. My third sentence in that paragraph gave you additional information that allowed you to revise your original estimate of probability. This happens all the time.

Probability, you gotta understand, is a polite fiction. Nothing is ever 66% likely to be true: it’s either 100% or 0% likely to be true. The only reason we play with probability is because we have incomplete information, and probability constitutes our best guess about the situation. As we gain new information, we revise this best guess.

Daniel

73

Sorry, Frylock, I thought I sent a response but I guess it didn’t post.

Okay, what I was trying to explain is that 2/3 couldn’t be right. I took the problem one step further and went ahead and chose (or guessed) the opposing because it was assumed to be 2/3 likely. If you chose the opposing child in all 100 families you would be right only 50% of the time, wrong (or lose) 50% of the time. The result would have been the same if you guessed the other child was the same sex as the given child.
As your random generated sample showed 50% of the time a matched pair would appear in the children. If you had guessed with the assumed odds of 2/3, all the families with two girls or two boys, which is half the families, you would have guessed wrong, while the assumed “2/3 premise” says it should only be 1/3.
I spoke of the other 21 families in your example because that is what seems to be missing between our thinking here. They didn’t have any girls, but the same “2/3 rule for opposites” should work on them as well. But in guessing opposite in all 21 families you would be wrong 100% of the time because they have two boys. So using your original numbers, you would guess wrong on the 30 two girl families and 21 times on the two boy families…51 out of a 100 times. I would say that is close to 50/50. The 2/3 idea would seem to imply that out of 100 attempts you would guess correctly 66.67% of the time, but the reality is that every other attempt you are likely to have another “double” combination come up (though it may be the opposite the first).

74

If I understand what you’re saying, you’re completely ignoring a crucial piece of information in the question, which is THE FAMILY HAS AT LEAST ONE DAUGHTER.

You can’t choose the opposing child in all 100 families. There are only 79 families that can make the claim that they have at least one daughter. 49 of them have one son and one daughter.

75

Heck, I don’t even know what “choose the opposing child” means in this context.

Maybe you can help explain Wissdok’s meaning to me.

-FrL-

76

I do believe what wissdok is doing is trying to generalize the problem into:

There is a family with two children. You have been told this family has a child of a certain sex. What are the odds that the two children are of opposite sex?

In which case he’s confounding his generalization by not seperating the specific cases of “boy” and “girl.” Or, alternatively, he’s changing the problem into:

There is a family with two children. You have been told one of the children is of a certain sex. What are the odds that the other child is of opposite sex?

Which is a different problem (semantic quibbling aside).

77

wissdok, I see your point now.

The question is not “What percentage of families have two children of different sexes?” The answer to that is indeed 50% (49/100 in the actual data).

The question is: “Out of the families who have at least one daughter (the 79 families in the actual data), what percentage of them will have a son?” We can see from the data (49/79) that 66.6% is a credible answer, and the mathematics that others have done demonstrates this.

We could also ask: “Out of the families that have at least one son (78 in the actual data), what percentage of them will have a daughter?” Again, the maths shows that it should be 66.6% - and the data (49/79) support this.

78

It is true we are being asked about just a girl, but the 2/3 argument should hold either way. If we only do this one time, for a given girl, the results are misleading because simultaneously we could be doing this for a given boy. It doesn’t take 100 families, if we only had just two families, one with a given girl and one with a given boy… the odds of one of the family having a set of either boys or girls is 75%. In Frylock’s sample, if we kept score from the beginning on the success of choosing opposite of a given child in all the families we would have never even come close to 66%. You would always get the *mixed *families right, and always get the matched families wrong.

79

Wissdok,

In the actual Monty Hall case (with the cars and the goats) do you think Marilyn Vos Savant and Cecil’s answers are right–that once the non-goat door has been opened, you are right to “switch” 2/3 of the time?

-FrL-

80

If you flip around the boys and girls completely (replace male nouns with female nouns and vice-versa everywhere in the problem’s text), then yes, you will still get an answer of 2/3. Or, if you had been asked the complementary question, “What’s the probability the person has no son”, then the answer would have been 1/3.

But maybe you meant something else by “either way”.

I can’t quite make sense of this. Two families, with what makeup of children? By “given boy/girl”, do you mean “given there’s at least one boy/girl in the family”, or “given a particular child in the family has been identified as a boy/girl”? And by the final clause, do mean the odds that a family (either exactly one, or at least one) has a set (pair?) of both boys, or both girls?

Assuming you meant one family is given to have at least one boy, and the other given to have at least one girl (and both are two-child families, as usual), then the probability that exactly one of the families has same-sexed children is 4/9, and the probability that at least one of the famlies has same-sexed children is 5/9.

But if this is what you meant, or something like it, it’s a very different sort of problem than the original one. I’m not sure how it’s helpful.

There’s no “always” right or wrong for any guess you’d care to make. If you always said “yes, this person has a son”, you’d be right 2/3 of the time, and wrong 1/3. Likewise if you always said “no, he has no son”, you’d be right 1/3 of the time, and wrong 2/3.

Again, it’s possible I’m misunderstanding what you’re claiming.