Yes, I agree with the original Monty Hall answer. I do know that Monty Hall disagreed with Marilyn, so I would like to hear is argument; But as it stands, I would switch doors.
Yes, Bytegeist, I was talking about two families…one family that we are given a girl and another family that we are given a boy. The 75% is based on the fact that half the time you will get a matched pair on a single family, the likelihood on any two families is 75%.
In the original answer by Cecil, the child would be the opposite of the first 2/3 of the time. This is because Cecil believed that after you removed the one possibility that is now impossible, the other three are equally likely. His 2/3 came from the two mixed possiblities, with the remaining 1/3 possibility belonging to the only possible same sexed (matched) family. This is where I get the argument that you would always guess right by picking opposites in mixed families and always wrong in families that had matched children.
I can’t agree with that, unless I’m just not following you. “One family that we are given a girl” means what, exactly? That we are told this family has at least one girl? Or are we told about a specific child, that she is a girl? Something else?
For a two-child family of any type, picked randomly from the entire population, the probability is indeed 50% that the children will be of the same sex. But if you’re stipulating (as you seem to be) that the one family has at least one girl, and other at least one boy, then those probabilities are no longer 50% and 50%.
Well, who wants impossible possibilities left lying around?
Let’s go over the possibilities again in painful detail.
There are two ways that people in this thread have been breaking down the population of all two-child families. Both are fine; neither is especially more useful than the other. Here are the types of Breakdown #1:
[ul]
[li](1A) first child boy, second child boy (1/4)[/li][li](1B) first child boy, second child girl (1/4)[/li][li](1C) first child girl, second child boy (1/4)[/li][li](1D) first child girl, second child girl (1/4)[/li][/ul]
(The words “first” and “second” there do not mean there’s a favored ordering for the children.)
And here are the types of Breakdown #2:
[ul]
[li](2A) both children are boys (1/4)[/li][li](2B) one boy, one girl (1/2)[/li][li](2C) both children girls (1/4)[/li][/ul]
If you prefer Breakdown #1, then the problem is essentially asking you to give the probability that a random family is either of types 1B or 1C, after type 1A has been excluded. Types 1B, 1C, and 1D remain possible. This probability then is (1/4 + 1/4) / (1/4 + 1/4 + 1/4) = 2/3.
If you prefer Breakdown #2, then the problem is asking you to give the probability that a random family is of type 2B, after type 2A has been excluded. Types 2B and 2C remain possible. This probability is then (1/2) / (1/2 + 1/4) = 2/3.
How you choose to group the possibilities doesn’t matter too much, as long as you know the relative proportions of the groupings. (And as long as the groupings don’t violate some condition of the problem.) Cecil was making use of Breakdown #1, but equally probable categories are not an essential assumption of his answer.
Of course it’s true that if the children are mixed, then you should guess that they’re mixed, if you want to guess correctly. The trick would be in knowing when to do that.
But you’re not really being asked to come up with a decision strategy. You’re effectively being asked what proportion of the time you’d be correct, if you always guessed “yes, this guy has a son”, after enduring a large number of repeated trials.
I agree with the numbers in Frylocks sample, but I don’t think they are **mutually exclusive. **For both to be right at the same time you would need 149 families not the 100 families that were in the sample. If we added the 49 mixed families to back this possible we would also need to add the 51 matched families which would require adding even more mixed…then matched…etc.
The main part of this problem is what is given to you and what you can make of the information. As I have stated before, a pair of same sex kids are possible 50% of the time, so 2/3 are only possible when looking at it from just girls or boys not both. For this problem with a given child, this is a practical dilemma. For the Frylock’s sample the 49 mixed families allow for 79 ( with the 30 girl/girl families) because a “girl was given,” at the same time these 49 families cannot be “boy is given;” It is either one or the other. If all 49 are “girl is given” then the only families with “boys are given” are families with two boys. So if the 62% is correct then at the same time your chance with a given boy is 0%. If we assume that half of the 49 said “given girl” and the other half said “given boy” then the results would be 45% girls and 53% for boys.
As to the statistics Bytegeist, the real numbers for a standard family of two random kids is: 50% matched pair/ 50% mixed pair. From that we get
25% all girls
25% girl and boy
25% boy and girl
25% all boys
If we take any two random families with two kids we have a 75% for at least one matched family and a 75% chance for at least one mixed family. (Same logic as the girl/boy.)
25% matched and matched
25% matched and mixed
25% mixed and matched
25% mixed and mixed
You’re just asking different questions about the same data. You don’t have to “add back” and double count and so forth. Just examine the data as they are.
You’re making this more complicated than it really is. I suggest you state specifically what the problem is that you’re solving. What is the question?
This is a good example of how tricky these can be, and why phrasing the question precisely is important. If you meet a girl and guess she has a brother (and we assume you know she has exactly one sibling), the probability she has a brother is 50%, not 67%. Where the mistake is above is that if you meet one of the two siblings, for the case where they are sisters, the chance that the one you met is a girl is 100%. If they are brother and sister, then it’s only 50/50 you’ll happen to meet the girl.
In the sample, there are 109 girls, and 91 boys, and you’re equally likely to meet any of them.
If you meet a girl, and guess that she has a brother, you’ll be right 49 times and wrong 60 times. 49/109 = 0.45
If you meet a boy, and guess that he has a sister, you’ll be right 49 times and wrong 42 times. 49/91 = 0.54
I found this on the web but I haven’t made it to the library to verify the original source. As it appears, I didn’t have an epiphany after all, others got this idea well before me. You might want to checkout my online source for this article at: http://www.wiskit.com/marilyn/boys.html
It appears on this website that others have also focused on the differences between the source of the “given girl” and which answer is correct. Those that have read all my post can recall I argued that for Cecil’s answer to be right… that the source had to know of both children. I guess I didn’t invent anything new there either.
What is my argument? For Cecil to be right I believe that these events must be assumed:
The source of the “given child” must be knowledgeable about the sex of both children. The source could not know of only one child because that child would then be specific.
The source must convey to you about the given child in such a way that you cannot make a distinguishing difference between the two children in a family. The most likely source for this scenario would be a parent being asked, “Do you have any girls?” or “Do you have any boys?”
That whatever child is given, all families that have that type of child would always provide the same clue. If a girl were given, then all families with girls would always speak of girls, even if the family had boys. It is also necessary that we were on the side that these 75% of the families are leaning, not the other side (25%) that these families are answering against.
I think that this problem requires too many assumptions to validate Cecil’s 2/3 claim. The fact that Cecil and Marilyn agree doesn’t make their conclusions any more likely to be right.
Why must the source be so knowledgeable? Supposing that he only knows one of the children to be a girl, and has no idea what the other child is, he can still report truthfully, “This family has a daughter.” It doesn’t matter what else he might or might not know. You can’t tell the difference anyway, whether he knows any more than he says.
Agreed to that — except that it doesn’t really matter how or whether the parents were interrogated. They could even both be dead, and the math problem would still live on.
Well, obviously if the family has two sons, your information source couldn’t have said, truthfully, that they have a daughter. That would be a big fat lie. Whatever information the problem gives us is assumed to be a reliable reflection of the underlying reality.
Obviously I don’t agree with those percentages — but beyond that, you’re still kind of reversing the flow of the reasoning. Assuming that whatever information we’re provided is correct, and given that our information source said, effectively, “This two-child family has at least one daughter,” then we can safely conclude that the children can’t both be boys. That fact, plus the other information provided (that child sexes are to be treated like coin flips), lets us calculate the probability the problem asks for.
That’s true — since they’re already 100% likely to be right in this case.
The cite… the links… your argument… they’re just… wrong.
It simply doesn’t matter how the informant came to have his information. All that matters is what he tells you–the sentence he utters to you. That’s the given in the problem. There doesn’t even need to be an informant in the problem. (Is there one in the question as worded in the OP?) It’s just a bare, unuttered given–a condition of the problem–that “one of the children is a girl.” Its also an implied given that “you don’t know how this information was arrived at.”
If you did know how it was arrived at, and it was arrived at in the second way mentioned in your SA cite, then the answer to the problem would be different–because it would then be a different problem.
The third cite you provided claims that, since we don’t know how the info was arrived at, the problem does not have a determinate answer. That’s incorrect. I’m not sure how I could possibly argue for this. I would need it explained why anyone would think they need to know how the information was arrived at in order to be able to give a determinate answer, before I could explain to them why they are wrong.
Frylock, I’ll just link to my earlier post in this thread as an example showing that how the information was determined affects the probability you will calculate, and how it can lead to the exact same informative statement (cf. Chronos’s post). Although I’m not 100% sure you’re in disagreement with what I said.
Certainly the problem is not ‘indeterminate’ because the information is ambiguous. Properly you should make clear what you’ve assumed the likelihood of each case to be when stating an answer. If we assume, based on this particular problem statement, that the person telling you would know or not know with equal probability, then you’d get an answer of (1/21/2 + 1/22/3=) 5/12.
You could easily adjust that in either direction based on further assumptions. wissdok pushes it entirely toward the first probability, assuming the second case to have probability 0 (in statement #2 above), though I don’t really agree with the reasoning.
Right, information as to how the data were obtained can change the probability, because additional information in general can change the probability. But the argument here seems to be that information about how the data were obtained is necessary in order to even be able to calculate a probability. This is what I am claiming is incorrect.
Since the problem as stated doesn’t say anything about how the data were obtained, you just take as an understood “given” a fact that “you don’t know how the data were obtained” and go with that.
So all you have is the following true statements:
There exists a family (call it F) with two children, one of which is a girl.
In general, half of all children are boys, and half are girls.
I maintain that this set of statements suffices to calculate a probability that both children in F are girls–and that that probability is 1/3.
And I maintain that this set of statements constitutes the problem unambiguously expressed by the wording of the OP.
The chance for any specific child to be a boy is 50/50. If the source knows about one child (the daughter), and has no information about the other child, the chance for the other child to be a boy has to be 50/50. How could it be anything else? You have no information about him/her.
If your source knows only of one child that child is specific. Knowing of one child is no different that seeing one child. This isn’t an idea I made up, both Marilyn Savant and Martin Gardner (the guru of logic and probability puzzles) both have stated that seeing one child changes the dynamic of the puzzle. The difference between the two problems is where the focus of the problem is…either the family or the child. If the statement of the source is general enough then all families are equal; if the statement of the source is not general enough then all the children are equal. Frylock is right that we don’t even need a source in the problem, but in this case we did. Cecil’s problem said we were **told **of a girl, not just given a girl.
As I explained earlier with cards, knowing just one child makes this problem 50/50. Make two piles of cards each with one red card and one black card. Have a friend draw two cards and look at just one. As long as he looked at just ONE card, you have a 50/50 chance to guess the other card correct. It doesn’t matter which pile the card came from because the other pile had exactly one red and one black. The focus here, with knowledge of just one card, is on the cards and their color. If your friend draws again and looks at both cards and then tells you just the color of one card, your chances are 1/3 and 2/3 respectively. The focus here, with knowledge on both cards, is with the possible hands that have the given color. This example is just like the family in question. In my require assumptions #1 and #2, the focus is placed on a random family other than on a random child. If we are given a girl and the family is random then there are 3 possibilities {GG,GB, BG}. If we are given a girl and the girl is random then there are 4 possibilities related to that girl (GG,GB,BG,GG).
What I was saying here was that the parents didn’t tell you of a child through use of a statement that gave a distinguishing feature to one child. Other posts have spoken of things like older/younger and taller/shorter, but it doesn’t need to so obvious. A simple statement like “We have a daughter that is a cheerleader,” would be specific to all girls not to families. In the family with two girls, both girls would be equally likely to be the “cheerleader.”
This is the same requirement that Scientific American stated in 1959. In the families that have both a girl and a boy, we have to assume that they will always answer one way. All the families with girls (if we are looking for girls) will always say they have a daughter even if they also have a son. As pointed out in the 1959 article, if the families with mixed children answered randomly, then the answer for original question would be 50/50.
“Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says “at least one is a boy.” If both are girls, he says “at least one is a girl.” And if both sexes are represented, he picks a child at random and says “at least one is a …” naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases – BB, BG, GB, GG – and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.” Scientific American, October, 1959
I don’t know why you don’t agree with the 75%. If we choose boys, 75% of the families have boys. If we choose girls, 75% of the families have girls. If we assume, as I explained above, that all families that have girls speak only of girls, then we also can conclude that we might be given a boy, which the only possible outcome for the other child is another boy.
By the way, I found a link to Monty Hall’s reply to Marilyn if anyone is interested. This is from the article titled *Behind Monty Hall’s Doors: Puzzle, Debate and Answer? * New York Times, July 21, 1991.
Regardless of what the source knows, you only know what you’re given in the problem, and what you’re being asked to calculate. The meaning of, “You have been told they have a daughter,” (effectively: “they have a daughter”) doesn’t depend on any extra information the informant might or might not have. Unless you have telepathy, the knowledge in the informant’s head is unavailable to you anyway.
If either child is identified, by any means (being seen, met, pointed at, marked with a crayon, shaved bald), and we’re asked about the other, non-identified child, then yes, certainly, it changes the dynamic of the puzzle. Basically, that would be a different puzzle than the one given.
Certainly the language of this problem’s text can suggest, if read a little too hastily, that it’s the one kind of problem when it is instead the other.
Mmm? We don’t need a source, but in this case we do (need one)?! Or did you mean: we don’t need a source, but in this case we have one anyway? Either way I’m not sure what that says.
“You have been told this family has a daughter. What are the odds they also have a son … ?”
We’re not told the sex of a specific child, and we’re not asked the sex of a specific child.
Right. In fact, I don’t care whether he looked at one card, both cards, or even no cards. The chance that the one specific, pointed-at card is black is 50%, regardless of what the other card is.
… from a pair of re-shuffled piles, one assumes …
I would word that differently. “Your friend … tells you the color of one card” strongly implies that he’s singling out one specific card and telling you its color. What I think you meant to say is that he tells you there’s at least one card of some particular color, but you still don’t know which card or cards of the pair have that color.
Assuming that’s what you meant, then we’re also no longer being asked to guess the color of a specific card, as we were in the other scenario. Our odds are then 1/3 that both cards are the same color, and 2/3 that they are different colors. If that’s what you’re saying, then we agree so far.
You keep using the phrase “given a girl”, but it’s not clear what that means. I wish you’d say either “given that there’s at least one girl” or “given that one particular child is a girl”, depending on what you intend — even though that’s a lot more typing — because this might be part of the difficulty in communicating.
To rewrite that last sentence a bit: if we are given that there’s at least one girl among the two children, and this two-child family was picked randomly from the set of all two-child families, and the sexes of all children are equally likely and independent of one other, then indeed, the three possible, equally-likely familes we might be facing are {GG, GB, BG}. If this is what you meant, then there’s no disagreement here either.
Now you’ve confused me again. In addition to clarifying the phrase “given a girl” you’re also going to have to elaborate on what you mean by “the girl is random”. There is no “the girl” to be talked about if all we’re given is that there’s at least one girl among the pair.
You’re making a good point about the distinguishing features, which is why I’m puzzled that you reach the opposite conclusion. You’re correct, there is no identified girl.
One point though about ordering the children — by height, age, or whatever — is that it doesn’t matter at all how you do it, or if you do it. Yes, the children are probably distinguishable by some criterion or another, apart from their sexes (when those are different). But this doesn’t matter for the problem. There are two snurples, equally likely to be blork or non-blork, independently. You are told at least one of them is blork. What are the odds that at least one of them is non-blork?
It can help to distinguish the children somehow in the mind, as long as we get the same answer when we reverse the distinguishing features — because those are irrelevant to the sexes of the children, for this problem anyway.
If I follow you, you’re saying that if a family has two girls, they are twice as likely to come up to you and say, “I have a girl. Guess what my other child is.”
That’s not true. A family with two girls is only just as likely as one with a boy and girl to walk up to you and say that. The family with two girls won’t walk up to you twice.
Since there are twice as many families that are boy/girl, we get 2/3.
Sorry for the delayed response…I’ve been a little busy.
Original statement: The Mclaren F1 is the fastest production car.
The Mclaren F1 is the fastest car.
The Mclaren F1 is the vehicle.
The Mclaren is the fastest production car.
In the above, even if the original statement is true the other variations of the statement may not be true. Rewording a true statement may make the “new” statement untrue. This is why in word problems and logic puzzles, proper wording is critical.
1959 version: Mr. Smith had two children, at least one of whom was a boy.
Marilyn’s version: A woman has exactly two children, at least one of whom is a boy.
Cecil’s version: There is a family with two children. You have been told this family has a daughter.
If the 1959 version was already ambiguous, Cecil’s puzzle added to this ambiguity. Cecil’s version provided us with a source that tells us of a child, the original did not. One of two possibilities exists, the source knew of both children and enjoys being cryptic, or the source told us all he knew. If the source is just being cryptic then all three families with girls are equal. If the source told us all he knew, then he knew of a single child. Knowing just one child makes the two girls in the same family along with the two girls in mixed families equally likely to be that child.
Bytegeist, I think you missed something in this experiment. In fact, you DO care whether he looked at one card, both cards, or even no cards. What you seem to be missing is that you don’t need to be present for you to have a 50/50 chance. All you need to know is that he drew two cards and looked at just one. This experiment could be done over the phone or the Internet. As long as he looked at just one card in his hand and told you the color, you can deduce that the other is equally likely to be a black or a red card. Why? Because whatever card he told you about came from one pile, while the other pile had just one black and one red. As for your friend seeing two cards, this experiment will not guarantee a 50/50 chance, because your friend is free to tell you of either card in his hand.
Your reference to my statement above confusions even me. Yes, in another part of the same post(#94), I am explaining that for Cecil’s version of the puzzle to match his answer that we have to assume that source of knowledge of the girl knew of both children and maintained the focus on the families not the child known. For you example of “walking up to you,” a family with two daughters could in fact walk up to you with either daughter. The families with a boy and a girl could only come up to you one way, as they have only one girl each.
My post you reprinted was referring to the same factors that Scientific American explained in 1959. If all families that have girls (GG,GB,BG) or boys (BB,BG,GB) provide the same clue, then the three families are equally possible. Scientific American explained that it was just as likely that the families with both a boy and a girl (BG,GB) would randomly provide you with a child, not always boys, not always girls. I took this one step further and said that if you assume that the families with girls always provide a girl clue, and then it would also be possible that you were given a boy who only could come from the two-boy family.
I wrote a quick program in C to simulate the Monty Hall Question, which pares it down to its essentials. I can’t understand the results, but they fit pretty much perfectly with the 2/3 vs 1/3 idea.
Here’s the essential part of the code, for those interested:
win=0; loss=0;
for (i=0; i<10000; i++)
{
for (j=0; j<3; door[j++]=0);
money=rand() % 3;
door[money]=1;
choice=rand() % 3;
if (door[choice]!=1)
win++;
else
loss++;
}
For this run (ie, where the decision is changed; ie that if the first door picked was correct, the final answer is wrong, and vice versa), the results were for ten runs were as follows for ten thousand trials:
6749 6658 6687 6690 6690 6729 6720 6654 6684 6647
When the code was reversed to change to keeping the first door chosen (ie the relevant line was changed to ‘if (door[choice]==1)’, the results were as follows (again for 10,000 trials):
3361 3331 3362 3342 3352 3352 3329 3305 3299 3348
Averages: 6690.8 vs 3338.1 - looks pretty close to 2/3 vs 1/3 to me. Not what I would have expected.
This may be obvious, but your code shows exactly why it works. If the player is using the “switch every time” strategy, he will always switch to a winning door when he first picks a losing door, because Monty will only show a non-winning door. He will always switch to a losing door when he first picks the winning door. He has probability 2/3 of picking a losing door first, and thus probability 2/3 of winning.
I wrote a much more complicated simulation, in which Monty actually picks a door and the player decides whether to switch or not with probability .5. The result was the same: when the player switches, he wins 2/3rds of the time; when he doesn’t he wins 1/3rd of the time. And, by the way, as the number of doors increases, the probability of the player winning after switching approaches 1/2, since the probability of switch to a losing door increases.
