November 20, 2006
Answers to Last Week's CPCP Challenge
Last week I posted some math and logic problems. Here are the answers (I have chosen some of the succinctly-expressed answers submitted, rather than write them all out again myself):
- Light both ends of the first rope, and one end of the second. When the first rope is completely burnt, light the other end of the second rope. The 45 minute mark is when the second rope is completely burnt.
- Take X coins and flip them. These form one pile, the rest of the coins form the other.
- Take the chicken and drop it off at the finish. Come back and get the dog, drop it off at the finish and grab the chicken again. Come back with the chicken, drop it off at the start and grab the corn. Drop off the corn with the dog. Head back to grab the chicken, and return to the finish.
- 3 cuts. Cut each link in one chain. Separate them, and use the links to join the ends of the 3 intact chains.
- She has two children, one of which is a daughter. Here are the possibilities: Boy/Boy -- this is impossible. Boy/Girl, Girl/Girl, or Girl/Boy. So it's a 1/3 chance that both children are girls.(NOTE: This answer is wrong, but I am leaving it here to explain the extensive debate in the comments below. The correct answer is 1/2.)
- Pick a jellybean from the box labeled blue&red. You can be sure that the all the jellybeans in there are the same color. The box labeled with the other color actually contains both blue and red. The box labeled with the color of the picked jellybean actually contains jellybeans of the opposite color.
- Unchanged. The floating cube displaces its own weight in water.
- First weight three coins against three others. If the weights are equal, weigh the remaining two against each other. The heavier one is the counterfeit. If one of the groups of three is heavier, weigh two of those coins against each other. If one is heavier, it's the counterfeit. If they have equal weight, the third coin is the counterfeit.
- Same amount of water in wine as wine in water. Think about it: however much water is missing in the one gallon jug of wine has to be in the other container and vice versa.
- 1&2 cross in 2 minutes. 1 returns in 1 minute. 5&10 cross in 10 minutes. 2 returns in 2 minutes. 1&2 cross in 2 minutes. Total: 17 minutes.
- Anywhere 1 mile north of the line of latitude near the South pole which is 1 mile in circumference will do, as will an infinite number of points below that point, all around the earth.
- 3, 3, and 8. The only groups of 3 factors of 72 to have non-unique sums are 2, 6, 6 and 3, 3, 8 (both add to 14). The presence of a single oldest child eliminates 2,6,6.
- Let's say it takes 24 hours to circle the planet. So each plane can carry
12 hours of fuel. At midnight THREE planes set out with full tanks. By 3 AM they have gotten 1/8 of the way around. Each has 9 hours of fuel remaining. Plane 1 gives 1/4 tank to each of 2 and 3, filling them up; it has 1/4 left and turns around. By 6 AM #2 and #3 have gotten 1/4 of the way around; each has 3/4 tank (9 hours )remaining. 2 gives 3 3 hours of fuel, filling him up and leaving himself with 6. He heads for home. 1 arrives home and refuels. At noon 3 is half way around. He has 6 hours of fuel remaining. 2 arrives home and refuels. 1 and 2 set out in the other direction. At 3 PM 3 is 5/8 of the way around, with 3 hours remaining. 1 and 2 are 7/8 of the way around; 1 fills 2 up and heads for home with 6 hours remaining. At 6 PM 3 is 3/4 of the way around and running on fumes. He meets 2, who immediately gives him 3 hours of fuel, leaving himself with 6. 1 arrives home, refuels, and sets out again. At 9 PM 3 is 7/8 of the way around and running on fumes again, while 2 is down to 3 hours. Luckily here comes 1 with 9 hours of fuel; he gives 3 3 hours. At midnight they all arrive safely; plane #1 even has 3 hours of fuel left. - Flip the first switch and leave it on for ten minutes. Turn it off, turn on the second switch and go upstairs. Look at the lamp and feel its bulb if it's off. If it's on, the second switch controls the light. If it's off and warm, the first switch controls the light. If it's off and cool, the third switch controls the light.
- I repeated question number 9 here by mistake.
Oh, and I had promised a harder problem. Here it is (Jesse Mazer mentioned it in the comments to the original post as well, I believe):
You have 12 balls. One of them is either lighter or heavier than the others. You have a scale and can only use it three times to find out which ball is different, AND whether it is lighter or heavier. How will you do it? Good luck.
Don't post answers in the comments. Email them to me at s.abbas.raza [at] att.net
Posted by Abbas Raza at 12:00 AM | Permalink
Comments
I think you neglect the fact that, if there are two girls, then there are two ways to select a girl to accompany the mother. If your answer to #5 is correct, please point out the error in the following reasoning.
I assume: 1) the combinations (given as younger/older) boy/boy, boy/girl, girl/boy, and girl/girl are equally likely; 2) it is equally likely that the mother would choose the younger as the older to walk with. Then, ignoring possibilities ruled out by the observation of a girl walking with the mother, it seems the remaining, still equally likely possibilities are:
(boy home)/(girl walking), (girl walking)/(boy home), (girl walking)/(girl home), and (girl home)/(girl walking). Of these, 50% have a girl at home, 50% a boy.
Posted by: Steve Durbin | Nov 20, 2006 9:03:14 AM
I'm also confused by number #5, and need the error in my reasoning pointed out.
I prefer to equate it to flipping coins. If I flip a coin and get heads ninety-nine times, my next flip still has a one in two chance of being heads. Isn't your answer to #5 equivalent to saying that the previous flips have altered the probability of the next one?
Also, if someone could discuss this scenario: you have two coins, and have flipped them without showing me the results. Let's consider two variations. In the first, I ask you to show me one of the coins, randomly. You show me a heads. In the second, I ask you, "If one or both of your coins is a heads, show one of them to me." Now, are the odds that the other coin is heads not different for those two scenarios?
Finally, is the probability of the question not the odds of each child being a girl multiplied together (1/1 X 1/2 = 1/2)? I can understand that this would not be the case if the situation were equivalent to the latter coin example, but since it's equivalent to the first...?
Posted by: Adam | Nov 20, 2006 11:11:37 AM
Here's another way of articulating my confusion on #5:
Your answer is valid if she had prefiltered the child with her to make it seem more likely to be a girl. Her possibilities for children are BG, GG and GB, as you stated: 1/3.
However, since her choice of child accompanying her is random, we don't need to consider the order in which her children were born. We can consider them by the sequence in which we learn of them: GG or GB: 1/2. What I'm saying is that BB and BG have both been eliminated by the fact that the first child we're exposed to is a G.
Posted by: Adam | Nov 20, 2006 11:32:28 AM
I eventually got #14 (actually my answer was slightly different than the one Abbas gives; mine works too but the given one is better), but I would also like to share my first try, which Abbas rejected on the grounds that it requires other people:
Flip switch #1 on. Then off. Then on. Then off. Keep this up until someone comes downstairs and pounds on the door yelling "WILL YOU CUT THAT OUT???" If nothing happens for, say, 15 minutes, go on to switch #2. And if necessary, #3.
Posted by: Dave M | Nov 20, 2006 12:06:14 PM
Ah, the old twelve balls problem. Brings back such memories!!
(Man, I'm a nerd.)
Posted by: Asad | Nov 20, 2006 12:34:11 PM
Dear Steve Durbin and Adam,
Here is the reason your boy/home, girl/walking type scenario doesn't work: it requires distinguishing the two girls, which the problem doesn't give us enough information to do. Adam, it works exactly the same way for coins: if I tell you that I have already flipped a coin twice and tell you that on one of those flips I got a head, then the chances that I also got a head on the other are 1/3.
Luckily, in case you continue to be confused and disagree with me, I don't have to keep arguing with you because I have what are known as Dutch Book arguments available to me: I will bet money with you using real life data from, say a hundred families with two children (or a hundred pairs of flipped coins), where a trusted third party tells us the sex of one child, and we bet on the sex of the other. I will always bet the other child is the same sex, and you should take my bet since it doesn't really matter to you which sex you choose (you believe the probability in both cases is the same or 1/2). Trust me, I will soon take all your money. (But if you are truly obstinate, I suppose you will then blame your losses on "bad luck.")
Flip two coins a hundred times, then try it!
Posted by: Abbas Raza | Nov 20, 2006 12:54:23 PM
Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied. What is the probability that both of her children are girls? Be warned: this question is not as trivial as it may look.
This question is as trivial as it looks. The only question being asked is "What is the probability of THE OTHER child being a girl?" The gender of the observed child has no bearing on the gender of the unseen child.
Neglecting non-equal birthrates between boys and girls in the general population , there is a 50% chance the unknown child is a girl.
Posted by: Bryan | Nov 20, 2006 1:28:11 PM
Abbas,
I'll take your bet if we can define terms. First, there should be a way for me to gain if I am right, rather than just break even. Since you say 2/3 of the time the unrevealed one of a pair of tossed coins will match the revealed one, while I say 1/2 the time, we should set bets such that we both break even (on average) if they match 7/12 of the time. So let's say I give you 5 monetary units each time you're right, and you give me 7 each time I'm right. If this is agreeable to you, we can define the exact protocol more precisely.
Steve
Posted by: Steve Durbin | Nov 20, 2006 1:50:12 PM
This is very interesting, and I hope those more qualified than I will help us out. (For the record, I gave the answer Abbas endorses.) In particular, I see connections to two other problems:
1. The Monty Hall problem (in all its variations)
2. The doctrine of "restricted choice" in bridge.
Posted by: Dave M | Nov 20, 2006 2:08:09 PM
"Here is the reason your boy/home, girl/walking type scenario doesn't work: it requires distinguishing the two girls, which the problem doesn't give us enough information to do. Adam, it works exactly the same way for coins: if I tell you that I have already flipped a coin twice and tell you that on one of those flips I got a head, then the chances that I also got a head on the other are 1/3."
But you're prefiltering the data. The woman walking is not a similar scenario, it's as if you showed me one of the coins randomly.
If you flip two coins and show me one of them randomly, and it happens to be a head, there's a 1/2 chance that the other one is a head.
If you flip two coins and show me a head by choice, the chance that the other one is heads is 1/3.
The daughter situation is analogous to the first scenario, not the second one.
Posted by: Adam | Nov 20, 2006 2:09:09 PM
For #14 you didn't specify that the light was off at the start, so there was no way of being able to tell with just one chance to look in the room :)
For #5 I think the answer would be 1/2. P(A|B)=P(AnB)/P(B) = P(2 girls being born n 1 girl being born) / P(one girl being born) = P(2 girls being born)/P(one girl being born) = (1/4)/(1/2) = 1/2.
Posted by: James R | Nov 20, 2006 2:20:08 PM
"Luckily, in case you continue to be confused and disagree with me, I don't have to keep arguing with you because I have what are known as Dutch Book arguments available to me: I will bet money with you using real life data from, say a hundred families with two children (or a hundred pairs of flipped coins), where a trusted third party tells us the sex of one child, and we bet on the sex of the other. I will always bet the other child is the same sex, and you should take my bet since it doesn't really matter to you which sex you choose (you believe the probability in both cases is the same or 1/2). Trust me, I will soon take all your money. (But if you are truly obstinate, I suppose you will then blame your losses on "bad luck.")"
We have equal odds here. Possibilities are BB, BG, GB, and GG. You are covering two of the choices with your betting method, as am I. We both have odds of 1/2.
Posted by: Adam | Nov 20, 2006 2:26:09 PM
The probability of a mother of two having two daughters is 1 in 3. The probability of a mother of two, who is known to have 1 daughter, having two daughters is 1 in 2.
Posted by: Bryan | Nov 20, 2006 2:30:22 PM
Bryan is wrong. I have already explained it as best as I can, and unlike Dave M., no longer find this debate at all interesting. If you don't understand either the problem or the answer, please read an elementary textbook on probability.
And Steve, I will not seriously take candy from a babe. :-) But go ahead and play the game with someone else to convince yourself...
Posted by: Abbas Raza | Nov 20, 2006 3:37:13 PM
Abbas,
I'm happy to drop the subject. But I have played the game with a computer. Unfortunately, it has no financial resources.
Steve
Posted by: Steve Durbin | Nov 20, 2006 4:03:12 PM
"Bryan is wrong. I have already explained it as best as I can, and unlike Dave M., no longer find this debate at all interesting. If you don't understand either the problem or the answer, please read an elementary textbook on probability."
Wow! I had no idea you were such an asshole! Thanks for clearing one thing up.
Posted by: Adam | Nov 20, 2006 4:05:37 PM
I am not wrong. I am just as correct as you. Your question is ambiguous.
The answer is different depending on whether you mean mothers of two in general, or this particular mother of two.
Posted by: Bryan | Nov 20, 2006 4:20:09 PM
It seems I have now caused offense. This was not my intention, and I apologize for seeming arrogant (especially to Adam, who seems quite upset). I just have nothing more to say about the problem.
Posted by: Abbas Raza | Nov 20, 2006 4:25:17 PM
No offense taken here. Thanks very much for the problems -- I really enjoyed thinking about them and sharing them. (And if you change your mind about the bet...)
Posted by: Steve Durbin | Nov 20, 2006 5:04:29 PM
Thank you for the apology. I apologize in turn. I still think you are incorrect.
I'm at work, and have nothing better to do than look into this. It appears that there are two subtle variations of this puzzle that seem to me to reinforce what I'm saying.
One version is: "A man has two children. One of them is a boy. What are the odds that the other is a boy?" This is 1/3. This allows the "revealed" boy to occupy either position of the 1,2 pair.
The other version is: "A man has two children. The first is a boy. What are the odds that the other is a boy?" This is 1/2. The first position is a boy, the second can be a boy or a girl.
Your question is the second version: the "first" needn't designate the eldest, it need only designate one randomly selected child (i.e. the first exposed to me without preference for exposing me to a boy). If you don't believe this you are maintaining that the previous flip of a coin alters the odds of subsequent flips. It doesn't.
Ummm...perhaps talking about exposing myself to boys isn't the best choice of words.
Posted by: Adam | Nov 20, 2006 5:36:35 PM
Adam,
I think we agree at this point. Though I am not sure what you mean in your "other version," we needn't worry about it since the original problem as I stated it last week doesn't have the word "first" or any other word which could be misinterpreted as that in it. Therefore, the first version of the two that you give (and which I understand) clearly applies.
Posted by: Abbas Raza | Nov 20, 2006 6:00:35 PM
I've got it now. You're right. Thanks for waiting patiently while I came around ;)
Posted by: Bryan | Nov 20, 2006 6:32:45 PM
Thanks Adam for clarifying. I maintain that the problem as stated effectively does have a "first", namely the one child that was out walking with the mother -- the first we know of. The two children are distinguishable, because one is out, the other is at home.
Posted by: Steve Durbin | Nov 20, 2006 7:07:12 PM
Here's the issue with #5 as I see it--if you run into a mother with one child on the street, then it is natural to treat this as if the child you saw were *randomly selected* from among her two children. The 1/3 answer would only make sense if you assume that if you repeated the experiment many times, then any time you run into a mother of two who has at least one girl, you are guaranteed to see a girl. This assumption makes very little sense, given the statement of the problem; it is much more natural to assume that if you randomly bumped into many different mothers of two, then of the mothers who have one boy and one girl, on 50% of the meetings the mother would happen to have the boy with her and on 50% of the meetings the mother would happen to have the girl with her. Thus on occasions when you *do* bump into a mother with a girl, the conditional probability of GG becomes greater than the conditional probability of BG, and likewise greater than the conditional probability of GB.
As an analogy, suppose I flip two coins, and randomly uncover one of them to show you what it is (without specifying whether the one I uncover is 'first' or 'last'), and you see it came up "heads". Would you agree that the probability the other is heads is 1/2, not 1/3?
Posted by: Jesse M. | Nov 20, 2006 8:21:20 PM
I still think I'm correct. Here's what I've done, and the results are far closer to the predictions than I would have thought:
I googled for a website with a random number generator. I set it to generate 250 binary pairs. I used the same set of numbers for both methods of counting.
Method 1: Look for each pair that does not contain two zeroes. Count the number of those pairs, and the number of those pairs that contain two ones. Prediction: 1/3. Result: 62/190, or .3263.
Method 2: Look at the first number of each pair. Count how many pairs have the same number as their second number. Prediction: 1/2. Result: 122/250, or .488.
I maintain that the problem as stated is equivalent to the second method. Am I wrong? Is 250 not a significant enough pool? I swear I didn't make this up; I thought 250 wouldn't be a large enough sample, but the similarities to my predictions are much closer than I expected.
Posted by: Adam Barnett | Nov 20, 2006 8:46:41 PM
Incidentally, the page on "Nuances of probability theory" by computer science PhD and "Bayesian statistical inference" researcher Mike Minka at http://research.microsoft.com/~minka/papers/nuances.html agrees that the answer to this sort of problem should be 1/2, saying:
My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities---two boys or two girls---have probabilities 1/4 and 1/4.
Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl? By the above reasoning, it is twice as likely for him to have one boy and one girl than two boys, so the odds are 2:1 which means the probability is 2/3. Bayes' rule will give the same result.
Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl? Observing the outcome of one coin has no affect on the other, so the answer should be 1/2. In fact that is what Bayes' rule says in this case. If you don't believe this, draw a tree describing the possible states of the world and the possible observations, along with the probabilities of each leaf. Condition on the event observed by setting all contradictory leaf probabilities to zero and renormalizing the nonzero leaves. The two cases have two different trees and thus two different answers.
This seems like a paradox because it seems that in both cases we could condition on the fact that "at least one child is a boy." But that is not correct; you must condition on the event actually observed, not its logical implications. In the first case, the event was "He said yes to my question." In the second case, the event was "One child appeared in front of me." The generating distribution is different for the two events. Probabilities reflect the number of possible ways an event can happen, like the number of roads to a town. Logical implications are further down the road and may be reached in more ways, through different towns. The different number of ways changes the probability.
Posted by: Jesse M. | Nov 20, 2006 8:53:53 PM
Steve has got what I'm saying. "First" is in your statement of the question, in terms of "the first of the two children to which I was exposed".
Compare it to my "method 2" of counting binary pairs in my immediately previous post. They are equivalent.
Posted by: Adam Barnett | Nov 20, 2006 8:59:56 PM
i can't believe that you (abbas) are not more open to these arguements. i hope that you can understand what jesse m has copied out of an 'elementrary textbook of probabilities"
i agree that the answer is 1/2 because like adam and steve are saying the child you meet is considered the first of the probabilities (it isn't always which comes first, ie. the oldest)...
cheers
Posted by: patrick | Nov 20, 2006 9:39:37 PM
Sorry, Abbas, I can't agree with you here. As you hint at above, the issue is whether the two children are in some way distinguishable. But they are distinguishable -- one is standing in front of you, and one is not. So the question is not "I have two children, and one is a girl; what is the chance that the other is a girl?" The question is "I have two children, and this one is a girl; what is the chance that the other is a girl?" The answer is 1/2.
Put another way: there are two children, the one you see and the one you don't. Each has a 50% chance a priori of being a boy or a girl, leading to 4 equally-likely possibilities. But your observation that the one in front of you is a girl removes two of those possibilities, leaving just two equally-likely possibilities.
Now, if you had met a woman by herself on the street, and she had mentioned the existence of a daughter, and later said "In fact I have two children..." they would be truly indistinguishable, and the probability that both were girls would be 1/3.
Posted by: Sean Carroll | Nov 21, 2006 12:31:42 AM
For a liberally-minded blog, the probability of 0.5 should be adjusted to 0.499, to allow for the finite probability that one of the children may be deeply disatisfied with their current gender identity.
Posted by: aguy109 | Nov 21, 2006 5:03:21 AM
And one consulted expert agrees with Abbas:
"Hi Adam,
This is a classic, and confusing, example in conditional probability. As the question is stated, the answer is indeed 1/3.
Everything depends on how you interpret the information. In your friend’s interpretation, we know there is at least one daughter. The question is then: “Given that a family of two contains one daughter, what is the probability that it contains 2?” to which the answer is 1/3. In your interpretation, you say that we know the oldest (or first, etc.) child is a daughter, and the question becomes: “Given that the oldest child is a daughter, what is the probability that the younger child is a daughter?” with answer ½. However, this depends on extra information which does not seem to be in the question, namely that we know which daughter we met (the older or the younger, the first or the second).
The issue here is that prior information about this family affects the probability distribution for the sexes of the children. Ordinarily, two children would be BB, BG, GB or GG each with probability ¼. This is impossible in the given scenario, since the family is known to contain a girl. The knowledge that there is a girl causes us to assess the probabilities differently – for example, we now consider that the probability of BB is 0, instead of ¼. On the other hand, the remaining 3 possibilities (BG, GB and GG) are all equally likely since the given information does not distinguish between them.
Your interpretation is a slight variant, where we know that the oldest is a girl. This eliminates 2 of 4 possibilities (BB and BG), leaving 2 which are now equally likely.
The rephrased question “Of families with 2 children, what is the probability that the second is the same sex as the first?” is actually a subtly different question. It is equivalent to the following experiment: randomly choose a family having 2 children and see if the children are BB or GG. Note that here you are asking for an unconditioned probability: you are considering the entire sample space of possible two-children families. But the original question gives you information about the family: you must restrict your attention to only families having at least 1 daughter. The experiment would be as follows. Randomly select a family with two children, but if you selected a family with two boys, move on to a different family. When you eventually find a suitable family, see if the children are BB or GG (in fact, they would have to be GG). Here you can see why the result is different: the experimental procedure is stacked, and gives a different answer than the unconditioned experiment.
One last way to explain the difference. Let X be the number of girls in a randomly selected family of 2 children. Evidently, X can have the values 0, 1 or 2. However, these are not equally likely; in fact, Prob(X=1) = ½ while Prob(X=0) = Prob(X=2) = ¼. In particular X is twice as likely to be 1 as it is to be 2. If you are now told only that X is not 0, then X is still twice as likely to be 1 as it is to be 2. But now it is either 1 or 2, so because probabilities must add up to one, the new conditional probabilities can only be CondProb(X=1) = 2/3 and CondProb(X=2) = 1/3 respectively.
I hope this explanation makes sense. Good luck,
Jesse Goodman"
Mr. Goodman is a graduate student at the University of British Columbia in some probability field that might has well have been greek to me.
I'm still sceptical, based on the other quoted expert above, and the fact that it still seems to me that I'm not assuming one of the sisters is the eldest; we can designate between them rather in the sense of the "first" which I've encountered.
Yet another way to put it: If the person were "stacking" their choice of children with them to present a daughter to us, it would be less likely that they have another daughter at home - that is, chances are they have one daughter and we're looking at her. However, if she selected a random child to come with her, it seems to me it MUST be more likely (than if she chose a daughter deliberately) that she has another daughter at home - the difference being between 1/3 and 1/2.
Posted by: Adam | Nov 21, 2006 11:03:28 AM
A portion of my reply to Mr. Goodman:
"Sorry to presume so much as to argue with you; you are induspitably the expert, but I hope you appreciate my arguing comes solely from a desire to understand.
You stated: The question is then: “Given that a family of two contains one daughter, what is the probability that it contains 2?”
That seems false to me. That seems equivalent to:
My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities---two boys or two girls---have probabilities 1/4 and 1/4.
Suppose I ask him whether he has any boys, and he says yes.
When the question as stated seems equivalent to:
Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl? Observing the outcome of one coin has no affect on the other, so the answer should be 1/2.
Adam"
It seems to me that those who are maintaining that the odds are 1/3 are denying that a deliberately chosen daughter has different odds than a randomly chosen daughter. I'm fully willing to accept that I'm wrong, mind you, I'm not being pig-headed about this. I just can't see it yet.
Posted by: Adam | Nov 21, 2006 11:10:31 AM
I was always bothered by these sort of problems until I came across examples of them in Byron P. Roe, Probability and Statistics in Experimental Physics.
He gives the following 2 cases:
- "Suppose we are given that a family has two children and at least one is a boy. What is the probability that they are both boys? The answer is 1/3."
- "Choose a boy at random. Suppose he comes from a family of two children. What is the probability his sibling is a boy? The answer here is 1/2!!!"
These really expose the subtlety of probability and the importance of well-posed problems. In the first case the "unit" of permutation is a family, hence the permutations boy-boy, boy-girl, and girl-boy give one a probability of 1/3. In the second case you are permuting over siblings and since it's equally probable to have a boy or a girl, you have 1/2.
So which sort of problem is this:
Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied. What is the probability that both of her children are girls?
It's maybe not as unambiguous as Roe's first case, but to me it seems that here Abbas is talking about permuting over families, so the answer is 1/3. However, the "curse" of probability is that the answer to a poorly-worded or ambiguous problem can be argued about until the end of the universe. (Not to say that this is such a problem by the way)
Posted by: Zevatron | Nov 21, 2006 11:17:45 AM
I think, though, Zevatron, that his question is the latter, not the former. "At least one is a boy," requires that the pool of both children is examined to determine if one or both of them is a boy. The question as stated has given no indication of such filtering. That's the distinction between the two subtle variations, and the question has not given that distinction. We are to assume that the child chosen to accompany her was random.
Posted by: Adam | Nov 21, 2006 11:35:11 AM
Adam, I'm not sure what you mean by "filtering" or why it's important. To understand the distinction between the two cases and why Abbas' problem as stated is more the former you only have to think of it in terms of permutations.
What is the probability his sibling is a boy?: This means you permute over possible siblings. You can only have either a boy or a girl (B or G), so the answer is 1/2.
What is the probability that they are both boys?
and What is the probability that both of her children are girls?:
Here you permute over possible families (which have two children, and therefore BG, GB, BB).
It's subtle, and makes your head hurt (speaking from experience), but once you think if it this way, it should be clear.
Cheers.
Posted by: Zevatron | Nov 21, 2006 11:57:45 AM
How is Abbas question not, "What is the probability his sibling is a boy?" That's exactly what his question is.
If the question was stated as, "A woman is walking down the street. You ask her is she has a boy. She says, "I do. I actually have two children." What is the probability that they are both boys?" THEN it's 1/3. That is not the stated question.
The fact that you have been shown a boy at random, instead of the data being filtered by the question, "do you have a boy?" changes the odds. If it's random, that's the same as flipping a coin successively. If you ask them to choose, that's the same as asking them to flip a sequence of coins and then polling the results. They have different answers.
Posted by: Adam | Nov 21, 2006 12:16:56 PM
I think all refutations of my reasoning come down to the fact that you must maintain that my second method in my binary table experiment is not equivalent to Abbas' question. Could someone address that? In what way are they not equivalent?
Posted by: Adam | Nov 21, 2006 12:19:28 PM
Or, conversely:
If you ask someone with two children if they have any boys, they have two children to choose from. Thus, they are effectively increasing the likelihood that their other child is not a boy: BG, GB, or BB: 1/3.
However, of two children families, half of them are going to have two children of the same sex: BB, BG, GB, and GG.
Being exposed to one child at random DOES NOT CHANGE THAT FACT. It is NOT equivalent to asking them to filter through both "children slots" for a child of a particular sex.
The question as stated is exposure to a random child. It does not involve filtering towards a particular sex.
Posted by: Adam | Nov 21, 2006 12:26:31 PM
Sorry, by using the two examples in the same case in the previous post I introduced some possible confusion. For two girls the permutations should be GG, BG, GB. For two boys: BB, BG, GB.
Furthermore, these permutations assume that we don't know whether the girl is the older or younger sibling in the family. We don't. (Reading over the comments, this was all covered by Abbas and Jesse Goodman).
Adam, the key word is both. This means that you are referrring to a family. It's all in the posing of the question.
OK, my head is starting to hurt. No more...
Posted by: Zevatron | Nov 21, 2006 12:31:22 PM
A similar debate emerged over the Monty Hall problem, and it was concluded that an ambiguity in phrasing allowed two answers depending on interpretation.
"You're on a game show with three doors. One has a prize behind it, two have nothing. After you choose one door, the host opens one other, and shows you that it has nothing behind it. Do you change doors?"
IF the host chooses a door RANDOMLY, you have been told nothing about what is behind either remaining door, and you have no incentive to change.
IF the host chooses a door with a goat behind it intentionally, he has filtered the results (that's what I mean) and it is to your benefit to change doors.
It seems that it is always to your benefit to change because if we extend the number of doors to a hundred, and you pick one, and then he opens ninety eight and shows you nothing, it seems clear that it is to your benefit to change. HOWEVER this apparent benefit is dependent on him deliberately choosing (filtering) nothing-doors. Again, if he in fact chose randomly, and happened to choose 98 nothing-doors (which is exceedingly unlikely), then we've learned nothing about the remaining two doors except that one of them must contain the prize.
Being exposed to one of her children by chance is equivalent to randomly opening doors.
Posted by: Adam | Nov 21, 2006 12:39:27 PM
Zevatron, I am growing increasingly certain you guys are wrong. Consider this, and refute. Or, again, someone refute my binary-table experiment results.
Half of all two-child families have two children of the same sex.
If your answer to the question stated is correct, then only of third of them do, because learning the sex of a random child would not effect the likelihood of a specific two-child family having two children of the same sex.
----
If you walked down the street looking for parents walking with a child of a particular sex, and then asked them if they had two children, then the likelihood of the two children being the same sex would be 1/3. This is not the question as stated.
Posted by: Adam | Nov 21, 2006 12:52:00 PM
Abbas is correct.
What is desired is the probability of the event B, "both children are daughters", given the event A, "at least one child is a daughter", in symbols Prob(B|A). By definition, Prob(B|A) = Prob(B and A)/Prob(A). Since having two daughters implies having at least one daughter, in this case Prob(B|A) = Prob(B)/Prob(A). Prob(B) is (1/2)(1/2) = 1/4. Prob(A) is 3/4. This means Prob(B|A) = 1/3. Q.E.D.
This is related to the difference between exchangeable random variables, and strictly independent ones.
Posted by: Cosma | Nov 21, 2006 1:33:01 PM
My expert now agrees with me:
"Hi Adam,
I see your point. I may have been thinking of a slightly differently stated question where the answer is unambiguously 1/3: Choose at random a family having 2 children. Given that one child is a daughter, what is the probability that there are 2 daughters? As stated in the Minka article, the question would be formulated in a different way: Choose at random a family having 2 children, then choose at random one of the 2 children. Given that the child is a boy, what is the probability that the other is a boy? There the answer is ½. On balance, this may more closely reflect the spirit of the question you asked. Considering the statement of the question further, it seems any answer depends on the interpretation of the daughter. If she was chosen at random, then that is the second scenario above and the probability is ½. In my interpretation, I am assuming that she is on of the daughters of the family, if there is one. Looking at it now, this is a somewhat unlikely interpretation.
Note that the question itself does not really tell you enough. It is not clear that the child accompanying the parent (or the child running) should really be considered to be one of the children chosen at random, although this is a supposition one might well make. I would also point out that in real life, one would expect neither of these possibilities to be exactly correct. Parents do not choose a child at random when they go outside. More plausibly, if we condition on the sexes of the 2 children, the distribution of which child accompanies the parent would have some unknown distribution which is not simply a random choice of the 2 children. (Thinking somewhat stereotypically, we might suppose, for instance, that the other sibling is probably a son because if it was a daughter she would have been more likely to want to come along, or some such thing.) In any event, this is probably closer in spirit to the random choice of one child that occurs in Minka’s article.
Regards,
Jesse Goodman"
Posted by: Adam | Nov 21, 2006 1:57:24 PM
In any question about probability, it is essential that you have a clear idea of the "sample space" of possible alternate outcomes that would be seen if you were to repeat the same sort of experiment multiple times. I think in this case it's clear we're meant to assume the sample space consists of running into multiple different mothers of two kids, each of whom has a single child with them. In any question about conditional probability--the probability of X given Y--the idea is to look only at the subset of trials in which Y occurred, and look at how often X occurred within that subset. In this case, we are looking at only the subset of cases where the mother had a girl with her, but the full sample space should include both cases where she brought a boy and cases where she brought a girl.
So the critical question of interpretation is, what assumption should we make about the probability that a mother with a boy and a girl happens to bring along the girl in any given trial? If we assume that all mothers with a boy and a girl always take the girl along when they go outside for walks with a child, and keep the boy chained up in the basement or something, then this means that any time the mother has at least one girl in the family we are guaranteed to see it, so the problem becomes equivalent to the one where we ask the mother "is at least one of your children a girl?" and she says "yes", in which case the probability the other child is a girl is 1/3. On the other hand, if we assume that a mother with one boy and one girl takes each child outside with equal frequency, so that when we randomly bump into one we are equally likely to see the boy as the girl, then the problem is not equivalent to the one where we ask the mother if she has at least one girl, because in that case we are guaranteed to know she has a girl if she does indeed have one, while in this case we might see the boy and be unsure whether the other child is a girl or a boy. If we assume the mother randomly selects which child to go out with, then each of the following possibilities in the sample space should occur with equal frequency:
1. older child boy, younger boy, mother takes younger
2. older boy, younger boy, mother takes older
3. older boy, younger girl, mother takes younger
4. older boy, younger girl, mother takes older
5. older girl, younger boy, mother takes younger
6. older girl, younger boy, mother takes older
7. older girl, younger girl, mother takes younger
8. older girl, younger girl, mother takes older
If we see a girl, then we know we are dealing with a subset of the sample space containing cases 3, 6, 7, and 8; since each of these occurs with equal frequency, and 3 and 6 mean the other child is a boy while 7 and 8 mean the other child is a girl, the probability must be 1/2.
Again, the critical issue is what we think the outcomes of multiple trials would look like, and specifically whether or not we think that in multiple trials, we are guaranteed to see a girl in every case where the mother has both a boy and a girl (the 'boy chained in the basement' assumption); this seems like a very strange assumption to make, and not justified in any way by the statement of the problem. If you assume that a mother with one boy and one girl is equally likely to take the girl or the boy with her, then the answer is 1/2.
Posted by: Jesse M. | Nov 21, 2006 1:59:05 PM
To quote the problem exactly "Walking down the street one day, I met a woman strolling with her daughter." The information given Cosma's event A) is not "at least one child is a daughter", but essentially "this particular child (who happens to be walking with me) is a daughter".
Posted by: Steve Durbin | Nov 21, 2006 2:00:22 PM
Cosma, I'd say you're incorrect that the problem is equivalent to the probability of the event B, "both children are daughters", given the event A, "at least one child is a daughter". Rather, it is the probability of event B, "both children are daughters", given the event A, "you see a daughter". If there is a 50/50 probability that a mother with one girl and one boy happens to have the boy along when you bump into her, then the two assumptions are different, because in the first case you are guaranteed to know the mother has a daughter as long as one exists, in the second case you are not.
If you disagree, consider the following two problems:
1. I flip two coins, and randomly uncover one for you. You see that it is heads. What is the probability the other is heads?
2. I flip two coins, peek at both, and if at least one of the two came up heads, I uncover a heads for you, while if both are tails I uncover a tails. If you see a heads, what is the probability the other is heads?
I would say the answer to #1 is 1/2, while the answer to #2 is 1/3, and that Abbas' problem is more analogous to #1, since there's no reason to think mothers with 1 boy and 1 girl are guaranteed to always take the girl outside while keeping the boy locked away in the house at all times. Do you disagree?
Posted by: Jesse M. | Nov 21, 2006 2:07:50 PM
The question as stated is not equivalent to what you've stated, Cosma.
1) The probability of a two-child family having two children of the same sex is 1 in 2.
2) By your reasoning, a randomly determined child's chance of having a sibling of the same sex is 1 in 3.
3) Find out the sex of one randomly chosen child in each two sibling family in the world. By your reasoning, you magically now changed the odds of having two siblings of the same sex to 1 in 3 for all families globally. This is obviously false.
The only room for denial seems to me to come from denying that the sibling is randomly chosen. However, the question, as stated, indicated that the woman had no predisposition to take a daughter with her as opposed to a son, nor that you had a predisposition to approach women with daughters. As such, Cosma has restated the question with such a predisposition built in.
Conversely, consider that the woman's second child has not been born yet. The second child has a 1 in 2 chance of being either sex. Consider, then, that her child was born yesterday. Obviously the child's odds aren't effected by their birthdate.
You are going to argue that I'm insisting on their age being a factor by assuming that the first child is eldest. To the contrary, I'm insisting that their age is irrelevant. Whether the other child is not born, or was born yesterday, or was born before the child we now have before us, doesn't effect the odds. Would you say that a child yet to be born has a 1 in 2 chance of being a particular sex, but an older sibling has a 1 in 3?
I flip a coin. I show you heads. What are the odds of the next flip being heads? 1 in 2.
I flip two coins. I show you that one of them is a heads. What are the odds of the other being heads? 1 in 3.
You are confusing the temporality of my flipping coins with the temporality of birth order. The temporality of the coin flips is actually analogous to "the sex of the first revealed child" being random. The wording of the question presents no indication that the woman has a preference for showing you a daughter, which is what "flipping two coins and showing you that one of them is heads" indicates.
Posted by: Adam | Nov 21, 2006 2:16:32 PM
Ha! Jesse beat me to it!
Posted by: Adam | Nov 21, 2006 2:19:09 PM
Jesse M., if you put it that way, then I see the wrinkle (finally):
Walking down the street one day, I met a woman strolling with her daughter.
I believe you're right; this does add something else to the sample space. The question, as posed is not equivalent to either "Suppose we are given that a family has two children and at least one is a girl. What is the probability that they are both girls?" or "Choose a girl at random. Suppose she comes from a family of two children. What is the probability her sibling is a girl?"
Without this addition, the problem does reduce to the former case, and hence 1/3. With it, 1/2.
Posted by: Zevatron | Nov 21, 2006 2:29:29 PM
If you think this exchange has gone on too long, imagine what it's like on a large particle physics experiment...
Posted by: Zevatron | Nov 21, 2006 2:32:57 PM
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