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

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 | Monday, November 20, 2006 at 09:03 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 | Monday, November 20, 2006 at 11:11 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 | Monday, November 20, 2006 at 11:32 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 | Monday, November 20, 2006 at 12:06 PM

Ah, the old twelve balls problem. Brings back such memories!!

(Man, I'm a nerd.)

Posted by: Asad | Monday, November 20, 2006 at 12:34 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 | Monday, November 20, 2006 at 12:54 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 | Monday, November 20, 2006 at 01:28 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 | Monday, November 20, 2006 at 01:50 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 | Monday, November 20, 2006 at 02:08 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 | Monday, November 20, 2006 at 02:09 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 | Monday, November 20, 2006 at 02:20 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 | Monday, November 20, 2006 at 02:26 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 | Monday, November 20, 2006 at 02:30 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 | Monday, November 20, 2006 at 03:37 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 | Monday, November 20, 2006 at 04:03 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 | Monday, November 20, 2006 at 04:05 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 | Monday, November 20, 2006 at 04:20 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 | Monday, November 20, 2006 at 04:25 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 | Monday, November 20, 2006 at 05:04 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 | Monday, November 20, 2006 at 05:36 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 | Monday, November 20, 2006 at 06:00 PM

I've got it now. You're right. Thanks for waiting patiently while I came around ;)

Posted by: Bryan | Monday, November 20, 2006 at 06:32 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 | Monday, November 20, 2006 at 07:07 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. | Monday, November 20, 2006 at 08:21 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 | Monday, November 20, 2006 at 08:46 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. | Monday, November 20, 2006 at 08:53 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 | Monday, November 20, 2006 at 08:59 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 | Monday, November 20, 2006 at 09:39 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

aredistinguishable -- 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, andthis oneis 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 prioriof 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 | Tuesday, November 21, 2006 at 12:31 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 | Tuesday, November 21, 2006 at 05:03 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 | Tuesday, November 21, 2006 at 11:03 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 | Tuesday, November 21, 2006 at 11:10 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 | Tuesday, November 21, 2006 at 11:17 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 | Tuesday, November 21, 2006 at 11:35 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 | Tuesday, November 21, 2006 at 11:57 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 | Tuesday, November 21, 2006 at 12:16 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 | Tuesday, November 21, 2006 at 12:19 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 | Tuesday, November 21, 2006 at 12:26 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 | Tuesday, November 21, 2006 at 12:31 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 | Tuesday, November 21, 2006 at 12:39 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 | Tuesday, November 21, 2006 at 12:52 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 | Tuesday, November 21, 2006 at 01:33 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 | Tuesday, November 21, 2006 at 01:57 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 isnotequivalent 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. | Tuesday, November 21, 2006 at 01:59 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 | Tuesday, November 21, 2006 at 02:00 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, "youseea 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. | Tuesday, November 21, 2006 at 02:07 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 | Tuesday, November 21, 2006 at 02:16 PM

Ha! Jesse beat me to it!

Posted by: Adam | Tuesday, November 21, 2006 at 02:19 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 | Tuesday, November 21, 2006 at 02:29 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 | Tuesday, November 21, 2006 at 02:32 PM

Well, the experts I asked (Sean Carroll, Cosma Shalizi, Alon Levy) have given contradictory answers, but there is much to think about here, and I must admit that the problem now seems not as clear-cut as I had thought. Unfortunately, I need to get into a car and drive four hours to Maryland in a few minutes, and don't have the time right now to look at all the responses. I'll try to say more tomorrow...

Posted by: Abbas Raza | Tuesday, November 21, 2006 at 03:01 PM

Oh, here is Alon Levy's response:

Abbas,Adam and Steve are right. Look at it this way: suppose that there are n mothers of two. Of these, n/4 have two boys, n/2 have a boy and a girl, and n/4 have two girls. If they walk with one of their kids, then the n/4 mothers of two boys will walk with a boy, the n/4 mothers of two girls will walk with a girl, and the n/2 mothers of a boy and a girl will be split into n/4 who walk with a boy and n/4 who walk with a girl. So given the information you have, the probability is 1/2, not 1/3.

Incidentally, there's a related problem, which involves cards. There are three cards - one with two red sides, one with two blue sides, and one with one red side and one blue side. A card is lying on the table with a red face up. What is the probability that it's the red/blue card?Posted by: Abbas Raza | Tuesday, November 21, 2006 at 03:04 PM

Abbas,

I appreciate your following up on this. It's an interesting and subtle point and I think considering and clarifying it has helped all of us. Thanks to you and your experts, as well as the other contributors.

Posted by: Steve Durbin | Tuesday, November 21, 2006 at 06:05 PM

I was discussing it elsewhere and came up with this example:

:If the doctor performs a sonogram and says, "the first child I checked is a girl," then the chance of the second one being a girl is one in two.

If the doctor performs a sonogram and says, "I've checked them both and at least one is a girl," then the chance of them both being a girl is one in three.

First situation possible permutations:

BG BB GB GG

He only has a one in two chance of reaching the first condition ("The first child I checked is a girl.")

We are now left with the permutations:

GB GG

There is a one in two chance of the second child being a girl.

Second situation possible permutations:

BG BB GB GG

He has a three in four chance of reaching the first condition ("At least one child is a girl.")

We are now left with three permutations:

BG GB GG

Only one of the three permutations has the second child as a girl.

The question as stated suggests that we're considering the first situation, and that the first condition has been met.

Note that the first situation is has a chance of one in two of reaching the first condition, and a one in two chance of reaching the second. A one in two chance times a one in two chance gives you a net probability of one in four of having two girls. Add the one in four chance of having two boys, and you get a one in two chance of having two same-sex siblings.

The second situation has a three in four chance times a one in three chance: net probability three in twelve. Add the three in twelve chance of having two boys, and you get six in twelve, also giving a one in two chance of having two same sex siblings."

Posted by: Adam | Tuesday, November 21, 2006 at 07:24 PM

I believe #8 can be solved with 9 coins instead of 8. In fact, it looks nicer that way, as each weighting will yield a full 'trit' (trinary digit) of information, since then in all circumstances the scales can rise, drop or remain even.

Posted by: Bruno Mota | Wednesday, November 22, 2006 at 12:22 AM

Bruno, you're right. Think of it this way - you're building a decision tree, and some of the branches terminate early: they don't force you to make all measurements you could conceivably spend.

A filled trinary tree has 3^n leaf nodes.

Posted by: Ian McMeans | Wednesday, November 22, 2006 at 02:12 AM

Two more experts weigh in (both faculty at UBC):

"Hi,

You asked:

"You're walking down the street and encounter a woman with a daughter. =

You discover that she has two children; what are the chances that the =

second is also a daughter?"

I'm afraid that you and your friend may continue the argument

indefinitely, for the problem is *very* badly stated, and you may each

have different interpretations. It is not clear whether "woman with a

daughter" implies that the daughter is there walking with the woman, or

that it has simply come out in conversation that the woman has at least

one daughter. The answer is different in the two cases. If you meet the

woman while she is walking with her daughter, then---under the usual

assumptions that births are independent, that boys and girls have equal

probability, AND that the woman is equally likely to be walking with

either of her children---the probability the second child is also a

daughter is 1/2. If the woman simply mentions in conversation that at

least one of her two children is a girl, the probability that the second

child is a girl is 1/3. (In the first case, you have to choose between GG

and GB. In the second, between GG, GB, and BG.)

This problem is pretty well-known. Lots of assumptions go into it,

and the answer depends on exactly how the family was chosen. Google

"sister probability problem" for some more discussion.

Cheers,

John Walsh"

"I agree with you, the probability is 1/2.

Rick Kenyon"

Posted by: Adam | Wednesday, November 22, 2006 at 12:48 PM

Bruno, perhaps the reason the problem is given with 8 coins rather than 9 is to make it harder to see the solution, since 9 would naturally tend to lead people to think of dividing the coins into three groups of 3.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 12:59 PM

I would like to add a voice that perhaps speaks for others as well as myself. I am not well versed in fundamentals of probability theory, but I have encountered the Monty Hall problem before and thought about that problem until I felt I understood what was going on. Ever since this "Problem #5," however, was first posted I have doubted the "not as trivial as it may look" caveat. But I could never have articulated my reasons for doubt as clearly and thoughtfully as have the devoted commentators above. I am very glad for the persistence that's been shown here, and I would guess that other readers like myself are similarly grateful. Thanks.

Posted by: Ihsan D. | Wednesday, November 22, 2006 at 02:22 PM

I used this very interesting problem set as an excuse to write about George Polya and heuristic, here.

Incidentally, Archimedes and I are pretty sure you got problem 7 wrong.

Posted by: Aaron Haspel | Wednesday, November 22, 2006 at 03:43 PM

I think you're incorrect, Aaron.

"The floating ice cube displaces a volume "D" of pure water given by {g*D*ρGlassWater = M*g} or {D = M/ρGlassWater}, where "M" is mass of ice cube. When "C" grams of ice melt, displacement volume decreases by ΔD = C/ρGlassWater. Furthermore, additional liquid water volume (from melting ice) is added given by ΔW = C/ρCubeWater. Since {ρGlassWater = ρCubeWater} because both are pure fresh water, we have {|ΔD| = |ΔW|}, and the glass water level remains constant."

Posted by: Adam | Wednesday, November 22, 2006 at 04:07 PM

Possibly; I want to think it over before I reply. However, we can surely agree that it is incorrect to say that "the floating cube displaces its own weight in water"; it does not.

Posted by: Aaron Haspel | Wednesday, November 22, 2006 at 04:34 PM

Aaron, in your blog post you gave the answer 2/3 to #5, while most commentators on this thread (including a number of experts) have come to agree the answer should be 1/2--did you read through the arguments on both sides?

As for the ice cube problem, this page notes some small corrections to the idealized case where the ice cube displaces exactly its weight so the water level remains unchanged when it melts:

http://van.physics.uiuc.edu/qa/listing.php?id=1650

Posted by: Jesse M. | Wednesday, November 22, 2006 at 04:35 PM

By the way Aaron, why do you say that "Archimedes and I are pretty sure you got problem 7 wrong"? If we consider this problem solely in terms of archimedes' principle, ignoring issues like the temperature change of the water after the ice melts, then it's clear that the cube *should* displace precisely its own weight in water. After all, Archimedes' principle says the bouyant force on an object under water is equal to the weight of the water it displaces; for the ice cube, the bouyant force upward will be equal to the weight of the water needed to fill the volume of the portion of the ice cube that is submerged. In order for the ice cube to be at equilibrium, neither sinking nor rising, the buoyant force pushing up on the cube must be equal and opposite to the gravitational force pulling down (ie the weight of the entire cube, including the part above water), so the weight of the water displaced by the part of the cube that's submerged must be equal to the weight of the entire cube.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 04:48 PM

In Aaron's thread, he was speaking of a 2/3 chance of the second child being of the opposite sex. He reversed it.

Posted by: Adam | Wednesday, November 22, 2006 at 04:52 PM

On Problem 7, you are both right, and I am wrong, and I will update my post accordingly.

As to Problem 5, yes, I'm familiar with the arguments on both sides. As I said in my post, it depends on exactly how the question is put. I read it as equivalent to, "one of the two children is girl," but I grant that the problem is ambiguously stated and that the other reading is also plausible. But a long discussion of Bayesian probability would have been too far beside my point and added too much to an already overlong post.

Posted by: Aaron Haspel | Wednesday, November 22, 2006 at 04:55 PM

With regard to question number 5. I am not sure this is resolved or not or whether this has already been stated. Answering the question initially I agreed with Adam. But thinking on it 1/3 is the coreect probability because she already had the children. If she was pregnant, then the likelyhood of her having a girl would be 1/2, but the fact is the unknown child is there already. The probability breakout would be g/g b/b b/g and g/b. We know b/b is out so this leaves g/g g/b and b/g therfore 1/3 is the answer.

Posted by: Bart | Wednesday, November 22, 2006 at 05:24 PM

I had another solution for the first one: light both ends of the first and when it's burnt light both ends and middle of the second one. A bit less precise and a bit less elegant, I guess.

Posted by: gemp | Wednesday, November 22, 2006 at 05:27 PM

Aaron, I disagree that the other reading is plausible--while I think an accepatable response would be "the problem does not provide enough information about the conditional probability of seeing a girl given a boy-girl family to give a definite answer", I think those who give the answer 1/3 are making a clear-cut mistake about probabilistic reasoning, namely the argument that "I see one child is a girl" can be treated as equivalent to "I know at least one child is a girl" without explicitly making the assumption that the conditional probability of seeing a girl given that the family contains a girl and a boy should be 1 (the 'boys chained in the basement' assumption). To see why it is definitely a mistake to treat the two as equivalent in meaning, just consider the problem I gave earlier: "I flip two coins, *randomly* choose one to show you, and you see that it is a heads. What is the probability the other is heads?" If someone said, "well, the fact that I see a heads logically implies that at least one coin is heads, and HT, TH and HH are equally probable, so the answer is 1/3", would you regard this as anything other than a misunderstanding of the way logical implications can be used in arguments about probability? Of course, in this example the randomness of my choice makes it explicit that the conditional probability of seeing a heads given one heads and one tails would be 1/2. but if I altered the wording so that I didn't specify how I chose which coin to show you, it wouldn't somehow become legitimate to treat seeing a head as equivalent to knowing at least one coin is a head--in this case there would simply not be enough information to solve the problem (after all, if I don't specify the procedure, it's possible that I'm still using the procedure of picking randomly). It would *never* be legitimate to give an answer to this problem without having some definite assumption about the conditional probability of seeing a heads given one heads and one tails.

Incidentally, I'd also disagree that explaining the answer requires a discussion of Bayesian probability--the frequentist interpretation of probability works fine for understanding this problem, as I tried to show earlier with my list of the 8 possible equally-probable elements of the sample space.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 05:30 PM

Bart, I think you're missing the main point, which is that even if she already has a girl and a boy, that does not mean there is a 100% chance you'll see her with a girl when you meet her, she might have happened to be walking with the boy that day instead. Do you disagree that if the experiment were repeated many times with different mothers of two, the following possibilities would 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

And again, if you repeat the experiment many times and then just look at the subset of cases where the woman had a girl with her, then this subset must consist of cases 3, 6, 7, and 8; in 3 and 6 the other child is a boy, while in 7 and 8 the other child is a girl, and all four occur with equal frequency.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 05:36 PM

gemp, I don't think your solution to problem 1 works, because it was stated in the problem that the rope doesn't burn at an even rate--if you light one rope in the middle, it's possible the fire would take 13 minutes to burn to the left end and 47 minutes to burn to the right end, for example.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 05:40 PM

The first rope burns in 30 minutes. The second lighted on three places: both ends and the middle (or anywhere around it), where it will burn on both directions, will effectively burn it in 15 minutes whatever the burning rate.

Posted by: gemp | Wednesday, November 22, 2006 at 05:51 PM

Okay, since Jesse is still going, I'll add some additional thoughts that I had on the matter, originally posted elsewhere:

Scenario 1: What are the odds that a mother will have two children both be sons?

Possible outcomes: GG GB BG BB

The odds of any one outcome is 1/4.

So the answer to scenario one is 1/4. A mother of two children has a 1/4 chance of having two boys.

Scenario 2: A mother of two is walking down the street with one of her children, chosen randomly. Given that the child with her happens to be a boy, what are the odds that the child at home also happens to be a boy?

Her possible families are: GG GB BG BB

Let's assume, for our example, that the child is in the first position. I know you don't believe you can do this; I hope to illustrate why.

So, first condition: that child happens to be a boy. First position boys have odds of 1/2. There is a 1/2 chance that a randomly selected child will be a boy. We can then eliminate all families without a boy in the first position, reducing her possible families now to: BG BB.

What are the odds that the second child is also a boy? We can see from her possible families that the odds are 1/2.

Now, you want to argue that we cannot assume the child is in the first position. That is because you're equating "position" with "birth order". Birth order is irrelevant. Because the choice was random, we can call that child "first".

Consider: If you reverse our assumption, and assume the child with her is in the second position, you then get the same answer. Thus our assumption was irrelevant to our having calculated the odds.

What I would like to point out is that our odds of the first child being a boy (1/2) times the second child being a boy (1/2) gives you total odds of 1/4, which is exactly correct for the odds of a two child family having two boys, as Scenario 1.

So the answer to scenario 2 is 1/2, however, there's a given assumption that the first condition (that the child with her is a boy) has already been met.

Scenario 3: You are told that a mother of two has at least one boy. What are the odds that she has two boys?

Okay, same initial possible families: GG GB BG BB.

3 of the 4 permutations have at least one boy. The odds of the first condition being true (that she has at least one boy) are 3/4.

This strikes GG from the possible families, leaving GB, BG and BB. Unlike the first scenario, where we can assume a "slot" for the boy because we've been shown one slot at random, in this scenario we know they've looked in both slots for a boy. That's why they have greater odds for the first condition being met.

That's also why they have less odds of the second condition being met. Both slots were already polled for boys, so unless there ARE two boys, the only available boy has been "used up."

So, getting back to the question, what are the odds, now, of the second condition being met? Well, we have three permutations of families and only one of them meets the criteria. Therefore, the odds of the other child being a boy is 1/3.

If, in this scenario, we multiply the odds of the first condition by the second condition, like we did for the second scenario, we get 3/4 times 1/3 equals 3/12, which reduces to 1/4. Note, again, that 1/4 is exactly what we predict the odds of being in scenario 1. Scenario 3 requires lower odds for the second condition being met since it has higher odds for the first condition being met. If it didn't balance out that way, the odds of having two children both be boys wouldn't be the same for each scenario, and it must be the same.

Posted by: Adam | Wednesday, November 22, 2006 at 05:57 PM

Not so, gemp. The problem states the rope burns at an inconsistant rate. The middle could be 5 minutes from one end and 55 from the other.

Posted by: Adam | Wednesday, November 22, 2006 at 05:58 PM

Dammit. Thanks for your comments. That makes a second one I got wrong (and two I didn't get at all).

I'll come around here a bit more. I'm learning things :)

Posted by: gemp | Wednesday, November 22, 2006 at 06:13 PM

gemp, if a rope lit only in the middle would take 52 minutes for the fire to reach the left end and 8 minutes for it to reach the right end, then if you light the rope at the middle and the ends at the same time, then the section from the right end to the middle will take 4 minutes to burn, while the section from the left end to the middle will take 26 minutes to burn, so the entire rope takes 26 minutes to burn, not 15.

Posted by: Jesse M. | Wednesday, November 22, 2006 at 06:16 PM

I'm not a mathematician, although I understand that the odds in the Monty Hall problem are 1/3.

But I was so hoping that someone would propose this alternative answer, and nobody has, so here it is:

The odds of the second child being a daughter are nil, because if both her children were girls, she'd have said, "I have two daughters." Or "two girls". She wouldn't be likely to refer to her progeny as "children" unless they had a mixture.

Hey, I think it's equally as mathematical as the answer to question 14!

Posted by: Kirsten | Wednesday, November 22, 2006 at 06:26 PM

By "a mixture", I meant "representatives of each gender".

That's what you get for redrafting. Sigh.

Posted by: Kirsten | Wednesday, November 22, 2006 at 06:29 PM

Again, the confusion comes from a subtlety of language that we don't distinguish between in ordinary cases. Saying, "Someone has seen one of their children at random and I can conclude that at least one is a girl," carries different information than, "Someone has seen both their children and at least one is a girl." In everyday speech, we would just say, "At least one is a girl," and leave it at that.

Posted by: Adam | Wednesday, November 22, 2006 at 06:42 PM

Adam, I don't think Kirsten was saying anything about the "at least one is a girl issue", I think she was just saying you could treat it as a trick question where if the mother had two girls, she would have used the word "daughters" rather than "children". In my experience parents do use the phrase "my kids" when talking about two sons or two daughters though...

Posted by: Jesse M. | Wednesday, November 22, 2006 at 07:13 PM

Oh, sorry, misunderstood.

Posted by: Adam Barnett | Wednesday, November 22, 2006 at 08:33 PM

At the risk of flogging the shadow of the ghost of a long-dead horse, let me suggest one other way of looking at the "daughter" problem.

Consider the following table of possible values of a 2-bit (no, we're not talking about the quarters problem here) number:

0: 00

1: 01

2: 10

3: 11

where we will call the right-hand bit "Bit 0" and the left-hand bit "Bit 1".

Now let's assume that I've randomly generated a 2-bit number. If I tell you that Bit 0 is a One, then the possible values of the number are 1 and 3. If I now ask, "What is the probability that Bit 1 is a One?", I think it's clear that the answer is 1/2.

On the other hand, if, instead of telling you that Bit 0 is a One, I tell you that there is a One in the number (by which I expect you to understand that either Bit 0 is a One, Bit 1 is a One, or both are Ones), I think it's clear that this condition only rules out the value 0, so I've reduced the size of the set of possible choices from four to three. If I now ask, "What is the probability that both bits are Ones?", I think it's clear that of the three remaining possible values, 1, 2, & 3, only one of the three satisfies this requirement, hence the odds are 1/3.

So the question, "Bit X is a 1; what are the odds that Bit Y is also a 1?" is a very different question from, "Either Bit X or Bit Y is a One or both are Ones; what are the odds that both are Ones?"

As I said, just another way of thinking about it. How you choose to map this into the semantics of the original problem statement is up to you :-).

By the way, speaking of semantics, in one of the comments John Walsh misquoted the original problem in an interesting way. He said, "You're walking down the street and encounter a woman with a daughter."

The original statement was "... I met a woman strolling with HER daughter." As a native speaker of more-or-less Midwestern American English, I would interpret the phrase "woman with a daughter" more loosely than "woman strolling with her daughter". In the "woman with a daughter" case, it is less certain that the daughter is present, and that the daughter, if present, is the encountered woman's daughter rather than the daughter of, say, a friend. However, in the "woman strolling with her daughter" case, I find it difficult to interpret in any other way than that the daughter was indeed present and was indeed the daughter of the woman encountered.

Posted by: M. S. | Wednesday, November 22, 2006 at 08:52 PM

If we're going to talk about realism in logic puzzles, then there are a couple issues with these puzzles:

1) The odds of a child being a boy or girl are not even. (And a Stanford prof concluded a couple years ago that the odds of a coin flip are not 50-50 either - unpracticed flips are slightly biased to land in the same orientation as they started. Puzzle writers will need to find a new, truly 50-50 real-world event to use in their puzzles. :)

2) Wine consists primarily of water, so the TRULY correct answer to #9 is: the wine jug contains more water than the water jug contains "wine". It starts and ends that way. Perhaps oil & water would be better examples, except the reader might then wonder how the liquids were poured/transferred... I'd suggest replacing "water" in this puzzle with another liquid, like "beer" (always a good move).

Also no one seems to have mentioned that there are two valid answers to question 3. On the second trip either the dog or the corn can be taken.

Someone else brought up the issue with #14: the puzzle needs to state that the lamp is initially off.

Finally, puzzle 12 is not well-worded. I understood the intent and the solution immediately, but also immediately saw a problem. There will always be an "oldest" daughter because two children cannot be born at the same instant (it's theoretically possible with two mothers, but unlikely, and certainly not guaranteed). The statement given in the solution, "the presence of a single oldest child eliminates 2,6,6," is incorrect. One of the 6-year-olds will be older than the other. The problem here is really one of cultural convention: when we refer to people's "age" we normally round down to a whole number of years (or months for babies), but when we use the word "oldest" we do not normally mean that the ages cannot be the same - only that one was born before the other. I have a friend who's a twin, but he's considered the oldest because he was born a few minutes earlier.

I don't think this means the puzzle or solution are "wrong". It's just a bad puzzle because it's hard to express the final statement ("...oldest daughter...") in a correct way without giving away the answer. I've seen other logic puzzles with exactly the same flaw. Maybe it has something to do with how we programmers can get into integer-thinking-mode and round numbers without realizing it.

Posted by: Dave | Wednesday, November 22, 2006 at 09:15 PM

sorry if i'm being dense, but i still don't get #5. trying to express my troubles as succinctly as possible:

assume you have an endless supply of mothers of two, and the gender distribution across all children is 50:50. now disregard all mothers of two boys, we already know we didn't run into one of those. now 2/3 of the remaining mothers have a boy and a girl (henceforth group A), while 1/3 have two girls (group B). so, let's say that group A is twice as large as group B. (mathematicians will realize that in reality both groups are equally large of course, but still you're twice as likely to run into a mother from group A; you get the idea.) now all mothers pick a child and take it for a walk. mothers from group A have a 50% chance of picking the boy. we have to throw those out, too; we know we didn't run into one of them. this leaves us with group A': mothers of a boy and a girl, which took the girl for a walk. note that group A' is now equally large as group B. we don't have to disregard anyone from group B, because no matter which child was taken for a walk, it is always a girl anyways.

now when we meet our mother of two, it is equally likely that she is from group A' as it is likely that she is from group B. therefore, it is equally likely that she has a boy at home, as it is likely that she has a girl at home. the probability of her other child being a girl is 1/2.

or put in much simpler terms: you meet a stranger and the stranger tells you she has a child, which you don't know, can't see, have never heard about. what's the probability of that child being a girl? well, roughly 50%.

now please, someone put me out of my misery and show me the flaw in my reasoning.

--~~--

about the light-switch-problem: i agree it's true that you'd have to know the initial state of the light bulb. i can't think of a solution that would work if you don't know the initial state. the problem doesn't specify the initial state, therefore the correct answer would be: "there is no solution." right?

Posted by: nex | Wednesday, November 22, 2006 at 10:04 PM

I wasn't trying to nitpick in my last comment, but to hint at a wider problem.

Logic puzzles are often solved by BREAKING convention... getting past the way you normally think... seeing things in a different light.

But there also seems to be a body of conventions and assumptions used by most logic puzzle writers, e.g., that heads and tails or boys and girls are equally likely. If you read lots of logic puzzles, you will eventually figure out what these conventions are.

(Or you might have that well-rewarded intuitive ability to quickly put yourself in the writer's shoes... folks like us are good at taking tests and do better than average at quiz shows... I've noticed that a small but fairly steady percentage of Jeopardy questions almost give you the answer, and seem intended to test the contestants' abilities to think quickly more than to test the depths of their knowledge. Some game shows like Millionaire and 1vs100 seem to be loaded with such questions; though the simplest explanation is just that their question writers aren't very good, especially at coming up with wrong answers that are misleading/believable. I'm thinking specifically of Ogi Ogas' million-dollar question on Millionaire, "Which of these ships was not one of the three taken over by colonists during the Boston Tea Party?" The correct answer was William, which immediately jumped out at me as the kind of fake English ship that a question writer would make up. But I seriously digress.)

The problem lies in knowing which conventions you can break and which you can't. Good logic puzzles don't force you to make that choice.

Posted by: Dave | Wednesday, November 22, 2006 at 10:21 PM

quoting abbas:

"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. [...]

Flip two coins a hundred times, then try it!"

are you saying that if i flip two coins a hundred times, you'd expect them to show identical sides 2/3 of the time? and shouldn't you actually expect that they're identical only 1/3 of the time, as the revealed coin is the girl we saw, and you claim the other child to also be a girl (identical gender) 1/3 of the time?

if i owned a casino, you'd be my favourite customer.

Posted by: nex | Wednesday, November 22, 2006 at 11:00 PM

Nex, you are correct. The 1/3 answer is coming from people who are interpreting the question as if the mothers from your Group A have automatically picked a girl, instead of a %50 chance of picking the girl, as you stated. The question, to me, doesn't seem to state this.

If the mother had a predisposition to bring a girl, then the chance of two girls is 1/3. If the mother chose what child to bring randomly, then the chance is 1/2.

Posted by: Adam Barnett | Thursday, November 23, 2006 at 12:59 AM

The answer to #7 seems wrong, we aren't talking about the weight but about the volume and as far as i know water expands when it freezes?

Posted by: Stephan | Thursday, November 23, 2006 at 11:07 AM

Jesse M, you are quite right - I was indeed playing with linguistic, rather than mathematical possibilities.

Kind of a feeble joke, but hey.

Posted by: Kirsten | Thursday, November 23, 2006 at 12:00 PM

stephan, i can help you there: weight is relevant as the property you describe is one of density, and density links weight to volume. because the ice does indeed shrink when it melts, one might ask if this causes the water level to go down. but while the density changes, the weight stays the same. (of course!) and therefor the water level stays the same, because of buoyancy and the fact that the ice cube did not displace its entire volume in the water, but only its weight in water. that's why ice cubes float and their tops stick out a bit.

Posted by: nex | Thursday, November 23, 2006 at 12:16 PM

The answer for #13 is over-complicated. All you do is replicate the actions made on the way out. Plane 1 meets 3 at 6/8s and both head home while 2 flies out to meet them both at 7/8s gives them both a quarter tank and they all come in landing on fumes. It's just simpler this way, less fill ups and less excess fuel.

Posted by: James | Thursday, November 23, 2006 at 04:30 PM

Couldn't 3,4, and 6 work for puzzle #12? 3 times 4 times 6 equals 72. The sum adds up to 13, not 14, but where does it say it has to add up to 14? Can't house number be 13?

Posted by: Chris | Thursday, November 23, 2006 at 05:51 PM

Confusion over #5 seems to me to be a semantic question of whether to interpret the problem as:

Given a two child family in which one child chosen at random is female, what is the probability that both are female? (50%)

or

given a two child family in which there is at least one female, what is the probability that both are female? (33%)

it seems to me that the way the original question is posed, it is the first case. Had he met a woman who simply asserted that one of her two children was female, it would have been the second case.

The difference in probabilities can be reduced to whether an unseen child is female (50%), or whether one child of two is female given that either of two children is female (33%).

I validated these results in an excel spreadsheet I'd be happy to email to anyone interested. Remove the *nospam* from my listed email.

Posted by: jb | Thursday, November 23, 2006 at 06:39 PM

Chris, there has to be two sets of numbers that multiply to 72 and add to the same number, so that he wasn't sure what the correct answer was by looking at the house number. If the house number had been 13, he would have been certain without the strawberry shortcake fact.

Posted by: Adam | Thursday, November 23, 2006 at 07:11 PM

I have a University degree in math. There is a 1/2 probability both children are girls.

Posted by: Mark | Thursday, November 23, 2006 at 08:36 PM

The answer to the mother/daughter problem should be 1/12, because when "she" said she has two children, that could be either the mother or the daughter speaking. The odds that the daughter has two daughters is 1/4, and the odds that the mother has another child who is a girl is 1/3, hence an overall probability of 1/12.

Posted by: Orlie Omegamu | Thursday, November 23, 2006 at 09:46 PM

Fantastic debate!

I would like to thank all participants, even though they are to blame for me not getting any sleep tonight.

Cheers from Denmark

Posted by: Niels Kern Bertelsen | Thursday, November 23, 2006 at 10:03 PM

If I may join the explanation of the mother-daughter problem....

The correct answer to the problem as stated is 1/3.

The probability of 2 daughters in a 2 child family is 1/4, not 1/3, since the boy-girl scenario is 2ce as likely as boy-boy or girl-girl. Imagine all 2-child families: 1/4th will be girl-girl, 1/2 will be boy-girl, and 1/4th will be boy-boy. Seeing one daughter tells you that the family is not boy-boy, which means you can eliminate all boy-boy families from the pool of families under consideration. Of the remaining families, 2/3rd will be boy-girl, and 1/3rd will be girl-girl. Thus the correct answer is 1/3rd.

The reason this answer is counter-intuitive is because English language is not designed for communicating such questions. No one here would be confused if this question were expressed mathematically or as a diagram.

Or please compare the question above to the following. Someone tells you that his best fried has 2 children and they are not both boys. What is the probability the two kids are both girls? How is this probability different from the one referred to in the original question? It is not. It is 1/3rd.

Posted by: Clyde | Thursday, November 23, 2006 at 11:16 PM

Also, re: Q13 - if one of the planes has some fuel left, it is probably not the most optimal solution. IMO

Posted by: Clyde | Thursday, November 23, 2006 at 11:18 PM

I got the answer to the harder problem, but then all the balls rolled off the scale and went in the sewer. Drag.

Posted by: dmac | Thursday, November 23, 2006 at 11:18 PM