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
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 | Nov 21, 2006 3:01:23 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 | Nov 21, 2006 3:04:48 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 | Nov 21, 2006 6:05:28 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 | Nov 21, 2006 7:24:00 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 | Nov 22, 2006 12:22:52 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 | Nov 22, 2006 2:12:38 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 | Nov 22, 2006 12:48:45 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. | Nov 22, 2006 12:59:58 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. | Nov 22, 2006 2:22:53 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 | Nov 22, 2006 3:43:47 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 | Nov 22, 2006 4:07:30 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 | Nov 22, 2006 4:34:40 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. | Nov 22, 2006 4:35:03 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. | Nov 22, 2006 4:48:17 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 | Nov 22, 2006 4:52:25 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 | Nov 22, 2006 4:55:36 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 | Nov 22, 2006 5:24:52 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 | Nov 22, 2006 5:27:20 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. | Nov 22, 2006 5:30:10 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. | Nov 22, 2006 5:36:07 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. | Nov 22, 2006 5:40:42 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 | Nov 22, 2006 5:51:35 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 | Nov 22, 2006 5:57:25 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 | Nov 22, 2006 5:58:55 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 | Nov 22, 2006 6:13:54 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. | Nov 22, 2006 6:16:50 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 | Nov 22, 2006 6:26:41 PM
By "a mixture", I meant "representatives of each gender".
That's what you get for redrafting. Sigh.
Posted by: Kirsten | Nov 22, 2006 6:29:09 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 | Nov 22, 2006 6:42:50 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. | Nov 22, 2006 7:13:27 PM
Oh, sorry, misunderstood.
Posted by: Adam Barnett | Nov 22, 2006 8:33:36 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. | Nov 22, 2006 8:52:13 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 | Nov 22, 2006 9:15:11 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 | Nov 22, 2006 10:04:47 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 | Nov 22, 2006 10:21:38 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 | Nov 22, 2006 11:00:11 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 | Nov 23, 2006 12:59:05 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 | Nov 23, 2006 11:07:54 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 | Nov 23, 2006 12:00:46 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 | Nov 23, 2006 12:16:25 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 | Nov 23, 2006 4:30:36 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 | Nov 23, 2006 5:51:44 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 | Nov 23, 2006 6:39:27 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 | Nov 23, 2006 7:11:53 PM
I have a University degree in math. There is a 1/2 probability both children are girls.
Posted by: Mark | Nov 23, 2006 8:36:27 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 | Nov 23, 2006 9:46:48 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 | Nov 23, 2006 10:03:32 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 | Nov 23, 2006 11:16:12 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 | Nov 23, 2006 11:18:11 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 | Nov 23, 2006 11:18:47 PM
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