## Three princesses

A forum for good logic/math puzzles.

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aaronspook
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Possible Solution:

How about we ask daughter A "Is daughter B older than daughter C?" If daughter A is the middle daughter, it doesn't matter which of the other two we choose. If daughter A is the eldest, we marry the one she indicates is younger. If daughter A is the youngest, we want to marry the elder of the other two, which means marrying the one she says is younger. So if the answer is yes, we always marry daughter C, and if it's no, we always marry daughter B.

I'm probably missing something, but this seems like a full solution.

DaveFP
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By Jove, I think he's got it! That's very clever, and thinking in completely the other direction that everyone else has been going in. Well done!

Fleshpiston
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### Congratulations!

Congratulations aaronspook! I was starting to wonder if anyone was going to get it. Your answer is exactly what I had in mind when I started the thread. It came just in time, as it looked like people were starting to give up.

I bet few people will be able to resist the temptation to read your spoiler, but if they can, they will know that you can come up with a solution on your own, and they should consider more ways a solution could be possible.

Thanks to everyone for their participation.

xkcd
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Very, very nice.

Shoofle
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*Punch self in face for responding like this*
I thought about going down a path of "Yes indicates one sister, no indicates another" (which is effectively what aaronspork has done) but hadn't connected it with questions of age.

RealGrouchy
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I read aaronspook's solution, but still don't understand it.

Maybe some sleep will help.

- RG>
Mighty Jalapeno wrote:At least he has the decency to REMOVE THE GAP BETWEEN HIS QUOTES....
Sungura wrote:I don't really miss him. At all. He was pretty grouchy.

ZombieRoboNinja
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GAHH, so simple...

I was all hung up on trying to get a useful piece of information out of the middle daughter.

Pyrthas
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sniffels wrote:Correct me if I am wrong, but I don't think it is possible to generate a question which doesn't refer to itself which is capable of producing a logical conundrum.
Well, you can construct some, although the problem still seems to be some sort of circularity: "What is the least integer not namable in fewer than nineteen syllables?" Unless I've just proved that there is no such integer, something has gone wrong.

Another: The word 'short' is short, but the word 'long' is not long. Call words of the second sort (i.e., those words that express a predicate that does not apply to the word itself) 'heterological'. Is 'heterological' heterological?

Anyway, sorry about the hijack, and nice answer. I, like Shoofle, was working on something similar, but hadn't come up with asking whether one specific one was older than another. Fun puzzle.

wisnij
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http://www.partiallyclips.com/index.php?id=1008

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xkcd
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wisnij wrote:http://www.partiallyclips.com/index.php?id=1008

Hehe, that was one of my favorites; I'd forgotten it.

Shoofle
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HAH! I had never seen that one before.

RealGrouchy
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Ah, now I understand.

It also helps that I re-read the question.

I think I'd want to marry the youngest sister, then always ask her in public if she thinks I'm a good lover.

- RG>
Mighty Jalapeno wrote:At least he has the decency to REMOVE THE GAP BETWEEN HIS QUOTES....
Sungura wrote:I don't really miss him. At all. He was pretty grouchy.

Westacular
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Aha! I like this. I really, honestly managed to get the solution shortly after reading Fleshpiston's whited-out clue, and my formulation of it was almost identical to aaronspook's.

This is interesting, in the way that it skirts around the apparent impossibility of trusting any answer -- the goal of the problem is actually more relaxed than it seems.

Since this thread is pretty dead now, I'm going to post spoilery details in clear text. Beware!

IF you are talking to the lying sister, that means it's already safe to marry either of the other two, regardless of her answer. So condition your plan on not marrying whomever you ask the question to. (STRATEGY: Pose the question to the least attractive one!) What's then needed is a question that will unambiguously identify which of the other two is the middle sibling if posed to either eldest or youngest. Asking A, "Is B older than C?" provides that, thanks to the ordinal organization and the symmetry between youngest and oldest. In hindsight it should have been clearer to me that the correct solution would exploit that.

ulnevets
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how about "can you tell me 100 facts?"

if they all seem to be truths, then it's probably the oldest. if they all seem to be lies, then it's probably the youngest. if there's some of each, then it's probably the youngest.

but this method is not foolproof, you say? well, it is as the number of facts approach infinity, so a variation would be to ask her to keep telling you facts. when you've reached the margin of error you're happy with, you can tell her to stop and to marry you.

ain't nobody said they had to be yes-or-no questions.

xkcd
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but this method is not foolproof, you say? well, it is as the number of facts approach infinity, so a variation would be to ask her to keep telling you facts. when you've reached the margin of error you're happy with, you can tell her to stop and to marry you.

Nowhere does it say she chooses the truth or lying randomly. Today could be an all-truth day.

Anyway, "can you tell me 100 facts?" gets the answer "yes" (or "no") :)

ulnevets
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xkcd wrote:
but this method is not foolproof, you say? well, it is as the number of facts approach infinity, so a variation would be to ask her to keep telling you facts. when you've reached the margin of error you're happy with, you can tell her to stop and to marry you.

Nowhere does it say she chooses the truth or lying randomly. Today could be an all-truth day.

Anyway, "can you tell me 100 facts?" gets the answer "yes" (or "no")

also, infinity means it'll last a lot longer than a day or a month. supposing you ask until she dies eventually, she will have to tell at least one lie or she will have turned into a truth-teller!

(i'm actually just screwing with you. i think aaronspook got it.)

archgoon
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About the paradox dragon, it doesn't seem to be resolvable. The middle head says "I was lying", which seems to peg him as the alternator (the truth teller would never have lied in the past,and the liar would never admit to it). Yet, that would imply in the first panel that the liar had agreed with the truth teller, which isn't possible.

Am I missing something? Or is the entire point that the paradox dragon is horribly confused?

Shoofle
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The paradox dragon is confused - remember how he died. Also, he is a paradox dragon.

Vonkwink
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that is very impressive... props to you, fleshpiston, for the puzzle, and also to aaronspook and everyone else who got the answer.

Weka
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### Two Challenges, Three Princesses

I really liked the 'Three Princesses' puzzle , but one thing about it irked me. It seemed that solving it depended on specifically knowing the relative ages of the princesses. I spent some time wondering if it might be possible to solve this if you didn't know anything about their ages (after all, it is jolly poor form to ask a woman about her age , and even worse to ask about another woman's age - and remember we're dealing with high society princesses here...)

Musing on this leads me to pose two challenges:

1) I had an idea for a solution, which involved intimidation with a nasty sharp sword (which, while unpleasant, is quite acceptable courtship etiquitte for enchanted princesses in hypothetical castles) and a curly question that trapped the always-truthful and always-untruthful princesses in a paradox (similar to the "I am lying to you now" paradox). Only the princess who is not constrained to telling just truth or just lies could answer, thus giving her away. However, while it is easy to trap the liar, I couldn't for the life of me think of a yes/no question that traps an always-truthful person into a paradox .

Challenge 1 is therefore 'Find a yes/no question that the always-truthful princess could never answer' (meta-questions are OK). I'd also be interested to know if there is a more general paradox that any always-truthful person cannot answer (even ones who don't have lying sisters to frame meta-questions around)? Warning - I don't know if there is an answer to this problem! This way lies madness...

2) While doing this, I stumbled upon a completely different solution to the problem, so Challenge 2 is to find a question you could ask one of the Princesses to evade the sometimes-liar even if you didn't know anything about their ages.

Torn Apart By Dingos
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### Re: Two Challenges, Three Princesses

Weka wrote:It seemed that solving it depended on specifically knowing the relative ages of the princesses.

You don't have to. You could use some other property, or even define one yourself. You could ask: "if i the lying princess is princess 1, the truthtelling one is princess 3, and the other one is princess 2, then ...", and use the question as above, but use their numbers instead of their ages.

Weka wrote:a curly question that trapped the always-truthful and always-untruthful princesses in a paradox (similar to the "I am lying to you now" paradox)

I'm not sure that would work. For one, if they're given a paradox, then the lying princess could say anything she wants, because she knows her answer will be a lie. But assuming you could trap both the liar and the truth-teller, why wouldn't the middle sister be trapped too? She's defined by sometimes telling truth, and sometimes lies, so she even when she decided what mode to be in, she couldn't answer.

Sidestepping that issue, the problem would need an additional time aspect, since you'd need to know how long it would take a princess to answer, if she has an answer - maybe there isn't a time limit, then you'd be none the wiser.

Tropylium
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### Re: Two Challenges, Three Princesses

Weka wrote:Challenge 1 is therefore 'Find a yes/no question that the always-truthful princess could never answer' (meta-questions are OK). I'd also be interested to know if there is a more general paradox that any always-truthful person cannot answer (even ones who don't have lying sisters to frame meta-questions around)? Warning - I don't know if there is an answer to this problem! This way lies madness...

I think this works:

(However, I think the "paradox" approach instead of revealing the pricesses' identity rather proovs that "always truthful" and "always lying" are impossibilitiesâ€¦)

theneutralnewt
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For challenge 2, how about asking A, "Is B more likely to lie than C?" and take B if yes, else C? If you ask the middle one it doesn't matter, if you ask the oldest it would point to the youngest, if you ask the youngest it would point to the oldest.

Westacular
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I think theneutralnewt is onto something. It has the same parallelism as the standard answer, without requiring the ages setup. Nice. Is that what you had in mind, Weka?

Fierce
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I dont see why aaronspook's works and I have my own version if you're allowed to ask the one question to all three sisters:

aaronspook's seems to need to know if the one you are asking is oldest/youngest/middle to decide on choosing the one she calls oldest or the middle (might be wrong about this)

If you ask all of them if they are the middle sister, the oldest will say no, the youngest yes, and the middle could be either. Then you simply pick the one that is alone
IE: Oldest:No Youngest:Yes Middle:No so you marry the youngest
or
Oldest:No Youngest:Yes Middle:Yes so you marry the oldest (the only one who said No)
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EndofEternity
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Fierce wrote:I dont see why aaronspook's works and I have my own version if you're allowed to ask the one question to all three sisters:

aaronspook's seems to need to know if the one you are asking is oldest/youngest/middle to decide on choosing the one she calls oldest or the middle (might be wrong about this)

If you ask all of them if they are the middle sister, the oldest will say no, the youngest yes, and the middle could be either. Then you simply pick the one that is alone
IE: Oldest:No Youngest:Yes Middle:No so you marry the youngest
or
Oldest:No Youngest:Yes Middle:Yes so you marry the oldest (the only one who said No)

aaronspook has a 100% working answer, it assumes nothing... you're just asking an arbitrary sister about the relative ages of the other two. It works this way because by asking one sister about the other two, you effectively remove her from your possible mates.

ulnevets
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who cares?

go for the hottest one

Nex3
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### Another Solution

I totally registered so I could post my answer . I may stay on, though.

Mine is different from aaronspook's in that the mischievous girl will always answer differently than the other two. The question you must ask is:

Is the statement "You can be honest, and if you are mischevious, then your answer to this question is dishonest." true?

Note that this uses the definition of "if/then" used in logic; the statement "If p then q" is false when p is true and q is false, and true otherwise.

So, let's go over this. If the girl is honest, the statement will be true, because she can be honest and is not mischevious. Thus, she will answer "Yes." If the girl is dishonest, the statement will be false, because she cannot be honest. Thus, she will answer "Yes."

Then, we come to the mischievous girl. She can be honest, so we can ignore that part (true and true is true, true and false is false, so it depends on the truth of the second part). In addition, she is mischevious, so the statement will be false if and only if the last part is false. If she answers honestly, her answer is not dishonest, so the statement is false and she says "No." If she answers dishonestly, though, the statement is true and she also answers "No."

Thus, the mischievous girl will always answer "No" and the others will always answer "Yes."

This was a delightful puzzle, by the way .
Last edited by Nex3 on Tue Sep 05, 2006 8:52 pm UTC, edited 1 time in total.

Fierce
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I still dont see why aaronspook's works, and this may explane why better:

Assume the daughter we ask is daughter 1
She then tells us daughter 2 is older than daughter 3

Now we still have to choose between 2 and 3 but:
2 could be the middle and 3 the youngest, so we would want 3
And
2 could be the oldest and 3 the youngest, so we would want either

2 could be the oldest and 3 the middle, so we would want 2

So unless we know where the daughter we ask stands reletive to the others, we dont know which way to go: chosing the one she says is younger or older

I also dont see Nex3's working:

While the oldest and youngest would answer yes, the middle could also answer yes too because that also is a dishonest answer to the question.
Really you statment breaks down to "if you are the middle daughter, you're answer will be wrong" and that doesn't have a right answer in itself but expecting the middle one to respond with a "no" makes "yes" dishonest.

Another explination because it confuses me even:

Is the statement "You can be honest, and if you are mischevious, then your answer to this question is dishonest." true?

Middle:
Can be honest, so first part is true
Is mischevious, so weather her answer will be dishonest or not remains to be seen

If she were to answer yes, she is being honest about the first part
So that means her answer is not dishonest
And that means she she is being dishonest because she said her answer is dishonest when it isnt
So then her answer is dishonest and then the honest answer is yes
So saying yes is not dishonest
So she is being dishonest because she said she was not
So saying yes is dishonest
So she is being honest because she said she was
etc. (similar pattern for "no")
But expecting a no from the middle sister means that yes is dishonest and can be used to answer.

I still stand by the one i suggested (If you ask all of them if they are the middle sister, the oldest will say no, the youngest yes, and the middle could be either. Then you simply pick the one that is alone) because the problem just wants you to be sure to pick either the oldest or the youngest

Sorry if i seem rude for trying to break you're theories without adding more myself (and for making a huge post to try and explain things as I see them), but feel free to let me know if I made a mistake here
Last edited by Fierce on Tue Sep 05, 2006 9:23 pm UTC, edited 1 time in total.
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Nex3
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Fierce wrote:I still dont see why aaronspook's works, and this may explane why better:

Assume the daughter we ask is daughter 1
She then tells us daughter 2 is older than daughter 3

Now we still have to choose between 2 and 3 but:
2 could be the middle and 3 the youngest, so we would want 3
And
2 could be the oldest and 3 the youngest, so we would want 2

so unless we know where the daughter we ask stands reletive to the others, we dont know which way to go: chosing the one she says is younger or older
Ah, but
Me wrote:we can take either the youngest or the oldest daughter. The only daughter we're trying to avoid is the middle daughter. In your situation, if 2 is the oldest and 3 the youngest, it doesn't matter which one we pick, because both are acceptable.

Fierce wrote:I also dont see Nex3's working:

While the oldest and youngest would answer yes, the middle could also answer yes too because that also is a dishonest answer to the question.
Really you statment breaks down to "if you are the middle daughter, you're answer will be wrong" and that doesn't have a right answer in itself but expecting the middle one to respond with a "no" makes "yes" dishonest.

Another explination because it confuses me even:

Is the statement "You can be honest, and if you are mischevious, then your answer to this question is dishonest." true?

Middle:
Can be honest, so first part is true
Is mischevious, so weather her answer will be dishonest or not remains to be seen

If she were to answer yes, she is being honest about the first part
So that means her answer is not dishonest
And that means she she is being dishonest because she said her answer is dishonest when it isnt
So then her answer is dishonest and then the honest answer is yes
So saying yes is not dishonest
So she is being dishonest because she said she was not
So saying yes is dishonest
So she is being honest because she said she was
etc. (similar pattern for "no")
But expecting a no from the middle sister means that yes is dishonest and can be used to answer.
Me wrote:It's true that she cannot answer "Yes." In fact, the solution relies on her always anwering "No." But answering "No" does end up with a contradiction. "No" is true when she is honest because she can be honest, she is mischievous, and the answer to the question is honest. It seems a bit circular, but it does work.
Fierce wrote:I still stand by the one i suggested (If you ask all of them if they are the middle sister, the oldest will say no, the youngest yes, and the middle could be either. Then you simply pick the one that is alone) because the problem just wants you to be sure to pick either the oldest or the youngest
Me wrote:The problem with this is that the premise of the problem is that you can only ask one girl. You can't ask all three.
Fierce wrote:Sorry if i seem rude for trying to break you're theories without adding more myself (and for making a huge post to try and explain things as I see them)
Not at all. Critique is an essential part of verifying truth.

Note: In case anyone is interested, here's my solution expressed symbollically as a logical proposition. I'll use ^ to be conjunction, ~ to be negation, -> to be implication, and <-> to be biconditional.
Me wrote:Let p be the statement "You can be honest."
Let q be the statement "You can be dishonest."
Let r be the statement "Your answer to this question is honest."

If this proposition is true for a given p, q, and r, then the girl who matches those criteria will answer "Yes" when asked the question I detailed above. If it is false, she will answer "No."

( p ^ ( ( p ^ q ) -> ~r ) ) <-> r

For the honest sister, p is true, q is false, and r is true.
( T ^ ( ( T ^ F ) -> ~T ) ) <-> T
( T ^ ( F -> F ) ) <-> T
( T ^ T ) <-> T
T <-> T
T

For the dishonest sister, p is false, q is true, and r is false.
( F ^ ( ( F ^ T ) -> ~F ) ) <-> F
( F ^ ( F -> T ) ) <-> F
( F ^ F ) <-> F
F <-> F
T

For the mischievous sister, both p and q will always be true.
( T ^ ( ( T ^ T ) -> ~r ) ) <-> r
( T ^ ( T -> ~r ) ) <-> r
( T ^ ~r ) <-> r
~r <-> r
F
So, she will answer "No," regardless of what r is (whether or not she is telling the truth).
Last edited by Nex3 on Tue Sep 05, 2006 9:42 pm UTC, edited 2 times in total.

Fierce
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Responce to Above: (not quoting all that to save space)
1. Typo on my part. I ment 2 could be the oldest and 3 the middle. While what I had is a possible combonation, this is too and makes it impossible to know who to pick for sure
It may not be circular and its my bad, but she can lie which allows for a yes answer (the honest dishonest answer is no, but she can be dishonest about the answer (which is dishonest in itself) to say yes if that makes any sense). Probably should have had that instead.
3. True, but its the best I had and I wanted to add some solotion instead of just nit-picking
4. Yay! Non-snob person to have a fun debate with !
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Nex3
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Fierce wrote:Responce to Above: (not quoting all that to save space)
1. Typo on my part. I ment 2 could be the oldest and 3 the middle. While what I had is a possible combonation, this is too and makes it impossible to know who to pick for sure
Me wrote:If two is the oldest, and three is in the middle, then you're asking the youngest sister. This means that she will lie and tell you that two is the youngest. So, when you pick the person you think is the youngest, you get two, who is, in fact, the oldest.
It may not be circular and its my bad, but she can lie which allows for a yes answer (the honest dishonest answer is no, but she can be dishonest about the answer (which is dishonest in itself) to say yes if that makes any sense). Probably should have had that instead.
Me wrote:If the middle sister is dishonest, then she is mischievous but she is not being honest. This means that the statement "If you are mischievous, you're being dishonest" is true, as is the statement "You can be honest and if you are mischievous, you're being dishonest." Because this statement is true and she's being dishonest, she answers "No."
Fierce wrote:3. True, but its the best I had and I wanted to add some solotion instead of just nit-picking
It would be a good solution if the problem were slightly different.
Fierce wrote:4. Yay! Non-snob person to have a fun debate with !
Indeed .

GreedyAlgorithm
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Oh man, the distinct solution thread is *such* a good idea.
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Fierce
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Well I now see where my mistake was with aaronspook's and it looks like it would work to me now though I still say that since she can lie the middle sister is able to choose which she answers with regardless of the question which was mentioned early on by someone else. Though this definatly was one of the better uses of my time.

And its why i don't like the idea of seperate problem/solotion threads, its dosn't take too long to explane the problem and its the solutions that are fun to discuss.
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xandah
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I realise I've come to this problem somewhat late, but isnt a slightly simpler answer:

"Is the sister standing to your left/right (just to specify a sister in general) more dishonest than your other sister?" and marrying whichever sister is pointed out as being more dishonest?

If you ask the honest sister, she will point out the always-lying sister.
(it is true that the always-lying sister is more dishonest than the sometimes-liar)
If you ask the lying sister, she will point out the truth sister.
(as it is a lie that the truth sister is more dishonest than the sometimes liar)
If you ask the lying sister it doesnt matter who she points out.

Of course, this relies on the sisters knowing the truth telling habits of thier siblings.

Vaniver
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Xandah's works. Aaronspook's doesn't, since his requires knowledge of daughter A's age. Nex's doesn't, because he assumes that the mischievous girl will always do something- that's not how she works. She doesn't even listen to the question and answers randomly, so any logic about her thought process is flawed.

aaronspook
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Actually, my solution only relies on the sisters knowing their relative ages. Xandah's solution also works, as does any that effectively involves comparing an attribute that scales from the lying sister to the truth-telling sister. As far as I can think of, age and honesty are the only two such attributes we know of in this case.

Jack Mac
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I think the way to approach this is to ask a question which will point you to one of the other girls. Ie, if you ask the truth-telling or lying girl, their answer will identify their counterpart. And if you ask the middle girl, her answer will indicate your choice, but be irrelevant since changing your choice in this case results in your desired outcome regardless.

I'm not entirely sure, but I can't see any problem with
"Would <indicate first other girl> lie more often than <indicate remaining girl> if I asked her repeatedly if she always told the truth?"

If you ask the truth teller or the liar, either way "yes" means your first choice is their counterpart, and you choose that one.
"No" means the opposite.
And asking the middle girl is irrelevant, since either of your two remaining choices is acceptable.

EDIT: It's very late at night where I am, sorry. I didn't even see there was a second page to this discussion. My solution is similar to the above, so ignore it.
<3

nes
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Hi, registered just to comment on this one, though maybe I'll participate in future threads. Just in case anyone is still confused by aaronspook's solution, I think I have a nifty chart with all possibilities here; someone please correct me if I made a mistake! I had it formatted a little nicer than this, but the code tag makes it show up in green, even with the color tag

Ask {a} "Is sister {b} older than sister {c}?"
As you will see, as long as you stick with choosing {c} if the answer is "Yes," and {b} if the answer is "No," you'll never get the random girl.

{a} (truth, elder), {b} (random, middle), {c} (liar, young)
- {a} truthfully answers "Yes," so choose {c} (the liar).

{a} (truth, elder), {b} (liar, young), {c} (random, middle)
- {a} truthfully answers "No," so choose {b} (the liar).

{a} (liar, young), {b} (truth, elder), {c} (random, middle)
- {a} falsely answers "No," so choose {b} (the truth teller).

{a} (liar, young), {b} (random, middle), {c} (truth, elder)
- {a} falsely answers "Yes," so choose {c} (the truth teller).

{a} (random, middle), {b} (truth, elder), {c} (liar, young)
- {a} truthfully answers "Yes," so choose {c} (the liar).
- {a} falsely answers "No," so choose {b} (the truth teller).

{a} (random, middle), {b} (liar, young), {c} (truth, elder)
- {a} falsely answers "Yes," so choose {c} (the truth teller).
- {a} truthfully answers "No," so choose {b} (the liar).

no-genius
Seemed like a good idea at the time
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I guess it works cos the truth-telling sister is the inverse of the lying sister. ie, (f)(f^-1) = 1
I don't sing, I just shout. All. On. One. Note.
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