Riddles

The first five. They all would have answered "no" if they themselves did not want coffee, because the waitress used the word *all*. That's my guess, anyway.

That is a very good guess. I should have seen that. :D

In her mind, she would assumed they all turned it down, though. Right?

In her mind, she would assumed they all turned it down, though. Right?

That is probable, yes. :)

In that, they all said no very clearly.

Voice of Reason is correct.

The five answered "I don't know" because the waitress asked if they all wanted coffee. If you were asked that, and you wanted coffee, but you didn't know if everyone in the group wanted coffee, you'd say "I don't know." But if you didn't want any, you'd say, "No," because then not all in the group want coffee.

So the five got coffee, because they indirectly stated that they wanted some, while the sixth got none, according to his own wishes. ;)

Here we are: *Solution:* _Pick out two cards of the same suit. Select a card for Alex where adding a number no greater than six will result in the number of the other card of the same suit. Adding one to the Ace would cycle to the beginning again and result in a Two. E.g. if you have a King and a Six of Diamonds, hand the King to Alex. The other three cards will be used to encode a number from 1 through 6. Devise a system with Peter to rank all cards uniquely from 1 to 52 (e.g. the two of hearts is 1, the two of diamonds is fourteen etc...). That will allow you to choose from six combinations, depending on where you put the lowest and highest cards._

Okay, yeah, I was on the wrong track. :P I'll go ahead and post my almost-solution though.

So going off of Christian's hint, I thought of a few specific orientations of the suits that Peter could receive that would indicate at most two suits the card would have to be in:

1. If Alex gave me 2 cards of one suit and 1 of the other three, I would give him back the lone card from the same color as the pair. E.g. if he gave me two hearts and one of each of the other three I would give him back a diamond. If Alex gave me 3 of one suit and two other different cards I would give him back one of the threesome. Both of these would give Peter 2 of Suit 1, 1 of Suit 2, and 1 of Suit 3, indicating that Alex's card is the same color as Suit 1.

2. If Alex gave me 2 cards of Suit 1, 2 of Suit 2, and 1 of Suit 3, I would give him back the card in suit 3. This would give Peter 2 cards of one suit and 2 of another, indicating that the card is not in either of these two suits. (This, unfortunately, is the only part that doesn't always work with the rest of my solution; I'll explain in a bit. xP)

3. If Alex gave me 3 cards of Suit 1 and 2 of Suit 2, I would give him back a card of Suit 2; whereas if he gave me 4 of Suit 1 and only 1 of Suit 2, I would give him back a card of Suit 1. Both of these would give Peter 3 of one suit and 1 of another, indicating that Alex's card is one of the two suits present.

4. The only other possibility is Alex giving me 5 cards of all one suit. Obviously, I would give one of the same suit back to him, and 4 cards of the same suit given to Peter would indicate to him that the Alex's card is of the same suit.

This narrows it down to at most two suits. The rest can be determined by ordering the cards. Keeping in mind the conventional order of suits (clubs lowest, then diamonds, hearts, spades) we can always find which card is the smallest, largest, etc. I will label these S (small), M (medium), L (large) and X (extra-large :P). Peter knows that Alex's card belongs to one of two specific suits. Each possible card in Alex's hand is coded for by a specific ordering of the cards in Peter's hand. Suit 1 in the chart below is by convention the lower-ranking suit of the two Peter knows. (This chart is based on the binary system.)

@ Suit 1 Suit 2@
@A SMLX SMXL@
@2 SLMX SLXM@
@3 SXML SXLM@
@4 MSLX MSXL@
@5 MLSX MLXS@
@6 MXSL MXLS@
@7 LSMX LSXM@
@8 LMSX LMXS@
@9 LXSM LXMS@
@10 XSML XSLM@
@j XMSL XMLS@
@Q XLSM XLMS@

This only leaves the kings to work with. If Alex's card is a king of Suit 1 in the chart above, use the code which also applies to the lowest card (in one of the two known suits) in Peter's hand, since that card obviously cannot be the same card Alex has. If it is a king of Suit 2, use the code for the second-lowest card in Peter's hand.

There's a couple times where this breaks down; first, in scenario 1 above, it is possible that Peter could receive a king of one (or both) of the known suits of Alex's card (and therefore the second-lowest card has no code of its own). However, this does not matter much, because there could now only be one king in Alex's hand - the one that Peter doesn't have. Again, this king is coded for by the same code of the other non-king card in Peter's hand.
Now here's where I'm stuck. In scenario 2 above, if Alex hands me a king in Suit 3, I would be forced to give it back to him, if we are to keep with the system; but then Peter would have no way of knowing what Alex's card was because there are no other cards of the same suits in his hand to code for. I thought of keeping the king to give to Peter, giving another card to Alex; but this could be ambigious to scenario 1.

So yeah, it almost works. :P Oh well, I'm rather proud of myself anyways for thinking of all of that myself. xP JK

Next one. This one is slightly less complex, but no less confusing: _I ask Alex to pick any 5 cards out of a deck with no Jokers._ _He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter._ _Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?_ _The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more._

This is a hard one. The way I might do it is card to two-digit number to code. So you assign each card a two digit number; there are many easy ways to do this, like spade = 1, heart = 2, club = 3, diamond = 4. Then you take the number of the card, or if it's a royal, create a special number for each. Now you have four two-digit numbers <70 to decode a two digit number <70. I don't know precisely how you can decode it because I'm too lazy to figure it out, but there probably is a way… Also, can you make any agreement with Peter beforehand? Like you will always pick a royal or something?

Sounds good, except that you can't very well control the cards that you are given… so how would that work?

I like truth-and-lie riddles. Here's one:

Tom always tells the truth. Dick sometimes tells the truth and sometimes lies. Harry always lies.

They are standing together in a line. The leftmost man says "The guy in the middle is Tom." The middle man says "I'm Dick." The rightmost man says "The guy in the middle is Harry."

Which is which?

Tom is on the right, Harry's in the middle, and Dick's on the left.

Tom can't be on the left, since he'd be lying about his own whereabouts; he also can't be in the middle, since he'd again be lying about himself. Therefore, he must be on the right, and the guy on the right says that Harry's in the middle. And that leaves Dick as the guy on the left.

Left guy = Dick
Center guy = Harry
Right guy = Tom

Am I right?

Good, you're both right! :) Here's another one.

The inhabitants of Truth City always speak the truth, while those of the city of Liars, of course, always lie. On your way to visit Truth City, you arrive at the crossroads leading to the two cities. The sign is confusing, so you are forced to ask a man standing at the crossroads for the right direction. Unfortunately, you don't know whether the man is a citizen of the city of Liars or is from Truth City. You are allowed to ask the man only one question. What question can you ask him to be sure to learn how to get to Truth City?

Ahhh! I've heard this one before, but I can't remember the answer :P Hmm…

Same here!
I'm still working on it…

Ask the person "Which city do you live in?"

But he could lie, so that wouldn't help much…

Not exactly, though you're on the right track :) The problem with that is that he will always answer Truth City, but that doesn't get the information you need.

I need to be going soon, but here's one more:

The citizens of the cosmopolitan city Nooneknowstruth are of three types: those who always speak the truth, those who always lie, and those who alternately lie and tell the truth. Again, you meet one of the residents. This time you are allowed to ask two questions. His answers must be sufficient for you to determine to which of the three groups the man belongs. What are the two questions you will ask him?

Have fun! :)

I'm going to bed now. Goodnight guys! Thanks for the riddles, Matthew– I'll sleep on it ;)

Goodnight! I'm heading out now too. :)

Yeah… I realized that just after I went to bed.
Is this right? "Which road will take me to the city you're from?"

2+2=?
Probably not the answer you're thinking of, but I'm pretty sure it works :)

Agh, that doesn't say if they're in the middle group. I'll think about it some more.

Is there anyone else near him?

Since the middle group alternately lies, would this work?
"2 + 2 = ?; 5 + 5 = ?"

Is there anyone else near him? Since the middle group alternately lies, would this work? "2 + 2 = ?; 5 + 5 = ?"

I was thinking along those lines, but I dunno if it's correct. x)

Yeah... I realized that just after I went to bed. Is this right? "Which road will take me to the city you're from?"

Correct! You don't find out whether the man is a citizen of the city of Truth or Lies, but that was not the question. Whichever road he points to is the road to Truth City. :)

Is there anyone else near him? Since the middle group alternately lies, would this work? "2 + 2 = ?; 5 + 5 = ?"

That's one possibility. :) My answer is to ask the person twice "Are you one of the citizens who alternately lies and tells the truth?" Two "yes" answers indicates the man is a liar, while two "no" answers indicates the man tells the truth. Two different answers means the man is one who alternately lies and tells the truth.

Here's an interesting one:

Ten pirates are trying to decide how to divvy out ten gold coins. They are ranked from the captain (1) to the cabin boy (10). Being a democratic crew they decide upon the following process to distribute the coins. The cabin boy will first nominate a distribution scheme (two coins to pirates 1, 4, 6, 7, and himself, for example) and the pirates will take a vote. If 50% or more of the pirates vote "yes", the distribution scheme will prevail; otherwise the cabin boy is thrown overboard and pirate number 9 comes up with a new scheme. This continues up the ranks until they settle on a favorable vote.

Can the cabin boy survive and come out with anything? Assume that the pirates are rational thinkers and will not vote against a scheme if as a result they are thrown overboard, or end up with fewer coins. If there is no personal loss to the pirate, he will otherwise be happy to see his fellow shipmate thrown overboard.

Hint: Start with just two or three pirates and see what happens.

Dose he gets to vote?

Assume that the pirates are rational thinkers and will *not* vote *against* a scheme if as a result they *are* thrown overboard, or end up with fewer coins.

Simplified: A pirate will vote against a scheme if as a result he is not thrown overboard, or does not receive fewer coins.

That could be interpreted two ways: does the "result" result from his vote, or from the proposed scheme? If from his vote, then I still don't know the answer. But if the result is the result of the scheme (and not his individual vote), then the cabin boy should propose to give all ten coins to himself.

I never did such a fine job of confusing myself, and probably everybody else who reads this. Exploiting syntactical ambiguity can be hard, sometimes…

I'm thinking that his vote would be a factor rather than a problem solver.

from what you proposed (cabin boy should propose to give all ten coins to himself) the top 5 pirats (who's turns are further away) will be like "seriously? NO!" And the other 5 (including the cabin boy) are like "Yeah. maybe he won't share… at lest I'm alive." Its a draw.

Anyhow the hole problem is based of off who is happy with what.
I say they all get one except the captain (he gets two) and the next guy how will be thrown over board next if he cant come up with something really good.

The problem with that is, it's fifty fifty that at lest four will be happy with just one coin and the C. with only two.
(assuming the cabinboy can't vote)

Everybody gets one coin… :P JK! :P

Yeah... I realized that just after I went to bed. Is this right? "Which road will take me to the city you're from?"

Correct! You don't find out whether the man is a citizen of the city of Truth or Lies, but that was not the question. Whichever road he points to is the road to Truth City. :)

Ohhhh!! That makes so much sense!

Here's an interesting one: Ten pirates are trying to decide how to divvy out ten gold coins. They are ranked from the captain (1) to the cabin boy (10). Being a democratic crew they decide upon the following process to distribute the coins. The cabin boy will first nominate a distribution scheme (two coins to pirates 1, 4, 6, 7, and himself, for example) and the pirates will take a vote. If 50% or more of the pirates vote "yes", the distribution scheme will prevail; otherwise the cabin boy is thrown overboard and pirate number 9 comes up with a new scheme. This continues up the ranks until they settle on a favorable vote. Can the cabin boy survive and come out with anything? Assume that the pirates are rational thinkers and will not vote against a scheme if as a result they are thrown overboard, or end up with fewer coins. If there is no personal loss to the pirate, he will otherwise be happy to see his fellow shipmate thrown overboard. *Hint:* Start with just two or three pirates and see what happens.

I'm not going to even try to solve this one xP

Ah ha! I solved it! (If 1 is the captain, and 10 is the cabin boy who cannot vote) 1,2,3,and 4, get nothing. 5,6,7, and 8, get two. 9 and 10 (the cabin boy) get one.
Ofcorse the captain will be mad, but he wouldn't be throwing away men so easily unless they were close to land or had a really small boat.

The vote of each pirate is weighted the same, whether of captain or cabin boy, if that helps in any way. :)

@Roy - Your solution would work; but it's not the best for the cabin boy. He can actually land himself survival and six gold coins!

The vote of each pirate is weighted the same, whether of captain or cabin boy, if that helps in any way. :) @Roy - Your solution would work; but it's not the best for the cabin boy. He can actually land himself survival and six gold coins!

well, Since he gets to vote, yes, I can see that working. we can risk one fewer vote, and the other four can make due with one coin….But would they Seriously vote for him if he is soooooooo blatantly selfish? I think not.

The vote of each pirate is weighted the same, whether of captain or cabin boy, if that helps in any way. :) @Roy - Your solution would work; but it's not the best for the cabin boy. He can actually land himself survival and six gold coins!

well, Since he gets to vote, yes, I can see that working. we can risk one fewer vote, and the other four can make due with one coin....But would they Seriously vote for him if he is soooooooo blatantly selfish? I think not.

You have a point. You'd have to assume in this case that each pirate understands the reasoning behind the solution (which as of yet I have not given), otherwise, yes, they probably wouldn't stand for him being selfish. :P
Here's a hint. Say there are only two pirates. In this case, pirate 2 can propose that he gets all 10 coins, and vote for himself to do so. He thus garners 50% of the votes and succeeds. What about three pirates?

Assume that the pirates are rational thinkers and will *not* vote *against* a scheme if as a result they *are* thrown overboard, or end up with fewer coins.

Simplified: A pirate will vote against a scheme if as a result he is not thrown overboard, or does not receive fewer coins. That could be interpreted two ways: does the "result" result from his vote, or from the proposed scheme? If from his vote, then I still don't know the answer. But if the result is the result of the scheme (and not his individual vote), then the cabin boy should propose to give all ten coins to himself. I never did such a fine job of confusing myself, and probably everybody else who reads this. Exploiting syntactical ambiguity can be hard, sometimes…

Lol :P I think the intended meaning of "as a result" is "as a result of their vote succeeding".

Actually I see now that my last solution wouldn't work since 8 and 9 can vote no, brake the tie, then make the exact same deal only with more to them selfs.

1 and 2 have no reason to vote 'yes' to anything, and one coin is probably the sweetest deal 3 will see all day, 5 will never be able convince 4 to vote 'yes' for him, so……. hmm……….. I just cant see 4, 6, 7, and 8, being happy with much from the cabinboy).

(1 N. 2 N. 3 Y[1]. 4 N. 5 Y[1]. 6[ ]. 7[ ]. 8[ ]. 9 Y. 10 Y[6?])

Short one vote and with three extra coins….Maybe he could buy them over with some salted fish?

We've got some complicated stuff going on here! :P You're right, 9 could vote no, I don't think 8 would though. (It's difficult to look into the minds of these pirates! xP)

My solution: The cabin boy proposes one coin to pirates 2, 4, 6, and 8, and the remaining six to himself. Here's why it works - maybe a little too simple, but I think it works if all the pirates understand it. :P Going off of what I said earlier, if there are three pirates, pirate three reasons that if he dies, captain gets no coins, given the two-pirate scheme. He can therefore buy the captain's vote by offering him a single coin. He thus garners two out of three votes and wins. If there are four pirates, pirate four reasons that if he dies, pirate 2 gets no coins; and so he can buy his vote with a single coin and win with 50%.

Going off of this principle, working down the ranks, we see that the cabin boy can land himself six coins if he offers the right pirates one coin each.

@1: 10@
@2: 0 10@
@3: 1 0 9@
@4: 0 1 0 9@
@5: 1 0 1 0 8@
@6: 0 1 0 1 0 8@
@7: 1 0 1 0 1 0 7@
@8: 0 1 0 1 0 1 0 7@
@9: 1 0 1 0 1 0 1 0 6@
@10 0 1 0 1 0 1 0 1 0 6@

You are a camp counselor. You and your eight campers are lost in the woods. You finally come upon a four-way intersection. You know your campsite is 20 minutes from there, but you don't know which path to take. You have an hour more of daylight, after which traveling is very dangerous. So you cannot travel with all eight campers down one route at a time. It would take too long.
Instead, you must send small groups 20 minutes down each path and have them rendezvous at the intersection in 40 minutes. You will then decide which route to take. (The counselor may also participate in the search in the first 40 minutes.)
The problem is that two of the campers in your group sometimes lie. You do not know which ones they are. How do you divide up your group into search parties? At rendezvous time, how to you decide which way to go?

There are several fun variations to this one, but I'll just post the basic one for now :)

I thought of a very simple answer, but it has nothing to do with how many people are in each group:

After the 40 minutes, whichever group claims to have made it to the campsite must prove it by having brought back with them an item from the campsite. Does that solve the problem?

….
My head hurts.

.... My head hurts.

Haha, IKR? These pirates basically have to be logicians in order for them to actually vote the way the cabin boy wants them to. :P

I thought of a very simple answer, but it has nothing to do with how many people are in each group: After the 40 minutes, whichever group claims to have made it to the campsite must prove it by having brought back with them an item from the campsite. Does that solve the problem?

Lol, I never thought of that one! That would probably work! xP Though, as you probably guessed it's not what I have as the answer. Also, some versions of this problem require one of the possible routes to be untraveled. If the campsite happened to be that way, they couldn't bring anything back obviously. :P

I thought of a very simple answer, but it has nothing to do with how many people are in each group: After the 40 minutes, whichever group claims to have made it to the campsite must prove it by having brought back with them an item from the campsite. Does that solve the problem?

Lol, I never thought of that one! That would probably work! xP Though, as you probably guessed it's not what I have as the answer. Also, some versions of this problem require one of the possible routes to be untraveled. If the campsite happened to be that way, they couldn't bring anything back obviously. :P

Yeah x) So I guess there are several plausible solutions.

Is this the solution you're talking about?

I (as the camp counselor) send two groups of three people each down two of the paths. I go down a third path with the two remaining people. That way, even if the two liars end up together, there's still one other truthful person with them to keep them accountable.

If none of the groups had found the campsite after a 20 minute walk, we all know that the untraveled path leads to the campsite. :)

But suppose that two groups came back saying the campsite is not down their paths. Then suppose two campers from the third group say the campsite is down their path and one camper in the third group says it isn't. You wouldn't know whether the third path is the right one or whether the unexplored one is right. What makes the problem tricky is that liars may lie but need not.

Ohhhh, right. Duh :P Hmmm. Yeah, this is a tricky problem… which is why I still like my simple solution just using hard evidence x)

Hehehe! :P

Here's a hint: The counselor doesn't lie.