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Posts posted by ThunderCloud


By looking at the OP, you can infer that all the logician's are "perfectly logical".
That is, they along will follow proper logic, and they can expect the other two to do so as well.
In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.
OK, with that out of the way, here's the problem: By allowing the option of "giving up", we introduce the observer effect into play (Giving up might change the conditions that made one conclude giving up initially).
The answer provided still may be correct, but it would need improved justification.
Consider the statement,
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked.
The logic that determines C to give up relies on then scenario where no can give up, which isn't the case, and thus the argument breaks down.
Additionally, A and B wouldn't be able to name their number on the third time, only A would per the OP rules. So wouldn't B give up at some point?
I haven't attempted the deeper logic; to answer the original problem: as each logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323, and 332 would need to be solved and would reveal the answer to the OP as all logicians viewpoints would be considered.
A helpful tidbit: a logician will say "I don't know" if and only if there exists a situation (from his viewpoint) in which he will be able voice his number. So every "I don't know" you conclude would require such an example. What? Yeah. It seems the proof would be quite involved.
(1) The original problem states that the question is repeated until each of the three logicians has either named his number or given up. So B is not forbidden from winning after A; in fact, by the rules of the OP, all three could theoretically win. In this respect the problem is different from many other similar puzzles which permit only one winner, so I should probably have emphasized this more in the original wording.
(2) The "observer effect" is an interesting aspect, but it is not problematic in this case due to note (1) above. No one who gives up (assuming they are perfect logicians, of course) can ever regret doing so; at most, he may change which of the other logicians may name their number afterward (and we can assume the logicians aren't petty enough to care about that ). Having said that, you have me wondering how a similar problem would play out if only the first logician to name his number could win, since then hypothetically there could be a situation where the winner is indeterminate (whoever gets frustrated and gives up first helps the other person to win). It could get interesting.
(3) Aside from that, what I posted was only the "short version" of the solution, with many details left out. The logic of the argument is correct (despite relying on a hypothetical), it just needs backup. I left out the details deliberately, so readers could work the problem out for themselves and yet have a way to check their answer. I may post a more complete solution later, if no one else does.
From your simple proof:
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked. C will never be able to.
So would the following be an accurate summarization?
If C keeps saying "I don't know" then both A and B will know their number on the third round based solely on the conclusions from each other saying "I don't know" two times?
As a result of this, C's number is irrelevant and this same outcome (A and B know on the third round) would occur if he is a 2 or a 3?
Note that (if this isn't the case) the observer effect does come into play as C decided to give up based on the conclusions from a hypothetical where no one gives up, rendering his decision moot.
Yes, that's exactly right.
With a few notes...
C's number is not altogether irrelevant: if it were anything other than 3 or 2, then the game would play out differently. Also, were no one to give up, A and B would not learn their numbers strictly from each other's responses, but from each other's responses in combination with C's responses. In the end, only C would be left hanging if no one gave up. So in anticipation of this, it makes sense for C to give up, and by doing so he provides A and B with an additional clue that helps them deduce their numbers sooner than they otherwise would have.

I think this works
Answer
A Pass
B Pass
C I know
A I know
B I give up
Reasoning
General Facts:
1. Every player sees a sum of 6 and thus knows he has a 2 or 3.
2. Thus, every player knows the each other player sees at least one 3 and either a 2 or 3.
3. Since each player knows they can only have a 2 or a 3 they know each other player is limited to believing they have a 2, 3, or 4. (Thinking, If I have a 2 then each player sees 5 and other players could think they have a 3 or 4. & if I have a 3, the other players see 6 and could think they have a 2 or 3.)
4. A player can only wonder if he has a 4. He cannot believe other players have a 4. (a consequence of rule 2)
the above rules allow us to create the starting viewpoints of each player.
A’s perspective
A’s perspective from the start: He knows the answer is 2 3 3 or 3 3 3
A knows B can think the answers are 233; 243; 333; 323
Step 1: A sees 6 so he can’t know if he has a 2 or 3 and passes.
Step 2: As soon as A passes he eliminates the possibility that he has a 2 and B thinks he has a 4. Both A and B know that C has a 3. Both A and B know that A cannot have a 1. Thus, if B has a 4 and C has a 3 A must have a 2. Since A passed he cannot have seen B with a 4 and C with a 3.
Thus, A knows that B can think the only remaining possibilities are 2 3 3; 3 3 3; and 3 2 3.
Now, if B sees that A has a 2 he will answer that he knows on his turn. (because the only possible answer that B can believe where A has a 2 is when B has a 3). Thus, if B passes on his turn A immediately knows he has a 3. (because B's two remaining possible solutions both leave A with a 3) but A must wait for his turn to state that he knows the answer.
B’s perspective:
B knows the answer is either 3 3 3 or 3 2 3
B knows A thinks the answer is: 233; 333; 423; or 323
B knows C thinks the answer is: 333; 323; 324; 332
When A passes it does not eliminate any options for what B thinks A thinks because there are no unique solutions.
But it does tell B that C can no longer think the solution is 324 as described above.
B thus passes without having learned anything about A’s beliefs.
C’s Perspective
C knows the answer is 333 or 323
C knows B could think the answer is: 333; 323; 342; 332
As described above C knows that B can’t believe he has a 4 after A passes. Thus eliminating B's being able to believe the solution is 342.
As soon as B passes it eliminates the possibility that B thinks the answer is 332 because if B saw that C had a 2 then B would know he had a 3. (because the only remaining solution where C has a 2 is 332)
B's only remaining beliefs are the solutions 323 and 333 thus C knows he has a 3.
So C says I know on his turn.
Then A says I know
B gives up because he knows that the reason A and C know their number is because they know he thinks the answer is either 333 or 323. Thus B is left with no way to eliminate the possibility that he has a 2.
I think you made a jump in "Step 2" of A's reasoning. A knows he can't have a 1, and B knows that A can't have a 1... but A cannot infer that B is aware that A knows he doesn't have a 1 (confusing, no?). To elaborate: If A has a 2, then he knows B must be thinking he either has a 3 or a 4. In that case, B could not know that he didn't have a 4, and therefore, B could not expect A to know that he doesn't have a 1 (since then (1,4,3) would become a valid combination). Which means, so far as A's perspective on B's reasoning is concerned, if A has a 2 and passes, then B doesn't know whether A was unable to decide between a 2 and a 3 (because B has a 3), or between a 2 and a 1 (because B has a 4). A cannot expect that B has learned anything new at this point.

By looking at the OP, you can infer that all the logician's are "perfectly logical".
That is, they along will follow proper logic, and they can expect the other two to do so as well.
In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.
OK, with that out of the way, here's the problem: By allowing the option of "giving up", we introduce the observer effect into play (Giving up might change the conditions that made one conclude giving up initially).
The answer provided still may be correct, but it would need improved justification.
Consider the statement,
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked.
The logic that determines C to give up relies on then scenario where no can give up, which isn't the case, and thus the argument breaks down.
Additionally, A and B wouldn't be able to name their number on the third time, only A would per the OP rules. So wouldn't B give up at some point?
I haven't attempted the deeper logic; to answer the original problem: as each logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323, and 332 would need to be solved and would reveal the answer to the OP as all logicians viewpoints would be considered.
A helpful tidbit: a logician will say "I don't know" if and only if there exists a situation (from his viewpoint) in which he will be able voice his number. So every "I don't know" you conclude would require such an example. What? Yeah. It seems the proof would be quite involved.
(1) The original problem states that the question is repeated until each of the three logicians has either named his number or given up. So B is not forbidden from winning after A; in fact, by the rules of the OP, all three could theoretically win. In this respect the problem is different from many other similar puzzles which permit only one winner, so I should probably have emphasized this more in the original wording.
(2) The "observer effect" is an interesting aspect, but it is not problematic in this case due to note (1) above. No one who gives up (assuming they are perfect logicians, of course) can ever regret doing so; at most, he may change which of the other logicians may name their number afterward (and we can assume the logicians aren't petty enough to care about that ). Having said that, you have me wondering how a similar problem would play out if only the first logician to name his number could win, since then hypothetically there could be a situation where the winner is indeterminate (whoever gets frustrated and gives up first helps the other person to win). It could get interesting.
(3) Aside from that, what I posted was only the "short version" of the solution, with many details left out. The logic of the argument is correct (despite relying on a hypothetical), it just needs backup. I left out the details deliberately, so readers could work the problem out for themselves and yet have a way to check their answer. I may post a more complete solution later, if no one else does.
From your simple proof:
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked. C will never be able to.
So would the following be an accurate summarization?
If C keeps saying "I don't know" then both A and B will know their number on the third round based solely on the conclusions from each other saying "I don't know" two times?
As a result of this, C's number is irrelevant and this same outcome (A and B know on the third round) would occur if he is a 2 or a 3?
Note that (if this isn't the case) the observer effect does come into play as C decided to give up based on the conclusions from a hypothetical where no one gives up, rendering his decision moot.
Yes, that's exactly right.

By looking at the OP, you can infer that all the logician's are "perfectly logical".
That is, they along will follow proper logic, and they can expect the other two to do so as well.
In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.
OK, with that out of the way, here's the problem: By allowing the option of "giving up", we introduce the observer effect into play (Giving up might change the conditions that made one conclude giving up initially).
The answer provided still may be correct, but it would need improved justification.
Consider the statement,
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked.
The logic that determines C to give up relies on then scenario where no can give up, which isn't the case, and thus the argument breaks down.
Additionally, A and B wouldn't be able to name their number on the third time, only A would per the OP rules. So wouldn't B give up at some point?
I haven't attempted the deeper logic; to answer the original problem: as each logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323, and 332 would need to be solved and would reveal the answer to the OP as all logicians viewpoints would be considered.
A helpful tidbit: a logician will say "I don't know" if and only if there exists a situation (from his viewpoint) in which he will be able voice his number. So every "I don't know" you conclude would require such an example. What? Yeah. It seems the proof would be quite involved.
(1) The original problem states that the question is repeated until each of the three logicians has either named his number or given up. So B is not forbidden from winning after A; in fact, by the rules of the OP, all three could theoretically win. In this respect the problem is different from many other similar puzzles which permit only one winner, so I should probably have emphasized this more in the original wording.
(2) The "observer effect" is an interesting aspect, but it is not problematic in this case due to note (1) above. No one who gives up (assuming they are perfect logicians, of course) can ever regret doing so; at most, he may change which of the other logicians may name their number afterward (and we can assume the logicians aren't petty enough to care about that ). Having said that, you have me wondering how a similar problem would play out if only the first logician to name his number could win, since then hypothetically there could be a situation where the winner is indeterminate (whoever gets frustrated and gives up first helps the other person to win). It could get interesting.
(3) Aside from that, what I posted was only the "short version" of the solution, with many details left out. The logic of the argument is correct (despite relying on a hypothetical), it just needs backup. I left out the details deliberately, so readers could work the problem out for themselves and yet have a way to check their answer. I may post a more complete solution later, if no one else does.

A sees two 3's. Everyone knows 242 doesn't work.
Right... we know that everyone sees a 3, but we only know that as the readers of the puzzle. C does not know that A sees two 3's, because C thinks he might have a 2 on his forehead. And if C
did have a 2, then B could not know that A sees any 3's at all, since B might have a 4 on his forehead. So, since C doesn't know his own number yet, there is no reason for him to expect (nor is there reason for C to expect that B would expect) that A could deduce that his number was 3. The hard part is keeping track of who knows what in each circumstance.A sees two 3's he knows B and C see at least one. That works for all people.
This is true, but not enough...
C knows that A sees at least one 3 (since B has one).
B knows that A sees at least one 3 (since C has one).
However, C does not know that B knows that A sees at least one 3, because to know that, C would have to (already) know that his own number was a 3.
Therefore, C cannot expect B to infer his number from the fact that A was unable to deduce his (at least on the first time around...).

A sees two 3's. Everyone knows 242 doesn't work.
Right... we know that everyone sees a 3, but we only know that as the readers of the puzzle. C does not know that A sees two 3's, because C thinks he might have a 2 on his forehead. And if C
did have a 2, then B could not know that A sees any 3's at all, since B might have a 4 on his forehead. So, since C doesn't know his own number yet, there is no reason for him to expect (nor is there reason for C to expect that B would expect) that A could deduce that his number was 3. The hard part is keeping track of who knows what in each circumstance. 
C thinks B thinks, if A saw 42, he would either have a 2 or a 3. Everyone already knows that 242 is not an option. So, C thinks B thinks, that if it was 342, A would have said 3.
I'm not sure I follow you there.
If A saw 4 and 2, then A could not know that 2,4,2 is an invalid combination (as you mentioned, he would know that he had a 2 or a 3 but could not tell which).
Everyone knows that there are at least two 3's. Everyone also knows that everyone else can see at least one 3. 242 has no 3's.
This is not true if B has a 4 and C has a 2. To elaborate: C knows he has either a 3 or 2. As you mentioned, C knows that
if he has a 2, then B does not know whether he has a 3 or a 4. Furthermore, if C has a 2, then C knows that B knows that if he has a 4, A must be wondering whether he has a 2 or a 3. But in such event, A would see a 4 and a 2, and therefore A would not have any way of knowing for certain that someone has a 3. In summary: C knows that if he has a 2, then B cannot assume that A is aware that any of them definitely has a 3. 
C thinks B thinks, if A saw 42, he would either have a 2 or a 3. Everyone already knows that 242 is not an option. So, C thinks B thinks, that if it was 342, A would have said 3.
I'm not sure I follow you there.
If A saw 4 and 2, then A could not know that 2,4,2 is an invalid combination (as you mentioned, he would know that he had a 2 or a 3 but could not tell which).

Two questions:
(1) Do any of the following count as "words": nonEnglish words, acronyms, proper nouns?
(2) If a word is a palindrome (such as "eye"), does it count once or twice?

Ah, more twist and turns. Thanks for such a great puzzle.
I erroneously interpreted a 'win' as being the first to name the hat number. My mistake. Defining a win as being able to name a hat number changes the games, since now C is motivated differently. The game now goes as follows
Turn 1 A: I don't know
Turn 2 B: I don't know
Turn 3 C: I give up
Turn 4 A: My number is 3 (C would not have given up if A's number is 2)
Turn 5 B: My number is 3.
I think you've got it now.
If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked. C will never be able to. On the other hand, if one of (A, B) has a 2, then C's situation is much improved, and he can eventually deduce his number. So, the fact that C gave up informs both A and B that their numbers are identical (neither A nor B would know his number at this point in any other way).
Glad you enjoyed it.

Comments
So the question is "in what circumstance would C give up, and in what circumstances would he not?"
First, let's bring back the first answer, which we both agree on. In the case where giving up is not allowed, then A wins and the game goes as follows.
Turn 1 A: I don't know
Turn 2 B: I don't know
Turn 3 C: I don't know
Turn 4 A: I don't know
Turn 5 B: I don't know
Turn 6 C: I don't know
Turn 7 A: My number is 3.
Obviously, C only gets to answer twice. Once at turn 3 and once at turn 6. The rest of this post will show C has no strategy to win, therefore he has an incentive to give up.
Note that since 'Give up' means 'I don't know' + 'I forfeit my right to play further', an 'I don't know' is the same as 'Give up' on turn 5 or 6, since B and C don't get to play further in any instance.
I think we both agree so far that turn 1 and 2 should both be 'I don't know'. So, let's look at turn 3. C has no idea what his hat number is, so he can't win. If C chooses to give up, the the game goes as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
If C chooses not to give up but to pass on turn 3, then the game goes as follows
Turn 1 A: I don't know
Turn 2 B: I don't know
Turn 3 C: I don't know
Turn 4 A: I don't know
Turn 5 B: I don't know/Give up (identical choices at this turn)
Turn 6 C: I don't know/Give up (identical choices at this turn)
Turn 7 A: My number is 3.
So essentially, C got no winning options if all hats are 3, so technically he would be required by the rules to give up on turn 3. Meaning that the game would proceed as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
Almost there...
Save for two things. First, a clarification of the problem statement: the game continues until each logician has either named his number or given up. So more than one of them can potentially "win". Secondly: would C give up if A had a 2 instead of a 3?

Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7  c). Continues the reasoning until the 7th turn as above.
You are right that A would answer with his number the third time he is asked, assuming no one has given up by that point. But what about the other two logicians? Can they also answer, or should they give up? ...Or should they have given up sooner than this?
If the two logicians are required to give up if they figure they would never know it, then the game would go as follows
A: I don't know
B: I give up
C: I give up
A: My number is 3.
Same reasoning as above. B would give up at 2nd turn since he figures the (a,c) = (3,3) would not allow him to win at 5th turn anyways. C would also give up because given that B gave up and (a,b) = (3,3), his hat could still be 3 or 2. A then wins.
In fact B is in the most advantageous position. He will never give up, because in any case he can decide his number. He learns at least as much as A or C in each round of questioning.
I see what you're saying
If B knows that C would give up, then he shouldn't give up. The game would then be
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
Getting warmer! Consider: in what circumstance would C give up, and in what circumstances would he not? When C gives up, the fact is very informative... to both A and B equally by symmetry...

Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7  c). Continues the reasoning until the 7th turn as above.
You are right that A would answer with his number the third time he is asked, assuming no one has given up by that point. But what about the other two logicians? Can they also answer, or should they give up? ...Or should they have given up sooner than this?
If the two logicians are required to give up if they figure they would never know it, then the game would go as follows
A: I don't know
B: I give up
C: I give up
A: My number is 3.
Same reasoning as above. B would give up at 2nd turn since he figures the (a,c) = (3,3) would not allow him to win at 5th turn anyways. C would also give up because given that B gave up and (a,b) = (3,3), his hat could still be 3 or 2. A then wins.
In fact B is in the most advantageous position. He will never give up, because in any case he can decide his number. He learns at least as much as A or C in each round of questioning.

C knows that if B saw 3 and 2, he would say 3 because 342 was eliminated.
Careful to track who has eliminated what. The combination 342 was eliminated on the basis that C can see that A and B have 3's. But B only knows he doesn't have a 4 if neither A nor C has a 2; and C can't know that he doesn't have a 2. So, from C's perspective, if C has a 2, then B can't eliminate 342 as a possibility. Since C can't infer that B can eliminate that case (even though we know he really can, since C actually has a 3), C cannot proceed to deduce his number from there.

Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7  c). Continues the reasoning until the 7th turn as above.
You are right that A would answer with his number the third time he is asked, assuming no one has given up by that point. But what about the other two logicians? Can they also answer, or should they give up? ...Or should they have given up sooner than this?

Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
 1

Here is a fun puzzle a college room mate once gave to me:
You have two lengths of fuse, and a book of matches. You know that each fuse length has (somehow) been cut so that if the fuse is allowed to burn from end to end, it will take precisely 1 minute to burn down. However, the fuses will not necessarily burn at an even pace; all you know is the total combustion time of each fuse. Using nothing but the two lengths of fuse and the matches to ignite them with, how can you measure an interval of 45 seconds?
Numbered Foreheads
in New Logic/Math Puzzles
Posted