30 replies to this topic

[ThunderCloud]

 

Spoiler for I probably misunderstood

A sees two 3's. Everyone knows 242 doesn't work.

 

Spoiler for It's a bit tricky...

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.

Spoiler for amount of 3's

A sees two 3's he knows B and C see at least one. That works for all people.

Spoiler for

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...).

[vinay.singh84]

Ah, more twist and turns. Thanks for such a great puzzle.

Spoiler for

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.

Spoiler for solution - short version

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.

 

Spoiler for I think some deeper logic is needed

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.

[ThunderCloud]

 

Spoiler for I think some deeper logic is needed

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.

 

Spoiler for A few notes

(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.

[lans ellion]

I think this works

Spoiler for Answer

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 2-3-3; 2-4-3; 3-3-3; 3-2-3

 

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: 2-3-3; 3-3-3; 4-2-3; or 3-2-3
 

B knows C thinks the answer is: 3-3-3; 3-2-3; 3-2-4; 3-3-2

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 3-2-4 as described above.
 

B thus passes without having learned anything about A’s beliefs.

 

C’s Perspective
C knows the answer is 3-3-3- or 3-2-3

C knows B could think the answer is: 3-3-3; 3-2-3; 3-4-2; 3-3-2

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 3-4-2.

As soon as B passes it eliminates the possibility that B thinks the answer is 3-3-2 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 3-3-2)

 

B's only remaining beliefs are the solutions 3-2-3 and 3-3-3 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 3-3-3 or 3-2-3. Thus B is left with no way to eliminate the possibility that he has a 2.

[lans ellion]

Whoops, noticed an error in my answer

Spoiler for corrected error

My rule number 4 in my post above should read. A player can only believe that another player believes that he has a 4. i.e. A can believe that B thinks he has a 4. But A can't Believe that B thinks C has a 4.

[vinay.singh84]

 

Spoiler for I think some deeper logic is needed

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.

 

Spoiler for A few notes

(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.

Spoiler for Help me understand

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.

[ThunderCloud]

 

 

Spoiler for I think some deeper logic is needed

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.

 

Spoiler for A few notes

(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.

Spoiler for Help me understand

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. 

[ThunderCloud]

I think this works

Spoiler for Answer

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 2-3-3; 2-4-3; 3-3-3; 3-2-3

 

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: 2-3-3; 3-3-3; 4-2-3; or 3-2-3
 

B knows C thinks the answer is: 3-3-3; 3-2-3; 3-2-4; 3-3-2

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 3-2-4 as described above.
 

B thus passes without having learned anything about A’s beliefs.

 

C’s Perspective
C knows the answer is 3-3-3- or 3-2-3

C knows B could think the answer is: 3-3-3; 3-2-3; 3-4-2; 3-3-2

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 3-4-2.

As soon as B passes it eliminates the possibility that B thinks the answer is 3-3-2 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 3-3-2)

 

B's only remaining beliefs are the solutions 3-2-3 and 3-3-3 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 3-3-3 or 3-2-3. Thus B is left with no way to eliminate the possibility that he has a 2.

Spoiler for

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.

[ThunderCloud]

 

 

 

Spoiler for I think some deeper logic is needed

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.

 

Spoiler for A few notes

(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.

Spoiler for Help me understand

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...

Spoiler for

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.

[vinay.singh84]

 

 

 

 

Spoiler for I think some deeper logic is needed

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.

 

Spoiler for A few notes

(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.

Spoiler for Help me understand

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...

Spoiler for

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.

Spoiler for well...

My point is C's number is irrelevant with the given circumstances (that A and B are both 3, which limits it to 2,3).

More importantly, I do believe A and B determine their numbers based strictly on their responses if C always say "I don't know".

Spoiler for Here is (cryptic) thought chart to show it

This is all the thought process of A. He can deduce his number solely based on what he thinks B thinks he thinks..etc (without a "C thinks...").

 

C's 3.
A know's he's 2,3.
    (A)CASE 2 B:3 A-2,3
                         4 A-1,2            
             (B)CASE 4 A:1 B-4,5
                                  2 B-3,4                     
                      (A)CASE 1 B: 4 A-1,2
                                            5 A-1! ~ {A would have voiced this}
                                           (A would know A's 1)    _Round 1
                      (B would know B's 4 [!5])
             (A would know A's 2 [!1])                           _Round 2
    (B would know B's 3 [!4])    
(A knows A's 3 [!2])                                                _Round 3

 

(B) CASE x A:  <--> Suppose B considered he was x. Then B knows A's thinking would be:

[!x] <--> because he cannot be x (via proof by contradiction for the hypothetical posed directly above [equally spaced])

Edited by vinay.singh84, 08 February 2013 - 07:52 AM.