30 replies to this topic

[ThunderCloud]

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?

[bushindo]

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

Spoiler for

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.

 

 

[googon97]

Spoiler for quick logic

3
Everyone knows that everybody knows that there must be one 3.
That leaves us with:
233
323
332
333
234
243
324
342
423
432
Everyone knows that the other two have 3's and they either have a 2 or 3. Only the person with a 2 in the scenario, doesn't know that two 3's and a 2 is wrong. The other two will know each other know.A will say I don't know. B and C know that A either saw 2 and 3 or 3 and 3. This eliminates 234, 243, 324, 342. This makes the possibilities:
233
323
332
333
423
432
It becomes B's turn. B sees 3 and 3. C knows that B sees either 3 and 3 or 3 and 2. C knows that the possibilities are 332 and 333. C knows that if B saw 3 and 2, he would say 3 because 342 was eliminated. B doesn't know whether he has 3 or 2, so he says I don't know. C can then guess 3.

Edited by googon97, 03 February 2013 - 02:44 AM.

[ThunderCloud]

 

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

Spoiler for

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.

 

 

 

Spoiler for

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?

[ThunderCloud]

Spoiler for quick logic

C knows that if B saw 3 and 2, he would say 3 because 342 was eliminated.

Spoiler for

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.

[bushindo]

 

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

Spoiler for

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.

 

 

 

Spoiler for

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

Spoiler for

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.

[ThunderCloud]

 

 

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

Spoiler for

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.

 

 

 

Spoiler for

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

Spoiler for

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.

Spoiler for

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.

[bushindo]

 

 

 

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

Spoiler for

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.

 

 

 

Spoiler for

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

Spoiler for

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.

Spoiler for

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

Spoiler for

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.

[ThunderCloud]

 

 

 

 

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

Spoiler for

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.

 

 

 

Spoiler for

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

Spoiler for

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.

Spoiler for

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

Spoiler for

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.

Spoiler for

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...

[bushindo]

Spoiler for

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...

 

Comments

Spoiler for

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.