Numbered Foreheads

[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?

[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.

[ThunderCloud]

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 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.

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]

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.

[vinay.singh84]

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.

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".

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 February 8, 2013 by vinay.singh84

[ThunderCloud]

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".

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])

You are right -- for A, C's responses are indeed redundant. But not so for B; in fact, if A and B were the only players in the game (knowing that their sum is either 5 or 6), B would be unable to deduce his number.