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## Question

A third party writes a positive integer on two post-it notes and then places one each on the foreheads of two perfect logicians. The integers are consecutive. All of this is common knowledge. He then asks the logician with the lower of the two integers to begin the conversation (this is not known by the two logicians). The following conversation ensues.

Logician A "I don't know my number."

Logician B "I don't know my number"......this stement is repeated N times.

Finally, one of the logicians says "I know my number."

As it relates to N, what is the integer on the forehead of the logician who first identifies his number? Why?

## 7 answers to this question

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I = N if I is even

I = N +1 if I is odd.

It will always be the guy with the larger integer to say "I know"...

I think

Edited by maurice

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A third party writes a positive integer on two post-it notes and then places one each on the foreheads of two perfect logicians. The integers are consecutive. All of this is common knowledge. He then asks the logician with the lower of the two integers to begin the conversation (this is not known by the two logicians). The following conversation ensues.

Logician A "I don't know my number."

Logician B "I don't know my number"......this stement is repeated N times.

Finally, one of the logicians says "I know my number."

As it relates to N, what is the integer on the forehead of the logician who first identifies his number? Why?

Let

A = Logician A's name

B = Logician B's name

a = A's number

b = B's number

Per A: = Perspective of A

Per AB: = The assumed perspective of B from A's perspective (What A would guess B is thinking)

Cycle = A says something and then B says something

Logic:

Cycle 1:

Per A:

I can be either b-1 or b+1. I have no reason to bias one over the other so I assume probability 0.5 for each.

B will look at my number, and will make a similar guess.

Assume a=b-1:

(Per AB: My number can be a-1 or a+1 (Per A: b-2 or b ))

Assume a=b+1:

(Per AB: My number can be a-1 or a+1 (Per A: b+2 or b ))

No assumption:

What does B think I am thinking?

Per AB:

Assume b=a-1:

(Per ABA: My number can be b-1 or b+1 (Per AB: a-2 or a ) (Per A:(b-3 or b-1) or (b-1 or b+1))

Assume b=a+1:

(Per ABA: My number can be b-1 or b+1 (Per AB: a+2 or a (Per A: (b+1 or b+3) or (b-1 or b+1)))

Per A:

Of these possible cases, only (Per AB: a-2) i.e. (Per A: b-3 or b-1) has the potential that it cannot be allowed (numbers must be positive integers).

Assume a-2 <= 0 and therefore not allowed:

Per AB:

If A does not know his number, it must be because (Per ABA: b-1) is valid.

If A does know his number, it must be because (Per ABA: b-1) is invalid.

(Per ABA: b-1) corresponds to (Per A: b-3 or b-1).

If b-1 is invalid, then so is b-3.

Changing logic to consider b-3, no time to edit portion below.

-----------------------------------------------------------------

If (Per ABA: b-1) is invalid, then a-2 must be invalid and A would only come to this conclusion if my number were b=a-1 and b-1 = a-2 is invalid.

Therefore, if A knows his number, then my number must be a-1. If A does not know his number, then I do not know my number.

It is not yet my turn, so I cannot do anything yet. Once A makes his move, I will know my number if he knows his, otherwise I will not know my number yet.

Per A:

I cannot reason any further until I know whether or not B knows his number at this point.

Cycle 2:

Per A:

Since B does not know his number, this must mean that (Per ABA: b-1) is valid.

...

Edited by mmiguel1

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Assuming they can see each others numbers

say we have 1 and 2 N=2

1 - well I'm 1 or 3. No way of knowing yet.

2 - Of course I'm 2.

say 2 and 3 N= 2

2 - well I'm 2 or 4 since he's 3.

3 - well I'm 1 or 3. But he'd know he were 2 if I were 1. I'm 3!

say 3 and 4 N = 4

3 - well I'm 3 or 5. He thinks he's 2,4 or 6

4 - well I'm 2 or 4. Based on prior examples, if I'm 2 he'll know he's 3 this time.

3 - well he knows he's not 2. I could still be 5 though.

4 - Aha! If I were 2 he'd know he were 3. I'm 4!

This reasoning repeats regardless of the integers. The logician with the higher integer will always think he can be lower than the logician with the lower integer thinks he can be (ie guy with I sees I+1 and thinks he's I or I+2, but guy with I+1 sees I and thinks he's I-1 or I+1), so he will be able to rule out his incorrect option first.

I think

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the logicians also consider the other possibilities that occur when the higher integered logician speaks first. It does not alter the results but just gives them more to think about...

for example when 2 and 3

3 says I don't know (but thinks "I'm 3 or 1"

2 thinks I'm 2 or 4. don't know.

3 says I'm 3! If I were I he would know he were 2.

I think...

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Assuming they can see each others numbers

say we have 1 and 2 N=2

1 - well I'm 1 or 3. No way of knowing yet.

2 - Of course I'm 2.

say 2 and 3 N= 2

2 - well I'm 2 or 4 since he's 3.

3 - well I'm 1 or 3. But he'd know he were 2 if I were 1. I'm 3!

say 3 and 4 N = 4

3 - well I'm 3 or 5. He thinks he's 2,4 or 6

4 - well I'm 2 or 4. Based on prior examples, if I'm 2 he'll know he's 3 this time.

3 - well he knows he's not 2. I could still be 5 though.

4 - Aha! If I were 2 he'd know he were 3. I'm 4!

This reasoning repeats regardless of the integers. The logician with the higher integer will always think he can be lower than the logician with the lower integer thinks he can be (ie guy with I sees I+1 and thinks he's I or I+2, but guy with I+1 sees I and thinks he's I-1 or I+1), so he will be able to rule out his incorrect option first.

I think

I followed the same logic, but...

The logician with the higher number will identify his number first, but if N is the number of times the statement "I don't know my number" is repeated then the number on the forehead of the logician that first figured it out will be

N+1 if the number is even

N+2 if the number is odd

If the logician with the larger number starts first then he is still the one who will figure it out first, but his number is

N+2 if the number is even

N+1 if the number is odd

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I followed the same logic, but...

The logician with the higher number will identify his number first, but if N is the number of times the statement "I don't know my number" is repeated then the number on the forehead of the logician that first figured it out will be

N+1 if the number is even

N+2 if the number is odd

If the logician with the larger number starts first then he is still the one who will figure it out first, but his number is

N+2 if the number is even

N+1 if the number is odd

it should be

I=N if I is odd

I=N-1 if I is even

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Before making the first statement, Logician A must reason that "all" he knows is that his number is consecutive to that of B.

Thus, if he sees a 1 he knows he has a 2 but otherwise cannot be sure. So he'll only answer "I don't know" if he doesn't see a 1.

Let's call this iteration 1.

Before making his reply, Logician B likewise knows he has a 2 if he sees a 1 but he also knows from Logician A's statement he can't have a 1 himself.

Thus, if he sees a if he sees a 2 he knows he has a 3 but otherwise cannot be sure. So he'll only answer "I don't know" if he doesn't see a 1 or a 2.

Let's call this iteration 2.

Now, Logician A follows similar reasoning to eliminate 1 and 2 from his possible set of integers. If he sees a 3, he knows he has a 4; otherwise, he is unsure.

So, he'll answer "I don't know" only if he doesn't see a 3. Let's call this iteration 4.

The cycle will continue in a like manner - creeping upward with each iteration - such that by keeping track of previous iterations,

Logician A has eliminated all lower integers and will be sure (he has I + 1) only if he sees I where I is the number of iterations

(i.e., if in iteration 13, Logician A sees a 13, he will be "sure" that his number is a 14.)

Likewise, Logician B has also eliminated all lower integers and will be all sure he has I + 1 if he sees I or that he has I if he sees I - 1

(i.e., if in iteration 14, Logician B sees either a 13 or a 14, he knows he must have 1 more.)

Based on the wording of the problem Logician A will always have the smaller integer (although he won't know this) and Logician B will always have the greater (and he doesn't know this).

However, it does mean that only Logician B can ever first declare he knows his number (either I or I + 1) since A is always looking for a number less than his own.

To restate, let a be the integer on Logician A's forehead and b be the integer on Logician B's with a = b - 1.

The "I don't know statement" is made a times if a is odd and a - 1 times if a is even after which Logician B will declare he knows his number.

(Thus based on the definition of N in the problem N = a if a is odd and a-1 if a is even where a is the smaller of the two integers.)

All this said, I suspect there is an underlying falacy in this thinking.

Edited by Chuck C.

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