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#11 Noct

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Posted 31 March 2008 - 04:23 AM

You can assume three numbers and ask what each logician in turn thinks and knows.
Then you can substitute variables [a, b, c] and do the same.
To get started, try this single question.

You're the first to be questioned.
You see 5 and 13.
What do you know about your number?


Could either be 8 or 18
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#12 bonanova

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Posted 31 March 2008 - 04:29 AM

Could either be 8 or 18

Yes.
And you answer "I don't know my number."

Next step:
What does that tell the next logician.
What numbers might he see.
What does that permit him to conclude.

Can you take this to the place where one logician eventually knows his number?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#13 Noct

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Posted 31 March 2008 - 04:34 AM

Yes.
And you answer "I don't know my number."

Next step:
What does that tell the next logician.
What numbers might he see.
What does that permit him to conclude.

Can you take this to the place where one logician eventually knows his number?


No it would just continue infinitely. If you have an 8, he will see 8 and 5, and you will have said you don't know. He could either have 13, or 3. Or if you have 18, he will see 5 and 18, so he will know he either has 13, or 23. So he will have to say he doesn't know.

Then it will come to the third person, and he will see 13 and your number. If it's 13 and 8, he will know he either has 5, or 21 so he won't know. If he sees 13 and 18, he will either have 5, or 31. So he will have to say he doesn't know.

And it will be back to you and you will see 5 and 13, and will still not know your number.
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#14 imtcb

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Posted 31 March 2008 - 05:03 AM

Here is what I think you are going for:
a is me, b is 5 and c is 13.
I could have 8 or 18, so I don't know.
b sees 8/13 or 18/13 and knows he can have 5/21 for my 8, or 5/31 for my 18. Giving him 5, 21 and 31 to pick from, so he doesn't know.
c sees 8/5 or 18/5 and knows he can have 3/13 for my 8, or 13/23 for my 18. He also takes into account what b is thinking, and adds into the mix 13/29, 3/39, 23/39, and 13/49. Giving him 3, 13, 23, 29, 39, and 49 to pick from, so he doesn't know.
Since I realize that the number of possibilities will continue to increase indefinitely, I concentrate on the lowest possible number and work from there. If c thinks he can possibly have a 3, and b has a 5 (also being the smallest he believes possible for himself) then my number could be a 2 or an 8, so I don't know.
b starts to think the same way, and realizes that I think I could have a 2, and c thinks he could have a 3, making his number a 1 or a 5. He doesn't know.
c, also realizing it, gets a 2 from me and a 1 from b, giving him a 1 or a 3. He doesn't know.
Me, now, I get left with a 1 and a 1, making mine a 2 if everyone really had those low number, but alas, they do not, so I don't really know.
b, hearing that I don't know, knows that the whole "lowest number" thing didn't pan out.

I guess this could continue until you exhausted the possibilities up the ladder of numbers, but I'm not really seeing how...

Could you give a bigger hint?
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#15 Noct

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Posted 31 March 2008 - 05:17 AM

Here is what I think you are going for:
a is me, b is 5 and c is 13.
I could have 8 or 18, so I don't know.
b sees 8/13 or 18/13 and knows he can have 5/21 for my 8, or 5/31 for my 18. Giving him 5, 21 and 31 to pick from, so he doesn't know.
c sees 8/5 or 18/5 and knows he can have 3/13 for my 8, or 13/23 for my 18. He also takes into account what b is thinking, and adds into the mix 13/29, 3/39, 23/39, and 13/49. Giving him 3, 13, 23, 29, 39, and 49 to pick from, so he doesn't know.
Since I realize that the number of possibilities will continue to increase indefinitely, I concentrate on the lowest possible number and work from there. If c thinks he can possibly have a 3, and b has a 5 (also being the smallest he believes possible for himself) then my number could be a 2 or an 8, so I don't know.
b starts to think the same way, and realizes that I think I could have a 2, and c thinks he could have a 3, making his number a 1 or a 5. He doesn't know.
c, also realizing it, gets a 2 from me and a 1 from b, giving him a 1 or a 3. He doesn't know.
Me, now, I get left with a 1 and a 1, making mine a 2 if everyone really had those low number, but alas, they do not, so I don't really know.
b, hearing that I don't know, knows that the whole "lowest number" thing didn't pan out.

I guess this could continue until you exhausted the possibilities up the ladder of numbers, but I'm not really seeing how...

Could you give a bigger hint?


You have some logic off. there could only be two possible choices for numbers. b can never have that many choices. His is either the compliment of the smaller number he sees, or the sum of both of them.
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#16 imtcb

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Posted 31 March 2008 - 05:27 AM

You have some logic off. there could only be two possible choices for numbers. b can never have that many choices. His is either the compliment of the smaller number he sees, or the sum of both of them.

From MY point of view, he could have the 3 options, because I don't know what my number is. I'm saying that, being all logicians, they think the same, so each knows that the others are contemplating what my numbers are, as well as the others.
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#17 Noct

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Posted 31 March 2008 - 05:32 AM

oops i meant c not b, sorry.
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#18 imtcb

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Posted 31 March 2008 - 05:36 AM

oops i meant c not b, sorry.

I agree, it gets pretty convoluted after b. What I was saying is that if b is taking into account what I am thinking about my own numbers, and c does this for both me and b, you get the added possible numbers.
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#19 Noct

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Posted 31 March 2008 - 05:39 AM

I agree, it gets pretty convoluted after b. What I was saying is that if b is taking into account what I am thinking about my own numbers, and c does this for both me and b, you get the added possible numbers.


But c shouldn't have that many possibilities, just the same amount of possibilities as b (in your eyes).
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#20 imtcb

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Posted 31 March 2008 - 05:46 AM

But c shouldn't have that many possibilities, just the same amount of possibilities as b (in your eyes).

c isn't just taking into account that I don't know my number, he is taking into account that neither of us do.

Don't get me wrong, I still don't see how this could be solvable (not saying it's not possible, I just don't see it), I'm just saying that you could end up with an infinite list by continually thinking about what the others are thinking because they don't know their number.

That list of numbers isn't what c actually thinks his number is, it is what the OTHERS could see as possibly what c thinks is his number, given that nobody knows their own number.

Now my head hurts...

I'll try to check back on this one tomorrow...
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