On 10/11/06, David Hobby <[EMAIL PROTECTED]> wrote:
jdiebremse wrote:
...
> But how does this work for N(blue) = 4?
>
> The initial state is that each native has two cases:
>
> 1) There are three blue-dot natives, and each blue dot native sees two
> blue dot natives.
>
> 2) There are four blue-dot natives, including himself, and each blue dot
> native sees three blue dot natives.
>
> In this case, I don't see how the naturalist provides any additional
> information. In the initial state, every native knows that every other
> native knows that there is at least one blue dot.
>
> JDG
JDG--
Maru's original post didn't say this, but the puzzle has
an additional assumption: All the natives are expert
logicians, they all know that all are, they all know that
everybody knows that all the natives are expert logicians,
etc. Without this, nothing happens even for only two blues,
as each would say, "So, maybe the other guy sees only reds
but is dumb."
True, true, but remember this is a logic problem, after all. If we
wanted to specify all the assumptions, we'd get into silliness like
"there exists an objective reality" or "each native will succeed in
killing themselves should they try".
The role of the outsider is to make it clear to everybody
that any situation with only one blue leads to suicide.
Of course when N = 4 everybody knows there are blues, but
this is different.
But isn't the case of only one blue already clear without the outsider?
All the natives would eventually conclude that there could
not be just three blues, since each of the three would only
see two, and eventually wonder why those two hadn't killed
themselves, finally concluding that the reason was that
each of the two actually saw two blues, since the one thinking
all this was the third blue.
Etc!
---David
And this reasoning stands for all N equal to or greater than 3?
~maru
Mmm... sicilian...
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