The elder learns they do not have brown eyes — Philosophim
Again, fun puzzle. :) — Philosophim
here's the more tricky part - what new information did the Guru give them that they didn't already have? — flannel jesus
How do they learn that? The elder could easily have brown eyes, as far as she's concerned. — flannel jesus
they are all distinct facts, all the way up — hypericin
In order to get started, so that the failure of anybody to leave is meaningful, all this must be known. And, for Michael solution to work, all this must be known too. Only a truth telling guru communicating to everyone that indeed there is a blue can cut through this recursive epistemic conundrum. — hypericin
If you are missing the need, you are missing it. — hypericin
are you saying that (2) is false and should instead say:
2. If everyone knows that (1) is true and if ... — Michael
1. As of right now everyone has come to know, through some means or another, that everyone knows that #101 sees blue
2. If (1) is true and if I do not see blue then I am blue and will leave this evening
3. If (1) is true and if I see 1 blue then if he does not leave this evening then I am blue and will leave tomorrow evening
4. If (1) is true and if I see 2 blue then ...
...
And it bears repeating (if any reader missed the previous comment), that even though as a practical matter (1) is true in counterfactual scenarios (2) and (3) only if someone says "I see blue" isn't that someone must say "I see blue" in every counterfactual and actual scenario for (1) to be true and for this reasoning to be usable.
It's worse than your amended 2. It recurses endlessly. — hypericin
But what's the relevant difference between seeing a piece of paper with the words "there is at least one blue" written on it and seeing 99 blue? How and why is it that the former can "cut through this recursive epistemic conundrum" but the latter can't? — Michael
That's what makes this puzzle so interesting. Truly, that's one of the biggest points, and why people find it fascinating. It's weird. — flannel jesus
If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".
Given that there are 201 participants, there are 40,401 possible combinations, so it's unfeasible for us to list them all, although our perfect logicians will be able to.
But we can make a start.
I'm #1 and I see 99 blue (#2-100), 1 green (#101), and 100 brown (#102-#201).
I ask myself:
Does #2 know that #101 sees blue? Yes; #2 knows that #101 can see blue #3.
Does #2 know that #1 knows that #101 sees blue? Yes; #2 knows that #1 and #101 both see blue #3.
Does #2 know that #2 knows that #101 sees blue? Yes; #2 knowing what #2 knows is a tautology.
Does #2 know that #3 knows that #101 sees blue? Yes; #2 knows that #3 and #101 both see blue #4.
Does #2 know that #4 knows that #101 sees blue? Yes; #2 knows that #4 and #101 both see blue #3.
Does #2 know that #5 knows that #101 sees blue? Yes; #2 knows that #5 and #101 both see blue #3.
Does #2 know that #6 knows that #101 sees blue? Yes; #2 knows that #6 and #101 both see blue #3.
...
I'm fairly certain the answer is always going to be "Yes", and so that there is no X and Y such that #X does not know that #Y knows that #101 sees blue.
So, returning back to the previous steps, I deduce that (1) is true
Okay, well I think the answer is that there isn't a difference — Michael
I'm very interested in that number. — flannel jesus
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