Or, if Max is really fast, he could run across the bridge with two of the eyes, then come back for the third before the natives catch up to him.
To stop the hijack, though, here’s another of my favorites:
Different bridge, same rickety-ness, two miles long. The weight limit on this one is 2,000 lbs. Trucks have to weigh in at a station just before the bridge to ensure they don’t collapse it. Ernie’s huge truck weighs precisely 1,999.999. “Phew! Just made it. I’ll be home in time for Jeopardy!,” Ernie thinks.
Ernie floors the accelerator and darts across the bridge. Halfway across, however, a tiny bird (or whatever), weighing roughly half a pound, flies in front of the truck and SPLAT! It gets stuck in the grill.
The bridge groans, but doesn’t break. Why not?
You may assume everything I told you is true, the weight limit is accurate, and that there is no word trickery here.
Ernie’s truck is a gas-guzzler, easily burning a half pound of fuel over a mile. Adjust lengths and weights to taste
A man parks his car in the outdoor office parking lot and realizes he forgot his umbrella at home. He has no hat, nor a newspaper, nor a raincoat. He darts to his office but doesn’t get wet. How/why?
It’s not raining.
When the same man is about to leave from work, the storm knocks the power out. Everything goes dark. He fumbles his way outside and sees that the entire block’s power is out. He has no flashlight and the street lights aren’t on. Yet he’s able to easily spot his black car across the parking lot. How?
It’s still daylight. His office has no windows.
It’s now raining hard. He still has no umbrella, coat, etc, yet not a hair on his head gets wet. How?
He’s bald
But he realizes he locked is keys in the car. Unfortunately, he’s able to easily retrieve them. Why is this unfortunate?
He drives a convertible. He just unlocks the door by hand because, unfortunately, he left the top down
Setting the example back to 100 blue eyed people, and for simplicity everyone assumes people will consder the case that they’re brown-eyed:
Blue A sees 99 blues and knows that everybody on the island sees at least 98 blues.
Blue A knows that all the brown-eyed people see at least 99 blues and all know that everybody on the island sees at least 98 blues.
Blue A also knows all the blue-eyed people see at least 98 blues and all know that everybody on the island sees at least 97 blues.
That’s all the people on the island. So, Blue A knows that everybody on the island knows that everyone else on the island sees at least 97 blue eyes.
The place induction fails is it tries to set up an arbitrary heirarchy of villagers statcked at levels one above one another stacking all the way down to 0/1 - but that’s not the way it works. There are precisely two levels of villiagers; blue and brown. A doesn’t think B thinks C thinks D thinks E thinks F thinks G sees 93 blue eyes when A knows everyone sees at least 98.
And the really unfortunate thing is, he left the top down and it’s been raining.
Well yeah. That’s where I was going with that.
Yes, A does know what everyone can see. But that’s not the point. You’re failing to keep up with the imaginary situations as they cascade. You’re reverting back to what E, F and G actually think and substituting that in. You’re not suspending your disbelief for long enough. Sure, in reality, the blues see the same amount. But they wouldn’t if, as you say, “consider the case that they’re brown-eyed”.
Let’s stick with your example of an entire island of blues and consider the following interview with a visitor (V) and A, the chief. Let’s also drop the fraction part and just stick with the lowest number:
V - “What’s the lowest amount of blues there could be?”
A - “I see 99, so while there could be 100, the lowest would be 99.”
V - “What would your daughter, B, say about it?”
A - “Well, I don’t know. She’s blue. DON’T TELL HER! But if I’m blue, then she’d say ‘99’. But it’s possible that I’m brown, so the lowest she could say would be 98. That’s all the villagers except her and me, since she doesn’t see herself.”
V later asks the question to B. She answers “99”, but that’s not the point. The point is, the chief can’t possibly know she’d say that. Back to the interview:
V - “So if the lowest scenario from B is 98 what would C tell her if B were to ask C? What’s the lowest C could answer?”
A - “Well, we’re still assuming I am, in fact, brown. In that case, C would say ‘98’ just like B, but there’s no way that B could possibly reason that out on her own. She’d actually have to ask C. And as you know, she wouldn’t do that, or she’d know if their eye color matched. So that conversation has never taken place, evidently, since they’re still alive. But if I’m brown eyed, and B reasons that she could possibly be brown-eyed too, then the lowest **B thinks **C would answer would be ‘97’. That’s missing me, her, and since C doesn’t know what he is himself, he wouldn’t count himself. So ‘97’.”
V - “So it’s possible for C to actually say 97?”
A - “No, the lowest he’d say in real life is 98. But if I’m brown, and B assumes she’s brown, and she reasons that C assumes he’s brown, then he could say 97.”
V - “Ah, so if you assume you’re brown, and B assumes she’s brown, and she assumes and C assumes he’s brown, then in that imaginary world, there are only 97 blues?”
A - “Yes.”
V - “Then in that world, what would C reason to be the minimum that D would say?”
A - “Well, 96, not counting me, her, C, and D doesn’t know about himself. That’s assuming that they don’t look at me and see blue. If they do, then this scenario is impossible. But I don’t know what B sees when she looks at me, so it’s possible that she thinks D will say ‘96’ minimum.”
So there you go. We’ve broken through your 97 barrier. I think that at a certain point, you bring it back up to the surface level instead of staying in the “I think she thinks that he will think that the next guy would think X” method. The part you’re forgetting is that when they assume they’re brown, then in the imaginary world, this is as good as fact. It effectively reduces the number of blues on the island. Sure, it doesn’t happen for real, but it does in the heads of the people they’re imagining.
Chessic Sense is correct, but it’s possible that you will find it is easier to think about the thought process from the bottom up; if so, try post #96 and keep extrapolating.
A - “Of course, B will also know that C would be wrong about what D would think, since B can plainly see that C’s eyes are blue, and knows that D can see that too, counting 97 blue eyes himself. And of course B would never imagine that C thinks that D thinks E sees 95 blue eyes for the same reason - B is imagining a world with only two brown-eyed people; herself and me. In that imagined world C would know that E can see D’s blue eyes, and that D can see E’s. Everybody can see at least 96 pairs of blue eyes - in the imagined world where myself, B, and C all have brown eyes, which of course all three of us know isn’t the case, because we can all see at least one of B and C.”
V - “Ah, now I see! I was scared for a while that I had doomed you all when I accidentally mentioned seeing blue eyes around.”
A - “No, no, we’re all good; I mean, we all know that there are lots of blue eyes around. Though I will add that I’m very glad we don’t live in a villiage with only four people in it, or we might have had a problem.”
On reflection, I made an error - a four-person village is immune to the scourge as well, as is the three-person. In a three-person village, each person sees two blue eyes, and knows that each other person sees at least one, so being informed of that fact would tell them nothing new.
A village with only two blue-eyed people would be in trouble, though.
About that bridge thing.
Isn’t he screwed anyway. Much like how when one steps on a scale the small amount of acceleration force downward due your mass shifting causes the scale to read high in the beginning, isn’t the bridge going to read “high” the first time he steps on it? In other words, he’ll exceed its maximum force rating and crash down regardless?
Your problem is that you’re changing people’s eyes back to blue again. You said “which we know isn’t the case” but that’s beside the point. Through this reasoning, we know that the other thinkers down the line would be wrong. But that doesn’t break the illusion.
To help you see this, I have a surprise for you:
I’m the 101st islander!
Yes, that’s right! I’m actually Z! I was just roll playing A, the chief! So although there are 101 islanders, we’re down to 96 now. I also have no idea what my eye color is, so I guess I should just assume it’s brown. And if the next doper wants to roll-play #102 and pretend to be me, that’d just extend the example.
Answer me this:
Do you realize that when A assumes he’s brown and opens his mouth to speak, it’s the exact equivalent of the interviewer going and talking to B directly, in the case that A actually is brown? That is, A talking while assuming brown = B talking while A actually is brown?
Do you further understand that B assuming brown while knowing A is brown = talking directly to C while A & B actually are brown? You’ll get the same answers either way.
Yes, this is probably true. However, the problem is much more fun if we ignore it 
I’m a bit disappointed that I went through the trouble of deriving a kinetic model for juggling, and now have nothing to show for it.
I don’t have any idea what you’re talking about - I assumed that A was an islander all along. And I don’t know what you mean about A opening his mouth to speak - in the actual hypothetical, none of the islanders discuss their eye color with one another, and none have to discuss their thought processess, because they’re all so brilliant they already figured out what the others would think.
Here’s an example: three islanders - A unknown, seeing two blues B and C.
V: “Hey y’all, one of you has blue eyes!”
A (thoughts):
(Oh dear, who is he talking about? Is he talking about me? Well, maybe he is, but he might have been talking about B or C instead, so I don’t know if V was talking about me. That means I still don’t know if I’m blue, so I don’t have to die tomorrow.)
(What about B? What will he think? Well, I don’t know if I’m blue-eyed or not, but C definitely is, so B won’t know whether V was talking about him or C, and won’t kill himself tomorrow, regardless of my own eye color.)
(C sees the same thing B does, and so won’t kill himself either. Good thing too - he owes me money.)
(And since I know none of us would under any circumstances kill himself tomorrow, the fact it won’t happen isn’t going to tell me anything, so I don’t have to do any inductive reasoning.)
A (speaking): “Phew! We really dodged a bullet there!”
B (speaking): “No kidding!”
C (speaking): “You got that right!”
A (speaking): “Hey guys, let’s kill and eat this visitor quick before he says something really damaging, like whether any of us has brown eyes.”
Others: “Right!”
V: “Um, help!”
The reason induction works is that induction relies on a clear ordered hierarchy, where each level of iteration is a specific number of ‘levels’ from the base cases and all the intermediate levels can be resolved. But the villagers aren’t in any particular order so this approach doesn’t work here.
I don’t mean that I, A, am the 101st islander. I mean that I, Chessic Sense, am the 101st islander. I was just role playing the A character, who was roll playing his daughter, pretending she was roll playing C, who was roll playing D. And if Indistinguishable comes back, he can be the 102nd and roll play me. Yet we can still reason down to 96 blues minimum. From 102 down to 96. That’s the same as 100 down to 94. Want to keep going?
And you’re missing a step in your ABC example. Sure, no one kills themselves the next day. But that means that everyone sees someone with blue eyes.
If you have brown eyes, then when you go “Phew! It’s a good thing we all saw blue eyes at least once!” then B knows that C saw B’s blue eyes, cause he sure didn’t see brown-eyed you. C reasons that he must be the blue eyed person that B saw. So they would have to kill themselves the next day, right?
But they don’t. So the above scenario must not have taken place. So if you don’t have brown eyes…::hands begbert a knife::
But in your hypothetical, A doesn’t think as deeply about B’s thought processes as A could. So, let’s continue A’s line of reasoning:
(Ok, B either sees one blue-eyed person, or two blue-eyed people. Let’s suppose for the moment that B sees only one blue-eyed person, C. In this case, B will have the following thought: “Ok, C either sees that no one else has blue eyes, or he sees one other blue-eyed person. Let’s suppose for the moment that C does not see any blue eyed people. If this is the case, then C will realize that V is talking about him, and will kill himself tomorrow. On the other hand, if C sees another blue-eyed person, me, C will not kill himself tomorrow. Therefore, C will not commit suicide if and only if I, B, have blue eyes. Thus, if C does not kill himself tomorrow, I will be forced to kill myself the day after tomorrow.” Of course, C will have the same thought. Thus, if I, A, do not have blue eyes, I know that both B and C will commit suicide the day after tomorrow. If, however, I do have blue eyes, both B and C will have had this exact same thought as I have, and will not kill themselves the day after tomorrow. Therefore, I have blue eyes if and only if B and C do not commit suicide the day after tomorrow."
The result is, A can test whether he has blue eyes, and if A is blue-eyed, will commit ritual suicide the second day after tomorrow.
The induction is on the number of islanders with blue eyes. If there are n islanders with blue eyes, then they will all kill themselves on the nth day after the missionary arrives. Clearly, this is (vacuously) true when n = 0, and the induction step is easy to establish as well: supposing this is true for n, we have to establish that it is true for n + 1. Very well; imagine an island with n + 1 blue-eyed islanders: any blue-eyed islander looks around and thinks “Hm, either there are n blue-eyed islanders who will all kill themselves on day n, or there are n + 1 blue-eyed islanders and I’m one of them”. If day n passes with no one killing themself, well, that leaves only one possibility…
I’m missing something here. The mother is 21 when her son is born at age 0. In 6 years the mother is 27 and the son is 6. 6 isn’t a fifth of 27. 
The mother isn’t impregnated at the same time her son is born, but, rather, 3/4 of a year earlier…
I’m impressed that all 1000 islanders are running around daily counting up the eye colors. Especially if figuring out your own eye color means death. I know I wouldn’t be counting.
Maybe he can just take off some clothes.
Unca cecil did this long ago. He tells us to “stay off that bridge”.
http://www.straightdope.com/columns/read/1224/riddle-me-this-how-can-68-3x1-70