My hypotheticals are also incompletely specified. However, the point of them was that the very same assumptions which would lead you to wanting to switch normally would (mutatis mutandis) lead to 50-50 answers in the hypotheticals. For example, merely happening to learn that there’s a goat behind door number 2 doesn’t give me any reason to prefer door number 3 to door number 1; the situation is totally symmetrical (as emphasized by putting another contestant at door number 3). It increases my confidence that my original guess was correct and increases the chances that door number 3 is correct. In the exact same way that if I were to furthermore see another goat behind door number 3, my confidence in my original guess would rise even further to 100% probability. And in the exact same way that, if there was a lottery, I would originally assign very low probability to my ticket winning, but each time I heard the announcer say another digit matching my own, my confidence in my ticket winning would rise accordingly. I wouldn’t sit there, all digits matched except the as-of-yet unrevealed last digit, thinking “Oh shit, I better go find a different ticket now; maybe another potential winner with a different last digit will be willing to trade with me”. I’d sit there thinking “Oh shit, I have a very good shot at being a winner now”.
Oh well, there goes the thread.
Sorry, you’re right. I did at one point recognize that I was feeding a hijack and should stop, but somehow let myself be sucked back in, “Someone on the Internet is wrong!”-style. I’ll put a moratorium on my Monty Hall discussion in this thread.
Max the Magician has a hobby of plundering the temples of tribal natives deep in the Wherever jungle to “acquire” items for his act. Most recently, he’s stolen the golden eyes from a statue of the tribe’s three-eyed, all-seeing god.
As he dashes through the jungle to escape his pursuers, he comes to a very wide ravine- at least 400 ft across- and a rickety, unstable plank-and-rope bridge. A sign posted (in English…how fortunate!) says that the bridge has a max load of 170 lbs. Anything more than that and the bridge will snap, sending its passenger plunging to his death!
Max knows that he and his clothes weigh exactly 150 lbs (What? Man’s gotta be prepared!), but each of the eyes weighs 7 lbs! Max (150) + 3 eyes (7x3) = 171 lbs!
Max paces back and forth, trying to come up with a solution to his predicament before the natives and their spears close in. He’s got a head start, but it’s a few minutes at best. No time to build a catapult or build another bridge. Max was a sprinter in college, but not a pitcher or quarterback. There’s no way he can throw one of the eyes other the ravine.
Then a lightbulb goes off. Max gets across the bridge safely and escapes.
How?
Max is an accomplished magician. Any magician worth his salt can juggle! He gets an eye in the air and proceeds across the bridge without dropping a single one.
I don’t get it. So the visitor tells them he’s surprised to see blue eyed people on the island - this isn’t new information, as everyone there was aware there were blue eyed tribesmen, since anyone with blue eyes could see 99 other people with blue eyes. It only works if there’s one tribesman with blue eyes.
That doesn’t work either. Max + 3 juggling balls still weighs 171 lbs (and at times, even more), regardless if one is in the air at all times or not.
How do you figure? So long as he gets the first ball airborne before he steps onto the bridge, then that ball doesn’t count in the system. The force required to propel the second ball into the air isn’t 7 pounds of pressure just because the ball weighs that. When the second ball lands, it’s the same thing. By the time the first ball lands, the second is in the air. So Max never weighs more than 170. Sure, when he tosses the original ball, he weighs slightly more than 171, but he’s still on the ground so it’s cool.
Doesn’t work. You have to apply an average of 7lb upthrust to keep a 7lb weight at a constant height to start with. You want to throw it up, you have to apply more - how much depends on how high you want it to go. But if the three balls are staying at the same average height throughout the crossing, then they’re each needing an average of 7lb of upwards shove. Otherwise your potential energy isn’t conserved.
God bless xkcd. 
It works due to inductive reasoning. But the answer is still wrong, because all the brown-eyed will die right after all the blue-eyed kill themselves.
If there was only one blue eyed person on the island, he would see 999 brown eyed people. When the missionary mentions he sees someone with blue eyes, that person would know that he was the only one with blue eyes. So he kills himself.
With 2 blue-eyed people, they each see 998 brown-eyed people, and 1 blue-eyed person. Since no one dies the first day, they each conclude indpendently that there must be more than 1 blue-eyed person (otherwise that person would die), so they naturally conclude that they each have blue eyes, and kill themselves the 2nd Day.
And so-on and so-forth. 100 blue-eyed people means 100 Days later they all kill themselves.
However, the brown-eyed people soon realize this. And because they know that they don’t see any other blue eyed people, and all the blue-eyed people that could see other blue-eyed people have now died, they must all have brown eyes.
So upon that conclusion, the rest of them all kill themselves.
Fucking genocidal tourists!
So? You’ve already counted the 7 lbs when you came up with 171/164. If I toss a ball up, I temporarily weigh more than 171, but then I weigh only 164 while it’s in the air. Then I step onto the bridge and apply a total force of, say, 9 lbs to the second ball, sending it airborne. How much do I weigh when I accelerate this second ball?
164 + 9, so 171? No. You already counted the initial 7 lbs. The answer is 164 + 2, or 150+ 7 (the 3rd ball) + 9 = 166. After that ball goes airborne, I weigh only 157 lbs.
Then that first ball comes home and I decelerate it with 9 lbs of upward force. When it hits my hand, I weigh 150 + 7 (3rd ball still in my hand) + 9 = 166 lbs again. The ball comes to a stop and I weigh 164 again while that 2nd ball hangs in the air. Repeat.
At no point do I cross the 170 threshold. Again, I think you’re counting the weight of the ball twice. You’re adding it to Max’s weight, and then saying he needs to apply more than 7 to get the ball moving. He doesn’t. You can either say Max is 150 and he uses 9 lbs of force to get the ball moving, or that he weighs 157 and uses 2 more lbs to get the ball airborne. Either way, he’s putting 159 lbs of weight onto the bridge. Sure, Max can break the bridge by throwing the ball really hard, but he’s no fool.
Your turn. Please demonstrate at what point he weighs more than 170 after the first ball is in the air.
Indeed. Imagine you’re on an island where you only see one blue-eyed person, Jim. Then the tourist says “Wow, there’s a blue-eyed person here.” You’d say to yourself “Aw man. Jim’s the only person with blue eyes. Now he knows what color eyes he has.”
Then the next day, you see Jim walking around normally. You go “Hey, he should be dead…unless he sees another person with blue eyes! Who could it be?” Then you start wandering around the island looking for the other blue-eyed person. You check everyone. You can’t find them. Everyone else is brown! Everyone!
Then the other person with blue eyes must be…shit. You go home and make out your last will and testament. Unfortunately, you realize, after you kill yourselves, all the blue eyed people will be dead. So the living must be all brown-eyed people…and they know it.
Might as well chuck your life savings into the ocean.
Nope - because each brown eyed person still doesn’t know what color their eyes are. If I’m one of those left, I could very well think I have brown eyes, green eyes, red eyes, etc.
Yes. But we’re not dealing with a population of 1 v. 999. The blue eyed people already know there are blue eyed people on the island - the missionary isn’t offering any new information.
Specifically, the new information the islanders receive is not “There is at least one person with blue eyes on the island”. Everyone already knew that; that’s true. The new information the islanders receive is “Everyone knows that everyone knows that everyone knows that… there is at least one person with blue eyes on the island”, for every number of “everyone knows”. It was not previously the case that the islanders knew every statement of this form; in fact, the islanders originally knew the statement with n “everyone knows” in it precisely when n < the number of blue-eyed people on the island.
It takes 7 lbs of force to get one ball up. Right as you apply it, you weigh 178 lbs. (we’re assuming you’re REALLY fast at juggling, and don’t need to keep the ball more than an infitesimal height above your hands, and the balls don’t get in each other’s way - you are a really good magician, after all). Once it’s in the air, you weight 164 (150 + 7 + 7). You need to get rid of ball #2 before ball #3 lands, so you throw it: 150 + 7 + 14 = 171.
That’s very confusing. Could you hash it out in long form for an island of 10 with 3 blue eyeds?
I should say, small correction, islander X originally knew the statement with n “everyone knows that” preceding “there is at least one blue-eyed islander” precisely when n < the number of blue-eyed people islander X sees.
I’ll do it inductively:
Let S(0) be the statement that there is at least one blue-eyed person on the island, and let S(n+1) be the statement “Everyone knows S(n)”. So S(1) is “Everyone knows there is at least one blue-eyed person on the island”, S(2) is “Everyone knows that everyone knows that there is at least one blue-eyed person on the island”, and so on.
Suppose there are 0 blue-eyed people on the island. Then S(0) is false.
Suppose there is just 1 blue-eyed person on the island. Then there is at least one blue-eyed islander, but not everyone knows there is at least one blue-eyed islander (the sole blue-eyed islander might think no one is blue-eyed). Thus, S(0) is true, but S(1) is false.
Suppose there are 2 blue-eyed people on the island. Then, everyone on the island knows there is at least one blue-eyed islander, but not everyone knows there are two blue-eyed islanders; some people think it’s possible that there’s just 1 blue-eyed islander. In other words, some people think it’s possible that the above case holds, which means some people think it’s possible that S(1) is false. Therefore, not everyone knows S(1); in other words, S(2) is false.
Suppose there are 3 blue-eyed people on the island. Then some people think it’s possible there are just 2 blue-eyed people on the island, which means some people think it’s possible that the above case holds, and thus some people think it’s possible that S(2) is false. Thus, not everyone knows S(2); in other words, S(3) is false.
And so on and so on.
See, your problem is that you’re counting the weight of the ball as 7, then saying that it needs 7 more to get airborne. It doesn’t. It requires 7 to hold still. That’s the weight. In addition to that, it needs any force whatsoever to go airborne.
So you say “150+7+14” and I say “Why are you throwing so hard?”
Throw it at 9 instead. “150+7+9= 166”. 9 is greater than 7, so the ball goes airborne. Yes, it requires 7 lbs of force to get one ball up. So why are you throwing at 14?
Get it now?
Sure. You’re a blue eyed person, but don’t know it. You think there are 2 blue-eyes on the island, Jim and Bob. This causes you to think that each of those guys sees only 1 other blue-eyed person there. Jim thinks it’s Bob and Bob thinks it’s Jim. As far as they know, there’s only one other person on the island with blue eyes. On day 1, no one kills himself. That’s cool. You knew that would happen. But Bob and Jim didn’t (or so you think). In reality, it’s no shock to them either, because they see you. As far as Jim knows, you think there’s only one other blue-eye on the island (Bob).
Since no one killed themselves on the first day, then Jim and Bob must not be looking out into the world and seeing only brown eyes. So there must be at least 2 people out there with blue eyes. But that’s cool, cuz you already knew that.
Then no one kills themselves on day 2. Hey, wait a minute, you say. if Jim and Bob both know that there are at least 2 people on the island with blue eyes, and they know who one of them is (the other one), and they check all the rest of the tribespeople and don’t find any other blue eyes, they’d know it was them. So they should be dead. Right?
Why didn’t Jim and Bob reason that out? They’re smart guys. Again, they know there are two blue-eyes out there, and they only see one, right? Right? RIGHT?!
Oh no, you realize, they DID find another person with blue eyes besides the two of them. Who could it be? You run to a 4th person. Nope, brown eyes. You check a fifth person. Nope, brown eyes. A 6th, a 7th, , and 8th, a 9th person…all brown eyes! Who is the mysterious third person?!
You pensively knock on the hut door of the 10th person on the island. But no, brown eyes. Now you know who the third person is. It’s got to be you.
Simultaneously, Jim and Bob realize the same thing. You see, all along, they thought you and the other guy were going to kill yourselves on the second day, due to the same reasoning. But they now know what you know- there are three blue eyes out there are they can only account for two, so it must be them.
So on the third day, you all off yourselves.
As an addendum, the strange thing is, the brown eyed people are HOPING that they have blue eyes themselves. You see, once they see Jim, Bob, and you heading up the mountain, they get in line behind them. They don’t even have to wait until noon. Because if Jim, Bob, and you all realize that you make up the blue-eyed population, then they must be in the brown eyed population themselves.
But if they in fact had blue eyes, then the whole tribe lives til day 4 instead of day 3! So they WANT to have the blue eyes so as to live an extra day.
Ah - there’s the error! (Also - the fact that when juggling, you have TWO balls in the air at a time…)
This is how most of these riddles work out, with someone saying something along the lines of, “from this day forward, if any prisoner can tell me how many blue hats there are, everyone goes free…”, thus setting a starting point. But in this stituation, there is no “Day 1”. Nor is there a defined number of color of eyes (thus saving the brown eyeds).
Did you see my post Munch for why, on an island of 10 with 3 blue-eyeds, it is not originally the case that everyone knows that everyone knows that everyone knows that there is at least one blue-eyed islander?
By “Day 1”, Chessic Sense means the 1st day after the foreigner makes his announcement. The announcement changes the islanders’ knowledge because the announcement entails that there is at least one blue-eyed islander, and the announcement entails that everyone knows the announcement. Thus, the announcement entails that everyone knows that everyone knows that everyone knows that … everyone knows that there is at least one blue-eyed islander, which, as shown above, would be new information.