For 4 blue eyed people... the same again. You know there's a blue eyed person, you know that person 2 knows there's a blue eyed person, you know that person 2 knows that person 3 knows there's a blue eyed person, but you don't know that person 2 knows that person 3 knows that person 4 knows there's a blue eyed person. This is the crucial piece of information the guru provides.
So you don't know that 2 knows that 3 knows that 4 knows that there's a blue eyed person.
However, when 4 doesn't leave that same night, you now know that 2 knows that 3 knows that 4 knows that there's a blue-eyed person.
That's the information that each new day provides. Okay, I see it.
You can always imagine that you have brown eyes, and assume that the next person imagines that they themselves might have brown eyes, and so on and so forth, so you know there's a chain of imagination stretching all the way down to N=1.
Yes, you know that the third guy in line sees that the second guy has brown instead of blue. But the second guy doesn't know, and makes his own assumptions about the third guy, plus your own assumptions about yourself.
You see (L=blue, R=brown, ?=unknown):
?LLL
You are now considering two different versions of person 2 based on your own potential eye colors. R2 sees R?LL, L2 sees L?LL.
Your imaginary R2 considers two different versions of person 3, corresponding to different versions of their own (2's) state. R2R3 sees RR?L, R2L3 sees RL?L, etc. Only ONE person is important right now, and they are imaginary: R2R3. When the guru speaks up, they give information to R2R3's head-4 alone. R2R3R4 sees RRR? and will leave. R2R3L4 sees RRL? and will stay.
So, YOU don't get new information, exactly. But virtual, theoretical, recursive people that you're imagining do.
God, how confusing.
There's a huge pile of envelopes that contain various amounts of money, from one dollar up to 65 thousand (give or take). Someone gives you two envelopes and tells you to pick one. Before you open it, they tell you that the other one has either half as much or twice as much, and offers to let you change your selection. What do you do? Would your answer change if you could open the first envelope and look inside before you switched?
The only time the one you chose is relevant is if it's worth at least half the maximum value or less than twice the minimum. If the most you can win is 65k and your envelope holds 40k, you can't possibly win 80k so that changes things. The same applies if your first choice is worth less than twice the minimum value.
Otherwise, normalize the amount of your first pick to 1. The expected value for switching is [(2 + 0.5)/2] = 1.25. Assuming a 50-50 chance of either, it's in your best interest to switch.
Aaand...Earthquake Damage is right with his first paragraph, at least. If you can look inside, it's good to switch if you're below (min*2), and bad to switch if you're above (max/2). Otherwise, it makes no difference at all! It's kind of like a broken Monty Hall problem, because switching really doesn't make any difference. You got two envelopes. One of them, at random, has twice the value of the other. You have a 50% chance of starting with it, and a 50% chance of starting with the other one. Doesn't matter which one you keep, even after looking inside, unless you have advance knowledge of the distribution function.
Or to put it another way: If you switched envelopes once, and then someone asked you if you wanted to switch back again, wouldn't you want to take them up on it again? After all, your expected value is still ((2 + 1/2) / 2) * the one you're currently holding... You'd never stop.
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Topic: The questions, riddles and puzzles thread (Read 37461 times)