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Author Topic: The questions, riddles and puzzles thread  (Read 38208 times)

Darvi

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Re: The questions, riddles and puzzles thread
« Reply #210 on: February 17, 2011, 06:04:12 pm »

It's nothing to do with that.  Assuming no communcation is possible, there really is no way for them to work out their eye colours without the announcement.
They could just try leave the day after the one with the amount of blue eyed people they see. The problem being, that strategy is based on that recursive line of reasoning, and without the guru it won't work.

Which is why brown won't leave.
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Leafsnail

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Re: The questions, riddles and puzzles thread
« Reply #211 on: February 17, 2011, 06:25:17 pm »

That's not a strategy though.  That's just a random guess.
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Darvi

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Re: The questions, riddles and puzzles thread
« Reply #212 on: February 17, 2011, 06:46:50 pm »

Exactly. They need the guru to  turn the guess into a usable strategy.
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Leafsnail

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Re: The questions, riddles and puzzles thread
« Reply #213 on: February 17, 2011, 06:48:26 pm »

No, not even that.

The guru adds something completely new and different that allows them to leave.

Imagine you were in this situation.  You see 2 other people with blue eyes, and hear no announcements.  Are you ever going to work out what colour your eyes are?
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Darvi

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Re: The questions, riddles and puzzles thread
« Reply #214 on: February 17, 2011, 06:50:37 pm »

No. I just said you'd need the guru.
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Sowelu

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Re: The questions, riddles and puzzles thread
« Reply #215 on: February 17, 2011, 07:32:15 pm »

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.
« Last Edit: February 17, 2011, 07:34:10 pm by Sowelu »
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Leafsnail

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Re: The questions, riddles and puzzles thread
« Reply #216 on: February 17, 2011, 07:47:56 pm »

So, YOU don't get new information, exactly.  But virtual, theoretical, recursive people that you're imagining do.
Honestly, I think I need to quote this because it sums up this riddle so well.  I mean, man, it's weird.
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Earthquake Damage

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Re: The questions, riddles and puzzles thread
« Reply #217 on: February 17, 2011, 08:26:22 pm »

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.

Hmm.

If you switch once then get a chance to switch again without seeing the result of the first switch...

If your new envelope has twice the money of the previous, switching gives an expected value of 3/4.
If your new envelope instead has half the money of the previous, switching gives an expected value of 3/2.
Averaging the two yields 9/8.
I'm not sure I should be averaging them.
Maybe (3/4)-1+(3/2)-1 = (eval)-1?  So an expected value of 1/2?
That seems too low.

Anyone mind showing me how to do this correctly?

Derp.  The first case should yield 3/8.  I think the rest still comes out wrong, though.
« Last Edit: February 17, 2011, 08:28:58 pm by Earthquake Damage »
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Sowelu

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Re: The questions, riddles and puzzles thread
« Reply #218 on: February 17, 2011, 08:29:11 pm »

Your pair of envelopes has a low value, L, and a high value, 2L.

Both envelopes have expected value of (L + 2L) / 2, or 1.5L.

L is hidden; you do not know how it relates to the amount in the opened value you are looking at.  But when you switch, you go from an expected value of 1.5L to an expected value of 1.5L.
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Earthquake Damage

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Re: The questions, riddles and puzzles thread
« Reply #219 on: February 17, 2011, 08:32:39 pm »

That can't be right.  Before you choose any of the three, they contain three values:  x, 2x, 4x.  For convenience, let x = 1.

Your first pick was 2x.  This is given, as you are told the other two contain x and 4x.

If you switch to 4x, the two left to switch to are x and 2x.
If you instead switched to x, then you have 2x and 4x available.
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G-Flex

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Re: The questions, riddles and puzzles thread
« Reply #220 on: February 17, 2011, 08:34:05 pm »

Sowelu: What are you smoking?

You're about to open an envelope with money in it, and are told the other envelope has either half that amount or double.

Define the money in your envelope as X. The amount of money in the other envelope is either X/2 or 2X. The average of X/2 and 2X is 1.5X.

So if you stick with the envelope you have, your expected value is X (by definition). If you switch, your expected (read: average) value is 1.5X. The variation between the X envelope and the other one ranges from X-(0.5X) and X+X. You can see what's going on here.
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Earthquake Damage

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Re: The questions, riddles and puzzles thread
« Reply #221 on: February 17, 2011, 08:41:20 pm »

I think he miscommunicated the problem or something.  I think he intended something like this:

You have $X.  You may keep the money or flip a fair coin.  Heads doubles your money.  Tails halves it.  You can flip as many times as you want.  How many times should you flip the coin?
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G-Flex

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Re: The questions, riddles and puzzles thread
« Reply #222 on: February 17, 2011, 08:43:52 pm »

That's an interesting question. In each instance, on average you get more money if you flip the coin. However, over the long run, on average nothing happens at all, because money gets halved as often as it gets doubled.
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Sowelu

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Re: The questions, riddles and puzzles thread
« Reply #223 on: February 17, 2011, 09:07:57 pm »

Yeah, it's a problem that's designed to sound complicated when there's really nothing going on at all.

50% of the time, you have the L envelope.  If you switch, you'll change to the 2L envelope, and gain L in value.
50% of the time, you have the 2L envelope.  If you switch, you'll change to the L envelope, and lose L in value.

So yes, switching will either halve or double your amount of money, but you gain or loes the same amount no matter what happens.  Funny, isn't it?
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G-Flex

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Re: The questions, riddles and puzzles thread
« Reply #224 on: February 17, 2011, 09:09:05 pm »

So yes, switching will either halve or double your amount of money, but you gain or loes the same amount no matter what happens.

Not really. On an individual turn, you either gain L (where L is the amount you already have) or lose half of L.
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