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Rogan Josh
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Edit: Sorry - the answer I just gave was bullshit. Still thinking....
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Rogan Josh
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Hmmm... well it really depends on how many numbers a given number is divisible by. For example 50 is divisible by 1,2,5,10,25 and 50, so it light number 50 will be switched 6 times so it will be off. Similarly 51 is divisible only by 1 and 51, so it will be flicked twice and also be off. Conservely, 64 is divisible by 1,2,4,8,16,32 and 64 - an odd number of numbers, so light 64 will be on.
So I can see how to see if any particular light is on or off, but I still can't see the general pattern... 
You thought this one was easier than the last one? You must have a strange mind...
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Richard Bruns
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NC, USA
Nov 1999
time: 06:23
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This is a really great thread. I've had fun solving some of these.
The only lights that are on are the rooms with square numbers. But I didn't really solve that.
I have a question about the robbery liars, however. From Rogan Josh's correct answer:
quote:
Chris lying -> one of Brad or Dave must be telling the truth |
I don't see how that can be inferred from the problem. All of them could be lying about the robbery, which means that all of them could be truthful when they say that others are lying about the robbery. You can't infer anything from that information.
I suppose that their statements are meant to refer to who is lying about who is lying, and not about who is lying about the robbery. But that creates a paradox. How can Alex know who is lying until he has heard all the statements of the others? How can any of them make such a statement until they have heard the others? It seems to be an impossible situation, a self-referencing paradox.
Can someone enlighten me? 
Also, a couple of comments about the three-way duel. It is very clear that MLeonard gains by not firing. But consider the eventual outcome of the duel:
Once the initiative passes to Ming, he faces the same choice. There are three outcomes. If he plugs MLeonard, he dies when MtG shoots. If he frags MtG, he has a 1/3 chance of getting blasted by MLeonard next turn. If he misses or doesn't shoot, the choice of firing passes to MtG.
If MtG wastes MLeonard, then he has a 2/3 chance of getting inhumed by Ming. If he eliminates Ming, he has a 1/3 chance of kicking the bucket when MLeonard fires.
But MtG knows, at this point, that MLeonard will not fire if all three are left alive. So his best choice of action is to blow down a tree or something, so that everyone is left alive. That way nobody tries to gun him down next turn. If Ming knows that MtG knows this, his best chance of survival is also to not shoot at anybody, because that is the only way that nobody will fire at him next turn.
Basically, if all three are logical and they know that the other two are logical, then nobody will fire in the duel. They will all waste shots until they run out of ammo. Of course, if they were all logical they wouldn't have gotten into a duel in the first place. Three-power situations typically result in standoffs, since nobody gains by taking the forst shot.
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Dauphin
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Caught in a tuna net
Jan 1970
time: 05:23
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quote:
Originally posted by Richard Bruns
If MtG wastes MLeonard, then he has a 2/3 chance of getting inhumed by Ming. If he eliminates Ming, he has a 1/3 chance of kicking the bucket when MLeonard fires.
But MtG knows, at this point, that MLeonard will not fire if all three are left alive. So his best choice of action is to blow down a tree or something, so that everyone is left alive. That way nobody tries to gun him down next turn. If Ming knows that MtG knows this, his best chance of survival is also to not shoot at anybody, because that is the only way that nobody will fire at him next turn.
Basically, if all three are logical and they know that the other two are logical, then nobody will fire in the duel. They will all waste shots until they run out of ammo. Of course, if they were all logical they wouldn't have gotten into a duel in the first place. Three-power situations typically result in standoffs, since nobody gains by taking the forst shot. |
Depends on if MtG is playing to win, or to not lose.
Ming and ML will never target each other - that would be tantamount to suicide. Therefore MtG's odds will never improve. So if he wants to win he has to shoot someone first.
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Zero-Tau
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Elsewhere
Aug 2002
time: 06:23
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quote:
Originally posted by Sagacious Dolphin
A hotel has 1,000 rooms numbered 1 to 1,000. Jack is looking after the hotel for the winter and is going slightly crazy. He decides to flick every roomlight switch in the hotel. He then decides to flick the switch in every other room (starting with room 2). He then flicks the switch in every third room (starting with room three). He then flicks the switch in every fourth room (starting in room 4).
If all lights were originally off, which lights remain on once Jack has finished switching every 1,000th roomlight switch? |
All rooms with square numbers have their lights flicked on. Their prime factorization would be a1^e1*a2^e2*a3^e3... with all exponents being even. Using the formula for the number of divisors (e1+1)(e2+1)(e3+1)... we see that the square numbers have an odd number of divisors, meaning that they'll end up with the lights on. Conversely, if the number of divisors is odd, then each of e1+1, e2+1, e3+1,... would be odd, meaning that all the exponents would be even, making the number a square.
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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quote:
Originally posted by Zero-Tau
All rooms with square numbers have their lights flicked on. Their prime factorization would be a1^e1*a2^e2*a3^e3... with all exponents being even. Using the formula for the number of divisors (e1+1)(e2+1)(e3+1)... we see that the square numbers have an odd number of divisors, meaning that they'll end up with the lights on. Conversely, if the number of divisors is odd, then each of e1+1, e2+1, e3+1,... would be odd, meaning that all the exponents would be even, making the number a square. |
Woah, I didn't even follow all that! As soon as Sagacious Dolphin asked what property a number needs to have an odd number of factors, I just figured: "For every number, each factor will be matched with exactly one corresponding factor so that the two together multiply to the number. A factor will be matched with itself if and only if it is a root of the number, so square numbers and only square numbers will have an odd number of factors, since the factors will come in pairs, except if the number is square, in which case the root will be by itself."
My way seems a lot simpler to me. (My explanation is a bit more wordy than my actual reasoning is complicated, trust me.)
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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quote:
Originally posted by Urban Ranger
Balls are not replaced: twice
Balls are replaced:
There are 5 configurations (I use letters instead of colours, it's the same) -
ABCD (original, config 0)
AACD (config 1)
AAAC (config 2)
AACC (config 3)
AAAA (config 4, goal)
config 0 -> config 1 : 1 turn
config 1: 1/2 chance remaining config 1, 1/3 chance becoming config 2, 1/6 chance becoming config 3
config 2: 1/2 chance no change, 1/3 chance becoming config 4 (done), 1/6 chance becoming config 3
config 3: 1/3 chance no change, 2/3 chance becoming config 2
Then I forgot how to convert this into expectations  |
I think config 2 should be: 1/2 chance no change, 1/4 chance becoming config 4 (done), 1/4 chance becoming config 3. The rest appear to be right.
Assuming that's correct...
The expected number of turns to get to config 4 from config 0, 1, 2, or 3 is 1 plus, for each possible next state, its probability of coming next times the expected number of turns to get to config 4 from that new state. Representing the expected number of turns to get to config 4 from config x by cx, we get
c0 = 1 + c1
c1 = 1 + (1/2)c1 + (1/3)c2 + (1/6)c3
c2 = 1 + (1/2)c2 + (1/4)c3 + (1/4)c4
c3 = 1 + (2/3)c2 + (1/3)c3
c4 = 0
After a lot of algebra, we can conclude that c0 = 9, c1 = 8, c2 = 5.5, c3 = 7, and of course c4 = 0.
So the expected number of turns after the first draw is 8.
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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OK, most of you should find this one fairly easy:
Three missionaries and three canibals are traveling together and come to a river. None of them can swim, but luckily, there is a boat tied to their side. It can only hold two people. For obvious reasons, the missionaries would prefer that no group of canibals ever outnumber a group of missionaries (where a group can be one, but not zero) at any one place. Figure out a way that they can maintain that condition and get everybody across the river.
If that's too easy for you, you might want to try the more general case of x missionaries and x canibals (I don't actually know whether that can be solved or not).
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28-10-2002 20:07
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