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Old 06-16-2010, 08:56 PM   #31
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Here it goes:

Spoiler:
On the night of the dinner they are told for the first time that there are green and blue marks. So that means that there is at least one of each color. For argument's sake let's say there was only one person with a green mark. Then that person would look around and see only blue marks and say aha I must have a green mark. The next morning he can leave.
Now if there were 2 people with green marks they each would see that on the first day neither leaves. Then they would assume that that green marked person saw at least one other person with a green mark and deduce that it must be them and leave the second day.
Now remember everybody here is logical then if there were 3 green marks they would wait 2 nights and see that each saw two others with green marks and leave the 3rd day.
On the 45th day all the green marked people will leave. On the 46 day all the blue marked people will leave
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Old 06-17-2010, 03:56 AM   #32
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Here it goes:

[/SPOILER]
Spot on. Exactly right. Congratulations.
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Old 06-17-2010, 04:45 AM   #33
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OK I give up. I started 3 different solutions and got lost in the reasoning in all 3 of them.

I even began again and tried to go super simple,
Spoiler:
1-- (on the day before 100th prisoner)

B sees 53 B & 45 G = 98 prisoners

2-- (prisoner #100 comes in with a B mark)

B now sees 54 B and 45 G = 99

but went blank on where to go from there.


Maybe, I'm not very logical?
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Old 06-17-2010, 04:49 AM   #34
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OK I give up. I started 3 different solutions and got lost in the reasoning in all 3 of them.
It is difficult to work out what happens in the case presented without considering simpler cases first. I refer you to my (small) clue here: http://www.mobileread.com/forums/sho...9&postcount=13
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Old 06-17-2010, 06:00 AM   #35
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Thoughts towards a solution, but I get lost quickly:
Spoiler:

When there's only two prisoners, and they know BOTH colours are present,
it is obvious to both of them that they have the colour they can't see, and both go.

When there's three prisoners, and they know both colours are present,
one of them sees only one colour, so he knows his is the other one, and goes. When he goes, the next day the other two go.

When there's four prisoners, and they know both colours are present, there are two possible cases. If they are divided in 1-3, again the one who sees only one colour leaves. If they are 2-2... Say you're blue, and you see one blue and two greens. You don't know your colour. You wait one day. If the blue one doesn't go away, you are not green. So tomorrow you can go away. Ok so far.

If there are five, 2 have one colour, 3 have the other. Say you are blue, you see one blue and 3 greens. You wait one day, no one leaves, which means you must be blue. Tomorrow you leave too.

If there are six, and say they are divided equally, because with groups of two it's simple. So you are green, and see two greens and three blues. If you are blue, you see three greens, 2 blues. Now what?

If I get stuck at six, how can I feel confident in a 100?
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Old 06-17-2010, 06:01 AM   #36
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Obs20,
I won't look at your solution just yet, but you are seriously starting to intimidate me
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Old 06-17-2010, 06:20 AM   #37
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Originally Posted by omk3 View Post
Thoughts towards a solution, but I get lost quickly:
Spoiler:

When there's only two prisoners, and they know BOTH colours are present,
it is obvious to both of them that they have the colour they can't see, and both go.

When there's three prisoners, and they know both colours are present,
one of them sees only one colour, so he knows his is the other one, and goes. When he goes, the next day the other two go.

When there's four prisoners, and they know both colours are present, there are two possible cases. If they are divided in 1-3, again the one who sees only one colour leaves. If they are 2-2... Say you're blue, and you see one blue and two greens. You don't know your colour. You wait one day. If the blue one doesn't go away, you are not green. So tomorrow you can go away. Ok so far.

If there are five, 2 have one colour, 3 have the other. Say you are blue, you see one blue and 3 greens. You wait one day, no one leaves, which means you must be blue. Tomorrow you leave too.

If there are six, and say they are divided equally, because with groups of two it's simple. So you are green, and see two greens and three blues. If you are blue, you see three greens, 2 blues. Now what?

If I get stuck at six, how can I feel confident in a 100?
I think you are on the same track I had tried, then got stuck.
Good Luck, I'm off to bed as the alarm goes off in about 4 hours.
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Old 06-17-2010, 09:12 AM   #38
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I gave in and looked at OBS's answer. I had initially started in that direction but stopped, my problem with it was

Spoiler:
I got wrapped up in human nature and asked myself "what if one of the prisoners did not attempt to leave at the earliest possible time?" That would have thrown everything out of wack. I didn't make the assumption, and maybe I should have, that logic implied they would leave as soon as they could - anything else would be illogical.
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Old 06-17-2010, 09:32 AM   #39
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Originally Posted by omk3 View Post
Thoughts towards a solution, but I get lost quickly:
You are very much along the right lines. Where you get stuck,

Clue:
Spoiler:
Imagine that you're that green prisoner (although you don't know you're green). Now imagine that you're that prisoner considering what would happen if you were a blue prisoner. Image what one of the greens you're looking at would see if you were blue. What would they do then? What can you conclude if they don't do that?
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Old 06-17-2010, 09:34 AM   #40
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Originally Posted by dsvick View Post
I gave in and looked at OBS's answer. I had initially started in that direction but stopped, my problem with it was

Spoiler:
I got wrapped up in human nature and asked myself "what if one of the prisoners did not attempt to leave at the earliest possible time?" That would have thrown everything out of wack. I didn't make the assumption, and maybe I should have, that logic implied they would leave as soon as they could - anything else would be illogical.
I should say that I don't think your spoiler tags are necessary, but I won't remove them. I will answer though.

This is really a logical puzzle given embodiment. It's not a logical analysis of a real situation. So bringing in human variation and fallibility is the wrong way to approach a solution.
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Old 06-17-2010, 09:46 AM   #41
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Okay, thanks to your clue I took it one step further. But this is making my head hurt!

Spoiler:
We are six, equally divided.
I wonder what my colour is. I see 3 green, 2 blue.
IF I am blue, the blue ones see 1 blue, 4 green.
The blue wonders what his colour is.
If he is green, the other blue should see 5 greens, and leave.
1 day passes, the other blue doesn't leave.
So it would follow that the second blue is actually blue.
The 2nd day passes, and none of the blues leave.
Which means I am blue too!

If I try to do this with bigger numbers, I think my head will break


But even so, all this takes into account that the prisoners only just realized that there are two colours only, and both are represented. But they knew that from the start, didn't they? And if so, shouldn't they have left sooner? What changed?
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Old 06-17-2010, 09:59 AM   #42
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But even so, all this takes into account that the prisoners only just realized that there are two colours only, and both are represented. But they knew that from the start, didn't they? And if so, shouldn't they have left sooner? What changed?
OMK has a point - and not on top of his head either
In the set up you said that the prisoners were all told they were marked with either Green or Blue. So actually, as soon as there were at least 2 prisoners with each color, they should have been able to figure it out in X number of days, where X is the number of people marked with the least represented color.
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Old 06-17-2010, 10:02 AM   #43
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Quote:
Originally Posted by dsvick View Post
OMK has a point - and not on top of his head either
In the set up you said that the prisoners were all told they were marked with either Green or Blue. So actually, as soon as there were at least 2 prisoners with each color, they should have been able to figure it out in X number of days, where X is the number of people marked with the least represented color.
aha !!!!

(just thoughts. it's not *that* big an "aha".)
Spoiler:
maybe the "new information" that pdurrant mentioned is the fact that so far, no-one has left. because maybe prisoner N°2 didn't know that he was only the second one ; for all he knew, there might have been 50 before him of whom 49 had managed to leave. right ?

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Old 06-17-2010, 11:34 AM   #44
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OMK has a point - and not on top of his head either
In the set up you said that the prisoners were all told they were marked with either Green or Blue. So actually, as soon as there were at least 2 prisoners with each color, they should have been able to figure it out in X number of days, where X is the number of people marked with the least represented color.
Do the thought experiment - pretend you're the first prisoner to arrive, and you're given a green mark (although you don't know it's green). Subsequently one more green and two blue prisoners arrive. Every prisoner can now see at least one prisoner of each colour. So every prisoner knows that the prison contains prisoners of both colours. Can any of them conclude anything about their own colour from this? I think you will see that they cannot, especially if you try to consider the situation from the viewpoint of any one of the prisoners.

Spoiler:
And yet if the governor gathers all four prisoners together and gives his speech to them, I think you'll find that there is a now a difference, even though he appears to have only told them something that they all already know. In fact, he's told them something else as well. Work out what that is, and you'll probably be able to see the answer.
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Old 06-17-2010, 11:37 AM   #45
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There's a country that has a special prison. It's not a prison for people who've broken the usual criminal laws, it's a prison for people who've irritated the rest of the population beyond endurance by making up (and insisting on explaining) far too many logical puzzles.
Despite the danger of going to this prison, I will give the answer, and insist on explaining the answer (as best I can) in about another day. Perhaps two, if Friday's busy for me.
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