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Yat wrote:If not :Spoiler:
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Yeah, I guess they could come up with a strategy to avoid the zero survivors scenario, but without having additionnal information from what the others say or what the cannibals do to them, I don't really see how they could individually escape the 2/3 dying probability...jestingrabbit wrote:I think the day to strategise means that they're okay actually.
Yat wrote:Yeah, I guess they could come up with a strategy to avoid the zero survivors scenario, but without having additionnal information from what the others say or what the cannibals do to them, I don't really see how they could individually escape the 2/3 dying probability...jestingrabbit wrote:I think the day to strategise means that they're okay actually.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:Yat wrote:If not :Spoiler:
I think the day to strategise means that they're okay actually.
This one is an easier case of the same problem with n people and n possible colors... having already seen the answer I can't really comment on how difficult it is.
jestingrabbit wrote:Yat wrote:Yeah, I guess they could come up with a strategy to avoid the zero survivors scenario, but without having additionnal information from what the others say or what the cannibals do to them, I don't really see how they could individually escape the 2/3 dying probability...jestingrabbit wrote:I think the day to strategise means that they're okay actually.
Yeah, that seems right. It needs to be clear that they hear each othe'rs guesses, or at least the first, to get a better than 1/3 survival.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:I did indeed think that one correct guess was enough to save them all. Reading comprehension fail on my part. Yat solves for the case where each individual decides their own fate, and they can hear the other respondents. In the case where they can't hear each other though, a 1/3 survival is possible, if they arrange forSpoiler:
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
tomtom2357 wrote:A version of this problem is similar to another thread, since I can't remember which one, I will explain it here...
Gwydion wrote:tomtom2357 wrote:A version of this problem is similar to another thread, since I can't remember which one, I will explain it here...
The thread you're looking for is viewtopic.php?f=3&t=76500 . I recall an older thread (which I can't find right now) which dealt with the same problem, yet no villain.
tomtom2357 wrote:Yes, that's the one! I am working on a better solution to the problem than above, but I can't figure out how to get past the worst-case scenario. Maybe if I try for a non-symmetric solution then that will work.
Edit: Eureka! If the person sees more than 32 red hats, guess red, if not, then if the person sees more than 32 blue hats, then guess blue, if not, then if the person sees more than 32 white hats, guess white. This works for all cases with a minimum of 33 prisoners let free!
Trebla wrote:tomtom2357 wrote:Yes, that's the one! I am working on a better solution to the problem than above, but I can't figure out how to get past the worst-case scenario. Maybe if I try for a non-symmetric solution then that will work.
Edit: Eureka! If the person sees more than 32 red hats, guess red, if not, then if the person sees more than 32 blue hats, then guess blue, if not, then if the person sees more than 32 white hats, guess white. This works for all cases with a minimum of 33 prisoners let free!
I'm not sure if you've digressed to the problem in the linked thread (which is actually quite different from the one in this thread). The solution presented by Yat in the third post is the "best" solution and guarantees 99 prisoners are let free in every case.
tomtom2357 wrote:Sorry, I haven't digressed, I'm working on the assumption that none of the prisoners can hear what the others are saying. Also, my solution need a little twerking, but I have now found the solution that saves at least 33 prisoners in the worst case. If a prisoner sees at least 32 red hats, then guess red, if not, then if the prisoner sees at least 32 blue hats, then guess blue, otherwise, they will definitely see at least 32 white hats, and therefore guess white. This will save at least 33 prisoners!
Edit, actually, this won't save at least 33 prisoners, the 32-32-34 case doesn't work (only that case doesn't work, it seems that its just 1 case every time that doesn't work!), I will get back to you when I have a better solution.
Trebla wrote:tomtom2357 wrote:Sorry, I haven't digressed, I'm working on the assumption that none of the prisoners can hear what the others are saying. Also, my solution need a little twerking, but I have now found the solution that saves at least 33 prisoners in the worst case. If a prisoner sees at least 32 red hats, then guess red, if not, then if the prisoner sees at least 32 blue hats, then guess blue, otherwise, they will definitely see at least 32 white hats, and therefore guess white. This will save at least 33 prisoners!
Edit, actually, this won't save at least 33 prisoners, the 32-32-34 case doesn't work (only that case doesn't work, it seems that its just 1 case every time that doesn't work!), I will get back to you when I have a better solution.
Now I follow you...
There are more cases that don't work for this plan.
Any case where the largest number of colored hats is the same.
E.g., 40 red, 40 blue, 20 white. Every Red will see 39 red and 40 blue, therefore they will guess blue. And every blue will see 40 red and 39 blue and therefore guess red. Every white will see an even number of blue/white and randomly guesses either blue or white. Number saved : 0/100
burgie12 wrote:Trebla wrote:tomtom2357 wrote:Sorry, I haven't digressed, I'm working on the assumption that none of the prisoners can hear what the others are saying. Also, my solution need a little twerking, but I have now found the solution that saves at least 33 prisoners in the worst case. If a prisoner sees at least 32 red hats, then guess red, if not, then if the prisoner sees at least 32 blue hats, then guess blue, otherwise, they will definitely see at least 32 white hats, and therefore guess white. This will save at least 33 prisoners!
Edit, actually, this won't save at least 33 prisoners, the 32-32-34 case doesn't work (only that case doesn't work, it seems that its just 1 case every time that doesn't work!), I will get back to you when I have a better solution.
Now I follow you...
There are more cases that don't work for this plan.
Any case where the largest number of colored hats is the same.
E.g., 40 red, 40 blue, 20 white. Every Red will see 39 red and 40 blue, therefore they will guess blue. And every blue will see 40 red and 39 blue and therefore guess red. Every white will see an even number of blue/white and randomly guesses either blue or white. Number saved : 0/100
No, by his rules, all 100 tourists will guess red. It's not that they guess whichever they see the most of, it's if they see some number (or greater) of red, then guess red. Otherwise, perform the same test with blue. If they still have not guessed, then they are guaranteed to guess white.
Slight nitpick, it would be a 32-32-36 case (there are 100 tourists, not 98). And, I'm not readily coming up with a strategy similar to your originally proposed solution that doesn't fall into the same trap that your current one does.
burgie12 wrote:No, by his rules, all 100 tourists will guess red. It's not that they guess whichever they see the most of, it's if they see some number (or greater) of red, then guess red. Otherwise, perform the same test with blue. If they still have not guessed, then they are guaranteed to guess white.
krogoth wrote:What if we get 3 correct guess's in a row? even if they are guess's would that be enough information to work out k?
krogoth wrote:Spoiler:
rokkshark wrote:I can think of a solution that saves at least half of the group (if they can hear what others are saying, and they are eaten immediately)
The first person asked responds with the color of the hat of the person to their right. That person remembers the color.
For everyone else, if the person to your left has already been asked, you respond with that color, if not you respond with the color of the person on your right.
Ultimately, Once they've asked 50% of the people, even assuming they were all wrong, the other 50% now know their hat color.
existential_elevator wrote:It's like a jigsaw puzzle of Hitler pissing on Mother Theresa. No individual piece is offensive, but together...
If you think hot women have it easy because everyone wants to have sex at them, you're both wrong and also the reason you're wrong.
Alfonzo227 wrote:I'm afraid I'm having a hard time following your logic in your spoiler, krogoth. What do you mean when you saykrogoth wrote:Spoiler:
Maybe you've just worked something out that I haven't seen yet, but surely our own hat is completely independent of what we see?
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