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Topic: forcefield detainment (Read 2474 times) |
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A group of prisoners are trapped in a forcefield. These prisoners are perfectly brave, meaning that they would attempt an escape on any positive probability of success. The prisoners are monitored by a guard who has only one bullet in his gun, but who also has perfect marksmanship skills (he never misses). A maintenance technician needs to tune up the forcefield generator, and so for one second, the forcefield is released. How can the guard still keep all the prisoners detained?
i dont get this puzzle much, can someone help me? im really lost here
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towr
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Re: forcefield detainment
« Reply #1 on: Apr 4th, 2005, 4:23am » |
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The guard has to say or do something which convinces the prisoners that individually they have no chance of escaping.
(Obviously after one has been shot, the rest can escape safely, but none of the prisoners wants to be the one shot.)
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| « Last Edit: Apr 4th, 2005, 4:25am by towr » |
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Grimbal
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Re: forcefield detainment
« Reply #2 on: Apr 4th, 2005, 4:59am » |
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I think you have to suppose the prisonners will prefer a low probability to be killed than a high one.
If not, if everybody tries his best to escape, he still has only one chance in N to be killed. That won't stop him.
But let's try. Put the prisonners in a circle around a pillar. Everybody holds his left and right neighbour. The first to release either his left or right neighbour gets shot.
But what if one wrestles out of his neighbour's grip? The neighbour gets killed and everybody else escapes.
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Icarus
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Re: forcefield detainment
« Reply #3 on: Apr 4th, 2005, 5:59pm » |
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Landslides arise when soil or rock is on the brink of falling. Though the potential is there, the slide itself is often set off by some small event that is insignificant by itself. Perhaps by a single small stone shifting. You may not be able to stop the landslide once it starts, but stopping that single stone from shifting is fairly easy, if you can only find a way to identify it before the slide starts.
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SWF
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Re: forcefield detainment
« Reply #4 on: Apr 4th, 2005, 10:05pm » |
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There is already a thread for this problem here.
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