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Topic: who is most possible to survive (Read 2478 times) |
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iloveriddle
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who is most possible to survive
« on: Jul 28th, 2004, 3:11pm » |
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I heard of this riddle from a friend. It's interesting.
5 prisoners are going to take green beans from a bag with 100 green beans. They will do it one by one. No communication is allowed between them. but they can count the green beans left in the bag. They must take out at least one bean each time. Those who take the largest and the smallest number of beans will die.
Question: who is most possible to survive.
hints: 1. they are all smart people.
2. they will try to survive first and then try to kill more people.
3. they donot need to take out all 100 beans
4. all prisoners who have the same largest or smallest number will die.
The riddle doesnot mention the condition when there is no bean left in the bag. I assume the prisoner get 0 beans in that case.
I had my answer. But I would like to hear yours first~
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towr
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Re: who is most possible to survive
« Reply #1 on: Jul 29th, 2004, 1:39am » |
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::If the first prisoner would die anyway, he'll just take all 100 beans at the start just so everyone else will die as well.
So either prisoner one doesn't die, or everyone dies.
If prisoner one can survive he'll take x < 50 beans, and prisoner two can take equally many. Of course if they are to survive more then x beans must be left, so prisoner 3 also takes x, and 4 and 5.
Suffice it to say, eventually they all die..::
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Eigenray
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Re: who is most possible to survive
« Reply #2 on: Jul 29th, 2004, 5:56am » |
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Some observations:
If prisoner 1 takes any more than 20 beans, then prisoner 2 lives.
If prisoner 4 finds himself with between 1 and 39 beans left to pick from, then he lives.
And if prisoner 5 finds himself with no more than 20 left, he should take them all.
It does seem like they all die though, because each would probably want to take the average of however many beans the prisoners before him took.
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beancounter
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Re: who is most possible to survive
« Reply #3 on: Jul 29th, 2004, 7:11am » |
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It does seem like player one will just kill them all. Maybe someone can suggest a rule variation that will save this puzzle. For instance, multiple rounds until the beans are gone and no ties allowed. You can probably think of some rule that will give this one a reason to be in Hard.
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Leo Broukhis
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Re: who is most possible to survive
« Reply #4 on: Jul 29th, 2004, 10:40am » |
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Are prisoners 2 to 5 told their ranks? I don't think so, therefore they could not reliably take the average of however many beans the prisoners before them took.
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Nigel_Parsons
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Re: who is most possible to survive
« Reply #5 on: Aug 1st, 2004, 10:08am » |
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First, “They must take out at least one bean each time”, this seems to preclude the possibility that the later prisoners will find the bag empty.
If (with no beans in the bag) remaining prisoners take no beans, does this count as “taking the smallest number of beans”? i.e. can taking no beans be taking a number of beans?
If a lack of beans is not “a number of beans” then prisoner 1 taking 100 beans will only mean his own death. But apart from survival (the first aim of the puzzle) the aim is to try to kill more people, so prisoner 1 will take less than 100 beans to increase the death toll.
Just semantics, but semantics count...1,2,3...
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Nigel_Parsons
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Re: who is most possible to survive
« Reply #6 on: Aug 1st, 2004, 10:46am » |
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Allowing for prisoners to die by taking No Beans:
Clearly any prisoner taking 50 or more beans dies.
So, Prisoner 1 will take less than 50 beans.
If P1 takes between 21 & 49 beans P2 is left with 51 to 79 beans and can take one bean less than P1, thus ensuring his own survival as at least one of the remaining prisoners cannot take as many beans as he has. P2 will not take the same number of beans as P1 as this gives him the chance of dying with the equal greatest number of beans. Of course, as the others are also smart, P2 will not wish to make his actions too obvious and so can take any number of beans which is greater than ¼ of the remainder and still guarantee his own safety.
As they are all smart people, P3 will have already worked out the above possibilities. He can now aim to take the number of beans matching P2’s take (thus not being the highest, nor the lowest, which must be P4 or P5 or Both). However, at this point P3 does not need to maximise the beans taken as he wishes to increase the chance of killing both P4 & P5. P3 can take any number of beans greater than 1/3 of the remainder and less than ½ of those already taken. He must, of course take an odd number (to return the total to even) to try to kill both.
If P1 takes 20 or less beans then P2 can match this exactly: taking more than P1 will allow P3 to get an average and survive quite possibly condemning P2 as having the most: similarly taking less than P1 allows the other prisoners to take the average condemning P2 with the least. So all prisoners take the same number of beans & die.
So prisoner1 has no survival strategy, and as already stated by others will condemn all 5 prisoners.
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BEANCOUNTER
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Re: who is most possible to survive
« Reply #7 on: Aug 1st, 2004, 6:47pm » |
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If no one knows his position, then the first prisoner counts 100 beans and knows he's first. There's no perfect choice ( or else every player will be able to figure out the whole situation, and we're back to a game where they know their position and they all die). Let's suppose he takes 10. #2 can guess he's probably #2, possibly #3. He wants an average choice. Let's say he goes between the two and takes 8. #3 finds 18 missing. He wants the average, but is that 18, 9, 6, or 4.5? He takes a mid number, 9 eg. #4 finds 27 missing, #5 will find somewhere around 35-40 missing. There's a good chance one of these guys will take more than 10, in case he's the second or third, and all the picks are going to be high. So #1 does seem to have a chance to survive.
Let's say #1 takes 20. #2's not so sure of his position now, but if he guesses he's late in the rotation, he'll go low, and no one will end up going over 20 and #1 dies.
Let's say #1 takes 4. #2 is sure he's #2--you wouldn't have 2 people taking 2. If I were him I'd take 5 or 6, to get the numbers rising fast enough to encourage the late guesses to go up into double digits. So #1 dies.
I'm thinking #1's strategy is to be unpredictable, but to be just high enough to make #2 unsure of where he stands. sounds like 8-12 might be the middle of the range to me. That's not really enough to keep #2 guessing, but it's better than certain death.
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