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Topic: Who will survive among these prisoners (Read 2441 times) |
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wonderful
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Who will survive among these prisoners
« on: Mar 26th, 2008, 7:48pm » |
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5 very smart prisoners play the following survival game. There are 100 diamonds in a box. Each prisoner will in turn pick up a certain number of diamonds once. The prisoner can pick as many as he wants. However, he would be killed if his number of diamonds is the greatest, or smallest, or equal to other(s).
The survived prisoner will keep all the diamonds he picked. There is no communication among the prisoners. The prisoner can count the number of remaining diamonds in the box before picking his own diamonds. Assume also that the prisoners are selfish.
Who will survive?
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| « Last Edit: Mar 27th, 2008, 10:34pm by wonderful » |
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Grimbal
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Re: Who will survive among these prisoners
« Reply #1 on: Mar 27th, 2008, 3:28am » |
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Suppose the first 3 take 20 and the fourth takes 19.
Then the last has the choice of taking 19, 20 or 21.
In every case he will die. But by choosing 21 he will save 3 people. What will he do?
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gotit
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Re: Who will survive among these prisoners
« Reply #2 on: Mar 27th, 2008, 5:17am » |
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I think the first prisoner will always survive if he makes sure that he picks 1/n of the diamonds, where n is the number of prisoners alive.
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towr
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Re: Who will survive among these prisoners
« Reply #3 on: Mar 27th, 2008, 5:41am » |
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Note in the opening post: "However, he would be killed if his number of diamonds is the greatest, or smallest, or equal to other(s)."
So if the first three people pick 20, all three die.
If the first person picks 20, he may still die if anyone else picks 20 (or if all the others pick less.)
To survive people need a unique number of diamonds which is neither the maximum nor minimum picked.
I'm wondering to what extent the prisoner scare whether the others live. If they are doomed to die, will they drag as many with them as they can?
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| « Last Edit: Mar 27th, 2008, 5:43am by towr » |
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gotit
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Re: Who will survive among these prisoners
« Reply #4 on: Mar 27th, 2008, 7:18am » |
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My assumption is that that if two or more prisoners pick up the same number of diamonds, then the one who was the first to pick would not be killed. If my assumption is wrong, then there may be a case when all prisoners pick up the same number of diamonds and all die. In such a situation, the question that who will be the survivor in the end does not make any sense.
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tohuvabohu
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Re: Who will survive among these prisoners
« Reply #5 on: Mar 27th, 2008, 7:49am » |
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I think "equal to the others" means all that are equal will die. Are they forced to play by an evil prison warden? Or is this just how they decided to divide the diamonds. Because if it's the latter, then they're not all going to kill themselves. At least one has to get the diamonds.
Anyway, if the first person takes 20, then second will count 80 and won't pick 20. He'll pick either 19 or 21, I think. He won't choose a number already taken. He won't leave a gap that later players could fill in, unless, maybe, he know's he's going to die and he's going to trick someone into picking an equal. But if later guessers correctly assume all earlier guessers have chosen adjacent numbers, then they know what numbers to avoid, and you won't have any equal guesses.
So we need to know if a doomed prisoner would take the rest down with him. We need to know if there's any guaranteed survivor.
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temporary
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Re: Who will survive among these prisoners
« Reply #6 on: Mar 27th, 2008, 7:31pm » |
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Anyway, why would each take 20? Maybe they can take more or they need to take less. Think of it like the pirate puzzle, the 1st pirate took 98 of the 100 coins and lived. Not that it applies in exactly the same way either, since that would be death.
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| « Last Edit: Apr 27th, 2008, 12:52pm by temporary » |
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wonderful
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Re: Who will survive among these prisoners
« Reply #7 on: Mar 27th, 2008, 10:34pm » |
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Great discussion as always! Thanks temporary for pointing that out. To answer tohuvabohu question regarding whether the prisoners are forced to play the game, one can think of the following story:
There was a king with many prisoners. One day he asked the five smartest prisoners to play the game. The rule of the game is mentioned above.
Have A Great Day!
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rmsgrey
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Re: Who will survive among these prisoners
« Reply #8 on: Mar 28th, 2008, 10:08am » |
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Looking at the 3 prisoner, 60 diamond variant, the outcome depends on what a doomed third prisoner is expected to do.
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Ajax
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Re: Who will survive among these prisoners
« Reply #9 on: Apr 7th, 2008, 12:56am » |
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I don't know if it has already has been written (I didn't notice) but it is not certain that the sum of the diamonds taken by all five prisoners must equal 100. Even if they are very greedy, they don't want to lose their heads.
One or two thoughts:
Prisoners 1,2,3 will take such numbers of diamonds that they will be consecutive (1st -> a, 2nd -> a-1 or a+1, 3rd -> a-2, or a+2), so as not to allow any of the rest two prisoners take diamonds and be spared.
Thus for the three first prisoners, they will take 3a diamonds <100
Moreover, none of them would like to lose his head for being greedy (taking the most diamonds), so they will let more than the most taken by anyone of them.
And one last thing, if the first prisoner realized that he had no chance surviving, he would take all 100 of the diamonds and doom all the group 
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tohuvabohu
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Re: Who will survive among these prisoners
« Reply #10 on: Apr 7th, 2008, 11:02am » |
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The only reason for anyone selecting a number outside the range of previous guesses is if he believes someone else will go farther outside the range. So the last player has no reason to do so; he might as well pick a central number and just hope it wasn't picked. That philosophy then extends to each earlier player, making the whole game hopeless.
Unless: if a player knows he is doomed and he is no longer motivated by any attempt at self-preservation, maybe greed will simply make him take all the remaining diamonds. Then, as long as the first three players make choices that give player D no winning strategy, then player D will take all the remaining diamonds, player E will get 0, and the first 3 live.
Assume, D knows E can't win under any circumstances, so E's strategy will always be to take all diamonds if there are any left. If the remaining diamonds was low enough so that D could take more than half and still be under the first three guesses, he would have a winning play. If the first three were 22, 21, and 20, or anything higher than that, D could take 19, leaving only 18 or less. So the first three guesses should be 21, 20, and 19 (with A taking the most). D can't win, so he'll take 40 and E gets 0.
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pex
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Re: Who will survive among these prisoners
« Reply #11 on: Apr 7th, 2008, 11:43am » |
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What's the point of being greedy when you know you won't live to enjoy what you took?
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Ajax
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Re: Who will survive among these prisoners
« Reply #12 on: Apr 7th, 2008, 11:46pm » |
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on Apr 7th, 2008, 11:43am, pex wrote:| What's the point of being greedy when you know you won't live to enjoy what you took? |
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Maybe a last "diamond-shower!?" 
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| « Last Edit: Apr 7th, 2008, 11:47pm by Ajax » |
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Grimbal
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Re: Who will survive among these prisoners
« Reply #13 on: Apr 8th, 2008, 12:51am » |
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To buy your freedom.
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temporary
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Re: Who will survive among these prisoners
« Reply #14 on: Apr 9th, 2008, 9:46pm » |
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It means that if he can survive with x to y gems, he will take y gems. Other than that, their strategy is assumed to be optimal. I think they are all doomed.
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pscoe2
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Re: Who will survive among these prisoners
« Reply #15 on: Apr 12th, 2008, 10:49am » |
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acc to me whtever the first picks...the second will know(as he counts the remaining and knows thr wr a total of 100)
the second will definitely not pick the same as the first one...
say the first one picked 20 and the second picked 18...
when the third prisoner comes and finds tht 62 diamonds are left he knows the others hv picked up at an avg of 19 but certainly not 19 each as he knows the others are smart...
he picks 19 to be sure of survival...
but if the case wud hv been tht the 2nd prisoner picks up 19...the avg becomes 19.5 and the third prisoner knows he has to pic 21 or 18...this is bcoz he knows the other 2 are smart and the diff btw the diamonds of the other 2 must be min....
now its his luck...if the fourth or fifth does not pick 21 but pick smthing bigger...
if nyone cud think on these lines and suggest a proper ans...
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Chekov
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Re: Who will survive among these prisoners
« Reply #16 on: Apr 16th, 2008, 4:57am » |
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In total there can be 3 prisoners who can survive
If we look at the first pick
say 10
2nd 1 will not get 10 but go 1 above or below. then the 3rd will see and count. the average can show him its either higher or lower.
the 4th then can get the middle again. he will try to goad the 5th into goin outside the range so he survives
the 5th will always die. whether 2 high or 2 low.
this is all in a situation that they will not pick some1 else to kill him.
if they do want to kill some ppl. then the situation can be that the choice of the 4th is binding. he is either 2 high or 2 low.
let me put it this way:
1 10
2 11
3 9
4 either 12 or 8
5 kills 1 because he picks 10
in that way either 2 or 3 will survive with that kind of number. depends on the choice of the 5th.
if the 1st guy gets 20
it can be like this:
1 20
2 21
3 19
4 either 22 or 18
5 depends if he wants to kill so number 1 wont get diamonds. then he picks 20. or he can kill 3 with 19 or 18 to "possible kill" 4
but thats up to chance
anyway
those were my 5 cents
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rmsgrey
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Re: Who will survive among these prisoners
« Reply #17 on: Apr 19th, 2008, 11:37am » |
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If we assume that the first four prisoners pick consecutive numbers, then the fifth prisoner will have no way to survive. If the fifth prisoner, knowing he can't survive, plays in the middle of the run to try to take out as many other prisoners as possible, then the fourth prisoner can't survive either. This also gives the fifth prisoner his best chance of survival - if the other 4 prisoners happen not to have made a solid run after all, then finding one of the gaps in the middle will save him - so it's not an unreasonable strategy for him.
If the fourth prisoner knows that the fifth prisoner will play that way, then he also knows that, if the other three have formed a consecutive run, then he has no chance of survival, which affects his optimal strategy.
On the other hand, if it's known that the fifth prisoner, when facing certain death, will play to minimise the total body count, it is in everyone else's best interest to play for a consecutive run.
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temporary
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Re: Who will survive among these prisoners
« Reply #18 on: Apr 27th, 2008, 12:56pm » |
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Yet if any one of them was doomed, they could take anyone(before or after them) down.
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