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Pirates and Laptops


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A friend told this to me, and I thought I'd share it here:

100 pirates need to allocate 100 identical laptops among them. Their democratic system works as follows:

All pirates are ranked by their seniority (all pirates have different ranks). First, the most senior pirate proposes a plan that states exactly how many laptops each pirate gets. The 100 pirates vote on the plan and it passes at least half of the pirates vote for it. If it passes, all pirates take their laptops and go home. If it fails, the one who proposed the plan (the most senior pirate in this case) is killed, and the second most senior pirate takes his place and proposes his plan.The same process is repeated in the order of seniority until someone's plan is passed.

Assume every pirate makes his decision based on the following priorities:

1. He doesn't want to die.

2. Given he's not going to die, he would prefer to get as many laptops as possible.

3. Given he's going to get the same number of laptops, he would prefer as many other pirates to die as possible.

Also assume every pirate is logical, rational, and selfish (wants as many laptops as possible and doesn't care what anyone else gets as long as it doesn't affect him) and knows everyone else is the same. What will happen? i.e. whose proposal will be passed and what is the proposal?

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I would say that the second to last pirates offer is the one who is accepted and he gets 100 Laptops. a 50% vote passes. The last 2 pirates are the only ones that can assure that outcome. So lets start at the begining... The first pirate could offer 2 laptops to the top 50 pirates but it is assumed that the other 49 pirates in the top 50 want as many laptops as they can get and with pirates ahead of them out of the way they can get more. During the vote only one pirate is in immediate danger so he is forced to make something that sounds reasonable but still gets him more than 1 laptop because he is greedy. Every pirate below him is forced to think that they can get more because there are less pirates.... Alll the way down to the last two pirates. The second to last pirate knows that the 50% vote passes and he knows how he will vote so he takes 100 laptops and the lonelyest pirate can't do anything about it.

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I would say that the second to last pirates offer is the one who is accepted and he gets 100 Laptops. a 50% vote passes. The last 2 pirates are the only ones that can assure that outcome. So lets start at the begining... The first pirate could offer 2 laptops to the top 50 pirates but it is assumed that the other 49 pirates in the top 50 want as many laptops as they can get and with pirates ahead of them out of the way they can get more. During the vote only one pirate is in immediate danger so he is forced to make something that sounds reasonable but still gets him more than 1 laptop because he is greedy. Every pirate below him is forced to think that they can get more because there are less pirates.... Alll the way down to the last two pirates. The second to last pirate knows that the 50% vote passes and he knows how he will vote so he takes 100 laptops and the lonelyest pirate can't do anything about it.

I feel like not posting the solution, if only because I'm curious to see what people invent. Here's a hint, though...

If you (and, by extension, all of the pirates) know what will happen when there are only 2 pirates left, what should the third-to-last pirate do?

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I would of thought that if the pirates knew what happened to the last 2 pirates, if you were the third-to-last pirate, surely you would order the death of one of the other pirates or kill him yourself and take it from there!

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If there was only 2 pirates were, first one proposes 100:0 and he gets all laptop. If there were 3

pirates, first pirate proposes 99:0:1 third one will accept it (if no this proposal couldn't be approved and first will be killed and second one proposes 100:0 and he gets all laptop leaving third with nothing). If there were 4 pirates first pirate proposes 99:0:1:0 and third support him (or first one dies and he gets nothing). If there were 5 pirates first one proposes 98:0:1:0:1 and gets third and fifth's support. And so on; if 100 pirates were present, first pirates propose 51:0:1:0......................:1:0:1:0 and this will won half the vote. (this is my first attempt and may have errors )

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Question: This is only my second topic posted, am I supposed to say when someone gets the answer?

I don't think there is a hard set rule on this but most posters do monitor, sometimes giving additional clues, answering questions and giving credit to progress toward the answer as well as congtaatulations to the solver.
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@Blavek...

Blavek, In light of Amiab's solution, which SeaCalMaster's hint addresses without providing the complete solution, I must concede that MY solution misses the essential quality of the greed factor.

However, I disagree that the final two pirates are left with a 100-0 solution. Again, see SeaCalMaster's hint (which I considered but did not pursue to the final logical solution presented by Amiab). The last pirate (call him number 1) would see what would happen if there were only two remaining, and would therefore accept #3's offer of 99-0-1. Similarly then, we begin climbing the logic ladder to #4, #5, etc. I think SeaCalMaster deserves a nod of recognition for the hint which clearly showed he knew the answer.

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I think the solution is: only 3 pirates left, the elder one will get 0 laptops, because his priority is not to die over having Laptop, then he propose all the 100 laptops to the second pirate and 0 to the third, that way he gets the most votes and doesn't die. This could not apply for 4 or more because they would have to split the laptops and if they split them then they would prefer to kill the first pirate. What do you Think?

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I think the solution is: only 3 pirates left, the elder one will get 0 laptops, because his priority is not to die over having Laptop, then he propose all the 100 laptops to the second pirate and 0 to the third, that way he gets the most votes and doesn't die. This could not apply for 4 or more because they would have to split the laptops and if they split them then they would prefer to kill the first pirate. What do you Think?

Actually I think 99 - 0 - 1 would be the solition

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On 5/4/2013 at 2:03 AM, petar said:

I think the solution is: only 3 pirates left, the elder one will get 0 laptops, because his priority is not to die over having Laptop, then he propose all the 100 laptops to the second pirate and 0 to the third, that way he gets the most votes and doesn't die. This could not apply for 4 or more because they would have to split the laptops and if they split them then they would prefer to kill the first pirate. What do you Think?

NO as 2nd pirate will also get 100 laptops if he kill 3rd one . so if he is getting same no of laptops then he will kill 3rd one. and finally will get 100 laptops.

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