可预测年薪超过8万美金的“趣味数学题”你能解吗？
- 主题发起人 Aeoliao
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最初由 Leo1976 发布
#1要利益最大化，就意味风险最大化，只要2345相互有一点信任或者仇恨，或者其他影响因素，#1死定了，无论怎么分配都不安全。
如果在现实生活中，不能完全把握关系和信息，如果#1能保持沉默或者弃权，最佳选择就是不表态了，等#2上来表演，等#2犯傻找死，或者看出2345之间的关系。如果#2也聪明，弃权，直到#5弃权，最后结果也就是都不愿意先表态，只好平均分配了。 好像目前中国的权利分配就是如此阿
You got my point.
It is a kind of wisdom to give up something.
要想得大便宜，必须要做好吃亏的准备。 :lol: :lol: :lol: :lol:
qwer
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应该是2号对。
典型的西方经济学问题。
每个人都理智的追求自己利益的最大化。
这题应这么想。
如果前4个海盗都死了，5号的分配方案是100个给自己。他希望前4个都死了才好。
如果前3个海盗都死了，4号的分配方案是都给5号，因为他不同意自己得死。
如果前2个海盗都死了，3号的分配方案是给4号1个，要好过自己死了，他1个都拿不着。不给5号，不怕他反对，给自己99个。
如果第1个海盗死了，2号的分配方案是先杀1个，当然是最富有的3号，1个都不给。
给5号1个，给4号2个，他们2个会同意，97个给自己。
所以头号海盗的分配方案是：97，0，1，0，2
给下一方案中最富有的2个（2号4号）0，
另外2个，稍微贿赂一下，每人加一个，免的他们算不清楚，5号2个，3号1个，
97个给自己。数数的时候要小心2号和4号，他们会造反。
对不对呀，8万快不要紧，命要紧。
典型的西方经济学问题。
每个人都理智的追求自己利益的最大化。
这题应这么想。
如果前4个海盗都死了，5号的分配方案是100个给自己。他希望前4个都死了才好。
如果前3个海盗都死了，4号的分配方案是都给5号，因为他不同意自己得死。
如果前2个海盗都死了，3号的分配方案是给4号1个，要好过自己死了，他1个都拿不着。不给5号，不怕他反对，给自己99个。
如果第1个海盗死了，2号的分配方案是先杀1个，当然是最富有的3号，1个都不给。
给5号1个，给4号2个，他们2个会同意，97个给自己。
所以头号海盗的分配方案是：97，0，1，0，2
给下一方案中最富有的2个（2号4号）0，
另外2个，稍微贿赂一下，每人加一个，免的他们算不清楚，5号2个，3号1个，
97个给自己。数数的时候要小心2号和4号，他们会造反。
对不对呀，8万快不要紧，命要紧。
无业游民
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您老兄很笨，但是没有把你的笨思路解释清楚。俺觉得老兄的解至少可以骗骗人的无数的解之中的一个。俺自己和老兄笨的相似（还有师长），所以越俎代庖帮你解释。
俺也没把这个问题当作纯数学题。不过如果把他当数学题做的的话，用以下这种繁琐罗嗦的方法来思考，可以使得所有的笨蛋包括俺都能懂：
1.假定俺是1号海盗，想要找到一种方案，使得俺的到的利益最大。
2.判定利益的大小看来是这样的：俺得100颗>得99>...1>0>死亡。（注：有少数聪明人认为死亡不在“利益”的可能性当中）
3.先定义死亡为-1（其实定义为任何负数实质都一样）。在从-1到100的可能值当中不管俺主观上希望得到多大的值，俺实际可能得到只有-1和0，这是因为：
a.其他4个海盗都没有感情而只又冷冰冰的理性的算计，试图通过各种手段使他们的利益最大化。俺知道这一点，这也是俺对他们的算计。
b.俺选任何除了0以外的其他期望值比如10颗，其他人投票让俺喂鱼就能使他们的利益更大一点点。换句话说，如果俺的方案中俺的所得不是0的话，俺实际能得的只会是-1。
4.俺也是个只懂得冷冰冰的算计的海盗，俺不在乎别人得到多少利益，只想让自己得益最多。在0和-1之中，不太显而易见的是，0要大一点。
所以出于追求俺的最大利益考虑，俺要选你的这种方案。聪明人会问即使得0，他们也可能投票要俺死。俺天真地认为他们只在乎自己的利益而没有感情(包括友情和仇恨）。俺已经得0了，他们无法通过让俺死来得到更大的利益。至于他们是否残忍仇恨俺是否要向先帝学习之类，都是另外的可以越扯越远的话题。无论如何即使有可能死，和必定死相比前者更好些。
师长，我们的差别在于：俺没有无我境界，只有出于理性的自私。以上3b是唯一还有漏洞的一条，但是解释起来太罗嗦没人要看
泛蓝
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ORINGINAL TEXT
(This is the original text before it was edited by Scientific American)
The logic of mathematics sometimes leads to apparently bizarre conclusions. The rule here is that provided the logicdoesn't have holes in it, then the conclusions are sound. If they conflict with your intuition, then either your intuition is okay in a different context --- but not the one underconsideration --- or you need to refine your intuition.In September 1998 Stephen M. Omohundro sent me a puzzle that falls into exactly this category. The puzzle was invented by Steven Landsberg (U Illinois, Urbana-Champaign) ,and Omohundro came up with a variant in which the logic becomes surprisingly convoluted and the conclusions are remarkable.
First, the original version of the puzzle.
Ten pirates have got their hands on a hoard of 100 goldpieces, and wish to divide the loot between them. They are democratic pirates, in their own way, and it is their custom to make such divisions in the following manner. The fiercest pirate makes a proposal about the division, and everybody votes --- one vote each including the proposer. If 50% or more are in favour, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard and the procedure is repeated with the next fiercest pirate.
All the pirates enjoy throwing people overboard, but given the choice they prefer hard cash. They dislike being thrown overboard themselves. All pirates are rational, know that the other pirates are rational, know that they know that... and so on. Unlike other puzzles with an apparently similar form (see Monks, blobs, and common knowledge, Scientific American August 1998, 96-97), this one does not hinge upon the revelation of some piece of 'common knowledge'. Any common knowledge is already commonly known. Moreover, no two pirates are equally fierce, so there is a precise 'pecking order' --- and it is known to them all. Finally: gold pieces are indivisible and arrangements to share pieces are not permitted (since no pirate trusts his fellows to stick to such an arrangement). It's every man for himself.
Which proposal will maximise the fiercest pirate's gain?
Omohundro's contribution is to ask the same question, but with 500 pirates instead of 10. (Still 100 gold pieces.) For convenience, number the pirates from 1 upwards in order of meekness, so that the least fierce is number 1, the next least fierce number 2, and so on. The fiercest pirate thus gets the biggest number, and proposals proceed in reverse numerical order from the top down.
Following Omohundro, I'm going to try to convince you that the answer to the 10-pirate version of the problem is this: Pirate 10 proposes to keep 96 gold pieces for himself, to give one gold piece to each of pirates 8, 6, 4, and 2, and none to the odd-numbered pirates.
In contrast, the answer to the 500-pirate version involves the first 44 pirates being thrown overboard, after which pirate number 456... but I'm in danger of giving too much away, so you'll have to wait for the rest.
The secret to analysing all such games of strategy is to work backwards from the end. At the end, you know which decisions are good and which bad. Having established that, you can transfer that knowledge to the next-to-last decision, then the next-to-next-to-last, and so on. Starting from the front, in the order in which the decisions are actually taken, doesn't get you very far. The reason is that strategic decisions are all about 'What will the next person do if I do this...?' so the decisions that follow yours are important. The ones that come before yours aren't, because you can't do anything about them anyway.
Bearing this in mind, the place to start is when (if!) the game gets down to just two pirates, P1 and P2. The fiercest pirate is P2, and (if the game ever gets this far) his optimal decision is obvious: propose 100 pieces for himself and none for P1. His own vote is 50% of the total, so it wins.
Now add in pirate P3. Pirate P1 knows --- and P3 knows that he knows --- that if P3's proposal is voted down, P1 will get nothing. So P1 will therefore vote in favour of anything that P3 proposes, provided it yields him more than nothing. P3 therefore uses as little as possible of the gold to bribe P1,leading to the following proposal: 99 to P3, 0 to P2, and 1 to P1. Let's write it out like this:
P1 P2 P3
1 0 99
The thought-processes of P4 are similar. He needs 50% of the vote, so again he needs to bring exactly one other pirate on board. The minimum bribe he can use is one gold piece, and he can offer this to P2 since P2 will get nothing if P4's proposal fails and P3's is voted on. So the proposed allocation here becomes
P1 P2 P3 P4
0 1 0 99
The thought-processes of P5 are slightly more subtle. He needs to bribe two pirates. The minimum bribe he can use is two gold pieces, and the unique way he can succeed with this number is to propose the allocation
P1 P2 P3 P4 P5
1 0 1 0 98
The analysis proceeds in the same manner, with each proposal being uniquely prescribed by giving the proposer the maximum possible subject to ensuring a favourable vote, until we get to the tenth pirate and find the allocation
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10
0 1 0 1 0 1 0 1 0 96
This is the proposal I indicated earlier, and it solves the 10-pirate puzzle.
Now comes Omohundro's twist: what if there are (a lot) more pirates? Clearly the same pattern persists --- for a while. In fact it persists up to the 200th pirate. P200 will offer nothing to the odd numbered pirates P1-P199, and one gold piece to each of the even-numbered pirates P2-P198, leaving one for himself.
However, we're trying to find the strategy for P500. At first sight, the argument breaks down at P200, because P201 has run out of bribes. However, he still has a vested interest in not being thrown overboard, so he can propose to take nothing himself:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201
1 0 1 0 ... 1 0 1 0 0
This opens up a new phase of the strategy, because pirate P202 knows that P201 has to accept nothing, or be thrown overboard. So he can count on P201's vote. However, P202 also is forced to accept nothing. He must use the entire 100 gold pieces to bribe 100 pirates --- and these must be among those who would get zero according to P201's proposal. Since there are 101 such pirates, P202's proposal is no longer unique. Let's use a star * to mark pirates who might get something from P202's proposal:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202
0 * 0 * ... 0 * 0 * * 0
At this point a further consideration comes into force. Pirates have to think about how a pirate who has some chance of receiving a gold piece will react if he is definitely offered a gold piece. This depends on how much money, on average, he is willing to sacrifice for the fun of throwing somebody overboard. The puzzle doesn't specify this, so it is reasonable to assume that the pirates don't know it. This means that the bribe might succeed, so it is rational to offer a definite piece to a pirate who only has a chance of one later. As it turns out, things are more satisfactory: there are enough definite 0's in each round for pirates from now on to offer the bribes only to them. Indeed, as Peter Norvig pointed out, there will always be 100 allocations of 0 among pirates P1-P200, so we may describe a solution in which bribes are offered only to (some of) them.
You may wonder why I'm bothering about all this, since clearly pirate 203 doesn't have enough cash available to bribe enough pirates to vote for his proposal, whatever it is. This is true, and P203 goes overboard whatever he proposes --- a result that we symbolize by a cross X. We write ? to show that his choices are irrelevant, giving the situation:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202 P203
? ? ? ? ... ? ? ? ? ? ? X
Even though P203 is destined to walk the plank, this doesn't mean that he plays no part in the proceedings. On the contrary, P204 now knows that P203's sole aim in life is to avoid having to propose a division of the spoils. So P204 can count on P203's vote, whatever P204 proposes. Now P204 just squeaks home --- he can count on his own vote, P203's vote, and 100 others from bribes of one gold coin: 102 votes in all, the necessary 50%. So P204 will propose
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202 P203 P204
* 0 * 0 ... * 0 * 0 * * 0 0
What of P205? He is not so fortunate! He cannot count on the votes of P203 or P204: if they vote against him they have the fun of throwing him overboard and can still save themselves. So P205 gets a X. So does P206 --- he can be sure of P205's vote, but that's not enough. In the same way, P207 needs 104 votes --- 3 plus his own plus 100 from bribes. He can get the votes of P205 and P206, but he needs one more and it's not available. So P207 gets a X as well.
P208 is more fortunate. He also needs 104 votes, but P205, P206, and P207 will vote for him! Add in his own vote and 100 bribes and he's in business. He must offer bribes to those pirates that would get 0 according to P204's proposal:
P1 P2 P3 P4 ... P198 P199 P200 P201 P202 P203 P204 P205 P206 P207 P208
0 * 0 * ... * 0 * 0 0 * * 0 0 0 0
Now a new pattern has set in, and it continues indefinitely. Pirates who make proposals (always to give themselves nothing and to bribe 100 of the first 200 pirates) are separated from each other by ever-longer sequences of pirates who will be thrown overboard no matter what proposal they make --- and whose vote is therefore ensured for any fiercer pirate's proposal. The pirates who avoid this fate are numbers P201, P202, P204, P208, P216, P232, P264, P328, P456,... and so on --- 200 plus a power of 2.
It remains to work out who are the lucky recipients of the bribes, just to make sure they will accept them. As I said, the solution is not unique, but one way to do this is for P201 to offer bribes to the odd-numbered pirates P1-P199, for P2 to offer bribes to the even-numbered pirates P2-P200, then P204 to the odd numbers, P208 to the evens, and so on, alternating from odd to even and back.
At any rate, we conclude that with 500 pirates and optimal strategy, the first 44 pirates are thrown overboard, and then P456 offers one gold piece to each of P1, P3, P5 ..., P199. Thanks to their democratic system, the pirates have arranged their affairs so that the very fierce ones mostly get thrown overboard, and at best can consider themselves lucky to escape with none of the loot. Only the 200 meekest pirates can expect anything, and only half of them.
Truly, the meek shall inherit the worth.
(This is the original text before it was edited by Scientific American)
The logic of mathematics sometimes leads to apparently bizarre conclusions. The rule here is that provided the logicdoesn't have holes in it, then the conclusions are sound. If they conflict with your intuition, then either your intuition is okay in a different context --- but not the one underconsideration --- or you need to refine your intuition.In September 1998 Stephen M. Omohundro sent me a puzzle that falls into exactly this category. The puzzle was invented by Steven Landsberg (U Illinois, Urbana-Champaign) ,and Omohundro came up with a variant in which the logic becomes surprisingly convoluted and the conclusions are remarkable.
First, the original version of the puzzle.
Ten pirates have got their hands on a hoard of 100 goldpieces, and wish to divide the loot between them. They are democratic pirates, in their own way, and it is their custom to make such divisions in the following manner. The fiercest pirate makes a proposal about the division, and everybody votes --- one vote each including the proposer. If 50% or more are in favour, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard and the procedure is repeated with the next fiercest pirate.
All the pirates enjoy throwing people overboard, but given the choice they prefer hard cash. They dislike being thrown overboard themselves. All pirates are rational, know that the other pirates are rational, know that they know that... and so on. Unlike other puzzles with an apparently similar form (see Monks, blobs, and common knowledge, Scientific American August 1998, 96-97), this one does not hinge upon the revelation of some piece of 'common knowledge'. Any common knowledge is already commonly known. Moreover, no two pirates are equally fierce, so there is a precise 'pecking order' --- and it is known to them all. Finally: gold pieces are indivisible and arrangements to share pieces are not permitted (since no pirate trusts his fellows to stick to such an arrangement). It's every man for himself.
Which proposal will maximise the fiercest pirate's gain?
Omohundro's contribution is to ask the same question, but with 500 pirates instead of 10. (Still 100 gold pieces.) For convenience, number the pirates from 1 upwards in order of meekness, so that the least fierce is number 1, the next least fierce number 2, and so on. The fiercest pirate thus gets the biggest number, and proposals proceed in reverse numerical order from the top down.
Following Omohundro, I'm going to try to convince you that the answer to the 10-pirate version of the problem is this: Pirate 10 proposes to keep 96 gold pieces for himself, to give one gold piece to each of pirates 8, 6, 4, and 2, and none to the odd-numbered pirates.
In contrast, the answer to the 500-pirate version involves the first 44 pirates being thrown overboard, after which pirate number 456... but I'm in danger of giving too much away, so you'll have to wait for the rest.
The secret to analysing all such games of strategy is to work backwards from the end. At the end, you know which decisions are good and which bad. Having established that, you can transfer that knowledge to the next-to-last decision, then the next-to-next-to-last, and so on. Starting from the front, in the order in which the decisions are actually taken, doesn't get you very far. The reason is that strategic decisions are all about 'What will the next person do if I do this...?' so the decisions that follow yours are important. The ones that come before yours aren't, because you can't do anything about them anyway.
Bearing this in mind, the place to start is when (if!) the game gets down to just two pirates, P1 and P2. The fiercest pirate is P2, and (if the game ever gets this far) his optimal decision is obvious: propose 100 pieces for himself and none for P1. His own vote is 50% of the total, so it wins.
Now add in pirate P3. Pirate P1 knows --- and P3 knows that he knows --- that if P3's proposal is voted down, P1 will get nothing. So P1 will therefore vote in favour of anything that P3 proposes, provided it yields him more than nothing. P3 therefore uses as little as possible of the gold to bribe P1,leading to the following proposal: 99 to P3, 0 to P2, and 1 to P1. Let's write it out like this:
P1 P2 P3
1 0 99
The thought-processes of P4 are similar. He needs 50% of the vote, so again he needs to bring exactly one other pirate on board. The minimum bribe he can use is one gold piece, and he can offer this to P2 since P2 will get nothing if P4's proposal fails and P3's is voted on. So the proposed allocation here becomes
P1 P2 P3 P4
0 1 0 99
The thought-processes of P5 are slightly more subtle. He needs to bribe two pirates. The minimum bribe he can use is two gold pieces, and the unique way he can succeed with this number is to propose the allocation
P1 P2 P3 P4 P5
1 0 1 0 98
The analysis proceeds in the same manner, with each proposal being uniquely prescribed by giving the proposer the maximum possible subject to ensuring a favourable vote, until we get to the tenth pirate and find the allocation
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10
0 1 0 1 0 1 0 1 0 96
This is the proposal I indicated earlier, and it solves the 10-pirate puzzle.
Now comes Omohundro's twist: what if there are (a lot) more pirates? Clearly the same pattern persists --- for a while. In fact it persists up to the 200th pirate. P200 will offer nothing to the odd numbered pirates P1-P199, and one gold piece to each of the even-numbered pirates P2-P198, leaving one for himself.
However, we're trying to find the strategy for P500. At first sight, the argument breaks down at P200, because P201 has run out of bribes. However, he still has a vested interest in not being thrown overboard, so he can propose to take nothing himself:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201
1 0 1 0 ... 1 0 1 0 0
This opens up a new phase of the strategy, because pirate P202 knows that P201 has to accept nothing, or be thrown overboard. So he can count on P201's vote. However, P202 also is forced to accept nothing. He must use the entire 100 gold pieces to bribe 100 pirates --- and these must be among those who would get zero according to P201's proposal. Since there are 101 such pirates, P202's proposal is no longer unique. Let's use a star * to mark pirates who might get something from P202's proposal:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202
0 * 0 * ... 0 * 0 * * 0
At this point a further consideration comes into force. Pirates have to think about how a pirate who has some chance of receiving a gold piece will react if he is definitely offered a gold piece. This depends on how much money, on average, he is willing to sacrifice for the fun of throwing somebody overboard. The puzzle doesn't specify this, so it is reasonable to assume that the pirates don't know it. This means that the bribe might succeed, so it is rational to offer a definite piece to a pirate who only has a chance of one later. As it turns out, things are more satisfactory: there are enough definite 0's in each round for pirates from now on to offer the bribes only to them. Indeed, as Peter Norvig pointed out, there will always be 100 allocations of 0 among pirates P1-P200, so we may describe a solution in which bribes are offered only to (some of) them.
You may wonder why I'm bothering about all this, since clearly pirate 203 doesn't have enough cash available to bribe enough pirates to vote for his proposal, whatever it is. This is true, and P203 goes overboard whatever he proposes --- a result that we symbolize by a cross X. We write ? to show that his choices are irrelevant, giving the situation:
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202 P203
? ? ? ? ... ? ? ? ? ? ? X
Even though P203 is destined to walk the plank, this doesn't mean that he plays no part in the proceedings. On the contrary, P204 now knows that P203's sole aim in life is to avoid having to propose a division of the spoils. So P204 can count on P203's vote, whatever P204 proposes. Now P204 just squeaks home --- he can count on his own vote, P203's vote, and 100 others from bribes of one gold coin: 102 votes in all, the necessary 50%. So P204 will propose
P1 P2 P3 P4 ... P197 P198 P199 P200 P201 P202 P203 P204
* 0 * 0 ... * 0 * 0 * * 0 0
What of P205? He is not so fortunate! He cannot count on the votes of P203 or P204: if they vote against him they have the fun of throwing him overboard and can still save themselves. So P205 gets a X. So does P206 --- he can be sure of P205's vote, but that's not enough. In the same way, P207 needs 104 votes --- 3 plus his own plus 100 from bribes. He can get the votes of P205 and P206, but he needs one more and it's not available. So P207 gets a X as well.
P208 is more fortunate. He also needs 104 votes, but P205, P206, and P207 will vote for him! Add in his own vote and 100 bribes and he's in business. He must offer bribes to those pirates that would get 0 according to P204's proposal:
P1 P2 P3 P4 ... P198 P199 P200 P201 P202 P203 P204 P205 P206 P207 P208
0 * 0 * ... * 0 * 0 0 * * 0 0 0 0
Now a new pattern has set in, and it continues indefinitely. Pirates who make proposals (always to give themselves nothing and to bribe 100 of the first 200 pirates) are separated from each other by ever-longer sequences of pirates who will be thrown overboard no matter what proposal they make --- and whose vote is therefore ensured for any fiercer pirate's proposal. The pirates who avoid this fate are numbers P201, P202, P204, P208, P216, P232, P264, P328, P456,... and so on --- 200 plus a power of 2.
It remains to work out who are the lucky recipients of the bribes, just to make sure they will accept them. As I said, the solution is not unique, but one way to do this is for P201 to offer bribes to the odd-numbered pirates P1-P199, for P2 to offer bribes to the even-numbered pirates P2-P200, then P204 to the odd numbers, P208 to the evens, and so on, alternating from odd to even and back.
At any rate, we conclude that with 500 pirates and optimal strategy, the first 44 pirates are thrown overboard, and then P456 offers one gold piece to each of P1, P3, P5 ..., P199. Thanks to their democratic system, the pirates have arranged their affairs so that the very fierce ones mostly get thrown overboard, and at best can consider themselves lucky to escape with none of the loot. Only the 200 meekest pirates can expect anything, and only half of them.
Truly, the meek shall inherit the worth.
