可预测年薪超过8万美金的“趣味数学题”你能解吗？

[LoveBaby]

The key point is 当且仅当超过半数的人同意时，按照他的提案进行分配，否则将被扔入大海喂鲨鱼。
Lets think reversely.

(1) If 4, 5 left. Then 4 will be thrown into the sea not matter what juicy offer he gave to 5, as he could not win more than half. So 4 will definitely accept 3's offer no matter how greediness he is.
(2) B/c of (1), 3 will survive anyhow even he claims 100 diamonds for himself and 4 will accept it for safety of his own life. So 4,5 expects 2 could provide better offer and will accept it.
(3) Then 2 will survive if he could give 4, 5 one diamond each, though 3 will object it loudly.
(4) What could smart 1 do? Simple, beat 2's offer and give 4,5 two diamonds each.

Final answer: 96,0,0,2,2

Any objection?

 

[LoveBaby]

最初由 fishman 发布


看我的解释吧

Sorry, actually I did not see anyone else answer. We have some minor difference here. Also I had some typo. My answer should be 96,0,0,2,2.

OK, the real difference is that 4,5 will object 1's offer for 98,0,0,1,1 as they could get it from 2's offer anyhow. Maybe 2 will give more than that. Who knows? Only could see it if throw 1 into the sea.

 

[泛蓝]

I answered the English version of this puzzle to my friend 3 years ago.It didn't mention anything about the 80k income by that time.

This Chinese version is a very bad one (lack of quite a few rules)

I will post the english version question so that everyone can get a

clear view of the question and rules.

Here is the oringinal test:

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?

 

[阿搜]

问题:
还是前面提过的,还有一个朋友也提了:

性命是第一利益,最多珠宝是第二利益:
1号必须考虑是否满足第二利益而失去了第一利益,也就是2,3,4,5可能有反对他的方案.

2,3,4,5就开始考虑1号的方案,比较的参考就是找一个比1号方案获得更多第二利益,而
又保证了第一利益的,很明显,四个人平分是一个可能实现得优于1号方案的方案.与1号
方案比较:首先大家还是都保存了性命,其次都比接受1号方案拿更多的珠宝.所以不管1
号出什么方案,他都得死.

 

[泛蓝]

I will give you guys some time to think.
And I will post the answer tonight.

 

[shoppingcart]

最初由 LoveBaby 发布


Sorry, actually I did not see anyone else answer. We have some minor difference here. Also I had some typo. My answer should be 96,0,0,2,2.

OK, the real difference is that 4,5 will object 1's offer for 98,0,0,1,1 as they could get it from 2's offer anyhow. Maybe 2 will give more than that. Who knows? Only could see it if throw 1 into the sea.

那就违背了"每个海盗都是很聪明的人，都能很理智的判断得失"的前提了。

 

[LoveBaby]

最初由 www.cs-cn.net 发布
I answered the English version of this puzzle to my friend 3 years ago.It didn't mention anything about the 80k income by that time.

This Chinese version is a very bad one (lack of quite a few rules)

I will post the english version question so that everyone can get a

clear view of the question and rules.

Here is the oringinal test:

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?

The condition in your puzzle is much different which include 50% for survial case comparing with the Chinese edition. Then my answer for that is:

96,0,1,0,1,0,1,0,1,0

Here is my revese deduction:

2 left; 100,0
3 left: 99,0,1
4 left: 99,0,1,0
5 left: 98,0,1,0,1
6 left: 98,0,1,0,1,0
7 left: 97,0,1,0,1,0,1
8 left: 97,0,1,0,1,0,1,0
9 left: 96,0,1,0,1,0,1,0,1
10 left: 96,0,1,0,1,0,1,0,1,0

The point here is that one has to make sure that there is no chance that he will get worse offer if he decide to vote to throw a pirate into the sea in the next round.

 

[qlgc]

最初由 LoveBaby 发布
The key point is 当且仅当超过半数的人同意时，按照他的提案进行分配，否则将被扔入大海喂鲨鱼。
Lets think reversely.

(1) If 4, 5 left. Then 4 will be thrown into the sea not matter what juicy offer he gave to 5, as he could not win more than half. So 4 will definitely accept 3's offer no matter how greediness he is.
(2) B/c of (1), 3 will survive anyhow even he claims 100 diamonds for himself and 4 will accept it for safety of his own life. So 4,5 expects 2 could provide better offer and will accept it.
(3) Then 2 will survive if he could give 4, 5 one diamond each, though 3 will object it loudly.
(4) What could smart 1 do? Simple, beat 2's offer and give 4,5 two diamonds each.

Final answer: 96,0,0,2,2

Any objection?

一直到2号，都对。
2号的方案：98-0-1-1
1号为什么要给多两颗出来给4，5？他们的成本现在是2，而3的成本是1，放弃4或5，收买3：
1号可以只花多一颗就保证自己的方案通过:
97-0-1-2-0

 

[qlgc]

最初由 LoveBaby 发布


The condition in your puzzle is much different which include 50% for survial case comparing with the Chinese edition. Then my answer for that is:

96,0,1,0,1,0,1,0,1,0

Here is my revese deduction:

2 left; 100,0
3 left: 99,0,1
4 left: 99,0,1,0
5 left: 98,0,1,0,1
6 left: 98,0,1,0,1,0
7 left: 97,0,1,0,1,0,1
8 left: 97,0,1,0,1,0,1,0
9 left: 96,0,1,0,1,0,1,0,1
10 left: 96,0,1,0,1,0,1,0,1,0

The point here is that one has to make sure that there is no chance that he will get worse offer if he decide to vote to throw a pirate into the sea in the next round.

This time you are 100% correct. Congratulations for U$80K contract!

 

[LoveBaby]

最初由 qlgc 发布


一直到2号，都对。
2号的方案：98-0-1-1
1号为什么要给多两颗出来给4，5？他们的成本现在是2，而3的成本是1，放弃4或5，收买3：
1号可以只花多一颗就保证自己的方案通过:
97-0-1-2-0

Good point!

Actually, I have used the same logic to solve the next puzzle in English version.

 

[Aeoliao]

最初由 淼 发布



原来题目是"当且仅当超过半数同意", 是大于50%, 不是>=50%.

小鱼的解法是最优的.

如果是>50%才能通过分配方案的情况

#1应该分为 97，0，1，2，0

Solution:
-----------1----2----3----4----5
Round 4:------------------0----100 (就算#4不要宝石，#5会留他小命么？还是扔下船最保险阿)
Round 3:-------------100--0----0 (所以#4为了100%的确保小命，这里会接受任何条件，所以怎么投票都不会超过ROUND3 )
Round 2:--------98---0----1----1( #4，#5，在这里是唯一赚钱的机会，有1个宝石就会赞成了。
Round 1:---97---0----1----2----0(#4在这里是唯一可以赚到比1个多的机会，能放过么？ #3是唯一可以赚到钱的机会，也不会放过。#2和#5倒希望#1能死就死)


>=50% 的话

Solution:
-----------1----2----3----4----5
Round 4:-----------------100---0 (#4自投 =50%通过)
Round 3:-------------99---0----1 (#5有1个就会赞成，最后机会呢，所以ROUND 4不会发生)
Round 2:--------99---0----1----0 (#4最后机会赚钱，#2也只会给他1个，以达到 50%通过)
Round 1:---98---0----1----0----1 (由于在ROUND2肯定会出现 #3和#5没钱赚就结束分配的情况， #3,#5有1个宝石就同意了）

答案 98-0-1-0-1

文字解释:

如果是>50%，不是>=50%，这题就没有那么复杂了。从出题的角度来说，应该是>=50%

>50%的解法

因为在第2轮，还有4人时，#2 要3张赞成票。这时最愿意投反对票的是 #3，因为#2一死，进入第3轮，#3可以开出 100，0，0的条件，如果#4不接受，进入第4轮，#5只要反对，#4就下海了，剩下只有1人，#5独吞。 所以，进入第4轮，#4死定了，#4保命的办法是：如果进入第3轮 必须在第3轮无条件支持 #3,也就没有宝石分，#4赚钱的办法是在第2轮以前投1个赞成票，过了第2轮就没有机会了。同样#5保命没问题，要赚钱，只有在第2轮以前投1个赞成票，所以只要有1个宝石就可以收买#5。#2在第二轮考虑到上面所有的发展趋势，肯定开出 98，0，1，1的条件，而且肯定 #4 #5赞成。

#1和#3预见到 第2轮的条件 会是98，0，1，1.而且肯定可行。#3就知道唯一赚钱的机会就是第一轮支持#1(如果有1个的话). #1也意识到只要给1个与#3就能获得支持。剩下就是拉拢#4,或#5。 对于#4来说，要赚钱必须在第2轮以前投赞成票，如果正好在第2轮投赞成票的话，#2只会给1个，否则1个都没有了。那么如果在第一轮有2个宝石的OFFER的话，#4就出手了。而#5开始的心理是巴不得全死完。#2巴不得到第2轮。 所以对于#1来说拉拢#4最合理。

 

[moose]

96 0 0 2 2

 

[moose]

if >=50%
then
97 0 0 2 1

 

[Aeoliao]

#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

 

[shoppingcart]

最初由 lfan_cn 发布
because there lacks one condition: does every pirate wants other to be thrown into the sea if there is neither harm nor benefit to him?

Yes