The Drunkard's Walk, How Randomnes Rules Out Lives, by Leonard Mlodinow

[hangchoi]

I think E will say NO to every suggestion unless he got quite a lot of it......because he will not die anyway (Just opposite to Kido's assumption) but, as all A, B, C, D know that, they will not give him any. E will ban all D's proposal in order to get the most.

C will be killed when D says NO (in principle), but if D really says NO, C will be killed and D will get nothing, so C may seek for majority i.e. C ~ 100%, D>0, E=0%

B, knowing that he will die if any two of C, D and E say NO, so he will have to make a proposal that any two of C, D and E would accept, It is clear that C would say NO anyway and B has to get the vote from both D and E to survive. If B dies, E must get nothing, so B has to offer E a little bit to make him say YES. So B~99%, D > 0%, E > 0%

A needs two votes only and notes that both D and E will get a bit even if he is killed, can offer them a bit more and keep the majority. So A ~ 99%, D > 0% and E > 0%

After I have finished writing the above, I check that my answer is quite like Kido's......but I have different view from yours.

1) E will not die in any case....Yes. He will accept whenever the offer is > 0, I think it is right too. He will absolutely get nothing if C makes proposal. So whenever A or B makes a proposal that he will have some share, he will say Yes.

2) The votes from D and E matter much. These two votes are the votes that A or B has to get in order to survive.

[wongyan]

you guys never question how E execute D if vote is not >50%???

[hangchoi]

What do you mean? If A, B and C are killed, E will ban D's proposal anyway because D will be killed after that and E will take all..

So D will alway let C's proposal pass no matter what C proposes.

[wongyan]

yeah, you are right, hang.  E will reject any proposal from D.  But how can he execute D?  (隻抽？）

[hangchoi]

假设这5个海盗都贪婪成性、残忍无比、绝顶聪明而又一诺千金，都想自己得到最多，都想看到别人死去而自己活。 <----that's the assumption.

So.....If E ban D's proposal, D will be 一诺千金 and let E kill him.......

[wongyan]

I simply do not believe 一诺千金can override survival.

anyway, model answer listed.

答案

要回答这个问题，一般人肯定会想到，1号必须先让另外两个人同意，所以，他可以自己得到32颗，而给2号3号各34颗。但只要仔细想想，就会发现不可能，

2号和3号有积极性让1号死，以便自己得到更多。所以，1号无奈之下，可能只有自己得0，而给2和3各50颗。但事实证明，这种做法依然不可行。为什么呢？

因为我们要先看4号和5号的反应才行。很显然，如果最后只剩下4和5，这无论4提出怎样的方案，5号都会坚决反对。即使4号提出自己要0，而把100颗钻石都给5，5也不会答应――因为5号愿意看到4号死掉。这样，5号最后顺利得到100颗钻石——因此，4的方案绝对无法获得半数以上通过，如果轮到4号分配，4号只有死，只有死！

由此可见，4号绝对不会允许自己来分。他注定是一个弱者中的弱者，他必须同意3号的任何方案！或者1号2号的合理方案。可见，如果1号2号死掉了，轮到3 号分，3号可以说：我自己100颗，4号5号0颗，同意的请举手！这时候，4号为了不死，只好举手，而5号暴跳如雷地反对，但是没有用。因为3个人里面有 2个人同意啊，通过率66.7％，大于50％！

由此可见，当轮到3号分配的时候，他自己100颗，4和5都是0。因此，4和5不会允许轮到3来分。如果2号能够给4和5一些利益，他们是会同意的。

比如2的分配方案是：98，0，1，1，那么，3的反对无效。4和5都能得到1，比3号来分配的时候只能得到0要好得多，所以他们不得不同意。

由此看来，2号的最大利益是98。1号要收买2号，是不可能的。在这种情况下，1号可以给4号和5号每人2颗，自己收买他们。这样，2号和3号反对是无效的。因此，1号的一种分配方案是：96，0，0，2，2。

这是不是最佳方案呢？再想一想，1号也可以不给4号和5号各2个，而只需要1个就搞定了3号，因为如果轮到2号来分配，2号是可以不给3号的，3号的得益只有0。所以，能得到1个，3号也该很满意了。所以，最后的解应该是：97，0，1，2，0。

好，再倒推。假设1号提出了97，0，1，0，2的方案，1号自己赞成。2和4反对。3∶2，关键就在于3号和5号会不会反对。假设3号反对，杀掉1 号，2号来分配，3自己只能得到0。显然，3号不划算，他不会反对。如果5号反对，轮到2号、3号、4号来分配，5号自己最多只能得到1。

所以，3号和5号与其各得到0和1，还不如现在的1和2。

正确的答案应该是：1号分配，依次是：97，0，1，0，2; 或者是：97，0，1，2，0。

人类如强盗，一直分钻石

出现这样的结果，相信是很多人没有想到的。
1、谁负责分配，谁的收益就最大！
2、看起来最强势的2号居然什么都得不到！他的反对是无效的！
3、次弱势的人如果位置恰当，稍微可以得到一些，如3号、4号、5号，但如果位置不当，也无所得，他们的反对票是根本无用的！如2号。
4、要使决策获得通过，只要收买个别次弱势群体就可以了。如果愿意多给他们一点，比如，给2号和4号各1个，而给3号和5号都增加1个，结果是：94，1，2，1，2。那么，他们4人也许都会感激涕零、高呼万岁、欢呼“青天大老爷”甚至于死心塌地地为决策者效力。
5、尽管1号的位置风险大，因为大家都会争先恐后去当1号。

[kido]

For those like counting, here is another interesting passage:

http://sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

Gary Foshee, a puzzle designer from Issaquah, Wash., posed this puzzle during his talk this past March at Gathering 4 Gardner, a convention of mathematicians, magicians and puzzle enthusiasts held biannually in Atlanta. The convention is inspired by Martin Gardner, the recreational mathematician, expositor and philosopher who died May 22 at age 95. Foshee’s riddle is a beautiful example of the kind of simple, surprising and sometimes controversial bits of mathematics that Gardner prized and shared with others.

“The first thing you think is ‘What has Tuesday got to do with it?’” said Foshee after posing his problem during his talk. “Well, it has everything to do with it.”

Even in that mathematician-filled audience, people laughed and shook their heads in astonishment.

When mathematician Keith Devlin of Stanford University later heard about the puzzle, he too initially thought the information about Tuesday should be irrelevant. But hearing that its provenance was the Gathering 4 Gardner conference, he studied it more carefully. He started first by recalling a simpler version of the question called the Two Children Problem, which Gardner himself posed in a Scientific American column in 1959. It leaves out the information about Tuesday entirely: Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?

Intuition would suggest that the answer should be 1/2, since the sex of one child is independent of the sex of the other. And indeed, had he been told which child was a boy (say, the younger one), this reasoning would be sufficient. But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth:

Boy, girl
Boy, boy
Girl, boy

Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.

He used this same method on the Tuesday birthday puzzle, enumerating the equally likely possibilities for the sex and birth day of each child and then counting them up.

If the older child is a boy born on Tuesday, there are 14 equally likely possibilities for the sex and birth day of his younger sibling: a girl born on any of the seven days of the week or a boy born on any of the seven days of the week. (This analysis ignores minor differences like the fact that slightly more babies are born on weekdays than on weekend days.)

Now suppose that the older child isn’t a boy born on Tuesday. The younger child then must be, of course. Now we count up the possibilities for the sex and birth day of the older child. If she’s a girl, she might have been born on any day of the week, generating seven more possibilities. If he’s a boy, he could have been born any day except Tuesday. (Otherwise this case would already have been counted in the first scenario: the older child a boy born on Tuesday). This second scenario generates just six, rather than seven, more possibilities.

Since each of these cases is (approximately) equally likely, we can compute the probability by dividing the number of cases in which there are two boys by the total number of cases. The total number of cases is 27: 14 if the older child is a boy born on Tuesday and 13 if the older child isn’t. In 13 of those cases both children are boys (7 if the older child is a boy born on Tuesday and 6 if he isn’t), yielding a probability of 13/27.

Devlin was astonished by this answer. As a mathematician, he had long been familiar with the Two Children Problem and its answer of 1/3. “Knowing the birth day is a Tuesday may (and does) make a difference, but it surely cannot make much of a difference, right?” he wrote in his blog, Devlin’s Angle. “Wrong.” After all, 13/27 is far closer to 1/2 than 1/3.

So why does intuition seem to lead us so astray? Both the intuitive and the mathematically informed guesses are wrong. Are human brains just badly wired for computing probabilities?

Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.

Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.

Gardner himself tripped up on his simpler Two Children Problem. Initially, he gave the answer as 1/3, but he later realized that the problem is ambiguous in the same way that Peres argues that the Tuesday Birthday Problem is. Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.

In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male.

The remarkable thing that Foshee’s variation points out is that any piece of information that affects the selection will also affect the probability. If, for example, you selected a family at random among those with two kids, one of whom is a boy who plays the ukulele and wants to become a dancer, the ukulele-playing and dancing ambitions would affect the probabilities about the sex of his sibling.

Peres says that we shouldn’t despair about our probabilistic intuition, as long as we apply it to familiar situations. The difficulty of these problems is rooted in their artificiality: In real life, we almost always know why the information was selected, whereas these problems have been devised to eliminate that knowledge. “The intuition develops,” he points out, “to handle situations that actually occur.”

Still, Gardner’s initial overly narrow interpretation warns of the dangers of over-hasty analysis of probability questions — and shows the wonder that can come from them.