toquebec

carson修正后我同意一半,因为NO2只要有机会就一定是98-0-1-1方案,那么NO1打出98-0-0-1-1方案98-0-1-0-1方案NO4及NO5都不会同意,因为放弃NO1的方案会得到多于一粒的希望(当然只是希望但至少保证有一粒).所以NO1应提出97-0-1-0-2方案争取NO3和NO5才能保证获胜.大家有意见吗?

test

很有意思。或许逆向思维会好理解一些。
如果由5分配，全部归5；
由4分配，4会死掉；(4,5没有机会分配啦）
由3分配，4会拿到一个，5没有（4不会让3死掉，所以给不给这一个看3够不够狠）；
由2分配，3没有，4会拿到1个，5拿一个；

最后由1分配，2没有（即使给了2，2也会为争取最大利益而不同意，也就是让1死掉），3拿一个，4没有，5拿2个（只拿一个，会有被陷害的可能）: 97,0,1,0,2；或者，3拿一个，4拿2个（如果拿一个的话，1就太冒险了），5没有: 97,0,1,2,0。

也就是说，找到比2分配能更好的满足其他2个人的方案，1就可以尽可能的捞钱拉。

如果1真敢冒险，争取做个奸商的话，答案应该是：
98，0，1，1，0（因为4最多只能拿到一个，若是不陷害1的话还可以赚个声望），不过这样上面的答案也可以改成：98，0，1，0，1

uwdanny

To 大山

   

对了,就是这么简单
当是老师上课就是举的这个例子,请2个同学上台表演
结果是,分配的一方以拿出2枚硬币成交.急得俺在下面喊,"ONE ENOUGH".

That's also Question 1 on my first assignment when I took Game Theory course. ^oo^

For this question, I agree with the answer ( 97, 0, 1, 0, 2 ) or ( 97,0,1,2,0 ). The key point is to ensure the satisfaction from #4 or #5, since #3 will agree anyways.

uwdanny

To rat, do we know whether the unbalanced individual weights less or more than the others? Just to make sure everything's absolutely clear.

rat

if you did it right, you will find that out as well, getting warmer..!!

uwdanny

Here's an attemp using enumeration-related knowledge.

Label 12 balls as

001, 112, 200,
010, 120, 201,
011, 121, 202,
012, 122, 220

I did this because there're 3^3 = 27 possible lables. We first need to eliminate 000, 111, 222, these 3 labels are useless; now there remains 27-3=24 labels. Since it is a circular list, we divide by 2 to eliminate the ordering. ( ie. 100 is same as 122 but with bit-flip, 1 -- 1, 0 --- 2 )

Now, we rename these labels as follows,

    A   001
    B   010
    C   011
    D   012
    E   112
    F   120
    G   121
    H   122
    I   200
    J   201
    K   202
    L   220

Notice that each digit has 4 "0"s, 4 "1"s and 4 "2's. Now we can start the 3 weighings.

First weigh: Weigh ABCD against IJKL, that is, weigh shots with first digit 0 against those with first digit 2. If Weight(ABCD) less than Weight(IJKL), we mark 0; else if Weight(ABCD) greater than Weight(IJKL), we mark down 2, else, (the different one will have first digit 1), we mark down 1

Second weigh: Weight AIJK against FGHL, that is, weigh shots with second digit 0 against those with second digit 2. If Weight(AIJK) less than Weight(FGHL), we mark 0; else if Weight(AIJK) greater than Weight(FGHL), we mark down 2, else, (the different one will have second digit 1), we mark down 1

Third weigh: similar to first, second weight, focus on the 3rd digit of the labels.

My claim is that, if we let x, y, z represent the numbers we mark down from weight 1, 2, 3 respectively, then the ball labelled xyz is the different one, which is lighter than the rest.

In the case if xyz doesn't appear on my labelling list, you could do a bit-flip operation, the label obtained after bit-flip is guaranteed to be on my labelled list, and that individual will be heavier than the rest.

Done. ^oo^

甑士隐

据统计，在美国，在一辈子内不能回答出这道题的人，平均年薪在8百万美金以上，象好莱乌演员,棒球篮球冰球运动员等等!!!!!!!

RedPath

It depends on how to understand the game rule. Especially for in case of only 4 and 5 left.
For only 4 and 5 left, If 5 votes no then 4 will die. If it is the rule,then 5 will always vote NO, and 4 always vote YES. As for 3, he expects 1 and 2 die since in his turn, 4 will vote yes, himesel will vote yes, then 3 will get all. Therefore, 3 will always vote NO. Therefore, if 1 die, 2 will die anyhow. Therefore, 2 will always votes YES. 1 can get 100 and gives others nil based on the rest won't risk to be voted die. Therefore, 100-0-0-0-0 is the deal.

freshland

参与一下：
原则：生命最重要，其次是财富。

5号：5号没有生命危险，他唯一要做的就是前面的人都死光，由他独吞。因此他会一直反对前面任何人的方案。

3号：3号的目的就是要1号和2号都死掉，由他来分配，这样一来因为4号要想活命就一定会同意3号的任何方案。

4号：生命非常危险，因为如果由3号分配（意味着1,2号已死），那么4号要想活命，无论3号提出什么方案他都会同意（此时3号肯定会给4号0）。而如果1号死了，由2号分配，由于3号和5号肯定投2号的反对票，所以此时意味着2号也死。所以，4号想活命的话有两个选择，一是支持1号，二是支持3号。但由于支持3号意味着4号仅能活命但颗粒无收，因此一但1号对4号有所表示，那么4号一定会支持1号。

2号：2号的生命应该说与1号绑在一起，1号死了，他也死了，因此，无论1号是什么分配方案，2号都会支持。

1号：由于已经有了铁定的自己的一票和2号的一票，那么1号要做的就是把可以有两个选择的4号拉过来。因此1号必须分一点利益给4号。对于1号来说，决定生杀大权的是4号。

综上所述，1号分配方案如下：1号-99；  2号-0； 3号-0； 4号-1；  5号-0

大山

TO RAT :

I think I know the answer now:

(1) 4--4, if level , go to (2),if not goto (5)

(2) 2 balls from step1--2 from the rest 4 ,if level, go to (3),if not goto (4)

(3) 1 from step(2)--one from the rest2 , if not level, this one ,if level , the very last one.

(4) if 2 from the rest 4 is heavier, 1--1 to find the heavier, or if they are lighter ,1--1 to find the lighter.

(5) remove 3 balls from either side ,say the lighter side (sideA)of the unleveled scale ,and put them aside, ; Move 2 balls from the other side(side B) to side A to form a 3-2 distribution, then add one ball from the rest balls to side B , to form a 3--3

(6)if the scale is level now , then we can tell:
   a)"the one" is one of the 3 balls removed in step 5

   b) "the one" is lighter ( remember we remove 3 balls from the lighter side)
  
   then 1--1 from the 3 balls , if level, the one is the very last, or if not the lighter one

7)if the scale change so that side A become heavier, then "the one" must be heavier and one of the 2 removed from sideB; Weight these 2 balls to find the heavier one .

8)if the scale remain unchanged,we can be sure:

a)of the 6 balls on the scale,3 balls ,one added to sideB and 2 removed from side b in step (5), are sure normal balls.

b)the ditribution of remaining 3 unresolved balls is depend on wether "the one" is heavier or ligher:

if heavier:  A:1 normal ball--B:1normal ball+"the one".

if lighter: A: "the one"--B:2 normal balls

so whether "the one" is heavier or lighter can be resolved by weighing the remaining 2 balls on side B to see if they are equal.

If equal , "the one" is the remaining unresolved ball on side A. If not it is the heavier of the 2 unresolved ball on side B.


What a waste!!!