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枫下沙龙 / 休闲娱乐 / （转贴）在20分钟内能回答出这道题的人，平均年薪在8万美金以上
-mafan(麻烦);
2002-10-17
{719}
(#802164@0)
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瞎猜, 有两个强盗得40颗钻石,自己拿20颗, 另两个人没有.
-decentboy(黄金重装甲骑士);
2002-10-17
(#802165@0)
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my answer is:第一个拿15个,第二个34,第三个51,第四个0,第5个0.,
-daisydaisy(士心);
2002-10-17
(#802869@0)
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100,0,0,0....在公布答案的一瞬间,打死其他四个强盗
-feijintom(恨我不聪);
2002-10-17
(#802949@0)
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可惜太老了.去年就有人贴过了.
-goodbaby(小宝);
2002-10-17
(#802166@0)
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第一个海盗死定了，基本上谁抽到最后一个签谁拿着宝石活着回去
-x86(爱谁谁);
2002-10-17
(#802170@0)
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叙叙旧
-supreguest(一枝狗尾巴花);
2002-10-17
(#802175@0)
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who's question is right?
-farmer(农夫®);
2002-10-17
(#802214@0)
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这问题不是去年那部片<<美丽人生>>的主角小约翰。纳什研究的博弈论吗？人家可是拿诺贝尔的。。。
-cocotea(可可茶-思考);
2002-10-17
(#802176@0)
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第一个拿10个,第二个30,第三个30,第四个20,第5个10.,对吗?
-vivianwang(归去来兮);
2002-10-17
(#802181@0)
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himself gets 20%, each of another 2 gets 40%. No diamond for the rest of them.
-birdswimming(feifei);
2002-10-17
(#802186@0)
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A--19;B--20;C--20;D--20;E--21ORA--19;B--20;C--20;D--0;E--41
-chinadragon(龙行天下);
2002-10-17
(#802194@0)
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Answer98---0---1---0---1
-yuanzidan(原子弹);
2002-10-17
{18}
(#802195@0)
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照您的分法: 2,3,4 都不会同意
-farmer(农夫®);
2002-10-17
(#802208@0)
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3 must agree, otherwise he will get nothing at the second turn(where 2 is the dealer at that time) The stradge of 2 is 99--0--1--0
-yuanzidan(原子弹);
2002-10-17
(#802216@0)
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no. if 1 dead, 3 will get 1 while 4,5 get nothing
-farmer(农夫®);
2002-10-17
(#802231@0)
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so, what's the strategy of 2?
-yuanzidan(原子弹);
2002-10-17
(#802241@0)
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99-1-0-0. butbefore 2nd turn, 3 still have chance to get more than one. For 4 &5, if they disagree after they get one for each, they will have chance to get nothing at 2nd turn
-farmer(农夫®);
2002-10-17
{163}
(#802254@0)
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Let's talk about 99-1-0-0 first. At this point, why 3 will agree? If 3 disagrees, he is the dealer next turn and he will get 99. (99-0-1)
-yuanzidan(原子弹);
2002-10-17
(#802268@0)
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that's why the reason 3 will disagree ur answer: 98-0-1-0-1
-farmer(农夫®);
2002-10-17
(#802276@0)
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呵呵，你脑子不大转
-xiaoma99(xiao ma);
2002-10-17
(#802286@0)
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who will agree at this point except 2?
-yuanzidan(原子弹);
2002-10-17
(#802279@0)
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the strategy of 3: 99-0-1
-farmer(农夫®);
2002-10-17
(#802270@0)
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why? 我没拿8万美金的年薪所以不明白
-fido(jumpingmonkey);
2002-10-17
(#802227@0)
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同意
-lilyba(Sunshine);
2002-10-17
(#802232@0)
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we are allies
-yuanzidan(原子弹);
2002-10-17
(#802272@0)
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那你就给我8万年薪吧。
-lilyba(Sunshine);
2002-10-17
(#802281@0)
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Hehe, you need ask mafan for that US80k, he offers that.
-yuanzidan(原子弹);
2002-10-17
(#802301@0)
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抓阄吧?60颗一堆,其余四堆每堆10颗,抓到头彩的拿大堆,其余四人安慰奖,嘿嘿.....接下来,有可能四人合伙干掉luckybone,然后再分?;-))
-mildkiller(M.K.);
2002-10-17
(#802196@0)
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1-98;2-0; 3-0;4-1;5-1
-farmer(农夫®);
2002-10-17
(#802201@0)
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4 may not argee with this. If he agrees, he gets only one. If he disagrees, he will get at least one.
-yuanzidan(原子弹);
2002-10-17
(#802229@0)
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if 4 disagree, he will get nothing
-farmer(农夫®);
2002-10-17
(#802234@0)
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why not:98 0 1 0 1?
-ra_95(小人-盼雪化!等草绿!);
2002-10-17
(#802235@0)
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because 3 has chance to get more than one if 2 made something wrong (giving 3 more than one) at 2nd turn by chance.
-farmer(农夫®);
2002-10-17
(#802239@0)
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I think there is a bug in the question. should be:只有超过半数以上的同意... or:if only 4 and 5 : 4 will raise: 100 0; so 5 will always agree 3's quotation if 3 give him more than 0
so 3 will raise: 99 0 1; so 4 will always agree on 2's decsion if 2 gives him more than 1;
so 2 will raise: 98 0 1 0 (Please notice: 4 and 2 agree this case) ...
so 1 will raise 98 0 1 0 1 and 3 and 5 will agree, then decision passed.
-ra_95(小人-盼雪化!等草绿!);
2002-10-17
{344}
(#802252@0)
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我看不对，同意小人的答案。case 1) 剩 #4, #5 ==> 100, 0
case 2) 剩 #3, #4, #5 => 99, 0, 1 (#4 disagree)
case 3) 剩 #2, #3, #4, #5 => (98, 0, 1,0) (#3, #5 disagree)
case 4) 剩 #1,#2, #3, #4, #5 => (97, 0,1,0,1) (#2, #4 disagree)
order is important.
-lanyi(蓝衣DX);
2002-10-17
{227}
(#802351@0)
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Himself get 32%, two other get 34%, the rest get zero.
-jeffrey815(Smartiecat);
2002-10-17
(#802205@0)
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1st get 100%.
-hhll26(likeplay);
2002-10-17
(#802215@0)
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1-98, 2-1, 4-1 或 1-97 2-1 4-2//答案剩余
5 100
5/4 （5）100， 4（0）4最多0个
5/4/3 （5）0， 4（1），3（99）4可多的1个，4，3同意
5/4/3/2 （5）0，4（1），3（99），2（0），（4）保证一个，3保持只剩3人的满意度，234同意
5/4/3/2/1 （5）0，（4）1，3（0），2（1），（1）98 //124同意
5/4/3/2/1 （5）0，（4）2，3（0），2（1），（1）97 //124同意
我想4如果满意度不变，不知道她会不会冒险搞掉1
-lilyba(Sunshine);
2002-10-17
{368}
(#802207@0)
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我同意你的分法, 但条件是把97粒钻石给我...
-decentboy(黄金重装甲骑士);
2002-10-17
(#802256@0)
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你就当no5吧。:P
-lilyba(Sunshine);
2002-10-17
(#802257@0)
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那你死定啦...我会许诺一粒钻石也不要而把你干掉...:-P
-decentboy(黄金重装甲骑士);
2002-10-17
(#802265@0)
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嘿嘿，我的假设条件里，压根就没有考虑no5是否开心。
-lilyba(Sunshine);
2002-10-17
(#802273@0)
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我觉的97,2,1,0,0 比 98,1,1,0,0 更诱人
-decentboy(黄金重装甲骑士);
2002-10-17
(#802259@0)
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安全些，风险收益取舍
-lilyba(Sunshine);
2002-10-17
(#802267@0)
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98-1-0-1-0更好些。因为2345存在时，4有一个钻石都拿不到的风险。
-lilyba(Sunshine);
2002-10-17
(#802288@0)
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不可行，我要是第三个人， 肯定不会同意的。因为我只有一个，太少， 如果把你去掉，第二个人为了拉拢我， 会至少给我1/3，就是33个，所以不可行。所以第三个人很重要， 他的最大收益是50 个。应该为16/34/50/0/0分配
-chenql26(真的不懂);
2002-10-17
(#803034@0)
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其实,1号的最大收益是不被喂鲨鱼,他给2,3每人50颗就行了.要设想自己完蛋后,2,3的最大收益是什么,只要满足,他们就会赞成,4,5的意见是废纸.所以,345时,3最多拿50,2345时,23分别最多也是50.1想贪,还是保命吧!
-mildkiller(M.K.);
2002-10-17
(#802237@0)
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This is my answer too.
-rudy2000(born in 1975);
2002-10-17
(#802298@0)
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太简单！要最大收益， 他肯定能拿到钻石
-chenql26(真的不懂);
2002-10-17
(#803039@0)
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如果只剩2个人他们怎么表决?
-lazybones(lazy);
2002-10-17
(#802243@0)
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2个人, 100,0; 3个人: 99,1,0; 4个人: 99,1,0,0; 5个人: ?????
-decentboy(黄金重装甲骑士);
2002-10-17
(#802249@0)
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5 gets nothing
-yuanzidan(原子弹);
2002-10-17
(#802250@0)
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98:0:1:0:1逆推,
如果最后剩两个人, 会是100-0, 因为自己已占一半的投票权
依此类推, 99:0:1(第五个海盗一个也会满意, 因为如果剩最后两个人的话, 一个都没有)
99:0:1:0
98:0:1:0:1
不过, 这和年薪有联系吗?!
-online(Online);
2002-10-17
{190}
(#802262@0)
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I think there is a bug in the question. should be:只有超过半数以上的同意 ... or too easy ...
-ra_95(小人-盼雪化!等草绿!);
2002-10-17
(#802299@0)
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哈哈, 答案会是 97:0:1:0:2同样是逆推
0:100
99:1:0
97:0:2:1
97:0:1:0:2
总要有点甜头才行呀. :D:D:D
-online(Online);
2002-10-17
{77}
(#802310@0)
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But the trick thing is: 0 : 100 will not MAKE 100% sure 5 will agree ... e.g. if I were 5, and as the style I played game, I will disagree .... then I can get all AND enjoy the happiness from killing.so, 5 will always disagree 4 will always agree to 3's proposal!
3: 100 0 0! so, 3 will disagree 1 and 2's proposal!
2: 98 0 1 1 how to make 5 happy? 1 is enough. But 5 knows if he disagree, 3 will give him nothing!
1: 97 0 1 2 0 ...?
97 0 1 0 2 ...?
Damn ... which is higher priority if "life is safe"? Killing or gem?
-ra_95(小人-盼雪化!等草绿!);
2002-10-17
{335}
(#802324@0)
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YES, 如果只剩两个人的时候, 第五个人可以ENJOY珠宝和把第四个扔下海. -- 但这可是一对一的较量 :P你的分法中有问题, 如果只剩3个人时, 他的方案会是99:1:0, 而不会是100:0:0, 除非它想去见鲨鱼.
-online(Online);
2002-10-17
{89}
(#802349@0)
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sorry, 没看清你写的, 你说的对, 100:0:0, 第四个人也会同意. 只是多重答案了:D
-online(Online);
2002-10-17
(#802354@0)
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如果向你所说，那就是另外一个问题了。
-lazycat(沧海一笑);
2002-10-17
(#802311@0)
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make sure other four earn at least 80K per year.
-mssg(mssg);
2002-10-17
(#802308@0)
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The first one is the boss of the other four. That's enough. :D
-online(Online);
2002-10-17
(#802321@0)
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同意，我在一次interview中被问过这题。被他提示了半天才答出来，所以不敢想八万年薪，呵呵。
-lanyi(蓝衣DX);
2002-10-17
(#802343@0)
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98,0,0,1,1 or 98,0,1,0,1
-falcon(令狐葱);
2002-10-17
(#802323@0)
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98, 1,0,10
-lazycat(沧海一笑);
2002-10-17
(#802333@0)
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let us think about this question in reverse.
-shanxiren(山西同乡会);
2002-10-17
{532}
(#802392@0)
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96,0,0,1,3
-debug(除虫);
2002-10-17
(#802461@0)
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应该加一条，海盗都是极端自私的人，并以杀到别的海盗伟目标。
-killer(流血五步伏尸二人);
2002-10-17
(#802925@0)
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98:0:1:1:0
-jianghongca(慎独);
2002-10-17
(#802994@0)
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16/34/50/0/0
-chenql26(真的不懂);
2002-10-17
(#803036@0)
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98-1-0-0-1
-eelvis(eelvis);
2002-10-17
(#803204@0)
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Final answer: 40-26-34-0-0
-mountainoutsider(outsider);
2002-10-17
(#803305@0)
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(1),5号最多期望是100枚,而此4号得0枚保住性命(其它全死);(2)3号最多得99枚(1,2号死掉),而此时4号得1枚,5号得0枚(3)如果要2号分则0:99:1:0
(4)如果让1号分, 3号和5号都难达到他们的期望值而又使自己最多,所以最终分法为97,1,0,2,0..
-seeu(冷冷冰雨);
2002-10-18
(#803543@0)
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98-0-1-0-1。理由——如果只有2个海盗，无疑会这么分：100，0。
如果有3个，他必须收买1个人，你认为他会收买老4还是老5呢？当然是老5，而且只需1块：99，0，1。
如果有4个，他必须至少收买1个人，他会去收买老4，如果老4不答应，轮到老3分，他就一个子儿都没了（上面的情况）：99，0，1，0。
如果有5个，必须收买2个人，他会去收买老3和老5，因为老2分的话，老3和老5都没戏（上面的情况）：98，0，1，0，1。
-simali(smile);
2002-10-18
{362}
(#803544@0)
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98-0-1-0-1
-auroraeos(北极光);
2002-10-18
(#803837@0)
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(1号)分法97:0:1:2:0或97:0:1:0:2.(2号)分法98:0:1:1(3号)分法:100:0:0.(4
号)分法0:100.
-wanderabout(驿动的心);
2002-10-18
(#804623@0)