【問題】一個有趣的邏輯推理問題

[-Hero-]

我覺得答案有點像對號入座, 且雖然各位高人也有所解釋,
但在下仍覺矛盾點太多, 這樣的答案與其證明, 恕在下難以信服。

[dominic]

好難理解....唉....領不到8萬美金的年薪了
嗚...................

[cookiess]

最初由 herozero 發表
我覺得答案有點像對號入座, 且雖然各位高人也有所解釋,
但在下仍覺矛盾點太多, 這樣的答案與其證明, 恕在下難以信服。

之前的解答我覺得寫的很清楚了…

至於大大難以信服的話，可以把你的疑點post上來，說不定正確解答就不再那麼正確了。

[Asa]

最初由 rainwen 發表


可以拿到100當然提100了......「最大利益原則」嘛！
1號提100也行，但提97最保險，保證3、4、5號一定投贊成！

1號不能提100
因為4號5號基於「最大利益原則」.......
會賭賭看2號可能不是全拿,所以一定會把1號丟下去餵魚.....
如果2號到時候也全拿100也沒差,賺了個機會

[linux_xp]

剩4-5時，4號不一定要死的：

[4]提議 5表決：5號只要一否決四號就會被丟去餵鯊魚，這時4號可以提100顆全部給5號，如果五號還是否決，他還要動手把4號丟到海裡去，反正他100顆都到手了，基於最大利益原則，何必浪費力氣動手，再說5號只剩自己一個人，他有十足把握可以打的過4號並把他丟入海中嗎？在這種情況下我認為5號會同意四號的提議--全部100顆給5號。

而這麼一來，[4-5] ，4號不一定會死，[3-4-5] 3號是死定了而且一顆都沒得分

關鍵還是[2-3-4-5] ，3號一定同意，4號不一定會死，所以5號也不一定否決

[rainwen]

最初由 linux_xp 發表
剩4-5時，4號不一定要死的：

[4]提議 5表決：5號只要一否決四號就會被丟去餵鯊魚，這時4號可以提100顆全部給5號，如果五號還是否決，他還要動手把4號丟到海裡去，反正他100顆都到手了，基於最大利益原則，何必浪費力氣動手，再說5號只剩自己一個人，他有十足把握可以打的過4號並把他丟入海中嗎？在這種情況下我認為5號會同意四號的提議--全部100顆給5號。

而這麼一來，[4-5] ，4號不一定會死，[3-4-5] 3號是死定了而且一顆都沒得分

關鍵還是[2-3-4-5] ，3號一定同意，4號不一定會死，所以5號也不一定否決

輪到3號、4號提議時是可以將所有寶石給5號以保命，
但卻會給5號一個殺自己的機會，反倒不如在2號提議時
讓2號過關。

[舒跑阿薩姆]

呃？今天天氣不錯！

[pete001]

這類問題很好玩, 然而也真的是不要多想
像是在算數學一樣, 想太多反而無解.
像是這一題, 所有可能的變局都卡在"條件"
跟" 目的" 這兩個限制之下.
看似很清楚, 要是仔細分析就變模糊了!
解題的關鍵點在於"倒推", 這也可能就是灰色地帶的開始.

由"目的" 得知, 每個人的起始需求應該是獨得100.
然而這樣將導致No.5 獨得的局面.
基於"條件" 得知, 每個海盜都是很聰明的人，
都能很理智的判斷得失, 所以"推論" 每個人都不想死.
且為了不死, 不惜犧牲任意數的寶藏.( 題目並沒有這樣寫,
只是合理的推論,所以說"推論").----------------------------**
解題到了這邊, 原本正面推演成為無解之法, 因為牽涉了
五個人會預測未來發展而影響現在之決策, 所以要確定
現在的決策必須從未來的發展來看, 解題演變成倒推.

正常倒推過程, 正如各位所說, 不過我試圖用沙盤推演倒推,
卻完全沒辦法進行, 因為有個頭痛問題, 最後到底是什麼情況?
沒辦法, 就用正常的想法來想吧
依照上面的解題過程, 順理成章的由(No.5 獨吞, No.4 預先知道
會有這種現象, 所以會在No.3 及以前之提議時改變原本的態度.)
這個路徑來倒推, 然而這裡出現疑問, 基於"條件", No.5 不可能
不知道No.4 改變決策的可能性, 同理, 每個人都了解這個現象.
所以說, No.4必同意No.3是有疑問的.

時間來到No.3 提案的時間:
一方面No.3 可以選擇支持No.2來免死,
另一方面No.3 也可以爭取No.5 的支持吧!
來看看另外一個故事:
No.3 基於不想死的原因, 現在No.3 提議0-0-100, No.5 必同意(***), No.4 該怎麼辦?
No.4 同意的話, 等於是讓No.5 獨吞一樣的局面, No.4 反對的話, 也是讓
No.5 獨吞的局面. 對於No.4 來說, 兩種選擇獲利皆為零. 所以為了避免這種局勢
No.4 必須改變在No.2及之前 的決定, 想辦法在前兩次投票獲得超過0 的利益.
對No.3 來說, 獲利0 還有性命之憂( 要看No.4,5 臉色), 所以也會改變前面的決定,
只要確定保命, 就不可能讓局勢發展到No.3 提案這個時點來.
對No.5 來說, 因為No.3 跟 No.4 的聯合行為, 使他無法達成獨吞的目的,

反推到No.2 提案的時間點, 可知的是 No.3 必同意,No.4 利益大於1則同意,
這時候No.2 只要提出99-0-1-0,
No.3 為求保命, 必同意
No.4 也因為可得超過0 的利益1, 必同意, 這個提案獲得兩票同意, 通過!

反推到No.1 提案的時間點, 可知的是 No.2 未滿99 必反對,
No.3 大於等於1 必同意, No.4 未滿1 必反對,
No.1 提出100-0-0-0-0, 2反對, 3未知, 4反對, 5未知
99-0-0-1-0, 2反對, 3未知, 4同意, 5未知
98-0-1-1-0, 2反對, 3同意, 4同意, 5未知
97-0-1-1-1, 2反對, 3同意, 4同意, 5未知****
0-99-0-1-0, 2同意, 3未知@@, 4同意, 5未知

這個裡面** 記號的部分, 我覺得可能是最大的問題, 這五位海盜到底是
命重要還是錢重要? 以上的推論假設命重要.
***記號的部分, 也是個好玩的地方, 題目並沒有說明當獲利相同的時候,
海盜會採取先獲利的決策, 還是後獲利的決策? 以及, 殺或者不殺?
****記號的部分, 困擾在5 沒辦法獨吞的時候, 不能斷言利益1 他會同意,
也不能斷言利益1 他會反對,No.5 變成一個大問號.
@@記號的部分, 說明No.3 在一樣可以保命, 一樣獲利0 的狀況下, 無法
得知他究竟要選擇在No.1 提案, 或者No.2 提案做出同意決定.

由此也證明, 我大概也拿不到高薪了...
會被老闆踢出應徵約談室.

[a650520]

這好像就是答案...哪位可以翻譯一下嗎

先跟大家說聲抱歉..小弟眼拙不知rainwen兄已經貼過...真是對不起...

The most common answer (and the one that lacks thought) is: the most senior pirate takes half the coins and divides up the rest among the least senior pirates. This, however, entirely misses the point. Why is this missing the point? Because that answer requires no logic at all.

How one should reason is that Pirate 5, being the most senior pirate, knows that he needs to get two other people to vote for his solution in order for him not to be executed. So who can he get to vote for him, and why would they choose to vote for him? Those are the questions that you are meant to ask. That is part of the point of the puzzle. If you start thinking that Pirate 4 will never vote for him, because he would rather have Pirate 5 die and then be in charge and take it all for himself, you are somewhat on the right track. Consider if there were only one pirate. In this case there is no problem: he takes all coins for himself since no one else is around to complain. What about two pirates? In this case, Pirate 2, being the most senior, would just vote for himself and that would be fifty percent of the vote right there, so obviously he is going to keep all the money for himself and not risk execution at all. What about with three pirates? Here Pirate 3 has to convince at least one other person to join in his plan. So who can he convince and how? Here is the leap that needs to be made to solve this problem and gets to the crux of the issue. Pirate 3 realizes that if his plan is not adopted he will be executed and they will be left with two pirates. He already knows what happens when there are two pirates (which is what we just figured out). So Pirate 3 proposes that he will take ninety-nine gold coins and give one coin to Pirate 1. He hopes that Pirate 1 thinks along the lines that one gold coin is better than none, and since Pirate 1 knows that if he does not vote for Pirate 3, he gets nothing, he should go along with Pirate 3's plan. (Remember all the pirates are very intelligent so we are assuming that Pirate 1 thinks along these lines.)

What about with four pirates? Here Pirate 4 has to convince one other person to join in his plan. He knows if he gets executed then Pirate 3 will get ninety-nine coins and Pirate 1 will get one coin. So Pirate 4 could propose giving Pirate 1 two coins, and surely Pirate 1 would vote for him, since two is better than one. But remember these pirates are extremely greedy, so Pirate 4 would rather not part with two whole coins if he does not have to. He realizes that if he gets executed, then Pirate 3's scenario will happen and Pirate 2 will get zero coins. So Pirate 4 proposes that he will give one coin to Pirate 2, and Pirate 2, seeing that one is better than zero, will obviously vote for this plan. Now, a common objection is that Pirate 2 is not guaranteed to vote for this plan since he might hope for the case when there are only two pirates and then he gets all the gold coins. But, again, remember the assumption: the pirates are all extremely intelligent. Pirate 2 realizes that Pirate 3 is smart enough to make the optimal proposal, so he realizes that there will never be two pirates left, because Pirate 3 does not want to die and we just showed that Pirate 3 has a winning proposal. So can we sum this up? Well, it helps to graph it out:

Pirate 1 2 3 4 5
5. ? ? ? ? ?
4. 0 1 0 99 -
3. 1 0 99 - -
2. 0 100 - - -
1.100
Seeing that pattern, it should all be quite clear. The key here is realizing that when a given pirate's plan does not succeed then that means you are in the same situation with one less pirate:

Pirate 1 needs zero other people to vote for him. So he votes for himself and takes all the gold.
Pirate 2 needs zero other people to vote for him. So he votes for himself and takes all the gold. Pirate 1 gets zero.
Pirate 3 needs one other person to vote for him. He gives one coin to Pirate 1 for his vote - if we are reduced to two pirates, Pirate 1 gets zero so Pirate 1 knows one is better than none. Pirate 3 takes ninety-nine. Pirate 2 gets zero.
Pirate 4 needs one other person to vote for him. He gives one coin to Pirate 2 - if we reduce to three pirates, Pirate 2 gets zero so Pirate 2 knows one is better than none. Pirate 4 takes ninety-nine. Pirate 3 gets zero. Pirate 1 gets 0.
Following that logic, Pirate 5 needs two other people to vote for him. It is clear now that the two people he needs to convince are the two who get shafted in the four pirate scenario - Pirate 3 and Pirate 1. So he can give them each one coin (which is better than zero - what they would get otherwise) and keep ninety-eight for himself. If you want to make sure you understand the logic, consider what happens if there are fifteen pirates. In this case, Pirate 15 needs seven other people to vote for him, so he recruits Pirates 13, 11, 9, 7, 5, 3, and 1 with one coin each and keeps ninety three coins for himself. Those pirates will all vote for him because they know that they get zero coins if he dies and Pirate 14 is in charge.

[darkwarrior]

這個答案與當初給的題目不合喔.....
答案是要每個人都加入投票，達一半即可
here's the answer...
首先由一個、兩個海盜推上來
會發現海盜2只要提0,100這個方案，最後一個根本不能選
然後三個時，海盜3只要有另一個支持即可
當然不是給2，因為2只要踢掉3就能拿到100個
所以3會給1一顆寶石，1一定會投給她(不然拿不到)
由此類推，4只要給1兩顆寶石，就能達到一半支持
可是若不給1呢...3是絕對不會支持的，那2呢
2知道3不會分給他，但3會分給1，所以只要4給他一顆就支持

重點是5個呢...他要另外找兩個人支持
當然5不會給4，因為4不會投她
那就是給誰呢...給3和1
因為他們都知道前面我們的推理過程
只要給1他一定會答應，反正最多1個寶石，誰給我就投誰
3也會答應，因為接下來4不會給他...2就不一定了，因為還有4會分給他
但是5只要另找2個支援，當然是給3和1