麻將的期望值(二) - 麻將
By Jack
at 2012-12-16T01:27
at 2012-12-16T01:27
Table of Contents
我決定放大絕了!
麻將總共144張牌對吧
所以起始牌局有144!種(大概10^250)
然後根據這麼多種牌局的先後次序大概會造成(10^250)!種 組合
[想簡單一點假設整副麻將只有
5張牌有5!=120種組合 根據打出來的順序
頂多有120!種組合]
以上可以扣掉4張雷同的 還有環狀排列 但是我只是示意所以不細算也無差
你把所有的資訊建檔好 得出的所有組合 就是所有麻將的事件數
然後根據手牌和桌面現出來的牌 總可以推斷出 哪樣好哪樣不好 進而得出必勝法
但是 為什麼做不到呢?
1 :據我所知宇宙原子大概10^75個 所以無法暴力破解
(題外話圍棋10^600就無法暴力破解了)
但是難保哪天找出新的演算法可以更快的處理此類問題
2 :建檔困難 (10^250)! 數據理想上人類科技做不到
3 :不可能每手牌給你想個5000年再打
以上是針對部份人認為數學不可靠的見解
其實不是不可靠 而是認識的太少....或許我上面的數字有所謬誤
但是我只是想把表達 確實是可以把所有牌局的資料進行統計來達到最佳策略的
--
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