migi自摸的台數問題 - 麻將

By Olga
at 2012-05-08T09:56
at 2012-05-08T09:56
Table of Contents
想請教版上各位版友一個有關migi自摸的問題:
在台數的計算上
migi 8台
自摸 1台
不求 1台
共計10台.
請問以上算法對不對?
因為實務上遇到有人覺得migi自摸就是8+1台=9台.
並沒有把"不求"算進去.
但是migi並不是"必定不求"
因為如果是胡別人的牌 就沒有"不求"了
一定要自摸 才有不求
所以小弟認為migi自摸該算不求人的1台.
可是有些朋友認為migi的8台就含門清及不求.
含門清我是認同的 因為migi不論胡牌或自摸 都必定門清
但不求的這一台 似乎不該含在那8台內 因為migi胡別人是沒有"不求"的.
不知道大家的看法如何呢 請賜教囉!
--
在台數的計算上
migi 8台
自摸 1台
不求 1台
共計10台.
請問以上算法對不對?
因為實務上遇到有人覺得migi自摸就是8+1台=9台.
並沒有把"不求"算進去.
但是migi並不是"必定不求"
因為如果是胡別人的牌 就沒有"不求"了
一定要自摸 才有不求
所以小弟認為migi自摸該算不求人的1台.
可是有些朋友認為migi的8台就含門清及不求.
含門清我是認同的 因為migi不論胡牌或自摸 都必定門清
但不求的這一台 似乎不該含在那8台內 因為migi胡別人是沒有"不求"的.
不知道大家的看法如何呢 請賜教囉!
--
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麻將
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