用Python分析石头剪刀布中的概率问题

　　2021-09-20_NebulaChien的博客-CSDN博客
　　两个人用猜拳的方式决定胜负，所有出拳的组合有3x3=9种，其中A胜、B胜和平局的组合各有三种，也就是说每一种结局对应的概率为1/3，那么这个过程到底是如何是实现的呢？下面我们用Python模拟这一过程。
　　代码如下，用到列表索引 、random库 #Possiblity of mora game from random import * mora_GamerA=[＂Rock＂,＂Paper＂,＂Scissor＂] mora_GamerB=[＂Rock＂,＂Paper＂,＂Scissor＂] count_A_win=0.0 count_B_win=0.0 count_Draw=0.0 Flag=1 while Flag:     round=int(input(＂Enetr rounds:(rounds>=10)＂))     ＂＂＂     use while of if to make sure their is correct index for range     ＂＂＂     if round<10:         print(＂The input bureau does not meet the requirements＂)         Flag=0         break     else:         for round in range(1,round+1):             print(＂Game round {}＂.format(round))             guess_A=mora_GamerA[randint(0,2)]             guess_B=mora_GamerB[randint(0,2)]             ＂＂＂             Attention the range of list index             ＂＂＂             if (guess_A==＂Rock＂ and guess_B==＂Scissor＂) or (guess_A==＂Paper＂ and guess_B==＂Rock＂) or (guess_A==＂Scissor＂ and guess_B==＂Paper＂):                 count_A_win+=1                 print(＂A is {},B is {}＂.format(guess_A,guess_B))                 print(＂Gamer A is winner＂)             elif guess_A==guess_B:                 count_Draw+=1                 print(＂A and B is {}＂.format(guess_A))                 print(＂Draw＂)             else:                 count_B_win+=1                 print(＂A is {},B is {}＂.format(guess_A,guess_B))                 print(＂Gamer B is winner＂)         print(＂In {} rounds,gamer A have won {} rounds and gamer B have won {} rounds,{} rounds is draw.＂.format(round,count_A_win,count_B_win,count_Draw))         print(＂The possiblity of A win is {}＂.format(count_A_win/round))         print(＂The possiblity of B win is {}＂.format(count_B_win/round))         print(＂The possiblity of draw is {}＂.format(count_Draw/round))         Flag=int(input(＂Enter 1 to continue and 0 to exit :＂))         if Flag!=1 and Flag!=0:             Flag=0             print(＂Force exit because mistake input＂)
　　可以看到，并不是每次输入一个rounds都对应1/3的结果，因为1/3是理想条件下根据大量实验统计得到的，而随着rounds逐渐增大，乃至于大于1000000，得出的结果会越来越逼近0.333333...
　　emmmm，其实写这段代码是放假无聊找点事做，因为计二报的是Python，虽然有比较强的基础，但还是看看书，顺带熟悉一下各种库的用法，random库是真的蛮有意思。
　　第一次在文章里面嵌入代码不太熟练，缩进什么的都惨不忍睹[捂脸]
　　各位小伙伴将就一下[谢谢]
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