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想了个办法计算这种牌的概率
概率一:
至少一玩家JJ,一玩家88, 一玩家77,一玩家9Ts且花色不跟对子重复
此概率为2.5e-7 (模拟了12,000,000次,出现了3次。误差肯定比较大,但是没时间做更多模拟了。语句见备注)
J high的这种情况有10个牌型,JT9+(KQ/Q8/87), JT8+(Q9/97), JT7+98, J98+(QT/T7), J97+8T, J87+T9
同理,K-5 high的均有10个牌型。
Ahigh的有6个 AKQ+JT,AKJ+QT, AKT+QJ,AQJ+KT, AQT+KJ, AJT+KQ
4high 有4个;432+(56/A5), 43A+52, 42A+35,
3high的有1个 32A+45
共90+6+4+1 = 101种
每种的概率都相等,所以一共 2.275e-5
概率二:当9家牌发好后,flop只有一种组合能满足要求。
概率是 1/ (34 choose 3) = 0.000167
总体的概率是二者相乘,
2.275e-5 * 0.000167 = 3.799e-9
或者每263209844手牌出现一次
列位看官您看准了,上面是263,209,844 也就是两亿多。比我上贴的5亿还更常见一些。
---------------备注---------------
语句为:
How often do(es)
( ( at least 1 player match hand range JJ!Jh AND at least 1 player match hand range 88!8h AND at least 1 player match hand range 77!7h AND at least 1 player match hand range Th9h) OR ( at least 1 player match hand range JJ!Jd AND at least 1 player match hand range 88!8d AND at least 1 player match hand range 77!7d AND at least 1 player match hand range Td9d) OR ( at least 1 player match hand range JJ!Jc AND at least 1 player match hand range 88!8c AND at least 1 player match hand range 77!7c AND at least 1 player match hand range Tc9c) OR ( at least 1 player match hand range JJ!Js AND at least 1 player match hand range 88!8s AND at least 1 player match hand range 77!7s AND at least 1 player match hand range Ts9s)) |
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