水-卯拱既树 發表於 2022-1-18 13:25

请教一段关于AK, AA/KK的计算

今天看到这个：If we are dealt AdKd, the number of combinations of AA and KK our opponent can have drops from 12 to 6. Since the number of possible combinations drop from 1326 -101= 1225 (there are 51 combinations each with Ad or Kd, which makes a total of 101 unique combinations containing Ad or Kd, since we count AdKd only once), the chance our opponent has AA or KK is 1/204 or 0.5%.如果是HU，我们拿到AK，对手拿AA or KK的机率是大约0.5%吗？

royalflush 發表於 2022-1-18 13:42

可以用spreadsheet软件的combin() 这个function 来计算combination=combin(52,2) = 1326如果我们拿了两张牌，Ad, Kd 那么只剩下 =combin(50,2) = 1225对手有 AA, KK的combination 是 =combin(3,2) * 2 = 6所以是 6/1225 = 0.49%如果是HU，我们拿到AK，对手拿AA or KK的机率是大约0.5%吗？ 对

水-卯拱既树 發表於 2022-1-18 14:01

royalflush 发表于 2014-2-23 09:40可以用spreadsheet软件的combin() 这个function 来计算combination=combin(52,2) = 1326如果我们拿了两张 ...ty~!!
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