抽5张牌获得three of a kind（三条）的概率计算及代码问题排查

• 未在循环内初始化three1/three2/three3变量，不仅会触发变量未定义报错，跨循环累计的计数也会导致判断完全失真
• 计数逻辑错误：any(hand[0][0] == x[0] for x in hand)的判断永远为真（元素会和自身比较），且单次匹配就加1的逻辑完全无法统计同面值的牌数
• 未排除四条、葫芦（三条+一对）的情况，会导致计算出的概率高于三条的真实概率
• 冗余操作：shuffle(deck)后再用sample抽牌完全多余，sample本身就会随机抽取不重复的元素，不需要提前洗牌

from random import sample
from itertools import product
from collections import Counter

# 生成牌库
suits = ["s","d","h","c"]
values = ["1","2","3","4","5","6","7","8","9","10","11","12","13"]
deck = list(product(values, suits))
sim = 100000

three_of_a_kind = 0
for i in range(sim):
    # 随机抽取5张手牌
    hand = sample(deck, 5)
    # 统计每个面值出现的次数，按降序排序
    value_counts = sorted(Counter(card[0] for card in hand).values(), reverse=True)
    # 判定三条标准：恰好一个面值出现3次，剩余两个面值各出现1次，排除葫芦、四条的情况
    if value_counts == [3, 1, 1]:
        three_of_a_kind += 1

prob_three = three_of_a_kind / sim
print(prob_three)

• 用collections.Counter统计手牌面值的出现次数，避免手动计数的逻辑错误
• 10万次模拟得到的概率约为2.1%，和官方公布的德州扑克三条出现概率（约2.11%）吻合，结果准确

内容的提问来源于stack exchange，提问作者The anime kid