基于Python 3.10：扩展卡组规模与抽取求和的概率对比

规则说明 #

• 基础卡组为[2,3,4,5,6,7,8,9,10]，定义为卡组规模1；
• 每提升一级卡组规模，需复制一份基础卡组加入；
• 卡组构建完成后，需额外添加[7,8,9,10]；
• 示例：规模2的卡组为[2,3,4,5,6,7,8,9,10,2,3,4,5,6,7,8,9,10,7,8,9,10]，规模3的卡组以此类推；
• 玩法：抽取次数等于卡组规模，结果为抽取卡牌的数值和，比如规模3卡组需计算3次抽取的和。

问题与实践 #

先编写了使用itertools生成符合规则卡组的代码，但不确定概率计算代码的正确性。最初的概率计算代码通过itertools.product遍历所有组合，对比不同卡组抽取和的获胜概率，但时间复杂度极高。随后尝试蒙特卡洛模拟方法，得到与全组合法相近的结果，最终问题已解决。

相关代码 #

卡组生成代码（版本1） #

import itertools

# define base deck
base_deck = [2, 3, 4, 5, 6, 7, 8, 9, 10]

# define deck sizes
deck_a_size = 2
deck_b_size = 3

# create decks
deck_a_cards = list(itertools.chain.from_iterable(itertools.repeat(base_deck, deck_a_size)))
deck_a_cards.extend(floater)
print(deck_a_cards)
deck_b_cards = list(itertools.chain.from_iterable(itertools.repeat(base_deck, deck_b_size)))
deck_a_cards.extend(floater)
print(deck_b_cards)

全组合概率计算代码 #

import itertools

# define base deck
base_deck = [2, 3, 4, 5, 6, 7, 8, 9, 10]

# define deck sizes
deck_a_size = 1
deck_b_size = 3

# create decks
deck_a_cards = list(itertools.chain.from_iterable(itertools.repeat(base_deck, deck_a_size)))
deck_a_cards.extend(floater)
print(deck_a_cards)
deck_b_cards = list(itertools.chain.from_iterable(itertools.repeat(base_deck, deck_b_size)))
deck_b_cards.extend(floater)
print(deck_b_cards)

deck_a_success = 0
deck_b_success = 0
total_combinations = 0

for i in itertools.product(deck_a_cards, repeat=deck_a_size):
    for j in itertools.product(deck_b_cards, repeat=deck_b_size):
        deck_a_sum = sum(i)
        deck_b_sum = sum(j)
        if deck_a_sum > deck_b_sum:
            deck_a_success += 1
        elif deck_a_sum < deck_b_sum:
            deck_b_success += 1
        else:
            deck_b_success += 1
        total_combinations += 1

deck_a_prob = deck_a_success / total_combinations
deck_b_prob = deck_b_success / total_combinations

print("Probability of success with {} cards drawn:".format(deck_a_size + deck_b_size))
print("Deck A:", deck_a_prob)
print("Deck B:", deck_b_prob)

蒙特卡洛模拟概率计算代码 #

import random

base_deck = [2, 3, 4, 5, 6, 7, 8, 9, 10]
deck_a_size = 2
deck_b_size = 3

num_trials = 1000000
deck_a_wins = 0
deck_b_wins = 0

for i in range(num_trials):
    deck_a_sum = sum(random.sample(base_deck, deck_a_size))
    deck_b_sum = sum(random.sample(base_deck * deck_b_size, deck_b_size))
    if deck_a_sum > deck_b_sum:
        deck_a_wins += 1
    elif deck_b_sum > deck_a_sum:
        deck_b_wins += 1

deck_a_prob = deck_a_wins / num_trials
deck_b_prob = deck_b_wins / num_trials

print("Probability of success with {} cards drawn:".format(deck_a_size+deck_b_size))
print("Deck A:", deck_a_prob)
print("Deck B:", deck_b_prob)

运行结果 #

全组合法 #

抽取5张卡牌的获胜概率：
Deck A: 0.1316926655146557
Deck B: 0.8683073344853444
运行时间：6.0s

蒙特卡洛法 #

抽取5张卡牌的获胜概率：
Deck A: 0.121357
Deck B: 0.837629
运行时间：3.8s

内容的提问来源于stack exchange，提问作者Thomas J Childers