Question

背景：今天我想我会开始一个建立扑克的小项目模拟器。我设定的第一项任务是从洗牌中处理卡片甲板，并检查各种数字生成的概率反对公认价值观。我检查的第一个这样的概率是单对概率 - 也就是说，生成（数字）作为输入数量给出的一对的概率发出的牌数和每手牌的牌数是从一个单独的洗牌甲板处理的。卡片是从甲板的顶部。下面我展示该程序的开头。我首先测试了数值生成的单对概率五张牌手。计算出的值就在里面接受单对概率的百分之十对于五张牌（但总是高约十分之一）：https://en.wikipedia.org/wiki/Poker_probability

但是，当我测试数值生成的单对概率时对于七张牌，我发现我离开了4％到5％可接受的值（例如，典型的计算值= 0.47828;上述接受值= 0.438）。我进行了数十次的数值实验百万手牌。计算出的单对概率为7 牌手稳定，并且与可接受的值保持4％至5％的关闭。目前尚不清楚为什么会这样。

问题：为什么会这样？我怀疑我的代码没有采取考虑到了什么，但我无法察觉到什么。 Python代码如下。。

注意：问题31381901与此类似。但是在下面的代码中，重复计数的问题是通过将处理的手转换为集合来解决的，这将消除重复值，从而将集合的大小（在7张牌的情况下）从7减少到6。减少表示一对。如果存在三种类型，则该集合的大小将为5，因为通过集合转换将消除三种类型中的三种卡中的两种。

from random import shuffle

def make_deck():
    '''Make a 52 card deck of cards.  First symbol
    is the value, second symbol is the suit.  Concatenate
    both symbols together.
    Input:  None
    Output:  List
    '''
    value = ['A', '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K']
    suit = ['C','H','S','D']
    deck = [j+i for j in value for i in suit]
    return deck


def shuffle_deck(deck, times_to_shuffle=7):
    '''Shuffle a deck of cards produced by make_deck().
    Default:  7 times.
    Input:  list, int
    Output:  list (modified in-place)
    '''
    for n in range(times_to_shuffle):
        shuffle(deck)


def test_for_single_pair(hand, cards_per_hand):
    '''Tests for presence of a single pair in 
    a dealt hand by converting the hand to a set.
    The set representation of a hand with a single
    pair will have one less member than the original
    hand.
    Input: list, int
    Output:  int
    '''
    hand_values_lst = [card[0] for card in hand]
    hand_values_set = set(hand_values_lst)
    set_size = len(hand_values_set)

    if set_size == (cards_per_hand - 1):
        return 1
    else:
        return 0


def deal_series_of_hands(num_hands,cards_per_hand):
    '''Deals a series of hands of cards and tests
    for single pairs in each hand.  Creates a deck
    of 52 cards, then begins dealing loop.  Shuffles
    deck thoroughly after each hand is dealt.
    Captures a list of the dealt hands that conform
    to the spec (i.e., that contain one pair each),
    for later debugging purposes
    Input:  int, int
    Output:  int, int, list
    '''

    deck = make_deck()
    single_pair_count = 0
    hand_capture = []

    for m in range(num_hands):
        shuffle_deck(deck)
        hand = deck[0:cards_per_hand]    #first cards dealt from the deck
        pair_count = test_for_single_pair(hand, cards_per_hand)
        if pair_count == 1:
            single_pair_count += pair_count
            hand_capture.append(hand)

    return (single_pair_count, num_hands, hand_capture)



cards_per_hand = 7   #User input parameter
num_hands = 50000    #user input parameter

single_pair_count, num_hands_dealt, hand_capture = deal_series_of_hands(num_hands, cards_per_hand)
single_pair_probability = single_pair_count/ num_hands_dealt
single_pair_str = 'Single pair probability (%d card deal; poker hands): '%(cards_per_hand)
print(single_pair_str, single_pair_probability)

Answer 1

如果手牌包含一对，但也包含更高价值的单位，例如直线或同花，那么您的代码仍会将其视为一对，而概率文章则没有。

处理七张扑克牌时计算单对概率的异常概率结果

1 个答案: