红心国王

King of Hearts

专题: Probability / 概率
难度: L5
来源: OpenQuant

个人笔记

题目详情

你拿一副标准的洗牌牌，不断翻转牌，直到出现第一张 A，即牌#14。下一张牌是红心K的概率是多少

英文原题

You take a standard shuffled deck of cards and keep flipping cards over until the first ace appears, which is card #14. What's the probability the next card is the king of hearts

解析

我们知道第 14 张牌一定是 A，而它之前的 13 张牌不可能是 A。因此，我们将 48 张非 A 牌中的 13 张排列为前 13 张，A 排列为第 14 位，并对其余 38 张牌进行排列以计算可能选择的总数。
${48\choose13}13!\cdot{4\choose1}\cdot38!$
为了求出第 15 张牌是 8h 的概率，我们需要求出 Kh 位于第 15 位且为有效选择的选择数。我们将 47 张非 A 非 Kh 牌中的 13 张排列为前 13 个位置，将 A 排列为第 14 个位置，将 8h 排列为第 15 个位置，然后排列其余 37 张牌以获得：
${47\choose13}13!\cdot{4\choose1}\cdot{1\choose1}\cdot37!$
将其与我们得到的可能选择总数相加：
$\frac{{47\choose13}13!\cdot{4\choose1}\cdot{1\choose1}\cdot37!}{{48\choose13}13!\cdot{4\choose1}\cdot38!} = \frac{35}{48\cdot38} = \boxed{\frac{35}{1824}}$

import random

suits = ['S', 'C', 'D', 'H']
ranks = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K']

class Deck:
    def __init__(self):
        self.cards = []
        for suit in suits:
            for rank in ranks:
                self.cards.append([rank, suit])

    def shuffle(self):
        # shuffles deck of cards
        return random.sample(self.cards, 52)

# number of decks where the first ace is on the 14th draw
num_valid_decks = 0
num_15th_is8h = 0

# create 10,000,000 shuffled decks
# see how many of the decks, where the first ace is on the 14th draw, have 8h as the next card
for i in range(1_000_000):
    deck = Deck().shuffle()

    # excluding all decks with an ace not being on the 14th draw
    if deck[13][0] != 'A':
        continue

    # exlcuding all decks with an ace in the first 13 draws
    for j in range(13):
        if deck[j][0] == 'A':
            break
    else:
        num_valid_decks += 1
        #checking if 15th card is 8 of hearts
        if deck[14]== ['K', 'H']:
            num_15th_is8h += 1

# should be expecting 35/1824 ~ 0.0191886
print(num_15th_is8h / num_valid_decks)

英文解析

We know the 14th card must be an ace and the 13 cards that precede it can not be an ace. Thus we permute 13 of the 48 non-ace cards to be among the first 13, an ace to be in spot 14, and permute the remaining 38 cards to count the total number of possible selections.
${48\choose13}13!\cdot{4\choose1}\cdot38!$
To find the probability the 15th card is 8h we need to find the number of selections with Kh being in spot 15 while being a valid selection. We permute 13 of the 47 non-ace non-Kh cards for the first 13 spots, an ace for spot 14, the 8h for spot 15, and permute the remaining 37 cards to get:
${47\choose13}13!\cdot{4\choose1}\cdot{1\choose1}\cdot37!$
Putting this over the total number of possible selections we get:
$\frac{{47\choose13}13!\cdot{4\choose1}\cdot{1\choose1}\cdot37!}{{48\choose13}13!\cdot{4\choose1}\cdot38!} = \frac{35}{48\cdot38} = \boxed{\frac{35}{1824}}$

import random

suits = ['S', 'C', 'D', 'H']
ranks = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K']

class Deck:
    def __init__(self):
        self.cards = []
        for suit in suits:
            for rank in ranks:
                self.cards.append([rank, suit])

    def shuffle(self):
        # shuffles deck of cards
        return random.sample(self.cards, 52)

# number of decks where the first ace is on the 14th draw
num_valid_decks = 0
num_15th_is8h = 0

# create 10,000,000 shuffled decks
# see how many of the decks, where the first ace is on the 14th draw, have 8h as the next card
for i in range(1_000_000):
    deck = Deck().shuffle()

    # excluding all decks with an ace not being on the 14th draw
    if deck[13][0] != 'A':
        continue

    # exlcuding all decks with an ace in the first 13 draws
    for j in range(13):
        if deck[j][0] == 'A':
            break
    else:
        num_valid_decks += 1
        #checking if 15th card is 8 of hearts
        if deck[14]== ['K', 'H']:
            num_15th_is8h += 1

# should be expecting 35/1824 ~ 0.0191886
print(num_15th_is8h / num_valid_decks)