Question

我的问题是：我有12名球员，其中3名分别被命名为A，B和C。 12名玩家被分成2组，每组6人。我需要计算出球员A和B在同一支球队中，而球员C在相反的一支球队中的可能性。数学不是我的强项，因为我很确定这不是一件难事，但是如果您能帮助我，我将不胜感激。这是我到目前为止写的：

import random

playersnumb = 12
players = list(range(12))

A = random.choice([x for x in range(12)])

B = random.choice([x for x in range(12) if x != A])

C = random.choice([x for x in range(12) if (x != A) and (x != B)])

random.shuffle(players)
team1 = (players[:6])
team2 = (players[6:])

if A in team1:
    print("Player A is in team 1")
else:
    print("Player A is in team 2")
if B in team1:
    print("Player B is in team 1")
else:
    print("Player B is in team 2")
if C in team1:
    print("Player C is in team 1")
else:
    print("Player C is in team 2")

感谢您的帮助。

Answer 1

填写1个列表（共六个）的方式数= 12！/（6！* 6！）foreach($value as $a) { $array = $a->VALUE; $ar = unserialize($array); echo $ar[0]; echo '</br>'; }

填充六个列表（包括A和B而不是C）的方式的数量= 9！/（4！* 5！）comb(12,6)

还要查找（不是A，不是B和C）= 9！/（5！* 4！）comb(9, 4)

comb(9, 5)

Answer 2

我根据您的代码写了一点。这个想法是要遍历您的测试代码多次，它不是100％准确，但是我认为对您来说足够了：

import random


def calculate(playercount: int = 12) -> bool:
    players = list(range(playercount))

    player_a = random.choice([x for x in range(playercount)])

    player_b = random.choice([x for x in range(playercount) if x != player_a])

    player_c = random.choice([x for x in range(playercount) if (x != player_a) and (x != player_b)])

    random.shuffle(players)
    team1 = (players[:playercount//2])
    team2 = (players[playercount//2:])

    # That are your "positive" events
    return (player_a in team1 and player_b in team1 and player_c in team2) or\
           (player_a in team2 and player_b in team2 and player_c in team1)


def calculate_all(runtimes: int = 100000) -> float:
    counter = 0
    # count all poyitive events
    for i in range(runtimes):
        if calculate():
            counter += 1

    # return how often they appeared, based on all tests
    return counter / runtimes


print("The probability is about {} %".format(calculate_all() * 100))

python中的概率计算

2 个答案: