如圖所示，求機率。
難度 ✩✩✩
最近看到羅智強在立法院質詢財政部「雲端種樹趣 e 起集點樹」抽獎的問題，有四個人竟然連中兩次大獎，這種幾乎不可能出現的機率，竟然發生！發生在某一個人就算，可是卻有四個人，告訴我抽獎過程完全公平沒問題，應該沒有人會相信，今年多項行政作為根本是在破壞人民對政府的信賴，尤其是大法官對死刑以及國會改革的釋憲。
尹倉20241229重大更新
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大家知道，尹倉是目前重碼最少的倉頡碼表。本次更新，重碼再降，一騎絕塵。 經過本次修改，「激淚、沉观」不再是重碼。
鶴飛四季一同發力，找到了「繏簈」這兩個遺留錯碼。 獲取地址 /ejcin 在線試玩 /ejime 重碼統計 2:1159 3:115 4:23 5:2
6:1 更新說明 「戾肩」將改取陸標 單純為...
1 天前
改为从9人中抽奖，概率大约是0.0636 :)
回覆刪除謝謝提供程式來估算機率，不過這方法在計算特定四個人
刪除# Revised approach: Reducing the number of trials and tracking only the necessary outcomes to speed up simulation
回覆刪除import numpy as np
num_people = 9
num_target_people = 4
prizes_per_round = [5,5,9]
target_wins = 2
# Adjusted parameters for a quicker estimation
num_trials = 10**6 # Reduced number of simulations for faster runtime
# Function to simulate one trial
def simulate_trial():
# Draw winners for each round
winners_round_1 = set(np.random.choice(num_people, prizes_per_round[0], replace=False))
winners_round_2 = set(np.random.choice(num_people, prizes_per_round[1], replace=False))
winners_round_3 = set(np.random.choice(num_people, prizes_per_round[2], replace=False))
# Track each target person's wins
wins = [0] * num_target_people
for i in range(num_target_people):
person_id = i # IDs 0, 1, 2, 3 for the target people
wins[i] += person_id in winners_round_1
wins[i] += person_id in winners_round_2
wins[i] += person_id in winners_round_3
# Check if exactly 4 people have exactly 2 wins
return sum(win == target_wins for win in wins) == num_target_people
# Run the simulation over the specified number of trials
target_people_wins = 0
for i in range(1,num_trials+1):
target_people_wins += simulate_trial()
# Calculate probability
probability_estimate = target_people_wins / i
print(probability_estimate,"")