我可以很容易地找到100個頭數和100個翻轉頭中硬幣翻轉頭的機率。該代碼如下。然而,我無法弄清楚如何輕鬆獲得連續連續10次擲硬幣的機率。連續10次硬幣翻轉頭的機率
import random
n = int(input("Enter the number of flips: "))
sum = 0
heads = 0
tails = 0
for n in range(1, n + 1):
number = random.randrange(0,2)
if number == 0:
heads += 1
else:
tails += 1
sum += number
odds = sum/n
print("No. of heads: ", heads)
print("No. of tails: ", tails)
print("Odds: ", odds)
您知道'(1/2)^ 10'嗎? – schwobaseggl
'odds = sum/n'應該不會縮進那裏。你用循環計數器覆蓋了'n'變量 –
@schwobaseggl:我認爲OP需要模擬翻轉,而不是計算* expected *值。 – Prune