unibiased flip：找出頭的次數和尾數的概率

Question

我正在嘗試開發一個函數，該函數將打印的頭的概率以h的次數打印出的次數除以h或t的總次數打印。

這裏是我的代碼 高清unbiasedFlip（N，P）：

for i in range(n+1): 
    p=Pr(Heads) 
    n=Totalflips 
    if num1>=p and num2<p: 
     print(Heads) 
    elif num1>=(1-p) and num2<(1-p): 
     print(Tails) 

NUM1和NUM2是它們應該被通過，如果函數生成的兩個隨機數。並提供可能性。當我運行程序時出現錯誤，我沒有定義公關或頭像。

Answer 1

注：此代碼可能不是你真正需要的，但我認爲它可以幫助你以某種方式......我希望如此......

反正看我的ptential前解決方案，我建議你嘗試學習Python（synthax，如何創建函數，創建隨機數等）。你會發現它很容易學習，你會完全喜歡它！ ：P

你可以找到幾種學習Python的方法（書籍，在線課程/文檔，嗜好Python XD的朋友等）。

檢查下面的鏈接，例如：http://docs.python.org/tutorial/

請記住，有一個明確的和可以理解的代碼可以幫助我們瞭解什麼是你的問題，並給出你得到一個更好的回答你的問題的最好機會;） 。

---

下面是一個簡單的代碼，我建議你集中在仔細閱讀註釋：輸出的

import random 

# The function "prob_head" below return the number of head divided by the number of coin toss 
# The input variable "number_toss" is number of times we toss a coin 
def prob_head(number_toss): 

    # "heads" is our number of heads. 
    # Initially it is equal to 0 
    heads = 0 

    # We toss a coin "number_toss" times... 
    for i in range(0, number_toss): 
     # We create a random number "flip" comprised in {0,1}   
     flip = int(random.random()*2) 

     # Let's say we follow the following rule: 
     # If "flip" = 0, then it's a head 
     # Else, if "flip" = 1, then it's a tail 

     if (flip == 0): 
      # "flip" = 0, so it's a head ! 
      # We have to increment the number of "heads" by 1: 
      heads=heads + 1 

    return float(heads)/number_toss 

# Here's a test of our function: "prob_head" 
my_number_toss = 100 
my_head_probability = prob_head(my_number_toss) 

print "Probability of heads = "+str(my_head_probability) 

例子：

概率頭= 0.41

---

上面的代碼爲您提供了模擬普通投擲硬幣的想法。

重新閱讀您的意見後，我覺得我的理解更多的是你真正想要的，所以我說這個額外的部分......

下面的代碼表示的方式來模擬一個「騙」 /「假「擲硬幣遊戲。

注重我提出的意見......輸出的

# The function "unbiasedFlip" returns the average probability of heads considering "n" coin 
# The variable "p" is a fixed probability condition for getting a head. 
def unbiasedFlip(n, p): 

    # The number of heads, initially set to 0 
    heads = 0 

    # We toss a coin n times... 
    for i in range(0, n): 

     # We generate "prob_heads": a random float number such that "prob_heads" < 1 
     prob_heads = float(random.random()) 

     # If "prob_heads" is greater of equal to "p", then we have a head 
     # and we increase the number of heads "heads" by 1: 
     if prob_heads>=p: 
      heads = heads+1 

    # We return the average probability of heads, considering n coin tosses 
    # Note: we don't need to return the average prob. for Tails since: 
    # it's equal to 1-Avg_Prob(Heads)    
    return float(heads)/n 

# An example for testing our function... 
# We consider 100 coin toss 
my_number_toss = 100 

# We want a Head only if our generated probability of head is greater or equal to 0.8 
# In fact, considering that the random number generator generates equally probability numbers 
# (which means that it provides as many chance to give a Tail or a Head) 
# it would be like saying: "we want a probability of 1-0.8 =0.2 chance of getting a head" 
my_defined_prob_heads = 0.8 

# We get our average probability of heads... 
average_prob_heads = unbiasedFlip(my_number_toss, my_defined_prob_heads) 
# We get our average probability of tails = 1-Avg_Prob(Heads) 
average_prob_tails = 1-average_prob_heads 

# We print the results... 
print "- Number of toss = "+str(my_number_toss) 
print "- Defined probability for head = "+str(my_defined_prob_heads) 
print "- Average P(Heads) for n tosses = "+str(average_prob_heads) 
print "- Average P(Tails) for n tosses = "+str(average_prob_tails) 

例子：

- Number of toss = 100 
- Defined probability for head = 0.8 
- Average P(Heads) for n tosses = 0.24 
- Average P(Tails) for n tosses = 0.76 

希望這有助於伴侶。

讓我知道你是否有問題，或者有什麼不清楚的地方。

Answer 2

首先，我們生成一個隨機coinflip序列：

import random 
n = 100 # number of flips 
p = 0.5 # P(Heads) - 0.5 is a fair coin 
flips = ['H' if (random.random() < p) else 'T' for flipnr in xrange(n)] 
print flips 

接下來，我們計算的頭和尾的號碼：

nheads = flips.count('H') 
ntails = flips.count('T') 

和計算的機會：

phead = float(nheads)/(nheads + ntails) 

請注意（在Python 2中），我們需要通過將其中一個變量轉換爲來強制進行浮點除法（這在Python 3中已修復）。