# Probability of getting at least K heads in N tosses of Coins

Given N number of coins, the task is to find probability of getting at least K number of heads after tossing all the N coins simultaneously.

Example :

Suppose we have 3 unbiased coins and we have to
find the probability of getting at least 2 heads,
so there are 23 = 8 ways to toss these
coins, i.e.,
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Out of which there are 4 set which contain at
HHH, HHT, HH, THH

So the probability is 4/8 or 0.5


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The probability of exactly k success in n trials with probability p of success in any trial is given by: So Probability ( getting at least 4 heads )=  Method 1 (Naive)
A Naive approach is to store the value of factorial in dp[] array and call it directly whenever it is required. But the problem of this approach is that we can only able to store it up to certain value, after that it will lead to overflow.

Below is the implementation of above approach

## C++

 // Naive approach in C++ to find probability of  // at least k heads  #include  using namespace std;  #define MAX 21     double fact[MAX];     // Returns probability of getting at least k  // heads in n tosses.  double probability(int k, int n)  {      double ans = 0;      for (int i = k; i <= n; ++i)             // Probability of getting exactly i          // heads out of n heads          ans += fact[n] / (fact[i] * fact[n - i]);         // Note: 1 << n = pow(2, n)      ans = ans / (1LL << n);      return ans;  }     void precompute()  {      // Preprocess all factorial only upto 19,      // as after that it will overflow      fact = fact = 1;         for (int i = 2; i < 20; ++i)          fact[i] = fact[i - 1] * i;  }     // Driver code  int main()  {      precompute();         // Probability of getting 2 head out of 3 coins      cout << probability(2, 3) << "\n";         // Probability of getting 3 head out of 6 coins      cout << probability(3, 6) <<"\n";         // Probability of getting 12 head out of 18 coins      cout << probability(12, 18);         return 0;  }

## Java

 // JAVA Code for Probability of getting   // atleast K heads in N tosses of Coins  class GFG {             public static double fact[];              // Returns probability of getting at least k      // heads in n tosses.      public static double probability(int k, int n)      {          double ans = 0;          for (int i = k; i <= n; ++ i)                      // Probability of getting exactly i              // heads out of n heads              ans += fact[n] / (fact[i] * fact[n-i]);                  // Note: 1 << n = pow(2, n)          ans = ans / (1 << n);          return ans;      }              public static void precompute()      {          // Preprocess all factorial only upto 19,          // as after that it will overflow          fact = fact = 1;                  for (int i = 2; i < 20; ++i)              fact[i] = fact[i - 1] * i;      }              // Driver code      public static void main(String[] args)       {          fact = new double;          precompute();                  // Probability of getting 2 head out          // of 3 coins          System.out.println(probability(2, 3));                  // Probability of getting 3 head out          // of 6 coins          System.out.println(probability(3, 6));                  // Probability of getting 12 head out          // of 18 coins          System.out.println(probability(12, 18));              }   }  // This code is contributed by Arnav Kr. Mandal

## Python3

 # Naive approach in Python3   # to find probability of  # at least k heads     MAX=21    fact=*MAX    # Returns probability of   # getting at least k  # heads in n tosses.  def probability(k, n):      ans = 0     for i in range(k,n+1):             # Probability of getting exactly i          # heads out of n heads          ans += fact[n] / (fact[i] * fact[n - i])         # Note: 1 << n = pow(2, n)      ans = ans / (1 << n)      return ans     def precompute():             # Preprocess all factorial       # only upto 19,      # as after that it       # will overflow      fact = 1     fact = 1        for i in range(2,20):          fact[i] = fact[i - 1] * i     # Driver code  if __name__=='__main__':      precompute()         # Probability of getting 2       # head out of 3 coins      print(probability(2, 3))         # Probability of getting       # 3 head out of 6 coins      print(probability(3, 6))         # Probability of getting       # 12 head out of 18 coins      print(probability(12, 18))         # This code is contributed by   # mits

## C#

 // C# Code for Probability of getting   // atleast K heads in N tosses of Coins  using System;     class GFG   {             public static double []fact;             // Returns probability of getting at least k      // heads in n tosses.      public static double probability(int k, int n)      {          double ans = 0;          for (int i = k; i <= n; ++ i)                     // Probability of getting exactly i              // heads out of n heads              ans += fact[n] / (fact[i] * fact[n - i]);                 // Note: 1 << n = pow(2, n)          ans = ans / (1 << n);          return ans;      }             public static void precompute()      {          // Preprocess all factorial only upto 19,          // as after that it will overflow          fact = fact = 1;                 for (int i = 2; i < 20; ++i)              fact[i] = fact[i - 1] * i;      }             // Driver code      public static void Main()       {          fact = new double;          precompute();                 // Probability of getting 2 head out          // of 3 coins          Console.WriteLine(probability(2, 3));                 // Probability of getting 3 head out          // of 6 coins          Console.WriteLine(probability(3, 6));                 // Probability of getting 12 head out          // of 18 coins          Console.Write(probability(12, 18));             }  }  // This code is contributed by nitin mittal.

## PHP

 

Output:

0.5
0.65625
0.118942


Time Complexity: O(n) where n < 20
Auxiliary space: O(n)

Method 2 (Dynamic Programming and Log)
Another way is to use Dynamic programming and logarithm. log() is indeed useful to store the factorial of any number without worrying about overflow. Let’s see how we use it:

At first let see how n! can be written.
n! = n * (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1

Now take log on base 2 both the sides as:
=> log(n!) = log(n) + log(n-1) + log(n-2) + ... + log(3)
+ log(2) + log(1)

Now whenever we need to find the factorial of any number, we can use
this precomputed value. For example:
Suppose if we want to find the value of nCi which can be written as:
=> nCi = n! / (i! * (n-i)! )

Taking log2() both sides as:
=> log2 (nCi) = log2 ( n! / (i! * (n-i)! ) )
=> log2 (nCi) = log2 ( n! ) - log2(i!) - log2( (n-i)! )

Putting dp[num] = log2 (num!), we get:
=> log2 (nCi) = dp[n] - dp[i] - dp[n-i]

But as we see in above relation there is an extra factor of 2n which
tells the probability of getting i heads, so
=> log2 (2n) = n.

We will subtract this n from above result to get the final answer:
=> Pi (log2 (nCi)) = dp[n] - dp[i] - dp[n-i] - n

Now: Pi (nCi) = 2 dp[n] - dp[i] - dp[n-i] - n

Tada! Now the questions boils down the summation of Pi for all i in
[k, n] will yield the answer which can be calculated easily without
overflow.


Below are the codes to illustrate this:

## C++

 // Dynamic and Logarithm approach find probability of  // at least k heads  #include  using namespace std;  #define MAX 100001     // dp[i] is going to store Log ( i !) in base 2  double dp[MAX];     double probability(int k, int n)  {      double ans = 0; // Initialize result         // Iterate from k heads to n heads      for (int i=k; i <= n; ++i)      {          double res = dp[n] - dp[i] - dp[n-i] - n;          ans += pow(2.0, res);      }         return ans;  }     void precompute()  {      // Preprocess all the logarithm value on base 2      for (int i=2; i < MAX; ++i)          dp[i] = log2(i) + dp[i-1];  }     // Driver code  int main()  {      precompute();         // Probability of getting 2 head out of 3 coins      cout << probability(2, 3) << "\n";         // Probability of getting 3 head out of 6 coins      cout << probability(3, 6) << "\n";         // Probability of getting 500 head out of 10000 coins      cout << probability(500, 1000);         return 0;  } 

## Java

 // Dynamic and Logarithm approach find probability of  // at least k heads  import java.math.*;  class GFG {         static int MAX = 100001;     // dp[i] is going to store Log ( i !) in base 2  static double dp[] = new double[MAX];     static double probability(int k, int n)  {      double ans = 0.0; // Initialize result         // Iterate from k heads to n heads      for (int i=k; i <= n; ++i)      {          double res = dp[n] - dp[i] - dp[n-i] - n;          ans += Math.pow(2.0, res);      }         return ans;  }     static void precompute()  {      // Preprocess all the logarithm value on base 2      for (int i=2; i < MAX; ++i)          dp[i] = (Math.log(i)/Math.log(2)) + dp[i-1];  }     // Driver code  public static void main(String args[])  {      precompute();         // Probability of getting 2 head out of 3 coins      System.out.println(probability(2, 3));         // Probability of getting 3 head out of 6 coins      System.out.println(probability(3, 6));         // Probability of getting 500 head out of 10000 coins      System.out.println(probability(500, 1000));  }     } 

## Python3

 # Dynamic and Logarithm approach find probability of   # at least k heads     from math import log2  MAX=100001    # dp[i] is going to store Log ( i !) in base 2   dp=*MAX    def probability( k, n):          ans = 0 # Initialize result          # Iterate from k heads to n heads       for i in range(k,n+1):             res = dp[n] - dp[i] - dp[n-i] - n           ans = ans + pow(2.0, res)                 return ans         def precompute():          # Preprocess all the logarithm value on base 2       for i in range(2,MAX):           dp[i] = log2(i) + dp[i-1]         # Driver code   if __name__=='__main__':      precompute()          # Probability of getting 2 head out of 3 coins       print(probability(2, 3))          # Probability of getting 3 head out of 6 coins       print(probability(3, 6))          # Probability of getting 500 head out of 10000 coins       print(probability(500, 1000))     #this code is contributed by ash264 

## C#

 // Dynamic and Logarithm approach find probability of   // at least k heads   using System;     class GFG   {          static int MAX = 100001;      // dp[i] is going to store Log ( i !) in base 2   static double[] dp = new double[MAX];      static double probability(int k, int n)   {       double ans = 0.0; // Initialize result          // Iterate from k heads to n heads       for (int i = k; i <= n; ++i)       {           double res = dp[n] - dp[i] - dp[n-i] - n;           ans += Math.Pow(2.0, res);       }       return ans;   }      static void precompute()   {       // Preprocess all the logarithm value on base 2       for (int i = 2; i < MAX; ++i)           dp[i] = (Math.Log(i) / Math.Log(2)) + dp[i - 1];   }      // Driver code   public static void Main()   {       precompute();          // Probability of getting 2 head out of 3 coins       Console.WriteLine(probability(2, 3));          // Probability of getting 3 head out of 6 coins       Console.WriteLine(probability(3, 6));          // Probability of getting 500 head out of 10000 coins       Console.WriteLine(Math.Round(probability(500, 1000),6));   }   }      // This code is contributed by mits 

## PHP

  

Output:

0.5
0.65625
0.512613
`

Time Complexity: O(n)
Auxiliary space: O(n)

This approach is beneficial for large value of n ranging from 1 to 106

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