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

• Difficulty Level : Hard
• Last Updated : 25 Mar, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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#includeusing 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 codeint 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 Coinsclass 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 codeif __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 Coinsusing 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

 

## Javascript

 

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#includeusing namespace std;#define MAX 100001 // dp[i] is going to store Log ( i !) in base 2double 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 codeint 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 headsimport java.math.*;class GFG {     static int MAX = 100001; // dp[i] is going to store Log ( i !) in base 2static 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 codepublic 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 log2MAX=100001 # dp[i] is going to store Log ( i !) in base 2dp=*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 codeif __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 headsusing System; class GFG{     static int MAX = 100001; // dp[i] is going to store Log ( i !) in base 2static 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 codepublic 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

 

## Javascript

 

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
This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.