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 least 2 Heads i.e., HHH, HHT, HH, THH So the probability is 4/8 or 0.5
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<bits/stdc++.h> 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[0] = fact[1] = 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 import java.io.*; 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[0] = fact[1] = 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[100]; 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=[0]*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[0] = 1 fact[1] = 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[0] = fact[1] = 1; for (int i = 2; i < 20; ++i) fact[i] = fact[i - 1] * i; } // Driver code public static void Main() { fact = new double[100]; 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
<?php // Naive approach in PHP to find // probability of at least k heads $MAX = 21; $fact = array_fill(0, $MAX, 0); // Returns probability of getting // at least k heads in n tosses. function probability($k, $n) { global $fact; $ans = 0; for ($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; } function precompute() { global $fact; // Preprocess all factorial only // upto 19, as after that it // will overflow $fact[0] = $fact[1] = 1; for ($i = 2; $i < 20; ++$i) $fact[$i] = $fact[$i - 1] * $i; } // Driver code precompute(); // Probability of getting 2 // head out of 3 coins echo number_format(probability(2, 3), 6) . "\n"; // Probability of getting 3 // head out of 6 coins echo number_format(probability(3, 6), 6) . "\n"; // Probability of getting 12 // head out of 18 coins echo number_format(probability(12, 18), 6); // This code is contributed by mits ?> |
Javascript
<script> // javascript Code for Probability of getting // atleast K heads in N tosses of Coins let fact; // Returns probability of getting at least k // heads in n tosses. function probability( k, n) { let ans = 0, i; for ( 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; } function precompute() { // Preprocess all factorial only upto 19, // as after that it will overflow fact[0] = fact[1] = 1; for ( let i = 2; i < 20; ++i) fact[i] = fact[i - 1] * i; } // Driver code fact = Array(100).fill(0); precompute(); // Probability of getting 2 head out // of 3 coins document.write(probability(2, 3)+"<br/>"); // Probability of getting 3 head out // of 6 coins document.write(probability(3, 6)+"<br/>"); // Probability of getting 12 head out // of 18 coins document.write(probability(12, 18).toFixed(6)+"<br/>"); // This code is contributed by shikhasingrajput </script> |
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<bits/stdc++.h> 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.io.*; 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=[0]*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
<?php // Dynamic and Logarithm approach // find probability of at least k heads $MAX = 100001; // dp[i] is going to store // Log ( i !) in base 2 $dp = array_fill(0, $MAX, 0); function probability($k, $n) { global $MAX, $dp; $ans = 0; // Initialize result // Iterate from k heads to n heads for ($i = $k; $i <= $n; ++$i) { $res = $dp[$n] - $dp[$i] - $dp[$n - $i] - $n; $ans += pow(2.0, $res); } return $ans; } function precompute() { global $MAX, $dp; // Preprocess all the logarithm // value on base 2 for ($i = 2; $i < $MAX; ++$i) // Note : log2() function is not in php // Some OUTPUT very in their decimal point // Basically log(value,base) is work as // this logic : "log10(value)/log10(2)" // equals to log2(value) $dp[$i] = log($i, 2) + $dp[$i - 1]; } // Driver code precompute(); // Probability of getting 2 // head out of 3 coins echo probability(2, 3)."\n"; // Probability of getting 3 // head out of 6 coins echo probability(3, 6)."\n"; // Probability of getting 500 // head out of 10000 coins echo probability(500, 1000); // This code is contributed by mits ?> |
Javascript
<script> // Dynamic and Logarithm approach find probability of // at least k heads let MAX = 100001; // dp[i] is going to store Log ( i !) in base 2 let dp = new Array(MAX).fill(0); function probability(k , n) { var ans = 0.0; // Initialize result // Iterate from k heads to n heads for (let i = k; i <= n; ++i) { var res = dp[n] - dp[i] - dp[n - i] - n; ans += Math.pow(2.0, res); } return ans; } function precompute() { // Preprocess all the logarithm value on base 2 for (let i = 2; i < MAX; ++i) dp[i] = (Math.log(i) / Math.log(2)) + dp[i - 1]; } // Driver code precompute(); // Probability of getting 2 head out of 3 coins document.write(probability(2, 3).toFixed(2)+"<br/>"); // Probability of getting 3 head out of 6 coins document.write(probability(3, 6).toFixed(5)+"<br/>"); // Probability of getting 500 head out of 10000 coins document.write(probability(500, 1000).toFixed(6)+"<br/>"); // This code is contributed by Amit Katiyar </script> |
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 neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
