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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
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:

{displaystyle inom{n}{k}p^k(1-p)^{n-k}=^{n}	extrm{c}_{k}(p)^k (1-p)^{n-k}=frac {n!(p)^k (1-p)^{n-k}}{(k)!(n-k)!}}


So Probability ( getting at least 4 heads )=

{displaystyle inom{6}{4}inom{1}{2}^4inom{1}{2}^2+inom{6}{5}inom{1}{2}^5inom{1}{2}^1+inom{6}{6}inom{1}{2}^6inom{1}{2}^0}

{displaystyle =frac {1}{2^6} left (  frac {6!}{4!2!}+ frac {6!}{5!1!}+ frac {6!}{6!0!}
ight )}

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;
}
  
// Drive code
int main()
{
    precompute();
  
    // Probability of getting 2 head out of 3 coins
    cout << probability(2, 3) << " ";
  
    // Probability of getting 3 head out of 6 coins
    cout << probability(3, 6) <<" ";
  
    // 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[0] = fact[1] = 1;
       
        for (int i = 2; i < 20; ++i)
            fact[i] = fact[i - 1] * i;
    }
       
    // Drive 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;
    }
      
    // Drive 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;
}
  
// Drive code
precompute();
  
// Probability of getting 2
// head out of 3 coins
echo number_format(probability(2, 3), 6) . " ";
  
// Probability of getting 3 
// head out of 6 coins
echo number_format(probability(3, 6), 6) . " ";
  
// Probability of getting 12
// head out of 18 coins
echo number_format(probability(12, 18), 6);
      
// This code is contributed by mits
?>


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];
}
  
// Drive code
int main()
{
    precompute();
  
    // Probability of getting 2 head out of 3 coins
    cout << probability(2, 3) << " ";
  
    // Probability of getting 3 head out of 6 coins
    cout << probability(3, 6) << " ";
  
    // 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];
}
  
// Drive 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] -
        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
  
  
# Drive 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]; 
  
// Drive 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)." "
  
// Probability of getting 3 
// head out of 6 coins 
echo probability(3, 6)." "
  
// Probability of getting 500
// head out of 10000 coins 
echo probability(500, 1000); 
  
// This code is contributed by mits
?>


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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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