# Probability of getting more heads than tails when N biased coins are tossed

Given an array **p[]** of odd length **N** where **p[i]** denotes the probability of getting a head on the **i ^{th}** coin. As the coins are biased, the probability of getting a head is not always equal to

**0.5**. The task is to find the probability of getting heads more number of times than tails.

**Examples:**

Input:p[] = {0.3, 0.4, 0.7}

Output:0.442

Probability for a tail = (1 – Probability for a head)

For heads greater than tails, there are 4 possibilities:

P({head, head, tail})= 0.3 x 0.4 x (1 – 0.7) = 0.036

P({tail, head, head})= (1 – 0.3) x 0.4 x 0.7 = 0.196

P({head, tail, head})= 0.3 x (1 – 0.4) x 0.7= 0.126

P({head, head, head})= 0.3 x 0.4 x 0.7 = 0.084

Adding the above probabilities

0.036 + 0.196 + 0.126 + 0.084 = 0.442

Input:p[] = {0.3, 0.5, 0.2, 0.6, 0.9}

Output:0.495

**Naive approach:** The naive approach would be creating all the **2 ^{n}** possibilities of heads and tails. Then calculating the probabilities for different permutations and adding them when the number of heads are greater than the number of tails just like the example explanation. This would give TLE when

**n**is large.

**Efficient approach:** The idea is to use dynamic programming. Let’s assume **dp[i][j]** to be the probability of getting **j** heads with first **i** coins. To get **j** heads at the **i ^{th}** position, there are two possibilities:

- If number of heads till
**(i – 1)**coins is equal to**j**then a tail comes at**i**.^{th} - If number of heads till
**(i – 1)**coins is equal to**(j – 1)**then a head comes at**i**position^{th}

Hence, it can be broken into its subproblems as follows:

dp[i][j] = dp[i – 1][j] * (1 – p[i]) + dp[i – 1][j – 1] * p[i]

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the probability when ` `// number of heads is greater than the number of tails ` `double` `Probability(` `double` `p[], ` `int` `n) ` `{ ` ` ` ` ` `// Declaring the DP table ` ` ` `double` `dp[n + 1][n + 1]; ` ` ` `memset` `(dp, 0.0, ` `sizeof` `(dp)); ` ` ` ` ` `// Base case ` ` ` `dp[0][0] = 1.0; ` ` ` ` ` `// Iterating for every coin ` ` ` `for` `(` `int` `i = 1; i <= n; i += 1) { ` ` ` ` ` `// j represents the numbers of heads ` ` ` `for` `(` `int` `j = 0; j <= i; j += 1) { ` ` ` ` ` `// If number of heads is equal to zero ` ` ` `// there there is only one possiblity ` ` ` `if` `(j == 0) ` ` ` `dp[i][j] = dp[i - 1][j] ` ` ` `* (1.0 - p[i]); ` ` ` `else` ` ` `dp[i][j] = dp[i - 1][j] ` ` ` `* (1.0 - p[i]) ` ` ` `+ dp[i - 1][j - 1] * p[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `double` `ans = 0.0; ` ` ` ` ` `// When the number of heads is greater than (n+1)/2 ` ` ` `// it means that heads are greater than tails as ` ` ` `// no of tails + no of heads is equal to n for ` ` ` `// any permuation of heads and tails ` ` ` `for` `(` `int` `i = (n + 1) / 2; i <= n; i += 1) ` ` ` `ans += dp[n][i]; ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// 1 based indexing ` ` ` `double` `p[] = { 0.0, 0.3, 0.4, 0.7 }; ` ` ` ` ` `// Number of coins ` ` ` `int` `n = ` `sizeof` `(p) / ` `sizeof` `(p[0]) - 1; ` ` ` ` ` `// Function call ` ` ` `cout << Probability(p, n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation of the above approach ` `class` `GFG ` `{ ` ` ` `// Function to return the probability when ` `// number of heads is greater than the number of tails ` `static` `double` `Probability(` `double` `p[], ` `int` `n) ` `{ ` ` ` ` ` `// Declaring the DP table ` ` ` `double` `[][]dp = ` `new` `double` `[n + ` `1` `][n + ` `1` `]; ` ` ` ` ` `// Base case ` ` ` `dp[` `0` `][` `0` `] = ` `1.0` `; ` ` ` ` ` `// Iterating for every coin ` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i += ` `1` `) ` ` ` `{ ` ` ` ` ` `// j represents the numbers of heads ` ` ` `for` `(` `int` `j = ` `0` `; j <= i; j += ` `1` `) ` ` ` `{ ` ` ` ` ` `// If number of heads is equal to zero ` ` ` `// there there is only one possiblity ` ` ` `if` `(j == ` `0` `) ` ` ` `dp[i][j] = dp[i - ` `1` `][j] ` ` ` `* (` `1.0` `- p[i]); ` ` ` `else` ` ` `dp[i][j] = dp[i - ` `1` `][j] ` ` ` `* (` `1.0` `- p[i]) ` ` ` `+ dp[i - ` `1` `][j - ` `1` `] * p[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `double` `ans = ` `0.0` `; ` ` ` ` ` `// When the number of heads is greater than (n+1)/2 ` ` ` `// it means that heads are greater than tails as ` ` ` `// no of tails + no of heads is equal to n for ` ` ` `// any permuation of heads and tails ` ` ` `for` `(` `int` `i = (n + ` `1` `) / ` `2` `; i <= n; i += ` `1` `) ` ` ` `ans += dp[n][i]; ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `// 1 based indexing ` ` ` `double` `p[] = { ` `0.0` `, ` `0.3` `, ` `0.4` `, ` `0.7` `}; ` ` ` ` ` `// Number of coins ` ` ` `int` `n = p.length - ` `1` `; ` ` ` ` ` `// Function call ` ` ` `System.out.println(Probability(p, n)); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 implementation of the above approach ` `import` `numpy as np ` ` ` `# Function to return the probability when ` `# number of heads is greater than ` `# the number of tails ` `def` `Probability(p, n) : ` ` ` ` ` `# Declaring the DP table ` ` ` `dp ` `=` `np.zeros((n ` `+` `1` `, n ` `+` `1` `)); ` ` ` `for` `i ` `in` `range` `(n ` `+` `1` `) : ` ` ` `for` `j ` `in` `range` `(n ` `+` `1` `) : ` ` ` `dp[i][j] ` `=` `0.0` ` ` ` ` `# Base case ` ` ` `dp[` `0` `][` `0` `] ` `=` `1.0` `; ` ` ` ` ` `# Iterating for every coin ` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `) : ` ` ` ` ` `# j represents the numbers of heads ` ` ` `for` `j ` `in` `range` `(i ` `+` `1` `) : ` ` ` ` ` `# If number of heads is equal to zero ` ` ` `# there there is only one possiblity ` ` ` `if` `(j ` `=` `=` `0` `) : ` ` ` `dp[i][j] ` `=` `dp[i ` `-` `1` `][j] ` `*` `(` `1.0` `-` `p[i]); ` ` ` `else` `: ` ` ` `dp[i][j] ` `=` `(dp[i ` `-` `1` `][j] ` `*` `(` `1.0` `-` `p[i]) ` `+` ` ` `dp[i ` `-` `1` `][j ` `-` `1` `] ` `*` `p[i]); ` ` ` ` ` `ans ` `=` `0.0` `; ` ` ` ` ` `# When the number of heads is greater than (n+1)/2 ` ` ` `# it means that heads are greater than tails as ` ` ` `# no of tails + no of heads is equal to n for ` ` ` `# any permuation of heads and tails ` ` ` `for` `i ` `in` `range` `((n ` `+` `1` `)` `/` `/` `2` `, n ` `+` `1` `) : ` ` ` `ans ` `+` `=` `dp[n][i]; ` ` ` ` ` `return` `ans; ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `# 1 based indexing ` ` ` `p ` `=` `[ ` `0.0` `, ` `0.3` `, ` `0.4` `, ` `0.7` `]; ` ` ` ` ` `# Number of coins ` ` ` `n ` `=` `len` `(p) ` `-` `1` `; ` ` ` ` ` `# Function call ` ` ` `print` `(Probability(p, n)); ` ` ` `# This code is contributed by AnkitRai01 ` |

*chevron_right*

*filter_none*

## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to return the probability when ` `// number of heads is greater than the number of tails ` `static` `double` `Probability(` `double` `[]p, ` `int` `n) ` `{ ` ` ` ` ` `// Declaring the DP table ` ` ` `double` `[,]dp = ` `new` `double` `[n + 1, n + 1]; ` ` ` ` ` `// Base case ` ` ` `dp[0, 0] = 1.0; ` ` ` ` ` `// Iterating for every coin ` ` ` `for` `(` `int` `i = 1; i <= n; i += 1) ` ` ` `{ ` ` ` ` ` `// j represents the numbers of heads ` ` ` `for` `(` `int` `j = 0; j <= i; j += 1) ` ` ` `{ ` ` ` ` ` `// If number of heads is equal to zero ` ` ` `// there there is only one possiblity ` ` ` `if` `(j == 0) ` ` ` `dp[i,j] = dp[i - 1,j] ` ` ` `* (1.0 - p[i]); ` ` ` `else` ` ` `dp[i,j] = dp[i - 1,j] ` ` ` `* (1.0 - p[i]) ` ` ` `+ dp[i - 1,j - 1] * p[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `double` `ans = 0.0; ` ` ` ` ` `// When the number of heads is greater than (n+1)/2 ` ` ` `// it means that heads are greater than tails as ` ` ` `// no of tails + no of heads is equal to n for ` ` ` `// any permuation of heads and tails ` ` ` `for` `(` `int` `i = (n + 1) / 2; i <= n; i += 1) ` ` ` `ans += dp[n,i]; ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver Code ` `static` `public` `void` `Main () ` `{ ` ` ` ` ` `// 1 based indexing ` ` ` `double` `[]p = { 0.0, 0.3, 0.4, 0.7 }; ` ` ` ` ` `// Number of coins ` ` ` `int` `n = p.Length - 1; ` ` ` ` ` `// Function call ` ` ` `Console.Write(Probability(p, n)); ` `} ` `} ` ` ` `// This code is contributed by ajit. ` |

*chevron_right*

*filter_none*

**Output:**

0.442

## Recommended Posts:

- Probability of getting at least K heads in N tosses of Coins
- Probability of getting two consecutive heads after choosing a random coin among two different types of coins
- Expected number of coin flips to get two heads in a row?
- Number of paths with exactly k coins
- Burst Balloon to maximize coins
- Collect maximum coins before hitting a dead end
- Find minimum number of coins that make a given value
- Probability of rain on N+1th day
- Probability of getting more value in third dice throw
- Probability that two persons will meet
- Aptitude | Probability | Question 1
- Aptitude | Probability | Question 10
- Aptitude | Probability | Question 2
- Aptitude | Probability | Question 1
- Aptitude | Probability | Question 3

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.