# 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.

For 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
#include<bits/stdc++.h>
using namespace std;
#define MAX 21

double fact[MAX];

// Returns probability of getting at least k
double probability(int k, int n)
{
double ans = 0;
for (int i=k; i <= n; ++i)

// Probability of getting exactly i
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) << "\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
public static double probability(int k, int n)
{
double ans = 0;
for (int i = k; i <= n; ++ i)

// Probability of getting exactly i
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
```

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 is C++ code to illustrate this:

```// Dynamic and Logarithm approach find probability of
#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

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) << "\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;
}
```

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.

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