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Probability of not getting two consecutive heads together in N tosses of coin
  • Last Updated : 19 Nov, 2020

Given a fair coin that is tossed N times, the task is to determine the probability such that no two heads occur consecutively. 

Examples: 

Input: N = 2 
Output: 0.75 
Explanation: 
When the coin is tossed 2 times, the possible outcomes are {TH, HT, TT, HH}. 
Since in 3 out of 4 outcomes, heads don’t occur together. 
Therefore, the required probability is (3/4) or 0.75

Input: N = 3 
Output: 0.62 
Explanation: 
When the coin is tossed 3 times, the possible outcomes are {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}. 
Since in 5 out of 8 outcomes, heads don’t occur together. 
Therefore, the required probability is (5/8) or 0.62 

Approach: The following observation on the number of favorable outcomes has to be made. 



  • When N = 1: The possible outcomes are {T, H}. There are two favorable outcomes out of the two.
  • When N = 2: The possible outcomes are {TH, HT, TT, HH}. There are three favorable outcomes out of four.
  • When N = 3: Similarly, the possible outcomes are {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}. There are five favorable outcomes out of eight.
  • When N = 4: Similarly, the possible outcomes are {TTTT, TTTH, TTHT, THTT, HTTT, TTHH, THTH, HTHT, HHTT, THHT, HTTH, THHH, HTHH, HHTH, HHHT, HHHH}. There are eight favorable outcomes out of sixteen.

Clearly, the number of favorable outcomes follow a Fibonacci series where Fn(1) = 2, Fn(2) = 3 and so on. Therefore, the idea is to implement the Fibonacci sequence in order to find the number of favorable cases. Clearly, the total number of cases is 2N
To calculate the probability, the following formula is used: 

P = favorable cases / Total number of cases 

Below is the implementation of the above approach:

C++

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// C++ implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
#include <bits/stdc++.h>
using namespace std;
 
float round(float var,int digit)
{
  float value = (int)(var *
                 pow(10, digit) + .5);
  return (float)value /
          pow(10, digit);
}
 
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
int probability(int N)
{
  // The first two numbers in
  // the sequence are initialized
  int a = 2;
  int b = 3;
 
  //  Base cases
  if (N == 1)
  {
    return a;
  }
  else if(N == 2)
  {
    return b;
  }
  else
  {
    // Loop to compute the fibonacci
    // sequence based on the first
    // two initialized numbers
    for(int i = 3; i <= N; i++)
    {
      int c = a + b;
      a = b;
      b = c;
    }
    return b;
  }
}
 
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
float operations(int N)
 {
  // Computing the number of
  // favourable cases
  int x = probability(N);
 
  // Computing the number of
  // all possible outcomes for
  // N tosses
  int y = pow(2, N);
 
  return round((float)x /
               (float)y, 2);
}
  
// Driver code
int main()
{
  int N = 10;
  cout << (operations(N));
}
 
// Thus code is contributed by Rutvik_56

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Java

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// Java implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
class GFG{
     
public static float round(float var, int digit)
{
    float value = (int)(var *
                   Math.pow(10, digit) + .5);
    return (float)value /
           (float)Math.pow(10, digit);
}
  
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
public static int probability(int N)
{
     
    // The first two numbers in
    // the sequence are initialized
    int a = 2;
    int b = 3;
     
    //  Base cases
    if (N == 1)
    {
        return a;
    }
    else if (N == 2)
    {
        return b;
    }
    else
    {
         
        // Loop to compute the fibonacci
        // sequence based on the first
        // two initialized numbers
        for(int i = 3; i <= N; i++)
        {
            int c = a + b;
            a = b;
            b = c;
        }
        return b;
    }
}
  
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
public static float operations(int N)
{
     
    // Computing the number of
    // favourable cases
    int x = probability(N);
     
    // Computing the number of
    // all possible outcomes for
    // N tosses
    int y = (int)Math.pow(2, N);
     
    return round((float)x /
                 (float)y, 2);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
     
    System.out.println((operations(N)));
}
}
 
// This code is contributed by divyeshrabadiya07

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Python3

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# Python3 implementation to find the
# probability of not getting two
# consecutive heads together when
# N coins are tossed
 
 
import math
 
# Function to compute the N-th
# Fibonacci number in the
# sequence where a = 2
# and b = 3
def probability(N):
 
    # The first two numbers in
    # the sequence are initialized
    a = 2
    b = 3
 
    # Base cases
    if N == 1:
        return a
    elif N == 2:
        return b
    else:
         
        # Loop to compute the fibonacci
        # sequence based on the first
        # two initialized numbers
        for i in range(3, N + 1):
            c = a + b
            a = b
            b = c
        return b
 
# Function to find the probability
# of not getting two consecutive
# heads when N coins are tossed
def operations(N):
 
    # Computing the number of
    # favourable cases
    x = probability (N)
 
    # Computing the number of
    # all possible outcomes for
    # N tosses
    y = math.pow(2, N)
 
    return round(x / y, 2)
 
# Driver code
if __name__ == '__main__':
 
    N = 10
     
    print(operations(N))
   

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C#

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// C# implementation to find the
// probability of not getting two
// consecutive heads together when
// N coins are tossed
using System;
 
class GFG{
     
public static float round(float var, int digit)
{
    float value = (int)(var *
                   Math.Pow(10, digit) + .5);
    return (float)value /
           (float)Math.Pow(10, digit);
}
  
// Function to compute the N-th
// Fibonacci number in the
// sequence where a = 2
// and b = 3
public static int probability(int N)
{
     
    // The first two numbers in
    // the sequence are initialized
    int a = 2;
    int b = 3;
     
    //  Base cases
    if (N == 1)
    {
        return a;
    }
    else if (N == 2)
    {
        return b;
    }
    else
    {
         
        // Loop to compute the fibonacci
        // sequence based on the first
        // two initialized numbers
        for(int i = 3; i <= N; i++)
        {
            int c = a + b;
            a = b;
            b = c;
        }
        return b;
    }
}
  
// Function to find the probability
// of not getting two consecutive
// heads when N coins are tossed
public static float operations(int N)
{
     
    // Computing the number of
    // favourable cases
    int x = probability(N);
     
    // Computing the number of
    // all possible outcomes for
    // N tosses
    int y = (int)Math.Pow(2, N);
     
    return round((float)x /
                 (float)y, 2);
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 10;
     
    Console.WriteLine((operations(N)));
}
}
 
// This code is contributed by chitranayal

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Output: 

0.14





 

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