Given three integers N, A, and B, the task is to calculate the probability that the sum of numbers obtained on throwing the dice exactly N times lies between A and B.
Examples:
Input: N = 1, A = 2, B = 3
Output: 0.333333
Explanation: Ways to obtained the sum 2 by N ( = 1) throws of a dice is 1 {2}. Therefore, required probability = 1/6 = 0.33333Input: N = 2, A = 3, B = 4
Output: 0.138889
Recursive Approach: Follow the steps below to solve the problem:
- Calculate probabilities for all the numbers between A and B and add them to get the answer.
- Call function find(N, sum) to calculate the probability for each number from a to b, where a number between a and b will be passed as sum.
- Base cases are:
- If the sum is either greater than 6 * N or less than N, then return 0 as it’s impossible to have sum greater than N * 6 or less than N.
- If N is equal to 1 and sum is in between 1 and 6, then return 1/6.
- Since at every state any number out of 1 to 6 in a single throw of dice may come, therefore recursion call should be made for the (sum up to that state – i) where 1≤ i ≤ 6.
- Return the resultant probability.
- Base cases are:
Recursion call:
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;
// Function to calculate the// probability for the given// sum to be equal to sum in// N throws of dicelong double find(int N, int sum)
{ // Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1) {
if (sum >= 1 && sum <= 6)
return 1.0 / 6;
else
return 0;
}
long double s = 0;
for (int i = 1; i <= 6; i++)
s = s + find(N - 1, sum - i) / 6;
return s;
}// Driver Codeint main()
{ int N = 4, a = 13, b = 17;
long double probability = 0.0;
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
cout << fixed << setprecision(6) << probability;
return 0;
} |
Java
// Java program for the above approach import java.util.*;
class GFG
{// Function to calculate the// probability for the given// sum to be equal to sum in// N throws of dicestatic double find(int N, int sum)
{ // Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1)
{
if (sum >= 1 && sum <= 6)
return 1.0 / 6;
else
return 0;
}
double s = 0;
for (int i = 1; i <= 6; i++)
s = s + find(N - 1, sum - i) / 6;
return s;
} // Driver codepublic static void main(String[] args)
{ int N = 4, a = 13, b = 17;
double probability = 0.0;
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
System.out.format("%.6f", probability);
}}// This code is contributed by code_hunt. |
Python3
# Python 2 program for above approach# Function to calculate the# probability for the given# sum to be equal to sum in# N throws of dicedef find(N, sum):
# Base cases
if (sum > 6 * N or sum < N):
return 0
if (N == 1):
if (sum >= 1 and sum <= 6):
return 1.0 / 6
else:
return 0
s = 0
for i in range(1, 7):
s = s + find(N - 1, sum - i) / 6
return s
# Driver Codeif __name__ == "__main__":
N = 4
a = 13
b = 17
probability = 0.0
for sum in range(a, b + 1):
probability = probability + find(N, sum)
# Print the answer
print(round(probability, 6))
# This code is contributed by chitranayal.
|
C#
// C# program for the above approach using System;
class GFG
{ // Function to calculate the
// probability for the given
// sum to be equal to sum in
// N throws of dice
static double find(int N, int sum)
{
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1)
{
if (sum >= 1 && sum <= 6)
return 1.0 / 6;
else
return 0;
}
double s = 0;
for (int i = 1; i <= 6; i++)
s = s + find(N - 1, sum - i) / 6;
return s;
}
// Driver code
static void Main()
{
int N = 4, a = 13, b = 17;
double probability = 0.0;
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
Console.WriteLine(Math.Round(probability,6));
}
}// This code is contributed by divyeshrabadiya07 |
Javascript
<script> // Javascript program for the above approach
// Function to calculate the
// probability for the given
// sum to be equal to sum in
// N throws of dice
function find(N, sum)
{
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1)
{
if (sum >= 1 && sum <= 6)
return 1.0 / 6;
else
return 0;
}
let s = 0;
for (let i = 1; i <= 6; i++)
s = s + find(N - 1, sum - i) / 6;
return s;
}
let N = 4, a = 13, b = 17;
let probability = 0.0;
for (let sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
document.write(probability.toFixed(6));
</script> |
Output:
0.505401
Time Complexity: O((b-a+1)6n)
Auxiliary Space: O(1)
Dynamic Programming Approach: The above recursive approach needs to be optimized by dealing with the following overlapping subproblems and optimal substructure:
Overlapping Subproblems:
Partial recursion tree for N=4 and sum=15:
Optimal Substructure:
For every state, recur for other 6 states, so the recursive definition of f(N, sum) is:
Top-Down Approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;
float dp[105][605];
// Function to calculate the// probability for the given// sum to be equal to sum in// N throws of dicefloat find(int N, int sum)
{ if (dp[N][sum])
return dp[N][sum];
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1) {
if (sum >= 1 && sum <= 6)
return 1.0 / 6;
else
return 0;
}
for (int i = 1; i <= 6; i++)
dp[N][sum] = dp[N][sum]
+ find(N - 1, sum - i) / 6;
return dp[N][sum];
}// Driver Codeint main()
{ int N = 4, a = 13, b = 17;
float probability = 0.0;
// Calculate probability of all
// sums from a to b
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
cout << fixed << setprecision(6) << probability;
return 0;
} |
Java
// Java program for above approachclass GFG
{ static float[][] dp = new float[105][605];
// Function to calculate the
// probability for the given
// sum to be equal to sum in
// N throws of dice
static float find(int N, int sum)
{
if (N < 0 | sum < 0)
return 0;
if (dp[N][sum] > 0)
return dp[N][sum];
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1) {
if (sum >= 1 && sum <= 6)
return (float) (1.0 / 6);
else
return 0;
}
for (int i = 1; i <= 6; i++)
dp[N][sum] = dp[N][sum] + find(N - 1, sum - i) / 6;
return dp[N][sum];
}
// Driver Code
public static void main(String[] args)
{
int N = 4, a = 13, b = 17;
float probability = 0.0f;
// Calculate probability of all
// sums from a to b
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
System.out.printf("%.6f", probability);
}
}// This code is contributed by shikhasingrajput |
Python3
# Python program for above approachdp = [[0 for i in range(605)] for j in range(105)];
# Function to calculate the# probability for the given# sum to be equal to sum in# N throws of dicedef find(N, sum):
if (N < 0 | sum < 0):
return 0;
if (dp[N][sum] > 0):
return dp[N][sum];
# Base cases
if (sum > 6 * N or sum < N):
return 0;
if (N == 1):
if (sum >= 1 and sum <= 6):
return (float)(1.0 / 6);
else:
return 0;
for i in range(1,7):
dp[N][sum] = dp[N][sum] + find(N - 1, sum - i) / 6;
return dp[N][sum];
# Driver Codeif __name__ == '__main__':
N = 4; a = 13; b = 17;
probability = 0.0
f = 0;
# Calculate probability of all
# sums from a to b
for sum in range(a,b+1):
probability = probability + find(N, sum);
# Print the answer
print("%.6f"% probability);
# This code is contributed by 29AjayKumar |
C#
// C# program for above approachusing System;
using System.Collections.Generic;
public class GFG
{ static float[,] dp = new float[105, 605];
// Function to calculate the
// probability for the given
// sum to be equal to sum in
// N throws of dice
static float find(int N, int sum)
{
if (N < 0 | sum < 0)
return 0;
if (dp[N, sum] > 0)
return dp[N, sum];
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1) {
if (sum >= 1 && sum <= 6)
return (float) (1.0 / 6);
else
return 0;
}
for (int i = 1; i <= 6; i++)
dp[N, sum] = dp[N, sum] + find(N - 1, sum - i) / 6;
return dp[N, sum];
}
// Driver Code
public static void Main(String[] args)
{
int N = 4, a = 13, b = 17;
float probability = 0.0f;
// Calculate probability of all
// sums from a to b
for (int sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answer
Console.Write("{0:F6}", probability);
}
}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for above approachvar dp = Array(105).fill().map(()=>Array(605).fill(0.0));
// Function to calculate the// probability for the given// sum to be equal to sum in// N throws of dicefunction find(N, sum)
{ if (N < 0 | sum < 0)
return 0;
if (dp[N][sum] > 0)
return dp[N][sum];
// Base cases
if (sum > 6 * N || sum < N)
return 0;
if (N == 1)
{
if (sum >= 1 && sum <= 6)
return (1.0 / 6);
else
return 0;
}
for(var i = 1; i <= 6; i++)
dp[N][sum] = dp[N][sum] +
find(N - 1, sum - i) / 6;
return dp[N][sum];
}// Driver Codevar N = 4, a = 13, b = 17;
var probability = 0.0;
// Calculate probability of all// sums from a to bfor(sum = a; sum <= b; sum++)
probability = probability + find(N, sum);
// Print the answerdocument.write(probability.toFixed(6));// This code is contributed by umadevi9616</script> |
Output:
0.505401
Time Complexity: O(n*sum)
Auxiliary Space: O(n*sum)
Bottom-Up Approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;
float dp[105][605];
// Function to calculate probability// that the sum of numbers on N throws// of dice lies between A and Bfloat find(int N, int a, int b)
{ float probability = 0.0;
// Base case
for (int i = 1; i <= 6; i++)
dp[1][i] = 1.0 / 6;
for (int i = 2; i <= N; i++) {
for (int j = i; j <= 6 * i; j++) {
for (int k = 1; k <= 6; k++) {
dp[i][j] = dp[i][j]
+ dp[i - 1][j - k] / 6;
}
}
}
// Add the probability for all
// the numbers between a and b
for (int sum = a; sum <= b; sum++)
probability = probability + dp[N][sum];
return probability;
}// Driver Codeint main()
{ int N = 4, a = 13, b = 17;
float probability = find(N, a, b);
// Print the answer
cout << fixed << setprecision(6) << probability;
return 0;
} |
Java
// Java program for above approachimport java.util.*;
class GFG{
static float [][]dp = new float[105][605];
// Function to calculate probability// that the sum of numbers on N throws// of dice lies between A and Bstatic float find(int N, int a, int b)
{ float probability = 0.0f;
// Base case
for (int i = 1; i <= 6; i++)
dp[1][i] = (float) (1.0 / 6);
for (int i = 2; i <= N; i++)
{
for (int j = i; j <= 6 * i; j++)
{
for (int k = 1; k <= 6 && k <= j; k++)
{
dp[i][j] = dp[i][j]
+ dp[i - 1][j - k] / 6;
}
}
}
// Add the probability for all
// the numbers between a and b
for (int sum = a; sum <= b; sum++)
probability = probability + dp[N][sum];
return probability;
}// Driver Codepublic static void main(String[] args)
{ int N = 4, a = 13, b = 17;
float probability = find(N, a, b);
// Print the answer
System.out.printf("%.6f",probability);
}}// This codeis contributed by shikhasingrajput |
Python3
# Python3 program for above approachdp = [[0 for i in range(605)] for j in range(105)]
# Function to calculate probability# that the sum of numbers on N throws# of dice lies between A and Bdef find(N, a, b) :
probability = 0.0
# Base case
for i in range(1, 7) :
dp[1][i] = 1.0 / 6
for i in range(2, N + 1) :
for j in range(i, (6*i) + 1) :
for k in range(1, 7) :
dp[i][j] = dp[i][j] + dp[i - 1][j - k] / 6
# Add the probability for all
# the numbers between a and b
for Sum in range(a, b + 1) :
probability = probability + dp[N][Sum]
return probability
N, a, b = 4, 13, 17
probability = find(N, a, b)
# Print the answerprint('%.6f'%probability)
# This code is contributed by divyesh072019. |
C#
// C# program for above approachusing System;
public class GFG
{static float [,]dp = new float[105, 605];
// Function to calculate probability// that the sum of numbers on N throws// of dice lies between A and Bstatic float find(int N, int a, int b)
{ float probability = 0.0f;
// Base case
for (int i = 1; i <= 6; i++)
dp[1, i] = (float) (1.0 / 6);
for (int i = 2; i <= N; i++)
{
for (int j = i; j <= 6 * i; j++)
{
for (int k = 1; k <= 6 && k <= j; k++)
{
dp[i, j] = dp[i, j]
+ dp[i - 1, j - k] / 6;
}
}
}
// Add the probability for all
// the numbers between a and b
for (int sum = a; sum <= b; sum++)
probability = probability + dp[N, sum];
return probability;
}// Driver Codepublic static void Main(String[] args)
{ int N = 4, a = 13, b = 17;
float probability = find(N, a, b);
// Print the answer
Console.Write("{0:F6}",probability);
}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript program of the above approachlet dp = new Array(105);
// Loop to create 2D array using 1D arrayfor(var i = 0; i < dp.length; i++)
{ dp[i] = new Array(2);
}for(var i = 0; i < dp.length; i++)
{ for(var j = 0; j < dp.length; j++)
{
dp[i][j] = 0;
}
} // Function to calculate probability// that the sum of numbers on N throws// of dice lies between A and Bfunction find(N, a, b)
{ let probability = 0.0;
// Base case
for(let i = 1; i <= 6; i++)
dp[1][i] = (1.0 / 6);
for(let i = 2; i <= N; i++)
{
for(let j = i; j <= 6 * i; j++)
{
for(let k = 1; k <= 6 && k <= j; k++)
{
dp[i][j] = dp[i][j] +
dp[i - 1][j - k] / 6;
}
}
}
// Add the probability for all
// the numbers between a and b
for(let sum = a; sum <= b; sum++)
probability = probability + dp[N][sum];
return probability;
}// Driver Codelet N = 4, a = 13, b = 17;let probability = find(N, a, b);// Print the answerdocument.write(probability);// This code is contributed by chinmoy1997pal</script> |
Output:
0.505401
Time Complexity: O(N * sum)
Auxiliary Space: O(N * sum)