# Probability of getting all possible values on throwing N dices

Given an integer N denoting the number of dices, the task is to find the probability of every possible value that can be obtained by throwing N dices together.

Examples:

Input: N = 1
Output:
1: 0.17
2: 0.17
3: 0.17
4: 0.17
5: 0.17
6: 0.17
Explanation: On throwing a dice, the probability of all values from [1, 6] to appear at the top is 1/6 = 0.17

Input: N = 2
Output:
2: 0.028
3: 0.056
4: 0.083
5: 0.11
6: 0.14
7: 0.17
8: 0.14
9: 0.11
10: 0.083
11: 0.056
12: 0.028
Explanation: The possible values of the sum of the two numbers that appear at the top on throwing two dices together ranges between [2, 12].

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use Dynamic programming and DP table to store the probability of each possible value.

• Store the probabilies of all the 6 numbers that can appear on throwing 1 dices.
• Now, for N=2, the probability for all possible sums between [2, 12] is equal to the sum of product of the respective probability of the two numbers that add up to that sum. For example,

Probability of 4 on throwing 2 dices = (Probability of 1 ) * ( Probability of 3) + (Probability of 2) * ( Probability of 2) + (Probability of 3 ) * ( Probability of 1)

• Hence for N dices,

Probability of Sum S = (Probability of 1) * (Probability of S – 1 using N -1 dices) + (Probability of 2) * (Probability of S – 2 using N-1 dices) + ….. + (Probability of 6) * (Probability of S – 6 using N -1 dices)

• Hence, in order to solve the problem, we need to fill dp[][] table from 2 to N using top-down approach using the relation:

dp[i][x] = dp[y] + dp[i-1][z] where x = y + z and i denotes the number of dices

• Display all the probabilities stored for N as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ Program to calculate ` `// the probabilty of ` `// all the possible values ` `// that can be obtained ` `// throwing N dices ` ` `  `#include ` `using` `namespace` `std; ` ` `  `void` `dicesSum(``int` `n) ` `{ ` `    ``// Store the probablities ` `    ``vector > dp(n + 1); ` `    ``// Precompute the probabilities ` `    ``// for values possible using 1 dice ` `    ``dp = { { 1, 1 / 6.0 }, ` `              ``{ 2, 1 / 6.0 }, ` `              ``{ 3, 1 / 6.0 }, ` `              ``{ 4, 1 / 6.0 }, ` `              ``{ 5, 1 / 6.0 }, ` `              ``{ 6, 1 / 6.0 } }; ` ` `  `    ``// Compute the probabilies ` `    ``// for all values from 2 to N ` `    ``for` `(``int` `i = 2; i <= n; i++) { ` `        ``for` `(``auto` `a1 : dp[i - 1]) { ` `            ``for` `(``auto` `a2 : dp) { ` `                ``dp[i][a1.first + a2.first] ` `                    ``+= a1.second * a2.second; ` `            ``} ` `        ``} ` `    ``} ` `    ``// Print the result ` `    ``for` `(``auto` `a : dp[n]) { ` `        ``cout << a.first << ``" "` `             ``<< setprecision(2) ` `             ``<< a.second ` `             ``<< endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 2; ` `    ``dicesSum(n); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program to calculate ` `# the probabilty of all the  ` `# possible values that can  ` `# be obtained throwing N dices ` `def` `diceSum(n): ` `     `  `    ``# Inititalize a 2d array upto  ` `    ``# (n*total sum possible) sum  ` `    ``# with value 0 ` `    ``dp ``=` `[[ ``0` `for` `j ``in` `range``(n ``*` `6``)] ` `              ``for` `i ``in` `range``(n ``+` `1``)] ` `                   `  `    ``# Store the probability in a  ` `    ``# single throw for 1,2,3,4,5,6 ` `    ``for` `i ``in` `range``(``6``): ` `        ``dp[``1``][i] ``=` `1` `/` `6` `         `  `    ``# Compute the probabilies  ` `    ``# for all values from 2 to N  ` `    ``for` `i ``in` `range``(``2``, n ``+` `1``): ` `        ``for` `j ``in` `range``(``len``(dp[i ``-` `1``])): ` `            ``for` `k ``in` `range``(``6``): ` `                     `  `                ``if` `(dp[i ``-` `1``][j] !``=` `0` `and`  `                    ``dp[i ``-` `1``][k] !``=` `0``): ` `                    ``dp[i][j ``+` `k] ``+``=` `(dp[i ``-` `1``][j] ``*`  `                                     ``dp[``1``][k]) ` `     `  `    ``# Print the result  ` `    ``for` `i ``in` `range``(``len``(dp[n]) ``-` `n ``+` `1``): ` `        ``print``(``"%d %0.3f"` `%` `(i ``+` `n, dp[n][i])) ` ` `  `# Driver code ` `n ``=` `2` ` `  `# Call the function ` `diceSum(n) ` ` `  `# This code is contributed by dipesh99kumar `

Output:

```2 0.028
3 0.056
4 0.083
5 0.11
6 0.14
7 0.17
8 0.14
9 0.11
10 0.083
11 0.056
12 0.028
```

Time Complexity: O(N2)
Auxiliary Space: O(N2)

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Improved By : dipesh99kumar, nidhi_biet