Let us call the two dice \(A\) and \(B\). Suppose that \(A\) shows an integer \(a\) and \(B\) shows \(b\). Then there are \(36\) combinations of \((a, b)\): \((1,1),(1,2),\ldots,(6,5),(6,6)\). Exactly one of these \(36\) events will happen with equal probability \(\dfrac1{36}\), so the problem can be solved by counting how many of the \(36\) pairs satisfies the condition.

The sample code is shown below.

Since your answer is considered correct if the error is at most \(10 ^ {-9}\), in some languages you may need to specify how many decimal places to print.

#include <iostream>
#include <iomanip>
using namespace std;

int main() {
    int X, Y;
    cin >> X >> Y;

    int count = 0; // Counter for conforming pairs
    for (int a = 1; a <= 6; ++a) // Number shown by the first die
        for (int b = 1; b <= 6; ++b) // Number shown by the second die
            if (a + b >= X || a + Y <= b || b + Y <= a) // If it satisfies the condition
                ++count; // increment the counter

    // The answer is obtained by dividing by 36
    cout << setprecision(10) << count / 36. << endl;
    return 0;
}

X, Y = map(int, input().split())

count = 0
for a in range(1, 7):
    for b in range(1, 7):
        if a + b >= X or a + Y <= b or b + Y <= a:
            count += 1

print(count / 36)