#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <climits>
#include <map>
#include <cmath>
using namespace std;

#define REP(i,n) for(int i=0;i<n;i++)
#define ALL(v)	(v).begin(), (v).end()
#define INF INT_MAX/3
#define MOD 1000000007
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef long long ll;

ll N, K;

ll dp1[100][100], dp2[100][100];
double per1[100], per2[100];
int main(){

	cin >> N >> K;
	dp1[0][0] = 1;
	dp2[0][0] = 1;
	//dp1[i][j]:太郎君がサイコロをi個投げて合計をjにする方法の個数
	//但し太郎君は最初のK個をイカサマなサイコロを投げる
	for (int i = 0; i < K; i++){
		for (int j = 60; j >= 0; j--){
			for (int k = 4; k <= 6; k++){
				dp1[i + 1][j + k] += dp1[i][j] * 2;
			}
		}
	}
	for (int i = K; i < N; i++){
		for (int j = 60; j >= 0; j--){
			for (int k = 1; k <= 6; k++){
				dp1[i + 1][j + k] += dp1[i][j];
			}
		}
	}

	for (int i = 0; i < N; i++){
		for (int j = 60; j >= 0; j--){
			for (int k = 1; k <= 6; k++){
				dp2[i + 1][j + k] += dp2[i][j];
			}
		}
	}


	double t = 1;
	for (int i = 0; i < N; i++) t *= 6;
	for (int i = 0; i <= 60; i++){
		per1[i] = dp1[N][i] / t;
		per2[i] = dp2[N][i] / t;
	}
	
	double ans = 0;
	for (int i = 0; i <= 60; i++){
		for (int j = i + 1; j <= 60; j++){
			ans += per1[j] * per2[i];
		}
	}
	printf("%.20f\n", ans);
	return 0;
}