#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>

using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef unsigned long long ULL;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(auto i: c)
#define SORT(c) sort((c).begin(),(c).end())

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()

const double EPS = 1e-10;
const double PI  = acos(-1.0);

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, K;
	cin >> N >> K;

	VVI dp1(N + 1, VI(70, 0)), dp2(N + 1, VI(70, 0));
	dp1[0][0] = dp2[0][0] = 1;
	int d1[6] = {1, 2, 3, 4, 5, 6};
	int d2[6] = {4, 5, 6, 4, 5, 6};

	REP(i, N){
		REP(j, 62){
			REP(k, 6){
				dp1[i + 1][j + d1[k]] += dp1[i][j];
				if(i < K) dp2[i + 1][j + d2[k]] += dp2[i][j];
				else dp2[i + 1][j + d1[k]] += dp2[i][j];
			}
		}
	}

	double ret = 0;
	int denominator1= 0, denominator2 = 0;
	REP(i, 70){
		denominator1 += dp1[N][i];
		denominator2 += dp2[N][i];
	}
	REP(i, 70) REP( j, 70){
		int numerator1 = 0, numerator2 = 0;
		if(i < j){
			numerator1 += dp1[N][i];
			numerator2 += dp2[N][j];
		}
		ret += 1.0 * numerator1 / denominator1 * numerator2 / denominator2;
	}
/*
	ret = 1.0 * (numerator1 * numerator2) / (denominator1 * denominator2);
	ret = 1.0 * numerator1 / denominator1 * numerator2 / denominator2;
	*/
	cout << ret << endl;

	return 0;
}