#include <bits/stdc++.h>
using namespace std;
// #include <atcoder/all>
// using namespace atcoder;

// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define irreps(i, m, n) for (long long i = ((long long)(n)-1); i > (long long)(m); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) (long long)(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;

// さいころは１０まで
// ６０まで出す確率を求め、勝つ場合の組の確率の積の和を求める

double dp1[11][100];
double dp2[11][100];

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  
  ll n,k; cin>>n>>k;
  
  dp1[0][0]=1;
  rep(i,n){
    if(i==0){
      reps(j,6) dp1[i+1][j]=dp1[i][0]/6;
    }else{
      reps(j,60){
        reps(k,6){
          dp1[i+1][j+k]+=dp1[i][j]/6;
        }
      }
    }
  }
  
  dp2[0][0]=1;
  rep(i,n){
    if(i<k){
      if(i==0){
        ireps(j,4,6) dp2[i+1][j]=dp2[i][0]/3;
      }else{
        reps(j,60){
          ireps(k,4,6){
            dp2[i+1][j+k]+=dp2[i][j]/3;
          }
        }
      }
    }else{
      if(i==0){
        reps(j,6) dp2[i+1][j]=dp2[i][0]/6;
      }else{
        reps(j,60){
          reps(k,6){
            dp2[i+1][j+k]+=dp2[i][j]/6;
          }
        }
      }
    }
  }
  
  double ans=0;
  reps(i,60) reps(j,60){
    if(i<j) ans+=dp1[n][i]*dp2[n][j];
  }
  printf("%.15lf\n", ans);
  
}