#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
#define rep(i, n) REP(i, 0, n)
#define REP(i, s, e) for (ll i = (s); i < (ll)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for (ll i = (ll)(s - 1); i >= (ll)(e); i--)
#define all(r) r.begin(), r.end()
#define rall(r) r.rbegin(), r.rend()

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;

template <typename T, typename U>
bool chmax(T& a, const U& b) {
    if (a >= b) return false;
    a = b;
    return true;
}
template <typename T, typename U>
bool chmin(T& a, const U& b) {
    if (a <= b) return false;
    a = b;
    return true;
}

void yes_no(bool f, string yes = "Yes", string no = "No") {
    cout << (f ? yes : no) << "\n";
}

void solve() {
    int n, k;
    cin >> n >> k;
    using D = long double;
    const int ma = 150;
    vector<D> dp(ma, 0), ep(ma, 0);
    dp[0] = 1;
    ep[0] = 1;
    D p = (D)1 / 6;
    rep(i, n) {
        vector<D> nxt(ma, 0);
        rep(j, ma - 6) REP(x, 1, 7) {
            nxt[j + x] += dp[j] * p;
        }
        swap(dp, nxt);
    }
    rep(i, n - k) {
        vector<D> nxt(ma, 0);
        rep(j, ma - 6) REP(x, 1, 7) {
            nxt[j + x] += ep[j] * p;
        }
        swap(ep, nxt);
    }
    rep(i, k) {
        vector<D> nxt(ma, 0);
        rep(j, ma - 6) REP(x, 4, 7) {
            nxt[j + x] += ep[j] * p;
            nxt[j + x] += ep[j] * p;
        }
        swap(ep, nxt);
    }
    D ans = 0;
    rep(i, ma) rep(j, i) {
        ans += ep[i] * dp[j];
    }
    cout << fixed << setprecision(15);
    cout << ans << "\n";
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t = 1;
    // cin >> t;
    rep(ti, t) solve();
    return 0;
}