#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")


/*
  (not a proof)
    fixed strategy: S \subseteq 3^N
    (coin <-> element of S)
    condition: for any x, #{ y \in S | dist(x, y) <= K } <= 3K+1
    {0,1} ~~> real weight p := S/3^N
    condition: p \sum[0<=i<=K] binom(N,i) 2^i <= 3K+1
    ans = (3K+1) / (\sum[0<=i<=K] binom(N,i) (1/3)^(N-i) (2/3)^i)
*/

constexpr double A = log(1.0L / 3.0L);
constexpr double B = log(2.0L / 3.0L);
double calc(Int N, Int i) {
  return lgamma(N + 1) - lgamma(i + 1) - lgamma(N - i + 1) + (N - i) * A + i * B;
}

// log(e^x + e^y)
double add(double x, double y) {
  if (x < y) swap(x, y);
  return x + log1p(exp(y - x));
}

int main() {
  for (int numCases; ~scanf("%d", &numCases); ) { for (int caseId = 1; caseId <= numCases; ++caseId) {
    Int N, K;
    scanf("%lld%lld", &N, &K);
    
    Int l = 0, r = K;
    chmax(l, N / 3 - 100);
    chmin(r, N / 3 + 100);
    double ans = calc(N, l);
    for (Int i = l + 1; i <= r; ++i) {
      ans = add(ans, calc(N, i));
    }
    ans = log((double)(3*K+1)) - ans;
    printf("%.7f\n", ans);
  }
#ifndef LOCAL
  break;
#endif
  }
  return 0;
}