// 想定解法: DP

#include <bits/stdc++.h>
using namespace std;
void rc(int v,int mn,int mx){if(v<mn||mx<v){cerr<<"error"<<endl;}}
#define REP(i, n)           for(int(i)=0;(i)<(n);++(i))
#define FOR(i, f, t)        for(int(i)=(f);(i)<(t);(++i))

const int MAX = 15;
double dp[MAX+1][(1<<MAX)+1];
int M, S[(1<<MAX)+1];

// m回戦に選手iが勝つ確率を返す
double dfs(int m, int i){
    if(dp[m][i] >= 0) return dp[m][i];
    int p = 1<<(m-1), p2 = p*2;
    int j = i/p2*p2, k = i/p*p;
    if(j == k) j += p;

    double t = 0;
    for(int l = j; l < j+p; l++){
        t += dfs(m-1,l) * S[i] / (S[i]+S[l]);
    }
    return dp[m][i] = t * dfs(m-1, i);
}


int main(){
    cin >> M; rc(M, 1, MAX);
    int n = 1 << M;
    REP(i, n){
        cin >> S[i];
        rc(S[i], 1, 30000);
        S[i] *= S[i];
        dp[0][i] = 1.0;
    }
    REP(i,M) REP(j,n) dp[i+1][j] = -1;
    REP(j,n) dp[0][j] = 1.0;

    cout << fixed << setprecision(16) << dfs(M,0) << endl;
    return 0;
}