#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

typedef long double Num;

inline Num sq(Num x) { return x*x; }

const Num A = 2.506628274631000502415765284811045253006986740609938316629923L;
Num pdf(Num mu, Num sigma, Num x) {
	return exp(-sq(x - mu) / 2 / sigma / sigma) / sigma / A;
}

Num cdf(Num mu, Num sigma, Num x) {
	return (1 + erf((x - mu) / (sigma * sqrt(2.L)))) / 2;
}

int main() {
	const int D = 6;
	
	long long N;
	while(~scanf("%lld", &N)) {
		long long L; long long R;
		scanf("%lld%lld", &L, &R);
		L -= N, R -= N;
		amax(L, 0LL);
		amin(R, N * (D - 1));
		Num ans;
		Num mu = (D - 1) / 2.L * N;
		Num sigma = sqrt(70.L / 24 * N);
		if(R < L) {
			ans = 0;
		}else if(N <= 10000) {
			int NN = (int)N;
			ans = 0;
			vector<Num> dp(NN * (D-1) + 1);
			dp[0] = 1;
			Num c = 1.L / D;
			rep(i, NN) {
				for(int j = i * (D - 1); j >= 0; -- j) {
					Num x = dp[j];
					if(x < 1e-99) continue;
					dp[j] = 0;
					rep(d, D)
						dp[j + d] += x * c;
				}
			}
			ans = 0;
			for(ll x = L; x <= R; ++ x)
				ans += dp[(int)x];
		} else if(R - L <= 100000) {
			ans = 0;
			for(ll x = L; x <= R; ++ x)
				ans += pdf(mu, sigma, x * 1.L);
		} else {
			ans = cdf(mu, sigma, R + .5) - cdf(mu, sigma, L - .5);
		}
		printf("%.10f\n", (double)ans);
	}
	return 0;
}