#include <bits/stdc++.h>
#define GET_MACRO(a, b, c, NAME, ...) NAME
#define rep(...) GET_MACRO(__VA_ARGS__, rep3, rep2)(__VA_ARGS__)
#define rep2(i, a) rep3 (i, 0, a)
#define rep3(i, a, b) for (int i = (a); i < (b); i++)
#define repr(...) GET_MACRO(__VA_ARGS__, repr3, repr2)(__VA_ARGS__)
#define repr2(i, a) repr3 (i, 0, a)
#define repr3(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define chmin(a, b) ((b) < a && (a = (b), true))
#define chmax(a, b) (a < (b) && (a = (b), true))
using namespace std;
typedef long long ll;
typedef long double ld;

ll N, L, R;

ld dp[101010];

void solve1() {
	dp[0] = 1;
	rep (i, N) repr (j, i, i * 6 + 7) {
		dp[j] = 0;
		rep (k, 1, 7) {
			if (j - k >= 0) dp[j] += dp[j - k] / 6l;
		}
	}
	ld pr = 0;
	rep (i, L, R + 1) pr += dp[i];
	printf("%.20f\n", (double)pr);
}

void solve2() {
	ld m = (7.0l / 2.0l) * N;
	ld s = sqrt((35.0l / 12.0l) * N);
	ld l = (L - m - 0.5l) / s;
	ld r = (R - m + 0.5l) / s;
	ld ans = (erfl(r / sqrt(2)) - erfl(l / sqrt(2))) / 2.0l;
	printf("%.20f\n", (double)ans);
}

int main() {
	cin >> N >> L >> R;
	chmin(L, N * 6);
	chmin(R, N * 6);
	if (N < 8000) {
		solve1();
	} else {
		solve2();
	}
    return 0;
}