// Arup Guha
// 10/12/2013
// Solution to UCF JV Practice Problem E: Archer

// Note: Here's a way to solve this without infinite series:
//       SmallR's chances of winning can be broken down into two parts:
//                      (a) He wins on turn one.
//                      (b) No one wins round 1, and the game starts over.
//       Let X equal probability of SmallR winning. Then
//       X = prob(win SmallR on one turn) + prob(SmallR misses, followed by Zanos misses)*X
//       Plugging in, X = (a/b) + (1 - a/b)(1 - c/d)X, then solve for X.

import java.util.*;

public class e {

	public static void main(String[] args) {

		// Read in input.
		Scanner stdin = new Scanner(System.in);
		int a = stdin.nextInt();
		int b = stdin.nextInt();
		int c = stdin.nextInt();
		int d = stdin.nextInt();

		// Probability of both players missing.
		double r = ((double)(b-a))/b*(d-c)/d;

		// Calculate probabilty first wins - infinite series with first term being chance SmallR wins on first turn
		// and the common ratio being the chance that both players miss, returning the game to SmallR.
		double prob = ((double)a)/(b*(1-r));
		System.out.println(prob);
	}
}