dp[x][y]:现在有x个白老鼠，y个黑老鼠，公主赢的概率。
那么：
如果公主直接拿到白老鼠，概率为x/(x+y),公主赢。
如果公主拿到黑老鼠，概率为y/(x+y),那么公主如果想赢，龙必须拿到黑老鼠，概率为(y-1)/(x+y-1);
那么逃跑的老鼠为黑老鼠的概率为(y-2)/(x+y-2),为白老鼠的概率为(x)/(x+y-2);
那么dp[x][y]=x/(x+y)+y/(x+y) * (y-1)/(x+y-1) * ( (y-2)/(x+y-2) * dp[x][y-3] + (x)/(x+y-2) * dp[x-1][y-2] );
记忆化深搜也行，直接递推DP也行。
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<stack>
using namespace std;
#define N 100010
#define LL long long
#define INF 0xfffffff
#define maxn 1100
double dp[maxn][maxn];
double dos(int x,int y)
{if(x<0||y<0)return 0;if(x==0)return 0;if(y==0)return 1;if(dp[x][y]>-0.5)return dp[x][y];double a,b;a=1.0*x;b=1.0*y;dp[x][y]=a/(a+b);if(x+y>=3){dp[x][y]+=(b*(b-1)/((a+b)*(a+b-1)))*(dos(x-1,y-2)*a/(a+b-2)+dos(x,y-3)*(b-2)/(a+b-2));}// cout<<x<<" "<<y<<" "<<dp[x][y]<<endl;return dp[x][y];
}
int main()
{int a,b;while(~scanf("%d%d",&a,&b)){memset(dp,-1,sizeof(dp));printf("%.10lf\n",dos(a,b));}return 0;
}