题意：36张牌分成9堆，没堆4张，每次可以拿走某两堆顶部的牌，但需要的点数相同，如果有多种拿法就等概率随机拿，问拿完所有牌的概率。
分析：直接用9元组表示当前状态，即每堆剩余的牌数，状态数为5^9=1953125。设d[i]表示状态i对应的概率，则根据全概率公式，d[i]为后继状态的成功概率的平均值，按照动态规划的写法计算即可。

代码如下：

//
//Created by BLUEBUFF 2016/1/11
//Copyright (c) 2016 BLUEBUFF.All Rights Reserved
//

#pragma comment(linker,"/STACK:102400000,102400000")
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/hash_policy.hpp>
//#include <bits/stdc++.h>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b) memset(a, b, sizeof(a))
#define MP(x, y) make_pair(x,y)
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
const int maxn = 100010;
const int maxm = 1e5+5;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9 + 7;
const int rev = (mod + 1) >> 1; // FWT
const double PI = acos(-1);
//head

double dp[5][5][5][5][5][5][5][5][5] = {0};
bool vis[5][5][5][5][5][5][5][5][5];
vector <char> q[10];
double dfs(int w1, int w2, int w3, int w4, int w5, int w6, int w7, int w8, int w9){
    if(vis[w1][w2][w3][w4][w5][w6][w7][w8][w9])
        return dp[w1][w2][w3][w4][w5][w6][w7][w8][w9];
    vis[w1][w2][w3][w4][w5][w6][w7][w8][w9] = 1;
    int tmp[10] = {w1, w2, w3, w4, w5, w6, w7, w8, w9};
    bool flag = 0;
    double &x = dp[w1][w2][w3][w4][w5][w6][w7][w8][w9];
    for(int i = 0; i < 9; i++){
        if(tmp[i]){
            flag = 1;
            break;
        }
    }
    if(flag == 0){
        return x = 1;
    }
    //else{
        int cnt = 0;
        for(int i = 0; i < 8; i++){
            if(tmp[i]){
            for(int j = i + 1; j < 9; j++){
                if(tmp[j] && q[i][tmp[i] - 1] == q[j][tmp[j] - 1]){
                    cnt++;
                    tmp[i]--;
                    tmp[j]--;
                    x += dfs(tmp[0], tmp[1], tmp[2], tmp[3], tmp[4], tmp[5], tmp[6], tmp[7], tmp[8]);
                    tmp[i]++;
                    tmp[j]++;
                }
            }
            }
        }
        if(x == 0) return x = 0;
        else{
            return x = x * 1.0 / (cnt * 1.0);
        }
    //}
}

int main()
{
    char s1[10], s2[10], s3[10], s4[10];
    while(scanf("%s%s%s%s", s1, s2, s3, s4) != EOF)
    {
        q[0].push_back(s1[0]);
        q[0].push_back(s2[0]);
        q[0].push_back(s3[0]);
        q[0].push_back(s4[0]);
        for(int i = 1; i < 9; i++){
            scanf("%s%s%s%s", s1, s2, s3, s4);
            q[i].push_back(s1[0]);
            q[i].push_back(s2[0]);
            q[i].push_back(s3[0]);
            q[i].push_back(s4[0]);
        }
        //cout << " ** " << endl;
        double ans = dfs(4, 4, 4, 4, 4, 4, 4, 4, 4);
        printf("%.6f\n", ans);
        CLR(dp, 0);
        CLR(vis, 0);
        REP1(i, 0, 10) q[i].clear();
    }
}