There are too many cards to simply try every set of five cards and see which ones form a full house. So we will need to use some maths to calculate the answer.

Firstly, count how many cards there are of each rank. Suppose there are ~C_i~ cards of rank ~i~ for each ~i\in\{\text{A},2,3,4,5,6,7,8,9,10,\text{J},\text{Q},\text{K}\}~.

Then, given two ranks ~i\ne j~, the number of full houses with 3 cards of rank ~i~ and 2 cards of rank ~j~ is ~\binom{C_i}3\binom{C_j}2=\frac1{12}C_iC_j(C_i-1)(C_i-2)(C_j-1)~. Sum over all 182 possible pairs of ranks to get the total number of full houses (which fits in a long long).

Finally, the number of hands of five cards is exactly ~\binom N5=\frac1{120}N(N-1)(N-2)(N-3)(N-4)~, which also fits in a long long. Divide one number by the other to get the desired probability.

Taking in the input is ~O(n)~ time complexity. While we have to check all possible pairs of cards, we are using a frequency counter to do so, and there is a limited amount (~13^2~) possible pairs so the overall time complexity is still ~O(n)~. The solution uses ~O(1)~ space as the map representing the card deck can only have up to 13 entries in it.

C++ solution:

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;

    unordered_map<string, int> freqs;
    string x;
    for (int i = 0; i < n; i++) {
        cin >> x;
        freqs[x]++;
    }

    long long numerator = 0;

    for (auto [card1, freq1]: freqs) {
        for (auto [card2, freq2]: freqs) {
            if (card1 == card2 || freq1 < 2 || freq2 < 3) continue;
            numerator += 1ll * freq1*(freq1-1)/2 * freq2*(freq2-1)*(freq2-2)/6;
        }
    }

    long long denominator = 1ll * n*(n-1)*(n-2)*(n-3)*(n-4)/120;
    cout << (double)numerator/denominator << endl;
}

Python solution:

from collections import Counter
from math import comb

n = int(input())

cards = input().split()
freqs = Counter(cards)

fullhouses = 0
for k1, v1 in freqs.items():
    for k2, v2 in freqs.items():
        if k1 == k2 or v1 < 2 or v2 < 3:
            continue
        fullhouses += comb(v1, 2) * comb(v2, 3)

print(fullhouses / comb(n, 5))