Algorithm

1. At each one of the $10$ tries, we keep filtering out the items from the list that does not fit the match number.
2. The actual probability could be computed using some INSANE analytic combinatorics. I cannot think of a better method than this. It is basically asking given a randomly generated $6n$ length strings where $n\geq10,$ what are the chances that the $6$ length cuts are quite different. To calculate the probability, one would need to work out all the patterns under which $6$ length cuts are quite similar. There are some easy patterns to start with, such that all $n$ cuts have the same number of matches. The probability of this is $$\mathbb{P}(n\text{ cuts have same matchs})=\sum_{i=1}^{6}\binom{6n}{6i}\left(\frac{1}{26}\right)^{6i}\left(\frac{1}{26}\right)^{6n-6i}.$$ But this does not help much. If you have a better idea, please PM me so that I can update it.

Code

class Solution {
 public:
  void findSecretWord(vector<string>& wordlist, Master& master) {
    std::random_device rd;
    std::mt19937 gen(rd());
    std::uniform_int_distribution<int> Unif(0, size(wordlist) - 1);

    auto match = [&](const string &x, const string & y) {
      int acc = 0;
      for (int i = 0; i < 6; ++i)
        if (x[i] == y[i])
          ++acc;
      return acc;
    };

    auto solve  = [&]()-> void {
      auto word_filter = vector<optional<string>>(begin(wordlist), end(wordlist));

      for (int i = 0; i < 10; ++i) {
        const auto candidate = [&] {
          const auto rand_id = [&](int self = 0) {
            while (not word_filter[self])
              self = Unif(gen);
            return self;
          }();
          return word_filter[rand_id].value();
        }();

        const auto response = master.guess(candidate);

        if (response == -1)
          continue;
        else {
          for (int i = 0; i < size(wordlist); ++i)
            if (word_filter[i] and match(word_filter[i].value(), candidate) != response)
              word_filter[i].reset();         
        }
      }
    };

    solve();

  }
};