As a good prob/stats student you recognize that this situation fits the requirements of the hypergeometric distribution with \(n_1 = 6, n_2 = 27, r=5\) and you want \(P(X \ge 2)\text{.}\) You determine that it would be easier to compute the complement
\begin{equation*}
1 - P(X \le 1) = 1 - f(0) - f(1).
\end{equation*}
Therefore,
\begin{equation*}
P (X \ge 2) = 1 - \frac{\binom{6}{0} \binom{27}{5}}{\binom{33}{5}} - \frac{\binom{6}{1} \binom{27}{4}}{\binom{33}{5}}
\end{equation*}
or after some simplification
\begin{equation*}
P (X \ge 2) = 1 - \frac{13455}{39556} - \frac{8775}{19778} \approx 0.21617
\end{equation*}
Therefore, you have about 1 chance out of 5 that the friend got 2 or more of your bars. You can be somewhat confident that plenty of dark chocolate bars remain hoarded for yourself.