One essential aspect of the Inclusion-Exclusion Principle (IEP) is the transformation of at least information to exact information.
If counting objects having at least a number of properties is simple, but counting objects having exactly a number of properties is difficult, than the IEP comes into play.
The objects are represented by the elements contained in $A_1,\dots,A_n$ and the properties of an element $x$ are the sets $A_j,1\leq j\leq n$, which contain $x$.
If we have this essence of IEP in mind and we look at the expression:
\begin{align*}
\left|\bigcup_{j=1}^{n}A_j\right|=\sum_{j=1}^{n}\left|A_j\right|
-\sum_{1\leq i \leq j \leq n}\left|A_i\cap A_j\right|\pm\cdots+(-1)^{n-1}\left|\bigcap_{j=1}^{n}A_j\right|
\end{align*}
we observe, that the right hand side (RHS) consists of summands with at least information.
Note, that the summand
$$\left|A_i\cap A_j\right|\quad \text{in}\quad\sum_{1\leq i \leq j \leq n}\left|A_i\cap A_j\right|$$
is not only the number of elements which are in $A_i$ and $A_j$ it is more precisely the number of elements which are at least in $A_i$ and $A_j$, since elements $x$ in $A_i\cap A_j$ may also be contained in other sets of $A_1,\dots,A_n$.
Whereas the LHS
$$\left|\bigcup_{j=1}^{n}A_j\right|$$
presents the number of elements which are exactly in $\bigcup_{j=1}^{n}A_j$.
We observe the IEP transforms counting information with at least properties into counting information with exact properties.
Note: This intuitive connection between at least and exact information has a formal represention. Using generating functions you will get a kind of eye-birds view at hand, which transforms the at least to an exact information as simple shift by one of the argument. For more information about this approach you may have a look at section 4.2 of H.S. Wilfs Generatingfunctionology.
My previous answer contains a misinterpretation: the question actually has no thing to do with derangements.
For the game, the probability of exactly $k$ person get correct is
$$P_k=\binom{6}{k}(\frac{5}{6})^{6-k}(\frac{1}{6})^k$$
So the answer is just $P_1+P_2+P_3 = 0.6564$
Best Answer
All singles - all pairs + all triples - all quadruples + all quintuples.