= Solution
\b[No. Strict room values do not remove the <allocation externalities> generated by friends.] Consider the following new instance, with the same friendship pairs $\{S_1,S_4\}$ and $\{S_2,S_3\}$ and friendship bonus $0.4$ for each student:
$$
\begin{array}{c|rrrr}
&r_1&r_2&r_3&r_4\\ \hline
S_1&0.50&0.30&0.02&0.01\\
S_2&0.01&0.50&0.30&0.02\\
S_3&0.01&0.70&0.02&0.50\\
S_4&0.01&0.50&0.02&0.03
\end{array}
$$
Each row is strictly ranked and every room value is positive. In picking order $S_1,S_2,S_3,S_4$, <backward induction> gives the following continuations. Room numbers in a tuple are listed in student order.
$$
\begin{array}{c|c|c}
S_1\text{ choice}&\text{resulting assignment}&U_1\\ \hline
r_1&(r_1,r_3,r_2,r_4)&0.50\\
r_2&(r_2,r_3,r_4,r_1)&0.70\\
r_3&(r_3,r_2,r_4,r_1)&0.02\\
r_4&(r_4,r_2,r_3,r_1)&0.01
\end{array}
$$
To check $S_2$'s incentives behind these continuations, the resulting <utilities> for their available choices, in ascending room order, are respectively $(0.50,0.70,0.02)$, $(0.01,0.70,0.42)$, $(0.41,0.50,0.02)$ and $(0.41,0.90,0.70)$. Student $S_3$'s own <best responses> determine those figures, and $S_4$ again takes the remaining room. All comparisons selecting these continuations are strict.
Hence $S_1$ chooses $r_2$ and the <subgame perfect equilibrium> outcome is
$$
a=(r_2,r_3,r_4,r_1),\qquad U(a)=(0.70,0.70,0.90,0.41).
$$
But the feasible assignment
$$
a'=(r_1,r_3,r_4,r_2),\qquad U(a')=(0.90,0.70,0.90,0.90)
$$
strictly improves $S_1$ and $S_4$ and leaves $S_2,S_3$ unchanged. Thus $a$ is not <Pareto optimal>. This is compatible with equilibrium: a <Pareto improvement> can require several players' actions to change. If $S_1$ alone takes $r_1$, the later <best responses> lead to $(r_1,r_3,r_2,r_4)$, not to $a'$; anticipating that reaction gives $S_1$ only $0.50$.
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