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Suppose each participant has at most one mutual friend, each friendship pair chooses consecutively, room values lie in $[0,1]$ and are distinct for each participant, and each friendship bonus exceeds $1$. Then every subgame perfect equilibrium outcome of serial dictatorship is Pareto optimal.

For a friend pair, if any remaining adjacent room pair exists, the leader chooses their uniquely best room having an available neighbor; the follower chooses their uniquely best available neighbor. Adjacency dominates every possible difference in own-room value for both friends. If no adjacent room pair remains, neither friend can obtain a bonus and each chooses their unique best available room. A participant without a friend also chooses their unique best available room.

For the proof, compare a purported Pareto improvement block by block, where a block is a single participant without a friend or one consecutive friendship pair. Earlier fixed blocks leave the same rooms available. A block receiving adjacency cannot lose it in a weak improvement. Its leader must retain the uniquely best eligible room, and its follower must retain the uniquely best neighboring room. Without available adjacency, both rooms are fixed by strict own-room rankings. Mathematical induction fixes every assignment, excluding a strict improvement.

A feasible change weakly increasing every participant's utility and strictly increasing at least one. An assignment is Pareto optimal exactly when no Pareto improvement exists. This comparison permits coordinated changes by several participants; it is different from a profitable unilateral deviation used to test a Nash equilibrium.

The assignment is Pareto optimal. Write $a_i$ for the room selected by the $i$th student. Suppose another assignment $a^{\prime }$ makes every student at least as well off. The first student already selected their uniquely most valuable room, so $a_1^{\prime }=a_1$. Once the rooms of students $1,\dots ,i-1$ are fixed, the rooms available to student $i$ in the original serial dictatorship are exactly the rooms that remain after those fixed assignments. Student $i$ chose the unique maximum among them, so weak improvement requires $a_i^{\prime }=a_i$.

Mathematical induction fixes every student's room. Thus $a^{\prime }=a$ and nobody can be strictly better off, contradicting the definition of a Pareto improvement. Strict room rankings and the absence of allocation externalities are the essential assumptions: a student's utility is determined solely by their own room.^[ref]

No: arbitrary ties can produce an assignment that is not Pareto optimal. A two-student counterexample suffices. Give the first student value $1$ for both rooms; give the second student value $1$ for $r_1$ and $0$ for $r_2$. If the first student breaks the tie by taking $r_1$, the second must take $r_2$, and the utility vector is $(1,0)$.

Swapping the rooms gives utility vector $(1,1)$, so it is a Pareto improvement. Both choices in the original run maximized the corresponding student's available room value. The strict-ranking argument from (a) fails because the first student's utility need not decrease when that student's room changes.

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: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.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 isBut the feasible assignmentstrictly 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 $a^{\prime }$; anticipating that reaction gives $S_1$ only $0.50$.

Every subgame perfect equilibrium outcome is Pareto optimal under the stated consecutive-friend and bonus conditions. Partition the choosing order into blocks: a single student without a friend, or two friends who choose consecutively. Their slots in the choosing order are adjacent; the proof must separately establish when their allocated rooms will be adjacent.

Let $R$ be the rooms remaining before a friend block, with leader $i$ and follower $j$. Because $0\le v_{i,r},v_{j,r}\le 1$ and $f_i,f_j>1$, each friend strictly prefers every adjacent-to-friend allocation to every nonadjacent one. After leader $i$ takes room $r$ with an available adjacent room, follower $j$ therefore takes their uniquely highest-valued room among the available neighbors of $r$. The leader consequently knows they will receive the adjacency bonus.

If $R$ contains at least one pair of adjacent rooms, let $E(R)$ be its nonisolated rooms: the rooms having an available neighbor. Leader $i$ must choose from $E(R)$, because choosing there guarantees utility greater than $1$, whereas an isolated choice gives at most $1$. Since the bonus is then fixed and room values are distinct, the leader chooses their unique maximum-value room in $E(R)$. The follower chooses their unique best neighbor. Later students cannot change either allocated room, so later play cannot alter these two students' utilities.

If $R$ contains no adjacent pair, neither friend can obtain the bonus under any allocation of rooms from $R$. Both simply choose their uniquely most valuable remaining room in turn. Removing rooms cannot create a new physical adjacency, so no hidden later move can change that conclusion. A one-student block likewise chooses their unique best available room.

Now compare the equilibrium assignment $a$ with any proposed Pareto improvement $a^{\prime }$. Induct over these blocks, assuming earlier blocks' rooms are already fixed. Thus $a$ and $a^{\prime }$ have the same remaining set $R$ at the present block. A lone student's room must stay fixed by strict preferences.

For a friend block obtaining adjacency in $a$, both students must retain adjacency in $a^{\prime }$: losing a bonus larger than $1$ cannot be compensated by any room-value increase. The leader's room in $a^{\prime }$ must therefore belong to $E(R)$. It cannot differ from the unique highest-valued room the leader selected there without reducing that leader's utility. Once the leader's room is fixed, the follower likewise must retain the uniquely best available neighbor. If $R$ has no adjacent pair, both rooms stay fixed by the ordinary strict-ranking argument.

Thus every block's rooms are unchanged. The only assignment weakly improving all students is $a$ itself, so no strict improvement is possible. This proves Pareto efficiency of serial dictatorship with consecutive friends. The proof applies to every subgame perfect equilibrium, since the required choices at each block are strict best responses determined by the available rooms.

An allocation procedure using a fixed priority order: each participant in turn selects one remaining object, which is permanently assigned to them. For participants whose utilities depend only on their own assigned objects, strict rankings give a unique best available choice and the resulting allocation is Pareto optimal. If a proposed Pareto improvement changes an assignment, its earliest changed participant loses their unique best available object. With ties or allocation externalities, that argument requires additional assumptions.