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Allocating group housing

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We study mechanisms for allocating objects to pairs of agents when agents may have nontrivial preferences over objects and pairings. In this environment, the mechanism may distort agents’ preferences over pairings. Compared to certain distortive mechanisms, a non-distortive one always has a stable allocation in our model, and selects stable outcomes that are ex ante preferred by all students under a regularity condition on the distribution of pair values.

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  1. Some examples may be business teams which are assigned projects/locations, or families with preferences over neighbors and schools.

  2. A survey of the top 25 National Universities and the top 25 National Liberal Arts colleges shows that 23 of 50 determine the housing priority before students decide on their roommates, and 16 of 50 determine the priorities after roommate groups form. The rankings come from U.S. News and World Report. The remaining schools either do not make their rules available or follow some other scheme.

  3. Preferences generated by distance in a metric space, first proposed by Bartholdi and Trick (1986), is another “no odd rings” type of preferences.

  4. This model borrows from Becker’s (1973) model of marriage.

  5. Although the common order assumption is strong, it allows us to avoid specifying a method for how roommates jointly decide on a room, given their possibly distinct preferences over rooms. We leave such a model as an avenue for further research.

  6. All of the results would follow if, rather than all roommates being acceptable and all students having common ordinal preferences over rooms, S could be partitioned into subsets such that for each subset \(S'\), for any student \(s\in S'\), \(S'\) contains all acceptable roommates for s, and within these sets students share ordinal preferences over rooms. For example, athletes may all prefer to live near the athletic facilities on campus, in contrast to other students. We require then that they also prefer to live with each other.

  7. In our examples \(>_S\) is often given, in which case we refer to “mechanism” \(\phi _L\), with the understanding that this is the \(\phi _L\) mechanism with respect to \(>_S\).

  8. Blocking pairs are one of \((s_1,s_2), (s_1,s_3), (s_1,s_4), (s_1,s_5), (s_2,s_4)\) and \((s_3,s_4)\).

  9. Bergstrom and Bagnoli (2005) discusses the key implications of this assumption and its use in economic models.

  10. With 4 students a match which is stable with respect to \(\phi _L\) always exists.

  11. Conditional on \(x_1\) being assigned the room corresponding to \(v_{x_1}\), the anonymity of the mechanism implies an equal probability that \(x_2\) will be assigned any of the remaining rooms, and similarly for \(y_1\) and \(y_2\) respectively.


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Correspondence to Francis X. Flanagan.

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We thank Thayer Morrill and an anonymous referee for helpful feedback.



Proof of Theorem 1

Construct match \(\mu _n\) as follows: let \(\mu _0=\emptyset \) and \(\mu _i=\mu _{i-1}\cup \{{(s,s')}_i\}\), where \({(s,s')}_i\in \arg \max _{\{s_j,s_k\}\subset S{\setminus }\mu _{i-1}}w(s_j,s_k)\), with any ties broken randomly if \((s,s')\) is not unique. We claim that \(\mu _n\) is stable with respect to \(\phi _R\). Under \(\phi _R\) the expected room value of each student is independent of the match, thus any blocking pair (st) must be such that \(f_s(w(s,t))>f_s(w(s,s'))\) and \(f_t(w(s,t))>f_t(w(t,t'))\), where \(s'=\mu _n(s)\) and \(t'=\mu _n(t)\). This implies that \(w(s,t)> \max \{w(s,s'),w(t,t')\}\), which is clearly a contradiction. \(\square \)

Proof of Theorem 2

Suppose that \(\mu \) is stable with respect to \(\phi _R\), that \((s_1,s_2)\in \mu \) is assigned to \(r_1\), and that \((t_1,t_2)\in \mu \) is assigned to \(r_2\). Without loss of generality, assume \(v_i(r_1)\ge v_{i}(r_2)\) for all \(i\in \{s_1,s_2,t_1,t_2\}\). For a contradiction, assume the allocation is not ex post stable, or that

$$\begin{aligned} f_{s_1}(w(s_1,t_1))+v_{s_1}(r_1)\ge & {} f_{s_1}(w(s_1,s_2))+v_{s_1}(r_1) {\implies } w(s_1,t_1) \ge w(s_1,s_2) \end{aligned}$$
$$\begin{aligned} f_{s_2}(w(s_2,t_2))+v_{s_2}(r_2)\ge & {} f_{s_2}(w(s_1,s_2))+v_{s_2}(r_1) \end{aligned}$$
$$\begin{aligned} f_{t_1}(w(s_1,t_1))+v_{t_1}(r_1)\ge & {} f_{t_1}(w(t_1,t_2))+v_{t_1}(r_2) \end{aligned}$$
$$\begin{aligned} f_{t_2}(w(s_2,t_2))+v_{t_2}(r_2)\ge & {} f_{t_2}(w(t_1,t_2))+v_{t_2}(r_2) {\implies } w(s_2,t_2) \ge w(t_1,t_2),\qquad \quad \end{aligned}$$

where at least one inequality is strict. First, suppose \(v_i(r_1)= v_{i}(r_2)\) for all \(i\in \{s_1,s_2,t_1,t_2\}\). Then all of the room values cancel and the strictness of one of the inequalities implies that a pair blocks \(\mu \) under \(\phi _R\), a contradiction. Therefore, we must have \(v_i(r_1)> v_{i}(r_2)\) for all \(i\in \{s_1,s_2,t_1,t_2\}\). Then (2) implies that \(w(s_2,t_2)>w(s_1,s_2)\), which in combination with (4) implies that \((s_2,t_2)\) is a blocking pair for \(\mu \) under \(\phi _R\), a contradiction. \(\square \)

Proof of Theorem 3

Keeping \(\phi \in \Phi _L\) fixed, let \(V_{s}(\phi ,\mu )\) denote the value of the room assigned to s under \(\phi \) when the match is \(\mu \). Recall that any \(\phi \in \Phi _L\) selects some priority order over pairs with probability one. Combined with the assumption of common ordinal preferences this implies, that given the mechanism \(\phi \) and match \(\mu \) completely determine the room assigned to each pair. In the rest of the proof, to simplify notation, we may denote \(V_{s_i}\) as \(V_i\).

Lemma 1

Suppose \(\phi \in \Phi _L\). For \(i\ne j\), \(i\ne k\) and \(j\ne k\), there exist four matches \(\mu _1\), \(\mu _2\), \(\mu _3\), and \(\mu _4\) such that: (i) \(\mu _1\leftrightarrow \mu _2\leftrightarrow \mu _3\leftrightarrow \mu _4\leftrightarrow \mu _1\), (ii) \(|\mu _1{\setminus }\mu _3|=3\), (iii) \((s_i,s_j)\in \mu _1 \cap \mu _4\), (iv) \((s_i,s_k)\in \mu _2 \cap \mu _3\), and (v) \(V_i\left( \phi ,\mu _2\right) - V_i\left( \phi ,\mu _1 \right) \ne V_i\left( \phi ,\mu _3\right) - V_i\left( \phi ,\mu _4\right) \).


To generate a contradiction, assume that the statement is not true. It is straightforward to find four matches satisfying conditions (i)–(iv) in the lemma. Therefore, we assume that for any such set (v) is violated, i.e., \(V_i\left( \phi ,\mu _2\right) - V_i\left( \phi ,\mu _1 \right) = V_i\left( \phi ,\mu _3\right) - V_i\left( \phi ,\mu _4\right) \). Let \(i=1\). Take a match \(\nu _1\) and three pairs \(\{(s_1,s_2), (s_3,s_4), (s_5,s_6)\}\subseteq \nu _1\). Holding all other pairs fixed, restrict attention to the subset \(\mathcal {M}'\subseteq \mathcal {M}\) of fifteen matches which differ only in how these six students are paired. Consider the following arrangement of members of this set:

$$\begin{aligned}&\{(s_1,s_2),(s_3,s_4),(s_5,s_6)\} \subseteq \nu _1&\{(s_1,s_2),(s_3,s_6),(s_4,s_5)\} \subseteq \nu _1'\\&\{(s_1,s_4),(s_2,s_3),(s_5,s_6)\} \subseteq \nu _2&\{(s_1,s_4),(s_3,s_6),(s_2,s_5)\} \subseteq \nu _2' \\&\{(s_1,s_3),(s_2,s_4),(s_5,s_6)\} \subseteq \nu _3&\{(s_1,s_3),(s_2,s_5),(s_4,s_6)\} \subseteq \nu _3' \\&\{(s_1,s_2),(s_3,s_4),(s_5,s_6)\} \subseteq \nu _1&\{(s_1,s_2),(s_3,s_5),(s_4,s_6)\} \subseteq \nu _1'' \end{aligned}$$

Note that they are arranged so that each match is adjacent to the one in the same row as well as to any match immediately above or below it. By the assumption that the lemma is false, and since \(V_1\left( \phi ,\mu _2\right) - V_1\left( \phi ,\mu _1 \right) = V_1\left( \phi ,\mu _3\right) - V_1\left( \phi ,\mu _4\right) \iff V_1\left( \phi ,\mu _4\right) - V_1\left( \phi ,\mu _1 \right) = V_1\left( \phi ,\mu _3\right) - V_1\left( \phi ,\mu _2\right) \), we have the following equalities:

$$\begin{aligned} V_1\left( \phi ,\nu _1'\right) - V_1\left( \phi ,\nu _1\right)= & {} V_1\left( \phi ,\nu _2'\right) - V_1\left( \phi ,\nu _2\right) \end{aligned}$$
$$\begin{aligned}= & {} V_1\left( \phi ,\nu _3'\right) - V_1\left( \phi ,\nu _3\right) \end{aligned}$$
$$\begin{aligned}= & {} V_1\left( \phi ,\nu _1''\right) - V_1\left( \phi ,\nu _1\right) \end{aligned}$$

This implies \(V_1\left( \phi ,\nu _1' \right) = V_1\left( \phi ,\nu _1''\right) \). By exchanging \(s_4\) and \(s_6\) in all of the above matches, we find that \(V_1\left( \phi ,\nu _1 \right) = V_1\left( \phi ,\nu _1''\right) \). Note that swapping two students consistently across all matches does not change any adjacency relationships. We therefore have that for matches in \(\mathcal {M}'\), \(s_1\) must receive a constant room value when paired with \(s_2\) independently of the arrangement of the other four students. By swapping \(s_1\) with \(s_i\) and \(s_2\) with \(s_j\) for any \(i,j\in \{1,\ldots ,6\}\) with \(i\ne j\), a similar argument establishes that the pair \((s_i,s_j)\) must receive the same room value for all matches in \(\mathcal {M}'\). Because the original match \(\nu _1\) was arbitrary, we conclude that the same is true of any such subset \(\mathcal {M}'\).

We show next that it is impossible to give a pair the same room value for all matches in some subset \(\mathcal {M}'\), and hence that there must exist four matches satisfying the conditions laid out in the lemma. We assume in the model that there exist three rooms with room values that differ from all others. We refer to these rooms as \(r_1\), \(r_2\) and \(r_3\). Without loss of generality, we assign \((s_1,s_2)\), \((s_3,s_4)\) and \((s_5,s_6)\) to rooms \(r_1\), \(r_2\) and \(r_3\) respectively. To derive a contradiction, assume that we can assign each pair, \((s_i,s_j)\), to the same room in every match that includes \((s_i,s_j)\). Consider assigning \((s_1,s_6)\) to \(r_1\). We deduce that the following assignments must be made, where the column of the pair corresponds to the assigned room,










By assumption





\((s_1,s_6)\) is in \(r_1\) and \((s_3,s_4)\) is in \(r_2\)





\((s_2,s_3)\) cannot be in \(r_3\) when \((s_5,s_6)\) is a pair





\((s_2,s_4)\) cannot be in \(r_3\) when \((s_5,s_6)\) is a pair





Using (1) and (3).

But this requires that we assign \((s_2,s_4)\) and \((s_3,s_6)\) to \(r_2\) in all matches, which is impossible because these two pairs appear in the same match with \((s_1,s_5)\). The assumption that \((s_1,s_6)\) is in \(r_1\) was not without loss, but a similar set of steps shows that assigning \((s_1,s_6)\) to \(r_3\) also creates a contradiction. Note that \((s_1,s_6)\) cannot occupy \(r_2\). We conclude that it is impossible to assign a pair \((s_i,s_j)\) with \(i\ne j\) to the same room for all \(\mu \in \mathcal {M}'\). \(\square \)

Take four matches, \(\mu _1\), \(\mu _2\), \(\mu _3\), and \(\mu _4\), satisfying conditions (i)–(iv) of the lemma with \(i=1\), \(j=2\) and \(k=3\). Then \(s_2=\mu _1(s_1)\) and \(s_3=\mu _2(s_1)\). Let \(s_4 =\mu _1(s_3)\) and \(s_5= \mu _3(s_2)\). Assume \(f_s(w(s,s'))=w(s,s')\) for all \(s,s'\in S\). Without loss of generality we may assume that \(V_1\left( \phi ,\mu _2 \right) > V_1\left( \phi ,\mu _1\right) \) and \(V_1\left( \phi ,\mu _2 \right) - V_1\left( \phi ,\mu _1\right) > V_1\left( \phi ,\mu _3\right) - V_1\left( \phi ,\mu _4\right) \). Whenever

$$\begin{aligned}&V_1\left( \phi ,\mu _2 \right) - V_1\left( \phi ,\mu _1\right) \nonumber \\&\quad> w(s_1,s_2)- w(s_1,s_3) \\&\quad > V_1\left( \phi ,\mu _3\right) - V_1\left( \phi ,\mu _4\right) \nonumber \end{aligned}$$

we have \(U_1(\phi ,\mu _2)>U_1(\phi ,\mu _1)\) but \(U_1(\phi ,\mu _3)<U_1(\phi ,\mu _4)\). Find some M such that

$$\begin{aligned} M>\max _{i,\mu \leftrightarrow \mu '}|V_i(\phi ,\mu )-V_i(\phi ,\mu ')|, \end{aligned}$$

and let \(w(s_1,s_2)=3M+\delta \) and \(w(s_1,s_3)=3M\) to satisfy (8). Let \(w(s_3,s_4)=w(s_2,s_5)=2M\). For all pairs \((s_i,s_j)\) remaining in \(\mu _1\cap \mu _3\) let \(w(s_i,s_j)=M\). For any other pair let \(w(s_i,s_j)=0\). We claim that these preferences are such that there does not exist a stable match under \(\phi \). First note that any match \(\mu \) in which there is some \(s\in S\) such that \(\mu (s)\notin \{\mu _1(s),\mu _3(s)\}\) is not stable. Next, since \(\mu _1\) and \(\mu _3\) differ by exactly three pairs, they are the only two matches in which every \(s\in S\) is such that \(\mu (s)\in \{\mu _1(s),\mu _3(s)\}\). Finally, note that \(\mu _1\) is blocked by \(\mu _2\), and \(\mu _3\) is blocked by \(\mu _4\). \(\square \)

Proof of Theorem 4

Suppose that \(\phi \) is such that \((s_1,s_2)\) receives \(r_1\), \((t_1,t_2)\) receives \(r_2\), \((s_1,t_1)\) receives \(r_3\), and \((s_2,t_2)\) receives \(r_4\). Let \(\mu \) be stable with respect to \(\phi \), and \((s_1,s_2),(t_1,t_2)\in \mu \). Consistent with \(\mu \) being stable with respect to \(\phi \), assume that \(w(s_1,t_1)> w(t_1,t_2)> w(s_1,s_2) > w(s_2,t_2)\) while

$$\begin{aligned} f_{s_1}(w(s_1,t_1))+v_{s_1}(r_3)< & {} f_{s_1}(w(s_1,s_2))+v_{s_1}(r_1) \le f_{s_1}(w(s_1,t_1))+v_{s_1}(r_1) \\ f_{t_1}(w(s_1,t_1))+v_{t_1}(r_3)< & {} f_{t_1}(w(t_1,t_2))+v_{t_1}(r_2) \le f_{t_1}(w(s_1,t_1))+v_{t_1}(r_1) \\ f_{s_2}(w(s_2,t_2))+v_{s_2}(r_4)< & {} f_{s_2}(w(s_1,s_2))+v_{s_2}(r_1)< f_{s_2}(w(s_2,t_2))+v_{s_2}(r_2)\\ f_{t_2}(w(s_2,t_2))+v_{t_2}(r_4)< & {} f_{t_2}(w(t_1,t_2))+v_{t_2}(r_2) < f_{t_2}(w(s_2,t_2))+v_{t_2}(r_2) \end{aligned}$$

with \(v_i(r_1)>v_i(r_2)>v_i(r_3)>v_i(r_4)\) for all \(i\in \{s_1,s_2,t_1,t_2\}\). For example, these inequalities are satisfied if for \(i\in \{s_1,s_2,t_1,t_2\}\) and \(j\in \{1,2,3,4\}\) \(v_i(r_j)=4-j\), \(f_i(x)=x\), \(w(s_1,s_2)=w(s_1,t_1)=1\), \(w(t_1,t_2)=2\), and \(w(s_2,t_2)=3\). They also provide a case where the match is stable with respect to \(\phi \) yet not ex post stable. To complete the example, for all \(s,s'\in S{\setminus }\{s_1,s_2,t_1,t_2\}\) and \(k\notin \{1,2,3,4\}\) set \(v_s(r_k)=v_{s'}(r_k)=0\), \(f_s(x)=f_{s'}(x)=x\), \(w(s,s')=10\) if \((s,s')\in \mu \) and \(w(s,s')=0\) if \((s,s')\notin \mu \). \(\square \)

Proof of Theorem 5

Let the roommate index for the pair \((s_i,s_j)\) be determined by the random variable \(X_{ij}\sim G\). Let x be a realization of roommate indices corresponding to a stable match under \(\phi \) and let y the realized roommate indices of an adjacent match that blocks x under \(\phi _R\). There need not be any match that blocks x under \(\phi _R\), but in this event there is no difference in roommate indices between the mechanisms.

Because they are adjacent, all but two elements each of x and y are identical. Let the four elements that differ with positive probability be \(y_1\), \(y_2\), \(x_1\), and \(x_2\). Denote the value of the rooms received by the pairs corresponding to \(x_1\) and \(x_2\) by \(v_{x_1}\) and \(v_{x_2}\). For \(y_1\) and \(y_2\), we need the values of the rooms received under \(\phi \) if these pairs form, and we denote those by \(v_{y_1}\) and \(v_{y_2}\) respectively. At this point we can only rule out that \(v_{x_1}\) and \(v_{x_2}\) (or \(v_{y_1}\) and \(v_{y_2}\)) refer to the same room.

The inequalities that characterize the event, E, that y blocks x under \(\phi _R\) but x blocks under \(\phi \) are

$$\begin{aligned} x_1\le & {} y_1\le x_1 + (v_{x_1} - v_{y_1})\nonumber \\ x_2\le & {} \min \{y_1,x_1+(v_{x_1}-v_{x_2})\} \end{aligned}$$
$$\begin{aligned} y_2\le & {} \min \{y_1,x_1+(v_{x_1}-v_{y_2})\}. \end{aligned}$$

These follow from observing that it must be that \(\max \{y_1,y_2\}\ge \max \{x_1,x_2\}\) and \(\max \{y_1+v_{y_1},y_2+v_{y_2}\}\le \max \{x_1+v_{x_1},x_2+v_{x_2}\}\), and assuming without loss of generality that \(x_1+v_{x_1}\ge x_2+v_{x_2}\) and \(y_1\ge y_2\). Note that it must be that \(v_{x_1}>v_{y_1}\) for this event to have non-zero probability.

We first sign the difference in the sum of the payoffs to all of the students when the match is y compared to when it is x. Since the students have the same values for each of the rooms and x and y differ for only four pairs, this difference is simply

$$\begin{aligned} \Delta = 2(y_1-x_1)+2(y_2-x_2). \end{aligned}$$

The difference \(y_1-x_1\) is positive in these events but the sign on \(y_2-x_2\) is ambiguous ex post. To sign \(\Delta \), we condition on E and \((x_1,y_1,v_{x_1},v_{y_1})\), taking expectations over all \(x_2\), \(y_2\), \(v_{x_2}\) and \(v_{y_2}\) in E.Footnote 11 Letting \(V{\setminus } \{v_{i}\}\) be the set of room values not including \(v_i\), we find that the value of \(\mathbb {E}[\Delta |E,x_1,y_1,v_{x_1},v_{y_1}]\) is

$$\begin{aligned}&2(y_1 - x_1) + \frac{2}{n-1} \sum _{v_j\in V{\setminus } \{v_{y_1}\}} \mathbb {E} [Y_2|Y_2\le \min \{y_1,x_1+v_{x_1}-v_{j}\}] \\&\qquad - \frac{2}{n-1}\sum _{v_j\in V{\setminus } \{v_{x_1}\}} \mathbb {E} [X_2|X_2\le \min \{y_1,x_1+v_{x_1}-v_{j}\}] \\&\quad = 2(y_1 - x_1) + \frac{2}{n-1}\mathbb {E} [Y_2|Y_2\le \min \{y_1,x_1\}] \\&\qquad - \frac{2}{n-1}\mathbb {E} [X_2|X_2\le \min \{y_1,x_1+v_{x_1}-v_{y_1}\}] \\&\quad = 2(y_1 - x_1) + \frac{2}{n-1}\left( \mathbb {E}[Y_2|Y_2\le x_1] - \mathbb {E} [X_2|X_2\le y_1]\right) > 0. \end{aligned}$$

The first equality follows from canceling terms across the sums, and the inequality follows from the assumption of log-concavity, which guarantees that \(y-\mathbb {E}[Y|Y<y]\) is a non-decreasing function of y.

This shows that when the stable match under \(\phi \) is blocked under \(\phi _R\) by an adjacent match the total payoff improves on average. If there is no such blocking pair, there is no change in the total payoff. Since students are identical ex ante, each student receives the same fraction of the total expected payoff from a mechanism and hence each student’s expected payoff improves with the formation of such a blocking pair.

We next show that forming subsequent blocking pairs under \(\phi _R\) improves the total expected payoff and hence each individual’s expected payoff. From Theorem 1, the stable match under \(\phi _R\) can be constructed by successively finding the pair with the highest roommate value among all remaining feasible pairs. Let the \(\phi _R\)-stable match’s roommate indices be given by z, which may refer to the same match as y. Comparing some z to y, sort the vectors in descending order so that \(z_1\) is the largest element of z. If z and y do not refer to the same match, with positive probability there is some i such that either \(i=1\) and \(z_1 > y_1\), or \(i>1\) and \(\forall j<i, z_j= y_j\) and \(z_i > y_i\). That is, they differ at some point in the process of selecting the best pair from the remaining feasible pairs. The pair corresponding to \(z_i\) blocks y under \(\phi _R\) and hence there is a match \(y'\) incorporating this blocking pair and the pair that results from this block, labeled \(y'_j\). Conditional on such a blocking pair forming, the expected difference in the total payoff between \(y'\) and y is \(z_i - y_i + y'_j - y_j\). Conditioning on \(z_i\) and \(y_i\), this difference is positive, since

$$\begin{aligned} z_i - y_i + \mathbb {E}[Y'_j|Y'_j\le z_i] - \mathbb {E}[Y_j|Y_j\le y_i] \ge 0, \end{aligned}$$

because \(\mathbb {E}[X|X\le \cdot ]\) is an increasing function. If \(y'\ne z\), then one can form a \(y''\) adjacent to \(y'\) in the same manner. This process can be continued until z is reached. At each step, there is a nonnegative change in the expected total payoff, and hence each student’s expected payoff improves ex ante. \(\square \)

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Burkett, J., Flanagan, F.X. & Griffith, A.L. Allocating group housing. Soc Choice Welf 50, 581–596 (2018).

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