## Abstract

We characterize security strategies and payoffs for three mechanisms for dissolving partnerships: the Texas Shoot-Out, the \(K+1\) auction, and the compensation auction. A security strategy maximizes a participant’s minimum payoff, and represents a natural starting point for analysis when a participant is either uncertain of the environment or uncertain of whether his rivals will play equilibrium. For the compensation auction, a dynamic dissolution mechanism, we introduce the notion of a perfect security strategy. Such a strategy maximizes a participant’s minimum payoff along every path of play. We show that while a player has many security strategies in the compensation auction, he has a unique perfect security strategy.

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## Notes

- 1.
McAfee (1992) characterizes equilibrium in the Texas Shoot-Out when partners have independent private values.

- 2.
- 3.
- 4.
These mechanisms also efficiently dissolve partnerships when all the players follow security strategy.

- 5.
See, for instance, Raiffa (1982) p. 297 which develops a numerical example.

- 6.
In an independent private values setting, the equilibrium in strictly increasing strategies of the \(K+1\) auction where \(K=0\) is also an equilibrium of the Tie-Favoring Auction.

- 7.
In the event that \(b_{\mathrm{f}}=b_{\mathrm{s}}\), the winner is selected randomly from among the high bids.

- 8.
This is the difference between bidder

*i*’s demand and the compensation \( p_{1}\) given to the rival who dropped. - 9.
Player 2’s equilibrium strategy coincides with his security strategy.

- 10.
See Cramton et al. (1987).

- 11.
The auction is not ex-post efficient since \(\beta (x)\ne \bar{\beta }(x)\). The probability that the allocation is inefficient is 1 / 6.

- 12.
See Van Essen and Wooders (2016).

- 13.
If at round \(k^{\prime }\) bidder

*i*bids \(p_{k^{\prime }}\), then he ties with another bid. We break the tie in the other bidder’s favor so that*i*does not drop out. This is justified since we consider*i*’s worst-case payoff.

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John Wooders is grateful for financial support from the Australian Research Council’s Discovery Projects funding scheme (Project Number DP140103566).

## 5 Appendix

### 5 Appendix

### Proof of Proposition 0

The proof is well known and is only included for completeness. It is easy to verify that for the given strategies the players each obtain a payoff of at least \(x_{1}/2\) and \( x_{2}/2\), respectively.

We show there is no strategy which guarantees player 1 more than \(x_{1}/2\). Consider a strategy \(\beta ^{1}\) such that \(\beta ^{1}(x_{1})>x_{1}/2\) for some \(x_{1}\). If player 1 with value \(x_{1}\) sets a price \(\beta ^{1}(x_{1})=p\) and player 2 chooses Sell, then player 1’s payoff is \( x_{1}-p<x_{1}/2\). Likewise, if \(\beta ^{1}(x_{1})=p<x_{1}/2\) for some \(x_{1}\), then player 1 obtains a payoff less than \(x_{1}/2\) if \(\beta ^{2}(x_{2};p)=\) “Buy”.

Likewise, if \(\beta ^{2}(x_{2},p)=\)“Buy” for some \(p>x_{2}/2\), then player 2 obtains a payoff \(x_{2}-p<x_{2}/2\) if player 1 offers price *p*. \(\square \)

### Proof of Proposition 1

For symmetric mechanisms, we can write \(v_{i}(\hat{x}_{i},x_{-i},\hat{\beta }^{i},\beta ^{-i})\) as \(v(\hat{x} _{i},x_{-i},\hat{\beta }^{i},\beta ^{-i})\).

Suppose to the contrary that bidder *i* with value \(\hat{x}_{i}\) can guarantee more than \(\hat{x}_{i}/N\), i.e., he has a (possibly mixed) strategy \(\hat{\beta }^{i}\) such that

Since the inequality holds for all \(x_{-i}\) and \(\beta ^{-i}\), then it holds in particular for \(\hat{x}_{-i}=(\hat{x}_{i},\ldots ,\hat{x}_{i})\) and \(\hat{ \beta }^{-i}=(\hat{\beta }^{i},\ldots ,\hat{\beta }^{i})\), i.e.,

Summing across the *N* bidders, the total payoff when all bidders have a value of \(\hat{x}_{i}\) and follow the same strategy \(\hat{\beta }^{i}\) is \(Nv( \hat{x}_{i},\hat{x}_{-i},\hat{\beta }^{i},\hat{\beta }^{-i})\), which is greater than \(\hat{x}_{i}\). This contradicts that the mechanism is payoff feasible. \(\square \)

### Proof of Proposition 2

We need to establish two facts: (i) \(\bar{\beta }^{i}(x_{i})=x_{i}\) guarantees bidder *i* a payoff of at least \(x_{i}/N\), and (ii) \(\bar{\beta }^{i}(x_{i})=x_{i}\) is the unique security strategy.

Part (i): Suppose that \(\bar{\beta }^{i}(x_{i})=x_{i}\). If bidder *i* wins, then he obtains a payoff of

where the inequality holds since \(x_{i}\ge b_{\mathrm{s}}\) as \(x_{i}\) is the winning bid. If bidder *i* loses, then he obtains

where the inequality holds since \(b_{\mathrm{f}}\ge b_{\mathrm{s}}\ge x_{i}\). His payoff, therefore, is at least \(x_{i}/N\). By Proposition 00, \(\bar{\beta } ^{i}(x_{i})=x_{i}\) is a security strategy.

Part (ii). Suppose \(\bar{\beta }^{i}(x_{i})=x_{i}+\varDelta \) for \(\varDelta >0\). Suppose all the other bidders bid \(x_{i}+\varDelta /2\). Bidder *i* wins and obtains a payoff of

since \(K(x_{i}+\varDelta /2)+(1-K)(x_{i}+\varDelta )>x_{i}\) for \(K\in [0,1]\) . Suppose \(\bar{\beta }^{i}(x_{i})=x_{i}-\varDelta \) for \(\varDelta >0\). Suppose all the other bidders bid \(x_{i}-\varDelta /2\). Bidder *i* loses and obtains a payoff of

\(\square \)

### Proof of Proposition 4

S \(\bar{\beta }^{i}\) that satisfies the conditions of the Proposition. Let \(x_{-i}\) and \(\beta ^{-i}\) be arbitrary.

We first show that the bids of bidder *i* are in the interval \([\frac{x_{i}}{ N}+p_{k-1},\frac{kx_{i}}{N}]\) for rounds *k* where he remains active. In round 1, \(p_{0}=0\) and \(\bar{\beta }_{1}^{i}(x_{i};\mathbf {p}_{0})=x_{i}/N\) is determined. Suppose *i* does not drop at round \(k-1\), i.e., \(p_{k-1}\le \bar{\beta }_{k-1}^{i}(x_{i};\mathbf {p}_{k-2})\). Then, \(\bar{\beta } _{k-1}^{i}(x_{i};\mathbf {p}_{k-2})\le (k-1)x_{i}/N\) implies

and thus the interval \([\frac{x_{i}}{N}+p_{k-1},\frac{kx_{i}}{N}]\) is non-empty and \(\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\in [\frac{ x_{i}}{N}+p_{k-1},\frac{kx_{i}}{N}]\) at round *k*.

Next, we show that bidder *i* obtains a payoff of at least \(x_{i}/N\) with \( \bar{\beta }^{i}\). Suppose bidder *i* drops at some round *k*. Then, \(\bar{ \beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\ge x_{i}/N+p_{k-1}\) implies bidder *i*’s payoff is

If bidder *i* wins the auction, then \(p_{N-1}\le \bar{\beta }_{N-1}^{i}(x_{i}; \mathbf {p}_{N-2})\le \frac{(N-1)x_{i}}{N}\) and hence bidder *i*’s payoff is

In either case, bidder *i*’s payoff is at least \(x_{i}/N\). By Proposition 0, a bidder can guarantee at most \(x_{i}/N\), and thus \(\bar{\beta }^{i}\) is a security strategy.

We now establish that every security strategy satisfies the conditions of the proposition. Suppose to the contrary that \(\bar{\beta }^{i}\) is a security strategy and \(\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})<x_{i}/N+p_{k-1}\) for some *k*, \(x_{i}\), and \(\mathbf {p}_{k-1}\) such that \(p_{m}\le \)\(\bar{\beta }_{m}^{i}(x_{i};\mathbf {p}_{m-1})\) for \( m=1,\ldots ,k-1\). We show that there are bids for the other bidders such that player *i*’s payoff is less than \(x_{i}/N\). Consider bids for the other bidders such that (i)\(\ \mathbf {p}_{k-1}\) is the sequence of dropout prices, and (ii) the bids at round *k* exceed \(\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})\). Then, bidder *i* drops at round *k* and his payoff is \(\bar{\beta } _{k}^{i}(x_{i};\mathbf {p}_{k-1})-p_{k-1}<x_{i}/N\), which contradicts that \( \bar{\beta }^{i}\) is a security strategy.

Suppose that \(\bar{\beta }^{i}\) is a security strategy and \(\bar{\beta } _{k}^{i}(x_{i};\mathbf {p}_{k-1})>kx_{i}/N\) for some *k*, \(x_{i}\), and \( \mathbf {p}_{k-1}\) such that \(p_{m}\le \)\(\bar{\beta }_{m}^{i}(x_{i};\mathbf {p }_{m-1})\) for \(m=1,\ldots ,k-1\). Consider bids of the other bidders such that (i)\(\ \mathbf {p}_{k-1}\) is the sequence of dropout prices, (ii) some bidder \(j\ne i\) drops at round *k* at price \(p_{k}=kx_{i}/N+\varDelta \) for \( \varDelta >0\) and \(p_{k}<\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\), and (iii) for rounds \(k^{\prime }>k\) the remaining bidders bid \(\frac{x_{i}}{N} +p_{k^{\prime }-1}\).

There are two cases. First, if bidder *i* drops at some round \(k^{\prime }>k\) at a price strictly less than \(\frac{x_{i}}{N}+p_{k^{\prime }-1}\), then his payoff is \(\bar{\beta }_{k^{\prime }}^{i}(x_{i};\mathbf {p}_{k^{\prime }-1})-p_{k^{\prime }-1}<x_{i}/N\), which contradicts that \(\bar{\beta }^{i}\) is a security strategy.^{Footnote 13} Second, if bidder *i* wins the auction, then he pays

and hence his payoff is \(x_{i}-p_{N-1}<x_{i}/N\), which contradicts that \( \bar{\beta }^{i}\) is a security strategy. \(\square \)

### Proof of Proposition 5

Using the same argument as in Proposition 1, it is immediate that each bidder *i*’s security payoff is at most \((x_{i}-p_{0})/n\). We show the given strategy achieves \((x_{i}-p_{0})/n\) , and therefore the given strategy is a security strategy in \(\varGamma (n,p_{0})\) and the security payoff is \((x_{i}-p_{0})/n\).

Let \(x_{-i}\) and \(\beta ^{-i}\) be arbitrary, and let \(p_{1},\ldots ,p_{n-1}\) be the sequence of dropout prices that results. Suppose that bidder *i* is not among the first \(\hat{k}-1\) bidders to drop. We show, by induction, that for \(k\in \{1,\ldots ,\hat{k}-1\}\) we have (i) \(p_{k}-p_{0}\le k(x_{i}-p_{0})/n\) and (ii) \(p_{k}-p_{k-1}\le (x_{i}-p_{k-1})/(n-k+1)\). Assume \(x_{i}>p_{0}\). Since bidder *i* is not the first to drop, then

i.e.,

Hence, (i) and (ii) hold for \(k=1\).

Assume that (i) and (ii) hold for some \(k^{\prime }<\hat{k}-1\). We show they hold for \(k^{\prime }+1\). By the induction hypothesis, \(p_{k^{\prime }}-p_{0}\le k^{\prime }(x_{i}-p_{0})/n\) and hence \(k^{\prime }<n\) and \( x_{i}>p_{0}\) imply \(p_{k^{\prime }}-p_{0}\le x_{i}-p_{0}\), i.e., \( p_{k^{\prime }}\le x_{i}\). Since bidder *i* did not drop at \(k^{\prime }+1\le \hat{k}-1\), then

which establishes (ii) for \(k=k^{\prime }+1\). This inequality can be rewritten as

By the induction hypothesis, \(p_{k^{\prime }}\le k^{\prime }(x_{i}-p_{0})/n+p_{0}\) and hence

and (i) holds for \(k=k^{\prime }+1\).

If bidder *i* drops in round \(\hat{k}\), then his payoff is \((x_{i}-p_{\hat{k} -1})/(n-\hat{k}+1)\). Since \(p_{\hat{k}-1}\le (\hat{k} -1)(x_{i}-p_{0})/n+p_{0}\), then

If bidder *i* is not among the first \(n-1\) bidders to drop, then \( p_{n-1}-p_{0}\le (n-1)(x_{i}-p_{0})/n\). He wins the auction and his payoff is

Hence, \(\bar{\beta }^{i}\) guarantee’s bidder *i* a payoff of \((x_{i}-p_{0})/n\) and is therefore a security strategy.

If \(x_{i}<p_{0}\), then bidder *i*’s payoff is negative if he wins the auction. We first show that \(\bar{\beta }^{i}\) guarantees bidder *i* a payoff of a least \((x_{i}-p_{0})/n\). Since \(\bar{\beta }_{1}^{i}\) calls for bidder *i* to drop immediately, his payoff is zero unless he wins the auction. The later occurs only if all \(n-1\) other bidders drop immediately and ties are broken in bidder *i*’s favor. In this case, bidder *i*’s payoff is \( x_{i}-p_{0}\). Since this occurs with at most probability 1 / *n*, his expected payoff is at least \((x_{i}-p_{0})/n\). \(\square \)

### Proof of Proposition 6

Write \(\bar{v} _{N-(k-1),p_{k-1}}(x_{i})\) for the security payoff of a bidder with value \( x_{i}\) in the subauction \(\varGamma (N-(k-1),p_{k-1})\). Suppose that \(\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})<(x_{i}-p_{k-1})/(N-(k-1))+p_{k-1}\) for some *k*, \(x_{i}\) and \(\mathbf {p}_{k-1}\) such that \(p_{0}\le p_{1}\le \cdots \le p_{k-1}\). We show that \(\bar{\beta }^{i}\) is not a perfect security strategy. In particular, we show that \(\bar{\beta }^{i}|_{\mathbf {p} _{k-1}}(x_{i})\) yields a payoff less than \(\bar{v}_{N-(k-1),p_{k-1}}(x_{i})\) for some \(x_{-i}\) and \(\beta ^{-i}\).

From Proposition 5, the security payoff of bidder *i* in \(\varGamma (N-(k-1),p_{k-1})\) is \(\bar{v} _{N-(k-1),p_{k-1}}(x_{i})=(x_{i}-p_{k-1})/(N-(k-1))\). Let \(x_{-i}\) and \( \beta ^{-i}\) be such that the bids of the other \(N-k\) bidders in round 1 of \( \varGamma (N-(k-1),p_{k-1})\) are greater than \(\bar{\beta }_{1}^{i}|_{\mathbf {p} _{k-1}}(x_{i})=\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\). Then, bidder *i* drops in round 1 and his payoff is

Hence, \(\bar{\beta }_{i}\) is not a perfect security strategy.

Suppose that \(\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})>(x_{i}-p_{k-1})/(N-k+1)+p_{k-1}\) for some *k*, \(x_{i}\) and \(\mathbf {p }_{k-1}\) such that \(p_{0}\le p_{1}\le \cdots \le p_{k-1}\). Let \(x_{-i}\) and \(\beta ^{-i}\) be such that (i) one of the other \(N-k\) bidders in \(\varGamma (N-(k-1),p_{k-1})\) has a dropout price \(\hat{p}_{k}\) satisfying

and (ii) the remaining bidders’ dropout prices are above \(\bar{\beta } _{1}^{i}|_{\mathbf {p}_{k-1}}(x_{i})=\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})\). Then, bidder 1 does not drop out in round 1 of \(\varGamma (N-(k-1),p_{k-1})\), but enters the subauction \(\varGamma (N-k,\hat{p}_{k})\). From Proposition 5, the largest payoff he can guarantee himself in this subauction is \(\bar{v}_{N-k,\hat{p}_{k}}(x_{i})=(x_{i}-\hat{p}_{k})/(N-k)\). We have that

Hence, \(\bar{\beta }_{i}\) is not a perfect security strategy. \(\square \)

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Van Essen, M., Wooders, J. Dissolving a partnership securely.
*Econ Theory* **69, **415–434 (2020). https://doi.org/10.1007/s00199-019-01177-9

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### Keywords

- Partnerships
- Maximin
- Auction

### JEL Classification

- D44
- D47
- D82