Say there are three roommates moving into an apartment with three rooms. Two of the apartment's rooms are identical, but the third one is valued higher by all three parties (say it's bigger and has a private bathroom). To decide which one gets the big room, the roommates decide to hold an auction.

If the apartment's monthly rent is $p$ dollars in total, the idea behind the auction is to find a number $\delta$ such that the winner of the auction pays $\frac{p}{3} + \delta$, while the two losers pay $\frac{p}{3} - \frac{\delta}{2}$. In other words, the winner of the auction (who gets the big room) pays a bigger share of the rent compared to the two losers.

Does this type of auction—where the two losers benefit from the value of $\delta$ being large—differ from traditional auctions, where it doesn't concern the losers what the winner pays? For example, does it still apply that if the price is decided using an English auction, that each bidder $i$ is incentivized to never bid higher than their value $v_i$? Intuitively, you might think that if the bidder $s$ with the second highest value $v_s$ for $\delta$ is reasonably confident that the to-be winner $w$ has a much higher value $v_w$ for $\delta$ they might bid higher than their own value $v_s$, in hopes of raising $\delta$ and lowering the rent that they have to pay.

  • $\begingroup$ Assuming all three reservation prices are drawn independently from a uniform distribution on $[0,1]$, and assuming also that my arithmetic is right, the Nash equilibrium bidding strategy is $\delta=v/2$, contrary to your expectation. $\endgroup$ Jun 11, 2020 at 8:48
  • $\begingroup$ @ÁrniDagur : Seems to me you've answered your own question. In the standard setting, bidding your true value is a weakly dominant strategy in an English auction. If you're going to be outbid, it doesn't help you to bid higher than your true value. But here, it does. So bidding your true value is no longer weakly dominant. $\endgroup$ Jun 11, 2020 at 13:23
  • $\begingroup$ You can also indirectly infer that an English auction does not easily solve this problem because the literature on rent division involves rather complicated protocols and not a simple auction. $\endgroup$ Jun 11, 2020 at 13:33
  • $\begingroup$ The business about the English auction makes no sense. In an English auction, by definition, the losers neither make nor receive any payments. $\endgroup$ Jun 12, 2020 at 2:22

1 Answer 1


In your auction, losing is not as bad as it is in an English auction. Therefore you should expect that in your auction, people bid lower, not higher.

To check this, I'll take the three reservation prices to be drawn independently and uniformly from $[0,1]$. (For other distributions, it will be easy to modify these calculations.)

1) If your reservation price is $x$ and your bid is $b$, your expected return is $$\pi(x,b)=Prob(Win)(x-b)+E(b)$$ where $E(b)$ is your expected gain if you lose the auction. ($E$ can't depend on $x$ because $x$ is not observable to anyone but you.)

2) Let $b=B(x)$ be a symmetric monotonic Nash equilibrium bidding strategy. (Symmetric means that all three players employ the same strategy.) Let $\pi_0(x)=\pi(x,B(x))$.

3) By the chain rule, $$\pi_0'(x)={\partial\pi\over\partial x}+{\partial\pi\over\partial b}B'(x)$$ But when you choose your bid optimally, the second partial is zero, so we have $$\pi_0'(x)={\partial\pi\over\partial x}=Prob(Win)=x^2$$ with the last equality following from the uniformity assumption.

Clearly $\pi_0(0)=1/3$, so we have $$\pi_0(x)={x^3\over 3}+{1\over 3}$$

4) Directly from the definitions, and continuing to assume we're in a symmetric monotonic Nash equilibrium, we have $$\pi_0(x)=\pi(x,B(x))=x^2(x-B(x))+E(B(x))$$ Combining this with (3), we get $${x^3\over 3}+{1\over 3}=x^2(x-B(x))+E(B(x))\hskip{2pc}(1)$$ where $$ 2E(B(x))=\int_0^1\int_0^1 max(B(y),B(z))dydz- \int_0^x \int_0^x max(B(y),B(z)))dz$$

Equation (1) is an integral equation for $B$. Solving it, I get $B(x)=x/2$. In particular, your optimal bid is always less than your reservation price.


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