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I'm dealing with a problem where Brown Bears are bidding on a pot of honey. Imagine that we have $n$ Brown Bears $B_1,B_2,...,B_n,$ and brown bear $B_i$ values the honey pot $v_i \in \mathbb{R}^{+}.$ Consider a bidding structure where each brown bear $B_i$ submits a bid $b_i \in \mathbb{R}^{+}\cup\{0\}$, and then the person with the second highest bid wins the auction and pays the fourth highest bid. Anyone who doesn't win does not pay a bid. Imagine that the payoff for bear $B_i$ is $v_i-b_i$ if bear $B_i$ wins and $0$ if bear $B_i$ loses. I want to figure out what is the optimal strategy for each bear. Is it possible that I should be using the revenue equivalence theorem in this scenario? Otherwise, I am not sure what theorems to potentially use here.

Brian
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  • I'm not sure I understand this auction. If there are $5$ bears, and they bid (in order) $2, 3, 5, 7, 11$, then the fourth bear wins the auction (having submitted the second-highest bid $7$) and pays only $3$ (the fourth-highest bid)? That seems strange but it looks like what you're describing. Can you clarify? – Brian Tung Jan 15 '16 at 00:10
  • I don't have time to check this myself, but perhaps you could examine the proof given in this article: https://en.wikipedia.org/wiki/Vickrey_auction and try to adapt it to this situation? – TomGrubb Jan 15 '16 at 00:11
  • Do the bears know what values the other bears place on the honey? – mjqxxxx Jan 15 '16 at 00:54
  • Assuming not (i.e., no collusion), the dominant strategy may well be for each bear to bid his true value, as in the Vickrey auction. I think you should clarify (if it's correct) that the payoff to the winning bear is $v_i$ minus the amount paid, not $v_i$ minus the amount bid. – mjqxxxx Jan 15 '16 at 00:58
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    Your payoff doesn't agree with your description. The payoff for the winning bear will be $v_i - b_k$, where $i$ is the winning bear, and $k$ is the fourth-highest bidding bear. – Paul Sinclair Jan 15 '16 at 03:50

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