I've been trying to work the following problem out but I can't quite seem to understand it, or the whole concept of first-place auctions. I don't understand how we get to the equilibrium.

The problem is the following: "Consider a first-price, sealed-bid auction in which a bidder’s valuation can take one of three values: 5, 7, and 10, occurring with probabilities 0.2, 0.5, and 0.3, respectively. There are two bidders, whose valuations are independently drawn by Nature. After each bidder learns her valuation, they simultaneously choose a bid that is required to be a positive integer. A bidder’s payoff is zero if she loses the auction and is her valuation minus her bid if she wins it. Determine whether it is a symmetric Bayes–Nash equilibrium for a bidder to bid 4 when her valuation is 5, 6 when her valuation is 7, and 9 when her valuation is 10."

In this case, we have discrete bids and discrete valuation, as well as different probabilities for each. Could someone help me out understand the problem?

  • $\begingroup$ What happens if both have a valuation of 7 and both bid 6? $\endgroup$
    – Henry
    Commented Nov 11, 2019 at 23:15
  • $\begingroup$ @Henry I guess it is implicitly assumed that the tie is broken randomly then. So both get the good with probability 1/2. $\endgroup$
    – Bayesian
    Commented Nov 12, 2019 at 13:31
  • $\begingroup$ "If you bid an integer less than 4, you lose for sure because the other player bids more regardless of their type." Could you explain why the other player must bid more? $\endgroup$ Commented Feb 14, 2022 at 6:25

1 Answer 1


Suppose you are one of the bidders and fix the strategy of the other one to be the one supposed. Your payoff will be $\mbox{Pr(winning the auction, i.e., b>b')}(v-b)$, with b being your bid and b' being the other's bid and v being your valuation (or type).

Now, imagine you are the type-5 player.

If you bid an integer less than 4, you lose for sure because the other player bids more regardless of their type. Hence, your expected payoff is 0.

If you bid 4, (we assumed that) you win with probabilty 0.5 if the other one is of type 5 (and hence bids 4) and lose in every other case. Hence, your expected payoff is $0.2*0.5*(5-4) = 0.1$.

If you bid 5, your payoff from winning is zero such that your expected payoff will be zero in any case.

If you bid more than 5, you make a loss when winning, making the expected payoff negative.

Hence, bidding 4 when being type 5 is a best response (with this restricted bid space!).

Similarly, imagine being a type-7 bidder. Bidding less than 3 makes you lose the auction, and bidding 7 or more does not seem to be a good idea (see above). Now calculate the winning probabilites for bidding 4 (same as above, but higher winning payoff), bidding 5 (winning against type 5 for sure, but losing against other types), bidding 6 (winning against type 5 for sure, but against type 7 only with probability 0.5). Next, you repeat the exercise for type 10. If you don't find a profitable deviation, you have got an equilibrium.


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