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Consider a sealed-bid second price auction with two bidders. Standard textbooks claim that bidding one's true valuation $v_i$ weakly dominates bidding lower than one's valuation $b_i<v_i$. But I am not sure of this.

If bidder 2 bids $b_2$, and bidder 1 bids $b_1<v_1$, the possible cases are

  • $b_2>b_1$: Then, $u_1=0$ because bidder 1 loses.
  • $b_2<b_1$: Then, $u_1=v_1-b_2$ because bidder 1 wins and pays the lower bid.
  • $b_2=b_1$: Then, $u_1=0.5(v_1-b_2)>0$, assuming that in the case of a tie, a winner is randomly chosen.

Now, if bidder 2 bids $b_2$, and bidder 1 bids $v_1$, the possible cases are

  • $b_2>b_1$: Then, $u_1=0$ because bidder 1 loses.
  • $b_2<b_1$: Then, $u_1=v_1-b_2$ because bidder 1 wins and pays the lower bid.
  • $b_2=b_1$: Then, $u_1=0$, assuming that in the case of a tie, a winner is randomly chosen. Since bidder 1 bids his valuation, even if he wins, he gets a payoff of $0$.

So, bidding one's true valuation gives the same payoff as bidding lower than the valuation in two cases, and $b_1=v_1$ gives a lower payoff in the case of a tie. I don't understand how do the textbooks claim this means $b_1=v_1$ weakly dominates $b_1<v_1$.

Could someone point out the flaw in my logic?

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1 Answer 1

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Consider the following game of chance:

  1. You pick a number $B_1$ in the interval $[0, V]$, where $V = 100$.
  2. A number $B_2$ in the interval $[0, 2V] = [0, 200]$ is randomly generated. The distribution is uniform.
  3. You win one dollar if $B_1 > B_2$. You lose one dollar if $B_1 < B_2$. Nothing happens if $B_1 = B_2$.

It is trivial that one should choose $B_1 = V = 100$ for the largest expected payout (which is zero dollars). Picking a lower number will lower the expected payout.

Now, here is a flawed argument that argues that picking any other number is equally good:

For any value that I pick for $B_1$, the rules state the following:

  1. If $B_2 > B_1$, lose one dollar.

  2. If $B_2 < B_1$, win one dollar.

  3. If $B_2 = B_1$, nothing.

The above statements are true regardless of the value of $B_1$. So, all values of $B_1$ are equally good.

What is the flaw?

Well, let's use a concrete example. Suppose someone uses that flawed argument to claim that picking $B_1 = 77$ is also optimal. Written down, they are claiming that the following two are equally good.

Incorrectly claimed "equally good" strategy:

  • If $B_2 > 77$, lose one dollar.
  • If $B_2 < 77$, win one dollar.
  • If $B_2 = 77$, nothing.

Obviously optimal strategy:

  • If $B_2 > 100$, lose one dollar.
  • If $B_2 < 100$, win one dollar.
  • If $B_2 = 100$, nothing.

Hopefully, it's easier to see the flaw now: The "lose one dollar", "win one dollar", and "nothing" outcomes are present in both, but the If clauses are different! For example, when $B_2 = 80$, the flawed strategy will result in a dollar lost (1st point: $B_2 > 77$) while the optimal strategy will bring a dollar to you (2nd point: $B_2 < 100$).


Could you now see what flaw is present in your second-price auction argument?

Here's the two auction strategies written down just like earlier. I'm reusing the numbers $77$ and $100$ too.

Incorrectly claimed "not weakly dominated" strategy:

  • If $b_2 > 77$, nothing.
  • If $b_2 < 77$, win $100 - 77$.
  • If $b_2 = 77$, win $0.5(100 - 77)$.

Optimal strategy:

  • If $b_2 > 100$, nothing.
  • If $b_2 < 100$, win $100 - 77$.
  • If $b_2 = 100$, nothing.

Notice that in all possible values of $b_2$, the optimal strategy is no worse than the other strategy.

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  • $\begingroup$ Ah ok, so I should fix $b_2>v_1$ (say), then compare $b_1<v_1$ and $b_1=v_1$ for bidder 1. Repeat for $b_2<v_1$ and $b_2=v_1$. $\endgroup$
    – PGupta
    Commented Aug 21, 2023 at 2:44
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    $\begingroup$ @PGupta Yes, that’s certainly one way to do it. $\endgroup$
    – VTand
    Commented Aug 21, 2023 at 10:55

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