# Weak dominance in second-price auction

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. But I am not sure of this.

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

• $$b_2>b_1$$: Then, $$u_1=0$$ because bidder 1 loses.
• $$b_2: Then, $$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.
• $$b_2: Then, $$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.

Could someone point out the flaw in my logic?

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.

• 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 $b_2=v_1$. Commented Aug 21, 2023 at 2:44
• @PGupta Yes, that’s certainly one way to do it. Commented Aug 21, 2023 at 10:55