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?
Best Answer
Consider the following game of chance:
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:
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.
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.
Notice that in all possible values of $b_2$, the optimal strategy is no worse than the other strategy.