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I'm trying to understand the basics of game theory and the topic of auctions has arisen. I understand the basic concepts of auctions but I'm struggling with second price sealed-bid auctions.

I understand that a weakly dominant strategy in a second price sealed-bid auction is to always bid the amount how much the item is worth to you. However I'm having trouble with some example questions that I've been given.

We have $n$ bidders, $n \ge 2$. There is only one object in the auction. Player $i, i = 1, . . . , n,$ evaluates the object by giving it a valuation $v_i$ , where:

$$v_1 > v_2 > v_3 > . . . > v_n > 0$$

Each player i submits a sealed bid $b_i \ , i = 1, . . . , n$. So, we can describe a bidding profile of all players as $(b_1, b_2, b_3, . . . , b_n)$.

Now what I what I want to understand is are both the listed bidding profiles below nash equilibrium?

A) bidding profile $(v_1, 0, 0, . . . , 0)$

B) bidding profile $(v_2, v_1, 0, . . . , 0)$

Surely both are a nash equilibrium because every bidder will bid there valuation for the item thus no one has any incentive to change there bid? E.G a weakly dominant strategy

I just wanted some clarification, any help on the matter would be greatly appreciated.

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In scenario $A$, only the highest bidder bids their true value and everyone else bids $0$. None of the other players can increase their payoff by changing their strategy alone, as no one values the item above $v_1$. This makes it a weakly dominant strategy.

In scenario $B$, the bidder valuing the item at $v_1$ bids $v_2$ and the bidder valuing the item at $v_2$ bids $v_1$, with $v_1 > v_2$. In this scenario, Player 2 is actually bidding $v_1 - v_2$ above their true value, making it beneficial for this player alone to switch to bidding lower than $v_2$ to obtain $0$ value instead of his current negative value. Thus this is not a weakly dominant strategy.

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