The question was the following:

Consider a setting with three bidders and complete information where bidders’ values are commonly known. There is a sealed-bid auction where the highest bidder wins the object and pays the AVERAGE of the all three bids. In the case of a tie, the lowest-indexed of the high bidders wins the auction and pays the average. Suppose v1 > v2 > v3. In any Nash equilibrium does player 1 win the object?

And the choices were: (a) For some v1, v2, v3 and (b) For all v1, v2, v3.

The use of “any” in this context confused me. I thought it meant that, for a given values v1, v2, v3, if there was any Nash equilibrium where player 1 wins the object, the condition is satisfied for the given values. Apparently any meant all! My professor says it means that given the values, player 1 must win the object in all the Nash equilibria (which is certainly not true).

I am more familiar with (and prefer) determining whether ∀i∃j such that P(i,j) is true or false. It avoids the confusion between any and all (apparently they mean the opposite of what I thought they meant), is more concise, and is language independent.