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Neither player has a weakly dominant strategy among its pure strategies. Against an opponent at site one, the unique best response is site two, yielding $2k-2$. Co-location yields only $k$, and moving further right reduces one's payoff. Against an opponent at site $k$, the unique best response is $k-1$, by reflection. For $k\ge 4$, these two sites differ. No single pure strategy can therefore be a best response to both opposing actions, as a weakly dominant strategy would require.

A weakly dominant strategy must be a best response to either opponent action. For X this requiresThe first is stronger because $t>0$. ThusFor Y the comparisons are $t/3-P\ge 0$ and $(t-P)/2\ge 0$, givingBelow each upper endpoint the corresponding strategy is strictly better against both opponent actions; at the endpoint one comparison ties and the other is strict. Thus the answers also satisfy the convention that weak dominance requires a strict advantage somewhere. Since expected payoffs are linear in the opponent's mixing probability, checking both opponent pure strategies suffices for every opponent mixed strategy.

At joint purchase, each player's gain from continuing to buy rather than deviating to $nb$ is $(t-P)/2$. Both actions are best responses exactly when this is nonnegative. ThereforeAt equality both players are indifferent to their individual unilateral deviations, which is permitted in a Nash equilibrium. The condition is less restrictive than requiring either player's action to be a weakly dominant strategy: only the opponent's actual action needs to be considered.