 Solution

For a fixed number choice, increasing one's own stake changes only the amount lost when one loses. It does not change the amount won, which is the opponent's stake, or the zero payoff of a tie. Thus doubling is either weakly dominated by retaining the original stake or payoff-equivalent to it. This establishes that there is no strategic advantage in doubling, but does not imply strict harm in every equilibrium.

For the first player, each of the two equilibrium choices loses with positive probability: choice one loses against the second player's four, and choice four loses against their one. Doubling either therefore gives a strictly smaller expected payoff than $-4/5$. By contrast, against the first player's support $\{1,4\}$, the second player's choices one and four either win or tie. Their own stake is never lost, so doubling it does not change their payoff.

More precisely, retain the first player's original-stake probabilities $(4/5,0,1/5)$. The second player may split their total probability $1/5$ on choice one, and $4/5$ on choice four, arbitrarily between original and double stakes. The first player's unused choice two has payoff at most $(1/5)2-(4/5)2=-6/5<-4/5$, even when the second player doubles their one stake. The doubled first-player choices are also worse. The second player's choice two, with either stake, is worse against the displayed first-player strategy. Hence these splits remain Nash equilibria of the enlarged zero-sum game.

The first player should not double; the second player is indifferent between doubling and retaining the original stakes on their equilibrium choices. This distinction is an example of weak domination does not exclude equilibrium strategies.