Fiddling for the win

Congratulations to Fiddler Nation for making it to the semifinals of the World Cup! All four teams that made it this far are equally matched in that they each possess the same total amount of “energy.” In advance of each semifinal game, teams must independently decide how much of their energy to allocate to the match; all remaining energy goes toward the finals. The team that spends more energy in any given game will win. The semifinals and finals occur so close in time that teams can’t recuperate any of their energy in between.

You’ve heard that the managers for the other three teams are abysmal and have no idea how to allocate their teams’ energy. Each of the other managers will independently pick a random percentage between 0 and 100 and allocate that portion of their team’s energy to the semifinal game; the rest of that team’s energy will go toward the final. Since you’re the cleverest manager of the bunch, you can choose an optimal strategy that will maximize Fiddler Nation’s probability of winning the World Cup. What is this optimal probability?

Let's define some terms, let's let $V$, $W$ and $X \sim U(0,1)$ be the i.i.d. random energy levels selected by the other three coaches to the first match. Without loss of generality, let's assume Fiddler Nation's first draw is against the team that devotes $V$ energy to the first match. Therefore, if we devote some level of energy $E \in [0,1]$ to the first match, then the probability of making the final is $\mathbb{P} \{ V \leq E \} = E.$

On the other hand since the other two coaches devoted $W$ and $X$ energies respectively to the first match, the coach that devoted $\max \{ W, X \}$ will have won the match, and the amount of energy remaining for the final is only $1 - \max \{W, X \}.$ Again, if our strategy was to devote $E$ to the first match that Fiddler Nation's fierce fiddling team would only have $1-E$ energy left for the final, which means that we will win the final only if $1 - E \geq 1 - \max \{W,X\},$ or equivalently, $\max \{W,X \} \geq E.$ So the probability of winning the final if you only have $1-E$ energy remaining is $$\mathbb{P} \{ \max \{W, X \} \geq E \} = 1 - \mathbb{P} \{ \max \{ W, X \} \leq E \} = 1 - E^2.$$

Putting these together we see that the probability of winning the final by devoting energy $E$ to the semifinal is $$f(E) = \mathbb{P} \{ V \leq E \} \mathbb{P} \{ \max \{W, X \} \geq E \} = E (1-E^2) = E - E^3.$$ In this case, we want to maximize this probability, which will occur when $f^\prime(E) = 1-3E^2 = 0,$ or at $E^* = \frac{1}{\sqrt{3}}.$ In this case, we see that the maximum probability of winning the World Cup is $$f(E^*) = \frac{1}{\sqrt{3}} - \left(\frac{1}{\sqrt{3}}\right)^3 = \frac{2\sqrt{3}}{9} \approx 0.38490017946\dots.$$