Less Gladiators

Time for the solution to the second gladiator problem. As before, I guess the thing to do is start with a simpler case.

Suppose Alice has a single gladiator of strength A, and Bob has two gladiators of strengths B₁ and B₂. Again, Bob needs only decide which of the two gladiators to send first and the whole match is set.

If Bob sends gladiator 1 first, he wins with chance B₁/(A+B₁) and loses with chance A/(A+B₁). If he loses the first game, he then has a B₂/(A+B₂) chance to win the second game.

This gives Bob's chance of winning as:

B₁/(A+B₁)+AB₂/(A+B₁)(A+B₂)
=1-A²/(A+B₁)(A+B₂)

unsurprising, its one minus Alices chance of winning both matches. This is symmetric in B₁ and B₂ again, so it is independent of Bobs strategy. Its true that the full game is also strategy-independent, but it is actually a bit weirder to prove this one. I was not able to come up with a proof myself, this was in the paper where I found this problem.

Let us consider to replace the gladiators with lightbulbs that have a fixed burn out time. The burn out time of one of these idealized lightbulbs is memoryless, and if the lightbulbs expected lifetime is x, then the probability that it is still burning after time t is given by e^-t/x.

If we have a lightbulb with lifetime x and a lightbulb with lifetime y both burning at the same time, the chance that y burns out first is given by x/(x+y). So two gladiators fighting can be appropriately replaced by a "lightbulb match" where both Alice and Bob must turn on one of their bulbs and whichever goes out first is defeated. Since the burnout time is memoryless, the chance that a given lightbulb will burn out in the next five minutes is the same as the chance it had to burn out in the past five minutes, so it is enough that the winner simply leave their light on and the loser must turn on a new bulb. In the end whoever runs out of lightbulbs is the loser of the match.

We can see this game could just be replaced by Alice and Bob showing up on separate days, Alice turns on one of her bulbs until it burns out, then she turns on another one and so on until she runs out of bulbs, when she is out of bulbs she records the total lifetime of her team. Bob then comes by and doe the same thing. Whoever has more total lifetime wins the gladiator match. Its pretty clear there is no strategy in this game, but it is the same as this gladiator match game.

This game is a bit less fun than the other one, since human intuition arrives at the "all strategies are the same" result much faster than the other game. I guess thats because in the simpler case its very obvious. If our opponent has only one gladiator, then it does not matter what order we send ours in, our opponent must defeat them all anyway. In spite of the result being more obvious though, the proof is actually trickier to think of, because the setup does not have as much of an obvious "zero expectation value" minigame.