## September 2011 solution

Virtually all solutions received this month were based on induction. The following solutions are not:

#### Part 1:

Consider the sum of the armament values in one of the teams. This value is a martingale. In particular, its expectation at the end of the competition must be the same as its value at the beginning.

Its value at the beginning is A, where A is the sum of all armament values for the gladiators in team "A". At the end of the competition it can be either zero (if the team loses) or A+B (if the team wins), where B is the sum of all armament values for the gladiators in team "B". For the two values to be equal in expectation, the probability of winning must be A/(A+B), regardless of the trainers' strategies.

#### Part 2:

Let us add an element of time to the competition. We say that a duel between a gladiator of strength a and a gladiator of strength b is an exponentially distributed random variable with expectation ab/(a+b).

Given this new time element, the probability density of the gladiator of strength a to die at any given point in time (given that he did not die earlier in the duel and that the duel is still ongoing) is a constant, 1/a, and is independent of the strength of the gladiator's opponent.

From this, we can calculate the exact distribution of the time it takes for all gladiators of team "A" to die: the probability that the last team member will die at time t is the probability that the sum of NA exponentially distributed random variables with expectations ai is t (where NA is the number of gladiators in the team and ai is the strength of the i'th one. The probability of team "A" winning is the probability that the random variable thus distributed is greater than the correspondingly defined random variable for team "B". It is independent of the trainers' strategies.