when playing with water guns ! It is not a reverse oedipal fantasy, it is simply a (stochastic) game. Actually, that game was mentioned here, in the context of a duel with laser guns, and the answer is there. Vincent (alias @Vicnent) presented the problem this way: consider a single combat between two single warriors (a duel) with laser guns. The probability to kill the other one is proportional to time (duel started at time 0). Initially, the have no chance to kill the other one, and after one hour, they are certain to kill the other one. At time , they kill the other one with probability (and miss him with probability ). What is the optimal time to shot ?
An alternative can be the following (yes, I prefer personal experience to science fiction): consider two warriors playing with water gun. We start with empty guns, and assume that both warriors fill their guns at two different tap water. The more water we have in guns, the more likely the other one will be wet. What is my optimal strategy to stop filling the gun, and start shooting at the other player ?
In order to formalize the game, assume that I will win 1 point if my son gets wet (and not me), I’ll get -1 if I am wet, and not him, and 0 if we are both wet (in Vincent’s game, we have the same payoff matrix, but I find odd to say that I have -1 if I die).
Let and denote time where players shoot. Vincent derived the following expected payoff function: for me (I shoot at time ) the payoff is
If we know that the other player shoot us, but missed us, it should be different: if I shot first, but missed my son, then he should wait until the end, and so, the payoff should be (thanks Jérôme)
Actually, we can consider a more general game, where probabilities are not proportional to , but functions for me ( stands for dad), and for the other player (here my son ). The expected payoff becomes,
Thus my best strategy is obtained by solving
If and are assumed to be continuous (we assume that water goes continuously into the tank of the gun), then there is such that
If , then and
and so . So
Similarly, if , then and
and so . So
So finally, the optimal strategy is to shoot at the same time, .
In the game considered initially by Vincent, . Hence, =1/2.
But it is also possible that I did not hear my son coming, shooting at me (and missing me). So in that case, the payoff is different, and closer to what Vincent proposed… but that will be in another post…