Following my previous post of yesterday, online here, assume now that I do not know if my son came when I turned my back at time, and missed me… Then the payoff function is the one propose by Vincent, i.e.
In that particular case,
If we draw those functions, on [0,1], the optimal value is solution of
i.e. (we focus only on solutions in [0,1]). Because here the game is symmetric, my son should also shoot at time
Thus, the payoff is then
Since we consider here a zero-sum game, this cannot be a solution of the game. So the game does not have pure strategy solution.
Assume that now I have a mixed strategy, i.e. a distribution of the optimal time to shot. My strategy has distribution , with density (we assume here that the density exists, or we seek only solution that are differentiable). Assume further that there exists >0 such that the support of my optimal shooting time (the time to shoot is now a random variable) is (,1] (or [,1] since we assume that is differentiable). There is a discussion at the end of Vincent’s post where he needs that assumption, at the end. Actually, I think we can make it now, since we can rationally assume that
I will not shot at time 0 (even on a neighborhood of 0 since I have zero chance to hit my son).
The expected payoff function, assuming that my son shoots at time y is
Since the zero-sum game is symmetric, again, the expected payoff should be zero. It comes that necessarily,
if . Hence, if we differentiate (with respect to y), we have
and if we differentiate one more time, it comes
i.e. a general solution should be of the form .
Here, we have the same solution as the one considered in Vincent’s blog. His solution is obtained as follows (with slightly different expressions) conditional to , my expected payoff is
With a simple integration by parts,
where , i.e.
So, if we want to be indifferent to ‘s strategy, , where
Consider solutions , then =1, i.e. either =1 and then is constant, or =-1. This means that is in proportional to .
If we substitute in the equation we had, initially, it comes that
If we consider =a and =1, it comes that =1/3 while =1/4 (but we don’t really care about that normalizing constant).
It means that we should not start shooting before 1/3 of the tank is fulled. Actually, it makes sense, since
if <1/3 (while =0 if >1/3).