 # When should I optimally shoot at my son ? (part 2)

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, becomes i.e. 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 i.e. With a simple integration by parts, where , i.e. Thus, So, if we want to be indifferent to ‘s strategy, , where with , 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 i.e. 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).

## One thought on “When should I optimally shoot at my son ? (part 2)”

1. Vicnent says:

C’est un peu fou, quand même, que nous nous cassions autant le c_l pour mettre au point puis résoudre et enfin présenter de tels sympathiques et intéressants problèmes et que in fine, personne ne dépose le moindre commentaire.

Peut être faudrait il relayer sur fr.sci.maths ? je l’avais fait à l’époque, et j’avais eu un bon retour. En tout cas, bien vu.

REPONSE: merci Vincent… personnellement ça ne me gêne pas de ne pas avoir de commentaires… je poste rarement des commentaires chez les autres, je n’ai pas grand chose à ajouter, ou à redire… Et je ne cherche pas spécialement à être relayé… Je suis vexé quand on pompe sans citer, mais sinon mon blog n’est destiné qu’à des geeks qui veulent refaire un peu de maths… Et être relayé par des bloggeurs qui se prennent trop au sérieux (comme il y en a dans la communauté mathématique), ça n’est pas un truc qui me fait tripper ! En tous les cas, ton problème m’a bien amusé !

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