There was (once again) a nice puzzle inhttp://www.futilitycloset.com/. Bayes and a good friend are playing Russian roulette. The revolver has six chambers. He puts two bullets in two adjacent chambers, spin the cylinder, hold the gun to his friend’s head, and pull the trigger. It clicks. So it is now Bayes’s turn: he can choose either to spin the cylinder again or leave it as it is. Which is better? Hopefully, Bayes knows his theorem: if he does spin it, the probability of getting killed is 2 out of 6 (four empty chambers out of six), but if he does not, since his friend is still alive, then the hammer should be next to one of the four cylinders in red, below
So here, there is 3 chance out of 4 to survive, i.e. the probability of getting killed is 1 out of 4 (while it was 1 out of 3 when spinning). So Bayes should not spin. And as always, it is possible to see it is a more general result: more generally, in a revolver with 
chambers, it there are 
bullets in adjacent chambers, if the first player survives, the probability of getting killed is k over , when spinning, while it would be 1 over 
if we don’t. Not spinning is better if and only if
i.e.
So you’d better not spin, unless there was one bullet in the revolver, i.e. 
… or 
(in that case, it might not be a good idea actually to play the game).
(we focus only on solutions in [0,1]). Because here the game is symmetric, my son should also shoot at time 
, 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 (
. Hence, if we differentiate (with respect to y), we have
.
, my expected payoff is
, i.e.
, where
,
, then 
=1, i.e. either 
=1 and then 
is constant, or 
is in proportional to 
.
=1/4 (but we don’t really care about that normalizing constant).
=0 if 
(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 
). What is the optimal time to shot ?
and 
for me (
stands for dad), and 
for the other player (here my son 
). The expected payoff becomes,
such that
, then 
and
. So
, then 
and
. So
. Hence, 
Ce week end, en sortant de l’exposition sur les mammouths à Rennes avec mon fils, on est passé à la fête de la science, voir si on pouvait faire quelques casses têtes à un stand tenu par des collègues de Beaulieu. Et là, un petit problème de théorie des jeux m’a été soumis, et je ne pouvais pas passer à côté de l’occasion (surtout quand on connaît la morale) d’en parler ici.