Russian Roulette variant, option to point away Two people are forced to sit at a table with a pile of money, and a six chambered revolver that has one bullet in a randomly spun chamber. The first player is given the option of pointing the gun at his own head or at the ceiling and pulling the trigger. If the first player points the gun at the ceiling and pulls the trigger, then he must point it at his own head and pull the trigger a second time.

After the first player has pulled the trigger with the gun to his own head once, then it becomes the second players turn and he is given the same option as the first player.

The first player to shoot the bullet at the ceiling wins, alternatively if the opposing player shoots the bullet at his own head then the player left alive obviously wins as well. In either case the game ends there and the winner gets the money.

We know that the first time the trigger is pulled there is a 1/6 chance of a bullet coming out. 

So its either:

OPTION A 1/6 chance of shooting the ceiling and winning followed by a 4/5 chance of aiming at his own head and pulling the trigger with no bullet coming out and surviving his turn to fire. 

Or 

OPTION B 1/6 chance of aiming at his own head and losing with the first pull of the trigger or a 5/6 chance of surviving his turn to fire.

Would it be better to try shooting the ceiling first, or aiming at your own head to try to end your turn faster?
Let us work backwards to see if there is a clear advance for either option.  

If there is only one chamber left, the person will shoot at the ceiling and win.  

If there are two chamber left, the person can shoot at his head, and survive with a $\frac{1}{2}$ chance and lose the game or lose and die with $\frac{1}{2}$ chance.
If  he shoots at the ceiling, he can win with $\frac{1}{2}$ chance or lose and die with $\frac{1}{2}$ chance.
So with 2 bullets left, you will shoot at the ceiling, chance of winning $\frac{1}{2}$, chance of losing $\frac{1}{2}$.

With three chamber left, you can shoot yourself in the head, and die with $\frac{1}{3}$ chance, or win with $\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$ chance.
If you shoot at the ceiling, you either win directly with $\frac{1}{3}$ chance, or you die with $\frac{2}{3}\cdot \frac{1}{2}=\frac{1}{3}$ chance, or you lose with $\frac{1}{3}$ as after this there is only one bullet left.
So with three bullets it doesn't matter what you do, unless you want the other person to do, in which case you shoot yourself, chance of winning $\frac{1}{3}$, chance of losing $\frac{2}{3}$.

With four chamber , you shoot yourself with $\frac{1}{4}$ chance of dying or $\frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$ of winning.
Or you shoot at the ceiling, which gives you a $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ chance of dying, and a $\frac{1}{4}$ chance of winning directly, and a $\frac{3}{4}\cdot \frac{2}{3} \cdot \frac{1}{2}=\frac{1}{4}$ chance of winning by surviving the round.
So with four bullets left, you shoot at the ceiling, with $\frac{1}{2}$ chance of winning, $\frac{1}{2}$ chance of losing.

The same kind of argument shows that you want to shoot at the ceiling if there are 5 chamber left.

And the final argument shows that you have a $\frac{1}{2}$ chance of winning by shooting yourself or the ceiling with 6 chamber left.
So it doesn't matter what you do if you only want to win.

Let us see what your chance of survival is.
If there is 1 chamber left, you both will survive.  

If there are 2 chambers left, you will survive with $\frac{1}{2}$ chance, and the other person will live. 

For 3 chambers, you die $\frac{1}{3}$ of the time, and the other person either survives or die $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ of the time.  

With 4 chambers left, shooting yourself results in $\frac{1}{4}$ or $\frac{1}{4} + \frac{3}{4}\cdot \frac{1}{3} = \frac{1}{2}$ chance of dying. The other person will have a $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ chance of dying.
Shooting the ceiling you die with $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$  chance. And the other person has a $\frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$ chance of dying. So you shoot at the ceiling.  

With 5 chambers left, shooting yourself result in a $\frac{1}{5}+\frac{4}{5} \cdot \frac{1}{4} = \frac{2}{5}$ chance of dying and $\frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5}$ chance of dying for the other person.
The ceiling let you die with either $\frac{4}{5} \cdot \frac{1}{4}  = \frac{1}{5}$ or $\frac{4}{5} \cdot \frac{1}{4} +\frac{4}{5} \cdot  \frac{3}{4} \cdot \frac{1}{3} = \frac{2}{5}$ chance of dying. The other person dies with $\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{5}$chance.
So you shoot the ceiling and die $\frac{1}{5}$ or $\frac{2}{5}$ and the other person $\frac{1}{5}$.

For 6 chambers, shooting yourself gives a $\frac{1}{6} + \frac{5}{6} \cdot \frac{1}{5} = \frac{1}{3}$ of dying. The other person has a $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ or $\frac{5}{6} \cdot \frac{2}{5} = \frac{1}{3}$ chance of dying.
Shooting the ceiling gives you a $\frac{5}{6} \cdot \frac{1}{5} + \frac{5}{6}\cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{3}$ chance of dying. The other person has a $\frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{6}$ chance of dying.

So for your own survival, it doesn't matter what you do. For the survival of the other person, you can double his/her chance to survive.
The answer I came up with is that if there are an odd number of chambers left in the gun to fire you want to aim at the ceiling, and if there are an even number of chambers left you want to aim at yourself. What I'm missing is the math to back up my answer, I just think its the right choice.