A six-shooter is loaded with three cartridges in consecutive chambers, before two persons start to play "Russian Roulette" until one is dead.

If the cylinder isn't spun after each attempt, would you rather be the first shooter, or the second?

What would be your answer if there are only TWO cartridges, of course in consecutive chambers?

Look at the possible combinations of the bullets' locations:

bbbeee (positions 1-6), ebbbee, eebbbe, eeebbb. 

I think I'd rather take the first shot if there were 3 bullets. 3 out of 4 chances that I've got an empty. Chances of being empty on #2 if #1 was empty are 2 out of 3. The thing is your chances of being alive on the third bullet are 1/2 if you shot first, but if you shot second and the first three were empty, you're one dead person on your second round.  I'd go first!

If there are only two consecutive bullets, this is what you'll have:

bbeeee, ebbeee, eebbee, eeebbe, eeeebb. 

Still the chances of getting an empty on the first bullet are 4/5 and the chances of getting an empty on the second bullet if the first was empty is 3/4. On the third shot, if the first two were empty, you have a 2/3 chance of survival and on the fourth shot, if the first three were empty you'd have a 1/2 chance of survival. That means that bullet #5 will kill the shooter, so you'd rather be the even numbered shooter.

Posted by Celie on 2004-09-11 15:11:39