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?

Label the empty slots 1, 2, 3 and the filled ones 4, 5, 6. For the first person to win/survive, he has to choose slot 1 or 3. So his probability is 2/6=1/3.

If it's 2 cartridges only, label the filled ones 5, 6. Then the first person will survive only if he chooses slot 2 or 4. Again his probability is 2/6=1/3.

Basically, the first person gets screwed for both cases. In fact, he's screwed for all cases except when there is only cartridge. When it's four or five cartridges, his chances of survival are 1/6. It's only fair 1/2 if it's one cartridge.

Posted by np_rt on 2004-09-11 14:26:58