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?

You would want to go second:

The first person will have a 3 in 6 chance of ending the game.  But if he succeeds in surviving his turn, based on the bullet placement you now have a 2 in 3 chance of surviving.  Only one of the three empty slots have a bullet after it.

If you survive, he then has a 1 in 2 chance of surviving at which point all the chambers are empty and you are out of luck.

Numerically

3/6 + 3/6(2/3 - 2/3(1/2)) = 4/6 probability to survive

Or to look at it another way, the person who goes first will only survive if he picks an empty chamber next to a full one.  2/6

Now if there are two consecutive bullets: The person who goes first again has only two chambers that will allow him to survive. So by going second the chance is 4 in 6.

Numerically

2/6 + 4/6(3/4 - 3/4(2/3 - 2/3(1/2))) = 4/6 prob to survive

 

 

Edited on September 11, 2004, 1:20 pm

Posted by Dan Blume on 2004-09-11 13:17:53