You are in a game of Russian roulette, but this time the gun (a 6 shooter revolver) has three bullets...

probability/roulette Added material

Problem

You are in a game of Russian roulette, but this time the gun (a 6
shooter revolver) has three bullets _in_a_row_ in three of the
chambers.  The barrel is spun only once.  Each player then points the
gun at his (her) head and pulls the trigger.  If he (she) is still
alive, the gun is passed to the other player who then points it at his
(her) own head and pulls the trigger.  The game stops when one player
dies.

Now to the point:  would you rather be first or second to shoot?

Show solution

All you need to consider are the six possible bullet configurations

    B B B E E E         -> player 1 dies
    E B B B E E         -> player 2 dies
    E E B B B E         -> player 1 dies
    E E E B B B         -> player 2 dies
    B E E E B B         -> player 1 dies
    B B E E E B         -> player 1 dies

One therefore has a 2/3 probability of winning (and a 1/3 probability of
dying) by shooting second.  I for one would prefer this option.

Additional material

[ Editor: a conditional-probability variant with the same gun (three
  adjacent loaded chambers, one spin), posted by Mike Naylor in 1996. ]

In one scene, Chains has a six-shooter and three hostages.  They are a
South American drug dealer who offended the Brotherhood, a girl named
Nancy, and Joe Huff.  For some unexplained reason, Chains empties his
revolver and then loads it with three cartridges, all next to each
other, so that there are three loaded chambers in a row and three
empty chambers.  He then gives the cylinder a fast spin and quickly
closes it.

He points the gun at the drug dealer and pulls the trigger.  As it
turns out, the gun fires and the drug dealer dies.  He then points the
gun at Nancy.

Question 1: What is Nancy's chance of getting shot (assuming Chains
plays fair)?

Question 2: What would Nancy's chance of getting shot be if the drug
dealer had not gotten shot?

As it turns out, Nancy gets shot too.  Chains then points his gun at
Joe Huff.

Question 3: What is Joe Huff's chance of getting shot?

Question 4: What would Joe's chance of getting shot be if the drug
dealer got shot but Nancy didn't?

Question 5: What would Joe's chance of getting shot be if the drug
dealer didn't get shot but Nancy did?

And finally question 6: What would Joe's chance of getting shot be if
the drug dealer didn't get shot and Nancy didn't get shot either?

-- Mike Naylor

[ Editor: Orrison's answers follow, numbered by question. ]

1.  2:3

The 3 bullets are loaded consecutively into chambers #1, #2, and #3.
The drug dealer was shot, his chamber was one of these three.  If it
was #1, or #2 then Nancy gets shot also, but if it were #3 she doesn't.

2.  1:3

The 3 empty chambers, also consecutive, are #4, #5, and #6.  The drug
dealer was not shot, his chamber was one of these three.  If it was #4,
or #5 then Nancy does not get shot, but if it were #6 she does.

3.  1:2

This means that the drug dealer, and Nancy were either chambers #1 and
#2, or #2 and #3.  If it was #1 and #2, Joe gets shot, but if it was #2
and #3 then he doesn't.

4.  0

This means that the drug dealer was #3 and Nancy was #4 so Joe must be
#5 which is empty.

5.  1

This means that the drug dealer was #6 and Nancy was #1 so Joe must be
#2 [Editor: was "#1"] which is loaded.

6.  1:2

This means that the drug dealer, and Nancy were either chambers #4 and
#5, or #5 and #6.  If it was #5 and #6, Joe gets shot, but if it was #4
and #5 then he doesn't.

-- Dwayne Orrison <dorrison@voicenet.com>

[ Editor: Orrison's "2:3", "1:3" and "1:2" are probabilities (2/3, 1/3,
  1/2), not odds: in answer 1, Nancy is shot in 2 of the 3 cases. ]