Is it better to go first in Russian Roulette?

Instead of stopping when the bullet is fired, keep the game going until the whole barrel is used. Each player will have fired three shots out of six; the one having the bullet loses. Clearly the odds of each player surviving is thus $\frac{3}{6} = \frac{1}{2}$ . This can be easily generalized for odd number of chambers, too; the second player gets less chambers, so the chances of surviving is higher.

Let's run a simulation of each of the 6 possible bullet positions and count the outcomes of wins and losses:

Python 2.7:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22

chambers = 6
def roulette_won(bullet_pos):
    first_player = True
    while bullet_pos != 1:
        bullet_pos -= 1
        first_player = not first_player
    return not first_player
wins = 0
losses = 0
# for each possible scenario...
for i in xrange(1, chambers + 1):
    # count wins and losses
    if roulette_won(i):
        wins += 1
    else:
        losses += 1
if wins == losses:
    print "They have equal odds."
elif wins > losses:
    print "The first player has better odds."
elif losses > wins:
    print "The second player has better odds."

This results in 3 wins as well as 3 losses for the first player, which means they have equal odds of survival. When the number of chambers is even, the first player's chance of survival is always $\frac{1}{2}$ .

However, something interesting happens when you have an odd number of chambers: The first player's chance of survival becomes $\frac{\left \lfloor{\frac{c}{2}}\right \rfloor}{c}$ where $c$ is the number of chambers; therefore, if the number of chambers is odd you have a better probability of survival as the 2nd player.