Assuming a perfectly fair revolver (unlikely, but let's roll with it) your cumulative odds for shooting yourself with no respin are:
n: [math]p = \frac{n}{6}[/math]
1: [math]p = \frac{1}{6}[/math]
2: [math]p = \frac{1}{3}[/math]
3: [math]p = \frac{1}{2}[/math]
4: [math]p = \frac{2}{3}[/math]
5: [math]p = \frac{5}{6}[/math]
6: [math]p = 1[/math]

And with respin:
n: [math]p = 1-{\left(\frac{n}{6}\right)}^n[/math]
1: [math]p = \frac{1}{6}[/math]
2: [math]p = \frac{11}{36}[/math]
3: [math]p = \frac{91}{216}[/math]
etc.

In terms of discrete pulls. Two is about equally risky with or without spins (basically a 1-in-3). Six with is about as dangerous as four without. Ten with is about as dangerous as five without.

>>17040327
why isn't the probability for respins:

p = 1/(7-n)

or even,

1: p = 1/6
2: p = 1/(6*5)
3: p = 1/(6*5*4)
...and so on.

>>17040365
The “empty chamber” chit is tossed back into the urn upon each spin.

>>17040365
Your probability for not getting shot is (5/6)^n with each reroll.
Thus your probability of getting shot is 1-(5/6)^n.
Your cumulative odds eventually approach 1, but it grows more slowly than without the reroll.