You have a 6-chamber revolver with multiple firing pins, containing 2 bullets in adjacent chambers. You put the gun to your head, pull the trigger once and get a click. If you are forced to pull the trigger again, should you re-spin the cylinder, or just fire again?
OK. assume chambers 1 and 2 contain the consec bullets.
Possible distribution of firing pins.
Consecutive pins. (x,x+1)
First pull of trigger - click. (firing pins were NOT 6 and 1, 1 and 2, or 2 and 3)
Firing pins were:
3 and 4
4 and 5
5 and 6.
On next pull,
3 and 4 becomes 4 and 5. (click)
4 and 5 becomes 5 and 6 (click)
5 and 6 becomes 6 and 1 (Bang)
Survival rate = 66.67% (when firing pins consecutive)=
EDIT: given that the opening options were a 50/50 chance of killing yourself, you are better off taking the next pull rather than a re-spin.
Pins with one space between them: ie, (x, x + 2)
Possible options: ( 1 and 2 have bullets)
1 and 3, = bang
2 and 4, = bang
3 and 5, = click
4 and 6, = click
5 and 1, = bang
6 and 2. = bang
Only (2 and 4) and (4 and 6) are viable first pull options. Since you survived, these are only valid options.
(2 and 4), leads to (3 and 5) and click.
(4 and 6), leads to (5 and 1) and bang,
Survival rate = 50%. (when firing pins 1 gap apart)
EDIT: default possibility is 2/6 of surviving with a re-spin, so taking the pull again is the best option.
Final option: with 2 spaces between them (x, x+3).
Firing pins were:
1 and 4 (bang)
2 and 5 (bang)
3 and 6 (click)
4 and 1 (bang)
5 and 2 (bang)
6 and 3. (click)
Since the first pull resulted in a click:
3 and 6 goes to (4 and 1) bang
6 and 3 goes to (1 and 4) bang
Unless I am very much mistaken, you get killed 100% of the time if the firing pins are 2 gaps apart!!
EDIT: the default chances are 2/6 of survival, so you would be better off taking the re-spin.
Unfortunately, I cant find an algebraic format for these, Since the numbers are manageable, its possible to go thru them manually.