Originally posted by talzamir
You're in a duel with n participants. On each round everyone who is still able to fires, in a descending order of probability of a hit, at the best marksman still standing. If more than one person still stands, another round then starts. The probabilities of your opponents are evenly spaced between 100% and 50%:
1 opponent: 3/4 chance of hit
2 opponents gun your opponent down.
How to get the best odds of winning against two or three opponents?
If 100% then you blast 5/6 but 4/6th gets you 4/6ths of the time, so you only have a 2/6ths chance
So, tweak your gun to 4/6ths - a tiny bit, then fire in the air until an opp gets shot, you then get first shot which gives you a better than 4/6ths chance (the other guy might miss, so you get another try). This looks like best.
Tweak your gun you just less than 5/8ths, then fire in the air until only 1 opp is left, its your go so you have a better than 5/8ths chance , beats 29% by miles.
I think tweak to just less than the worst and fire in the air until only on opp remains is the best strategy with any number of opps.