Free for All Duel

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: 4/6 and 5/6
3 opponents. 5/8, 6/8 and 7/8

On each round you can either fire in the air or at an opponent of your choice. You can also tweak your firearm so that it has a hitting probability of your choice, but you can't readjust between rounds.

Surviving against one opponent is simple enough. Tweak the gun to hit 100% of the time and to get the right to go first and gun your opponent down.

How to get the best odds of winning against two or three opponents?

I don't understand the "tweak your firearm" statement. Wouldn't you always want your gun to hit 100% of the time?

*edit* Also, have you seen this Thread 132170?

Oh my. That was before my time and I had not seen it. Mine seems a generalized version of that, and the the rotation is better to worse instead of worse to better.

As for 100% rate of hit, note that everyone aims at the most accurate person still standing, so if you have a perfect gun, all the other guns keep aiming at you. If there is only one opponent that does not matter as he won't ever get to fire. But if there are two, you would take out the second best, and the worst shot would take you out in two cases out of three. Who knows, maybe that's the best you can do, but being the only one with a choice of what to do and the privilege of adjusting your firearm might give better chances than 1/n where n is the number of people in the duel.

With two opponents the only way to avoid getting shot at on round 1.. and the only plausible alternative to a 100% gun is to have a gun with an infinitesimally worse chance to hit than 4/6, say, 3999999/6000000. The best marksman fires at the other guy, and whether he hits or not, you won't be shot at. Either the 4/6 shoots at the one with the best gun, or he won't shoot at anyone ever again. So.. is that better than a 100% gun? And how to get a max survival rate.. at the very least it need be above 25%.. in a 4-way duel? 100% would mean you get shot at three times so your survival chance is 6/8 x (5/8)^2 = 150 / 512 = 29%.. above 25%, but is that the best one can do?

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?

2 opps:
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.

3 opps:
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.