You and 3 others are being held captive by the Great Leader and will be forced to engage in a 4-way shootout. The last one standing will be set free and the top priority for all involved is to go free. If everyone stops shooting and there are 2 or more left then all will be executed. One person shoots at a time and has one shot for each round. The shooting order for each round is fixed and guns with fixed accuracies will be provided. Everyone knows that the accuracies for the guns (in order number) are .1 .3 .4 .6 and these are the probabilities that you will kill who you shoot at. You may choose not to shoot and you may choose who to shoot at. There is no communication between the 4 shooters. There is no limit to the number of rounds before 1 person is left. However luckily for you the Great Leader's daughter has taken a shining to you and has convinced Dad to let you pick your order number. So what do you pick? This problem is based upon an experience in a prior life where I learned a lot.
You should choose
to go second.
We start by considering cases where 2 players are left alive. Then the clear strategy is to shoot at the other remaining player. We can straightforwardly calculate the probability of each player winning (which will depend on who gets the first shot).
Then consider what happens when 3 players are left. I calculated the best strategy for each player under all possible combinations of (fixed) strategies for the other players. The only equilibrium occurs when the player with the least accurate gun skips their turn and the other two players shoot at each other. The advantage of skipping their turn is that the player skipping their turn is guaranteed to go first in the ensuing duel (but only because no one is shooting at them). But, this isn't the only possibility with three players. For example, the player with the least accurate gun is still incentivized to shoot if there is a big enough difference in the accuracy of the remaining players, as dueling against a less accurate opponent may still be preferable to going first. So, some calculation must be done to justify the choice of strategy. But, once the best strategy is known, we can easily calculate the chance of each player winning (again depending on who shoots first).
Finally, we can consider 4 players. We need to decide on a strategy, so we can proceed as in the 3 player case and consider the best strategy for each player under each combination of strategies for the other players. It turns out that it is best for players 1 and 3 to shoot at player 4, for player 2 to shoot at player 1, and for player 4 to shoot at player 3. (There is no advantage for player 1 to skip, as if they live, they are guaranteed to make it to the duel shooting first, so they want to eliminate the most accurate player. Player 2 wants to be the least accurate player with three remaining, so they shoot at player 1.)
The probabilities of winning with these strategies are $1045297985/6703211264$, $33497145/86522428$, $3193470/13446053$, and $1132299/5160704$ with the highest chance for player 2 (around 38.7%).
I have checked that these probabilities are close to the win rates I get when simulating with these strategies. This may be a bad answer since I haven't written out all the intermediate calculations, but this would be quite time consuming and, I think, not that interesting.
This is only a partial solution but I believe it shows how the general puzzle should be solved and is hopefully helpful for a complete solution. Namely I will only consider the case of 2 shooters with accuracies p and q, assuming p < q.
First if it is q's turn to shoot first, q always has a bigger chance to survive. So in that case choosing the higher accuracy is the better choice.
If it is p's turn first, things get more interesting. With probability p p's first shot will hit and p will win. With probabity (1-p)q p will miss and q will hit and win. With probability (1-p)(1-q) both will miss once and we are back at the initial situation of the game. That means we only need to compare p and (1-p)q to decide who is favored to win.
If p < (1-p)q then q is favored, if p > (1-p)q then p is favored.
Now plugging in the accuracies from the problem gives us the following. If 0.4 and 0.6 are the last two standing and it is 0.4 turn to shoot, then the 0.4 accuracy is favored to win. Similarly if 0.3 and 0.4 are the last two standing and it is 0.3 turn to shoot, then 0.3 accuracy is favored to win. In all other cases the higher accuracy is favored to win.
Edited Two more observations.
One observation for shooting with more than 2 players standing.
If you do shoot at someone, you should always shoot at the highest accuracy opponent. If you miss it doesn't matter anyway where you aimed but if you hit the remaining order will be the same but if you hit a weaker shooter you will be left with stronger opponents.
Now looking at 2 player constellations:
If one looks at a 3 player constellation with the weakest shooter to shoot first, this player should deliberately shoot in the air. If they shoot and hit they would get to a 2-way standoff with opponent to shoot first. If they deliberately miss, the 2 stronger shooters will try to shoot each other and if one of them hits, the weakest will be in a 2-way shootout but will be first to shoot which is clearly better.
From here one could work out all the different constellations but that would be super long and tedious. From the observation of shootouts with more than 2 player it follows that the weakest shooter will always make it to the final 2 and he will always be the first to shoot. This first made me think that one should pick the weakest gun, but the chance for the weakest gun to win a 2-way shootout is only 15.6%, 21.7% or 27% depending on which opponent they get. So their total chance of winning is somewhere in between these numbers, so it looks like its going to be smaller than 25% so can't be the best choice.
These general observation about the 3-player version maybe helpful to appreciate the complexity of the generalized game, but I think for given accuracies the exotic cases do not matter, and tehtmi already gave a good answer. So, this most likely will not become a full answer.
Chicken-loving suicidal perfect logicians playing a Mexican game.
In a subgame of 'chicken' as meant here, no one giving up before the end give the worst result, and winning is the only beneficial result. Perfect logicians would only start such a game if they 'know' they will win, i.e. if they are willing to go to the bitter end. If several of such suicidal perfect logicians get together, my conclusion is that each of them will start a game of chicken whenever possible, since the others will immediately decline. Therefor suicidal perfect logicians love the game of chicken.
This seems to apply to this question. This question implicitly assumes (perfect) logical decisions, and considering the value placed on freedom the 4 are at least a little suicidal.
A two player Mexican standoff needs little reasoning, shooting the other is obviously the best option.
A three-player game is already much more complicated, shooting the strongest opponent is often the best strategy. However being first in the duel phase is also beneficial, thus passing as number one (who no one cares to shoot) can also be best. So there are 2 reasonable strategies: (which 1,2 and 3 shooting at increasing efficiency)
A: If 3 is much better, 1 and 2 will team up and both shoot at 3.
B: If 2 and 3 are of comparable skill, 1 will pass until someone is dead. This makes sure 1 can fire the first shot in the duel.
2 will of course not be happy if 1 passes and can start a game of chicken: By passing! Assuming 3 misses 2, 1 now has 2 decent options. A+ Teaming up with 2 and start shooting.
C Pass until 3 kills 2 and facing the most skilled opponent
Switching to supporting 2 (A+) is the best of the 2 available options in some cases (see below)
3 will of course not be happy if 1 shoots at him, and this also allows 2 to start a game of chicken: By passing! 3 now has 2 decent options.
A+ Ignore the implicit alliance offer and shoot 2, obviously 2 will then start shooting 3
D: Accept the alliance and shoot 1, 2 will keep passing until someone is killed and get the first shot.
A+ may well be better than D in some cases (however: see below)
Final 3 player remarks
According to my calculation C is valid option if the accuracies are:
1: 1/12 2: 1/11 3: 1/9
strategy B yields 0.4737 / 0.2368 / 0.2895
strategy C yields 0.45 / 0.00 / 0.55
strategy A+ yields 0.4656 / 0.2987 / 0.2357
Since 2 has invalidated possibility B, player 1 will pick the best remaining: A+
Player 1 could have picked strategy A; one could thus reason that player 1 already started a game of chicken, and also being a chicken-loving suicidal perfect logician will continue to pass.
Regarding strategy D: Note that if 1 misses 3; 2 passes, and 3 misses 1, 1 still may have a trick: change strategy: by 'sub optimally' shooting 2 or passing an alliance with 3 is implicitly offered, which 3 is likely to take.
The four player (generalized) game
1 will shoot 4 (will remain weakest shooter until killed, thus wants strongest removed)*
2 may shoot 1 (to get 'weakest shooter' advantage)
2 may shoot 3 (sometimes if 1 will help him against 4 but pass against 3)
2 may shoot 4
3 will shoot 4*
4 will shoot 3
If 2 shoots at 1 or 3, as a chicken-loving suicidal perfect logician the targeted one should consider shooting 2 to 'force' 2 in another strategy.
Pick the forth order. You will get the highest accuracy in every round regardless of how many people have been killed. The plan is to kill the second highest accuracy, which is always the position before you. Since this is all based on luck, there is no guarantee that you will win this; however, it seems like this is the best strategy (has highest chance to win).
or you can just shoot the Great Leader, if that is even possible.
Given that everybody knows the probabilities and everybody keeps their gun for the next rounds, it would be best to be
Everybody(except for the last) would to try to get rid of the one with the highest accuracy in order to survive. So the first 3 would shoot the fourth person and the fourth try to get rid of the third one. With probabilities of hits being equal to .1, .3, .4 and .6, if the first 3 all shoot at number 4, his probability of survival would be: $(1-.1)*(1-.3)*(1-.4)=.378$. If he survives, he will most likely shoot the third one and chance of survival of number 3 is then $1-.6=.4$.
Starting from round 2
You will have a gun with the highest accuracy, so if the first 2 will try to shoot you, you will have $(1-.1)*(1-.3)=.63$ chance of surviving, but when you shoot the second guy, his chance of survival is only $(1-.4)=.6$.
And of course round 3(when there are only 2 people left)
You will have 4 times more chance of surviving, so statistically being the third one will give you the highest chance of survival