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In an adventure to a remote island, you get caught by the xenophobic locals and are about to be executed. But the tribe chieftain, known for his sadistic taste, suddenly changes his mind.

"I'll arrange you three into a truel (three way duel)," he says, pointing at you and two other prisoners, "the survivor of which will be pardoned and released."

"In front of you are three guns," he continued,"a malfunctioning gun that hits target 20% of the time, a lame gun that hits 40% of the time, and a golden gun that hits 90% of the time. You must choose your guns and take turns to shoot at each other until only one survives."

"But," he chieftain adds with a wryly smile, breaking the eerie silence, "he who's the last one to choose will enjoy the privilege to determine the shooting order, i.e. who gets to shoot first, who shoots second and last. One more point, whoever chooses the malfunctioning gun can pass his turn without shooting, but the lame and the golden guns must always shoot in their turns."

The chieftain then turns to you and asks "I know you like solving puzzles, so would you like to choose first? Or do you want to wait and be the 2nd/3rd one to choose?"

You all know the chieftain is a man of his word, and everyone wants to maximize their surviving probabilities. Is it better for you to choose first or be the 2nd/3rd one to choose? What's your best strategy?


Hint (updated):

Always keep in mind your two opponents are no fools. This puzzle seems like the same kind as the simple good old truel puzzle at first glance, but that is very beguiling. There're a couple of really hard-to-discern mindset traps for you to fall into and land on a suboptimal strategy. Reason and calculate ultra carefully!

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  • $\begingroup$ Can 2 opponents be killed in 1 shot? $\endgroup$ – Daniil May 10 at 14:43
  • $\begingroup$ @Daniil No. You can only shoot at one. Ditto for your opponents. $\endgroup$ – Eric May 10 at 14:47
  • $\begingroup$ I think you need to add the computer puzzle tag :) $\endgroup$ – Daniil May 10 at 14:49
  • $\begingroup$ @Daniil A calculator will suffice for this one. $\endgroup$ – Eric May 10 at 14:56
  • $\begingroup$ There are 10+ possibilities though... Each one chooses a different gun, whether malfunctioning one will skip or not etc. $\endgroup$ – Daniil May 10 at 14:57
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Suppose that the guns, and order of turns, have already been determined. There are only two plausible strategies for the player with the bad gun: either they always shoot at the player with the good gun, or they pass until someone else dies. (It is clearly wrong to pass when only two players remain, and every time it is your turn with three players left, the situation is exactly the same. Also if you shoot, no matter what gun you have, you should always shoot whichever opponent has the better gun, since it makes no difference if you miss but gives you the better chance if you hit.)

We can work out the probability of each player winning for a particular order of play, with bad gun always passing when three players are alive:

order  bad wins   mid wins   good wins
BMG    0.28855    0.26186    0.44958
MBG    0.28855    0.26186    0.44958
MGB    0.28855    0.26186    0.44958
BGM    0.22450    0.02618    0.74930
GBM    0.22450    0.02618    0.74930
GMB    0.22450    0.02618    0.74930

(note that, since we assume bad does not shoot until someone else dies, and then goes first, this only depends on the order of the other two)

We can also (although the calculations are more complicated) work out the probability of each player winning assuming bad always shoots.

order  bad wins   mid wins   good wins
BMG    0.27641    0.36845    0.35513
MBG    0.28933    0.35552    0.35513
MGB    0.28782    0.26826    0.44391
BGM    0.22582    0.18228    0.59188
GBM    0.22329    0.03684    0.73986
GMB    0.22458    0.03555    0.73986

The player with the bad gun will follow whichever strategy gives the best probability, that is should shoot in every even row and pass in every odd row. The probabilities with the optimal strategy are therefore

order  bad wins   mid wins   good wins
BMG    0.28855    0.26186    0.44958
MBG    0.28933    0.35552    0.35513
MGB    0.28855    0.26186    0.44958
BGM    0.22582    0.18228    0.59188
GBM    0.22450    0.02618    0.74930
GMB    0.22458    0.03555    0.73986

Now how does this information affect the choice of guns? Certainly you should never choose the bad gun if the good gun is still available, since the good gun always gives more than $35\%$ chance of survival and the bad gun always gives less than $30\%$. That means the good gun is not chosen last, so the last player has the bad or middle gun. The last player gets to choose the order, and the best order for either the bad or middle gun is middle, bad, good, so that is what will be chosen.

For this order, the middle gun has a slightly better chance of winning than the good gun. So the best option (marginally, and assuming I haven't made a mistake) is to go first and pick the middle gun. This assumes guaranteed perfect play by the other players.

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  • $\begingroup$ Congratulations! I think you got the correct answer :) $\endgroup$ – Eric May 11 at 13:33
  • $\begingroup$ Do you find it conceivable that for some other hit probabilities of the guns, it'll be best for you to be the second one to choose? $\endgroup$ – Eric May 11 at 14:33
  • $\begingroup$ @Eric Good question. My feeling is that that should be impossible although it seems difficult to prove. $\endgroup$ – Especially Lime May 11 at 15:50
  • $\begingroup$ I did some simulation for one million random triples of guns probabilities. I couldn't find a single case where the second player beats the first. But some came arbitrarily close as this one did. I wonder if there could be some explanation for this coincidence. $\endgroup$ – Eric May 11 at 16:16

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