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In the well known classic three way duel puzzle, 3 players Alice, Bob and Carol take turns to shoot each other until only one survives. In his/her turn, a player can either choose to shoot or pass$^{1}$.

In this variant, the players must choose their guns before the game starts. The three guns available have hit probabilities $g_1\lt g_2\lt g_3$. To add more drama, the last person to choose enjoys the privilege to determine the shooting order (i.e. who gets to shoot first, who's second and who's last). Alice is the first to choose her gun, Bob the second, Carol the last.

Can there exist $g_1, g_2, g_3$ such that Alice has the lowest surviving probability among the three, given that everyone act to maximize his/her own surviving probability? If no, is there a not-too-involved proof that explains why not? If yes, what is an example?

$^{1}$Technical note: to rule out the bug equilibrium where everyone just passes, we simply add a powerful enforcer who ensures that as soon as three consecutive passes occur, all players will be killed right on spot. This will justify the best and middle shooters' motive to always shoot instead of passing under any situation.


Update: after trying a few examples, I think a complete list of endgames would be necessary. How Carol determines the optimal shooting order for herself is relatively easy, the only ambiguity being the middle gun. Thus there're eight possibilities as shown below:

enter image description here

where I use w, m and b to denote the worst, the middle and the best guns respectively.

For each possibility, Alice faring the worst means

enter image description here

where m w b(underlined) is a shorthand notation to denote the surviving probability of the best gun b under the shooting order m w b. The colors are there to make verification easy, and are otherwise inessential. (Red for Alice, Blue for Bob and Green for Carol)

But how do we proceed from here? It seems we can add 8 more constraints similar to those (1)~(8) above for Bob and Carol to decide to arrive at each possibilities, and hopefully teasing out some obvious contradictions from these 16 constraints (such as (m w b $\gt$ m w b) && (m w b $\lt$ m w b)). But I'm not sure we can work up from there to Alice. And I'm not sure it's the right way to go.

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  • $\begingroup$ I don't know, but the "powerful enforcer" has to be named Ted. imdb.com/title/tt0064100 $\endgroup$ May 7, 2020 at 23:32
  • $\begingroup$ Do we assume that Carol chooses randomly between orderings that she is indifferent to? For example, if Carol has the b gun, then bmw and bwm are both optimal for her. Depending how Carol breaks ties, this could affect Alice and Bob's strategies. $\endgroup$
    – Tony Huynh
    May 9, 2020 at 8:10
  • $\begingroup$ @TonyHuynh If Carol has the b gun, bwm can be strictly better than bmw for her, because there're đť‘”1,đť‘”2,đť‘”3 for which w gun would optimally choose to shoot for bmw but to pass for bwm. If we adopt the reasonable assumption that for all equally optimal strategies Carol will choose the simplest one (the one requiring least amount of calculation), she would just choose to simply stick to bwm under all circumstances. – $\endgroup$
    – Eric
    May 9, 2020 at 13:29

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