The three way duel, reinvented

[lostkiwi]

This is a modification of the original three duel puzzle
http://en.wikipedia.org/wiki/Three_way_duel_(puzzle)

The puzzle is this:
Three men are in a pistol duel. Each man will shoot in turn. The three men are identified as A, B, and C:
A is a poor shot, and only hits 40% of the time
B is a reasonable shot, and hits 60% of the time
C is a good shot, and hits 80% of the time.

Each man knows each others shooting capabilities, and therefore in order to maximise their survival chances, they will always target the best shooter still standing. They will always aim for a legitimate shot, contenstants are unable to forfeit their shots by firing into the ground.

Once someone is hit, they are no longer able to fire any shots. (on account of being dead and all)

Each man shoots in turn, and the shooting order is determined before the fight, and that shooting order cycles until there is only one contestant standing.

Question 1)
Assume that you are one of these contestants, and that you are able to specify the shooting order. Which contestant would you be, and what shooting order would you choose, in order to maximise your chances of survival?

Question 2)
Alternatively, assume you are one of these contestants, however you do not know the shooting order until the duel starts (it is determined randomly from a third party source). Which contestant would you be in order to maximise your chances of survival?

[lostdummy]

I'm not sure about

[lostkiwi]

[ralphmerridew]

Given that lostdummy joined in June 2004 and that lostkiwi joined in January 2005, what is the probability lostdummy's name is mocking lostkiwi's?

[lostkiwi]

[lostdummy]

Well, if I correctly disabled intentional misses from my solution to last "truel" problem, it results in:


P[ 0.40 ; 0.60 ; 0.80 ] = (3,3,2) 37.916% 51.084% 11.001%
P[ 0.40 ; 0.80 ; 0.60 ] = (2,3,2) 35.744% 27.502% 36.754%
P[ 0.60 ; 0.40 ; 0.80 ] = (3,3,1) 43.123% 45.877% 11.001%
P[ 0.60 ; 0.80 ; 0.40 ] = (2,1,2) 32.508% 18.335% 49.158%
P[ 0.80 ; 0.40 ; 0.60 ] = (3,1,1) 45.837% 43.947% 10.217%
P[ 0.80 ; 0.60 ; 0.40 ] = (2,1,1) 45.837% 8.625% 45.539%

where P[order of shooting] = (their target order) their survival chances%

Also, average survival chances for players regardless of order are
A:43.03% B:30.39% C:26.59%


From that, it follows:
Q1) Shooting order A,B,C and best chance for survival for B (51%)
Q2) Best chance for survival regardless of shooting order has A (43%)

[lostdummy]

Also, related to "it is always best to target best shot", it may hold true for 3 players, but it is not true for 4 players or more, even when missing is not allowed.

Consider this same problem with 4 players: A=80%, B=60%, C=40% and D=20%

What would be targeting order (if shooting order is ABCD) that would maximize survival chance? Would it be BAAA (ie, target best shoot)?

[lostkiwi]

[lostkiwi]