3 gunmen with different accuracies

♠ Posted by GeekyFry in , at 5:45 AM

You are in duel with two other gunmen. First guy shoot with 100% accuracy, second person shoot with 50% accuracy and you shoot with 33% accuracy.
Everyone will get a chance to shoot in every round and shooting will start from the guy with worst accuracy. Who will you shoot in first round?

Possible Solution:
Since the shooting will start from worst accuracy so first you will get a chance, then guy with 50% accuracy and then guy with 100% accuracy.
First option is to shoot at 50% guy then death comes with 72% or Shoot at the 100% guy then death comes with 63.64%. Shoot into the air then death comes with 58.33%.
So, you can shoot into the air and hope for the best.
The p in the probabilities below represents your probability of dying at some point if you miss either guy in the first round. It’ll be filled in later.
If you shoot at the 50% guy and get him, you’re guaranteed to get shot the next round.
Probability of dying at some point by aiming at the 50% guy:
(=) (33.3%)(100%)+(66.67%)*(p)
(=) (33.3%)+(66.67%)(p)
If you shoot at the 100% guy and get him, you’re left with this geometric sum of probability of getting shot by the 50% guy at some point in the future. it converges.
(=) [(50%)] +
[(50%)(66.67%)(50%)] +
[(50%)(66.67%)(50%)(66.67%)(50%)] + …
(=) 50% * { 1 + (1/3) + (1/3)^2 + … }
(=) 50% * {3/2)
(=) 75%
So, your probability of getting shot if you shoot at the 100% guy is:
(=) (33.3%)(75%)+(66.67%)(p)
(=) (24.75%) + (66.67%)(p)
Still crummy, but it dominates the alternative, for positive p. What happens if you miss? If you miss, the second guy has a choice to make. Does he
shoot at you, or the other guy? His options:
Shoot you: if he gets you, he’s guaranteed to die on the next shot.
if he misses, he has q chance of dying (which we’ll get later).
(=) (50%)(100%)+(50%)(q)
(=) (50%)+(0.5)(q)
(2) Shoot at the 100% guy
If he gets the 100% guy, there some infinite sum representing his probability of getting
shot by you:
[(33.33%)] +
[(66.67%)(50%)(33.33%)] +
[(66.67%)(50%)(66.67%)(50%)(33.33%)] + …
(=) 33.33% * [ 1 + (1/3) + (1/3)^2 + … ]
(=) 33.33% * [(3/2)]
(=) (50.0%)
So, he chances of getting shot if he shoots at the 100% guy are:
(=) (50%)*(50%) + (50%)(q)
(=) (25%)+(.5)q
No matter what q is, he’d prefer shooting the 100% guy to shooting you.
Now, what happens if he misses (the 100%) guy? ie, what’s q? if he misses, then the 100% guy has to make a decision:
Shoot you:
he’s guaranteed to get you, so his chances of dying are just 50%. (game ends after the next round.the 50% gets only one shot)
prob of dying: 50%
Shoot the 50% guy:
By the same logic, he’d have a 33.33% chance of death (getting shot by you).
So, he prefers to shoot the 50% guy.
So, q is 100% (ie, if the 50% guy misses, the 100% guy shoots him immediately). From that, we know the 50% guy’s optimal move, if everyone’s around on his first shot. He’d prefer shooting at the 100% guy to shooting at you or purposely missing.
Now, we need to get p, which is our probability of dying if we purposely miss either guy
(shoot into the air).
We know the 50% guy shoots at the 100% guy if we miss.
With 50% probability, the 50% guy kills the 100% guy, resulting in a shootout b/w us and the 50% guy.
(=) (50%) * [ (66.67%)*(50%) + (66.67%)*(50%)*(66.67%)(50%) + …
(=) (50%) * [ (1/3) + (1/3)^2 + … ]
(=) (50%) * [0.5]
(=) (25%)
He misses him. Then, we get one shot at the 100% guy (after the 100% guy shoots the
50% guy).
Our chances of death are:
(50%)*(66.67%) = (33.33%)
So, p is (25%)+(33.33%) = 58.3%
So, again, our three choices are
Shoot at 50% guy. Death comes with:
(=) (33.3%)+(66.67%)(p)
(=) 72% likilhood
Shoot at the 100% guy. Death comes with:
(=) (24.75%) + (66.67%)(p)
(=) 63.64% likihood
Shoot into the air (purposely miss).
(=) p
(=) 58.33% likelihood.

0 comments:

Post a Comment