well, since it is not mentioned that everyone gets to shoot only once, C can
keep shooting A & B until both are dead... case solved :-))

On Sat, Jan 1, 2011 at 8:29 PM, Dave <dave_and_da...@juno.com> wrote:

> @Rahul: If shooter C shoots into the air, doesn't he still have only a
> 33% chance of hitting it? If so, then he would have a 67% chance of
> hitting B or C, or, I suppose, himself. :-)
>
> This brings up another alternative. In round 1 he shoots himself in
> the foot. If he is successful (33% of the time), he is certain to
> survive the duel because he is out as soon as he is hit. If he is
> unsuccessful (67% of the time), the result is the same as if he had
> shot into the air, i.e., he survives with probability 66/133 ~=
> 0.49624. Putting this all together, by shooting himself in the foot,
> his probability of survival increases to about 0.66248. :-)
>
> Dave
>
> On Jan 1, 6:58 am, RAHUL KUJUR <kujurismonu2...@gmail.com> wrote:
> > Suppose three gunmen are A, B, and C who have a probability of 100%, 50%
> and
> > 33% respectively. The shooting will start from C, then B and at last A.
> > Now there are several possibilities for C. If C shoots B, then A would
> shoot
> > C with an accuracy of 100% or in other case if C shoots A, then B would
> > shoot him with an accuracy of 50%. So he has a probability of getting
> > killed. We can see in either of the cases C will die.
> > So what C will do in first round is that it will fire the shot in air.
> Now
> > the scenario gets interesting. By doing this C has turned the battle
> among
> > three people into two people A and B. This will increase the chances of
> > survival of C. So now its B's turn of firing. So he can fire at either A
> or
> > C. If B fires at C, then A will shoot B with an accuracy of 100% and B
> knows
> > that he will surely die so B won't do that. If B shoots A, then C will
> shoot
> > B.
> > I think this is the solution. Please point out if there are any
> loopholes.
>
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A-414, Hostel 12
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