@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. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
