@Jai: C shoots at B and misses. Then B shoots at A and hits. Then C and B have a shootout with C getting the first shot. C kills B with probability 1/3 + 2/3 * 1/2 * 1/3 + (2/3 * 1/2)^2 * 1/3 + ... = (1/3) / (1 - 2/3 * 1/3) = (1/3) / (2/3) = 1/2 because it is an infinite geometric series. So your formula should be (2/3)*( (1/2 * 1/2 + 1/2 * 1/3) ) = 5/18.
Dave On Jan 3, 11:04 am, jai gupta <[email protected]> wrote: > @Dave: Sorry It was a typo but for the probability figures, > When C shoots B then if he is successful then A will shoot C > Hence he must be unsuccessful and then > if B is unsuccessful then A must will Kill B and then C must kill A > if B is successful then C must kill B now > Hence P(when C shoots at B)=(2/3 )*( (1/2)*(1/3) + (1/2)*(1/3) ) > =2/9 > > When C shoots at A i am similarly getting for the case when A is killed as > 1/12 > and when A is safe as 5/18 > Hence total 13/36 > > where am i wrong? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
