@Salil: The argument I was making is that A will choose to shoot at B rather than C on his first shot if both are standing, because doing so maximizes A's chances of survival. Perhaps we are using the notation P(A shoots at C) to mean different things.
Dave On Jan 3, 12:24 am, Salil Joshi <[email protected]> wrote: > @Dave > In point # 1, I had mentioned that P(A survives) + P(B survives) + P(C > survives) = 1 as the dual will be carried out till only 1 man is left. Do > you agree on this?... > > read more » > > By your calculations, P(A survives) = 0.17 * 1 + 0.5 = 0.67 > P(C survives) = 0.4962 > They add to more than 1. Please let me know your views. > > > > On Mon, Jan 3, 2011 at 11:34 AM, Dave <[email protected]> wrote: > > @Salil. The point that is incorrect is: > > 2. Now, by shooting in air C increases his probability by your > > argument, > > which is not good for A. > > Thus, P(A shooting at C) should NOT be 0 if A is intelligent. > > > If A shoots C, then B hits A with 50% probability, but if A shoots B, > > then C hits A with only 33% probability. Thus, A's probability of > > survival is higher if he shoots B rather than C. > > Hence, P(A survives) = .67 * P(A shoots B) + .5 * P(A shoots C). > > But since P(A shoots B) + P(A shoots C) = 1, it follows that > > P(A survives) = 0.67 * P(A shoots B) + 0.5 * [1 - P(A shoots B)] > > = 0.17 * P(A shoots B) + 0.5. > > Therefore, P(A survives) is maximized when P(A shoots B) = 1 and P(A > > shoots C) = 0. > > > Dave > > > On Jan 2, 11:07 pm, Salil Joshi <[email protected]> wrote: > > > @Dave, > > > Can you please point out which of the 3 points mentioned earlier sounds > > > incorrect? Because I think that this problem is much like the Three Body > > > problem, and we can not just maximize the likelihood for C, ignoring A & > > B. > > > They will also try to maximize their survival likelihood.... > > > > On Mon, Jan 3, 2011 at 1:34 AM, Dave <[email protected]> wrote: > > > > @Salil: If C shoots at A instead of into the air, he increases the > > > > odds that he will be shot by B, because if C hits A then B will shoot > > > > at C instead of A. > > > > > On the other hand, if C shoots at B instead of into the air, he > > > > increases the odds that he will be shot by A. > > > > > Thus, shooting at either A or B decreases his odds of survival. > > > > > Dave > > > > > On Jan 2, 12:25 pm, Salil Joshi <[email protected]> wrote: > > > > > @Dave > > > > > 1. In the end only 1 will survive (after max of 2 rounds). > > > > > i.e. P(A survives in end) + P(B survives in end) + P(C survives in > > end) = > > > > 1 > > > > > > 2. Now, by shooting in air C increases his probability by your > > argument, > > > > > which is not good for A. > > > > > Thus, P(A shooting at C) should NOT be 0 if A is intelligent. > > > > > > 3. If it is not 0, P(C's survival) decreases (doesn't remain > > 0.49624). > > > > > > Please let me know if this is clear enough. > > > > > > On Sun, Jan 2, 2011 at 11:15 PM, Dave <[email protected]> > > wrote: > > > > > > @Salil: Just to make sure we are on the same page, A hits with 100% > > > > > > probability, B hits with 50% probability, and C hits with 33% > > > > > > probability. C shoots first, then B, then A. Then the shooting > > > > > > continues among the survivors in that order until only one is > > > > > > standing. > > > > > > > If all three are alive and it is A's turn to shoot, he logically > > will > > > > > > choose to shoot at B rather than C since B has a greater > > probability > > > > > > of hitting him. Thus, your P(A shooting at C) = 0 if B is unhit > > when > > > > > > it is A's turn. > > > > > > > Similarly, if it is B's turn to choose between shooting at A or C, > > he > > > > > > rationally will choose A since A will shoot at and hit B if A gets > > a > > > > > > turn. So your P(B shooting at C) = 0 if A is unhit when it is B's > > > > > > turn. > > > > > > > If you dispute either of the above two paragraphs, please clealy > > state > > > > > > your objection. > > > > > > > Dave > > > > > > > On Jan 1, 11:19 pm, Salil Joshi <[email protected]> wrote: > > > > > > > @Dave, > > > > > > > Yeah, I had read those numbers on internet as this puzzle is well > > > > known. > > > > > > > However I am not convinced with the calculations because of > > following > > > > 2 > > > > > > > points: > > > > > > > > 1) If C shoots in air, the probability of survival is more for > > the > > > > > > > probabilities considered in the calculations with which A & B > > will > > > > shoot > > > > > > at > > > > > > > him. > > > > > > > Now, if A & B are intelligent, they will know that increasing > > > > survival > > > > > > > probability for C is bad for them (you can calculate survival > > > > probability > > > > > > > for A & B in each case), and therefore they will shoot at C with > > > > higher > > > > > > > probability than what they were planning earlier. > > > > > > > > 2) C's survival probability depends on P(A shooting at C) * 1 and > > P(B > > > > > > > shooting at C) * 1/2. > > > > > > > If C shoots at A, P(A shooting at C) is less by 33% and P(B > > shooting > > > > at > > > > > > C) > > > > > > > is more by 33%. So, if P(A shooting at C) dominates by logic in > > 1st > > > > > > point, > > > > > > > C's survival probability will be now more. > > > > > > > > On Sun, Jan 2, 2011 at 7:59 AM, Dave <[email protected]> > > > > wrote: > > > > > > > > @Salil: Working out the probabilities, we find that: > > > > > > > > > 1. If C initially shoots at A, C's probability of survival is ~ > > > > > > > > 0.35867. > > > > > > > > 2. If C initially shoots at B, C's probability of survival is ~ > > > > > > > > 0.27679. > > > > > > > > 3. If C initially shoots in the air, C's probability of > > survival is > > > > ~ > > > > > > > > 0.49624. > > > > > > > > > Dave > > > > > > > > > On Jan 1, 11:30 am, Salil Joshi <[email protected]> > > wrote: > > > > > > > > > @Rahul, > > > > > > > > > As per my understanding, > > > > > > > > > In any round P(C is dead) = P(A is alive * A shoots C * A's > > shot > > > > is > > > > > > > > > accurate) + P(B is alive * B shoots C * B's shot is accurate) > > > > > > > > > this is to be minimized. > > > > > > > > > by not shooting at either A or B in 1st chance, how is this > > > > > > probability > > > > > > > > less > > > > > > > > > for C? > > > > > > > > > > On Sat, Jan 1, 2011 at 10:43 PM, Salil Joshi < > > > > > > [email protected] > > > > > > > > >wrote: > > > > > > > > > > > @Rahul, > > > > > > > > > > What purpose is served by wasting the shot? If C shoots at > > A or > > > > B, > > > > > > at > > > > > > > > least > > > > > > > > > > some probability that C is dead in future will be reduced. > > > > > > > > > > > On Sat, Jan 1, 2011 at 10:14 PM, RAHUL KUJUR < > > > > > > > > [email protected]>wrote: > > > > > > > > > > >> @snehal: > > > > > > > > > >> will the shooting take place in increasing order of > > accuracy > > > > of > > > > > > > > hitting > > > > > > > > > >> the target and is that at a time only one person can take > > a > > > > > > shot??? > > > > > > > > > >> if yes then > > > > > > > > > >> @Salil: > > > > > > > > > >> my answer would be the same as above. what C will do is > > that > > > > it > > > > > > will > > > > > > > > first > > > > > > > > > >> let A and B kill each other first. > > > > > > > > > >> After C wastes his shot it will be B's turn. B can kill C, > > but > > > > in > > > > > > that > > > > > > > > > >> case the turn would go to A and he would surely kill B. If > > B > > > > goes > > > > > > > > after A, > > > > > > > > > >> then B may hit it or miss it(as its probability of hitting > > is > > > > 50%) > > > > > > > > > >> If B misses it > > > > > > > > > >> then > > > > > > > > > >> it depends on A whom to kill. A may kill B or C. A will > > try to > > > > > > kill > > > > > > > > one > > > > > > > > > >> who is better shooter i.e. B as C is less likely to hit A. > > > > > > > > > >> If B hits A then we are done. Round 1 is complete(as > > required > > > > in > > > > > > the > > > > > > > > > >> question) and C survives the first round. > > > > > > > > > >> Look the problem is not that who gets killed at last but > > > > rather > > > > > > what C > > > > > > > > > >> should fire in the first round obviously to survive(as I > > > > > > understood > > > > > > > > the > > > > > > > > > >> problem). It may happen that eventually C gets killed. But > > > > what > > > > > > should > > > > > > > > C > > > > > > > > > >> shoot in first round to survive. > > > > > > > > > > >> -- > > > > > > > > > >> 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]<algogeeks%2bunsubscr...@googlegroups.com> > > <algogeeks%2bunsubscr...@googlegroups.com> > > > > <algogeeks%2bunsubscr...@googlegroups.com> > > > > > > <algogeeks%2bunsubscr...@googlegroups.com> > > > > > > > > <algogeeks%2bunsubscr...@googlegroups.com> > > > > > > > > > >> . > > > > > > > > > >> For more options, visit this group at > > > > > > > > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > > > > > > > > -- > > > > > > > > > > > -------- > > > > > > > > > > Thanks & Regards > > > > > > > > > > Salil Joshi. > > > > > > > > > > CSE MTech II, IITB > > > > > > > > > > A-414, Hostel 12 > > > > > > > > > > +91.9819.442.865 > > > > > > > > > > > This is a confidential E-Mail. If it has reached you by > > mistake > > > > or > > > > > > if > > > > > > > > you > > > > > > > > > > are not the intended receiver, please send it back to me. > > > > > > > > > > -- > > > > > > > > > > -------- > > > > > > > > > Thanks & Regards > > > > > > > > > Salil Joshi. > > > > > > > > > CSE MTech II, IITB > > > > > > > > > A-414, Hostel 12 > > > > > > > > > +91.9819.442.865 > > > > > > > > > > This is a confidential E-Mail. 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