@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?

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.
> >
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Thanks & Regards
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A-414, Hostel 12
+91.9819.442.865


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