hah hah

Dear ALvin,

Once again your solution agrees with mine. I think I will present it as a tree diagram since that will be easiest for readers to follow.

Also, the result for the A-B duel can be gotten recursively, without using geometric series. If x is the probability that A survives, then x = P(A kills B) + P(A misses B)P(B misses A)(x), so: x = 1/3 +(2/3)(1/3)x, and thus x = 3/7.

Best,

MB

-----"Loo Chee Wee" wrote: -----

To:
From: "Loo Chee Wee"
Date: 12/28/2008 11:32PM
Subject: RE: Triple shootout: Charlie goof

Hi sir

yar sure you are free to post up my solution
Here's my try-out for the infinite shootout

everyone's intention is to be last man standing or survive as long as possible
they move in sequence A,B,C
shootings are independent of each other
B and C are still trying to kill each other as top priority

scenario 1: A shoots B
A hits: C then shoots A and C survives
A misses and B shoots C, if B hits then A and B engages in a 1-1 shootout with A making first move, probability of A surviving is 4/21 from a geometric progression...if B misses C, c kills B with his shoot and A must hit his next shoot at C for any chance of survival, probability of surviving is 2/27

summing everything: under this case A has 50/189 chances of surviving

Scenario 2: A shoots C
A hits, B and A then engages in 1-1 combat with B making first move, probability of A surviving is 1/21, from a geometric progression
A misses, B shoots C, if B hits, B and A engages in 1-1 combat with A making first move, probability of A surviving is 4/21 using geometric progression
if B misses, then C kills off B with his shot and A must then hit C to survive, probability of A surviving is 2/27

summing everything: A has a 59/189 chance of surviving

Therefore: A should start with shooting C in the infinite case

I'm not so confident of this solution though but it's certainly more fun
Hope you can advise on this

Regards
ALvin