hah

This is correct -- at least, it agrees with what I calculated for a 1-shot triple shootout.

The case of an "infinite" shootout (participants continue shooting until someone wins) is even more interesting. Why not give it a "shot"?

With your permission, I will post your solution for a the 1-shot case.

Thanks for writing.

M. Bridger

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

To:
From: "Loo Chee Wee"
Date: 12/28/2008 06:13AM
Subject: Triple shootout: Charlie goofs

Hi sir
Here's my try-out for the question you posted
Hope you can advise

P(A misses)=2/3
P(B misses)=1/3
P(C misses)=0

suppose they shoot in the sequence of A,B,C and each only has 1 bullet
Assume that the shootings are independent, that is C's probability of hitting target doesn't change whether A,B hits or not

scenario 1: Suppose A aims B

(1) :P(A misses)=2/3 and A will survive as B will aim for C and vice versa....(A)
(2): P(A hits)=1/3 and A dies and only C remains and he/she never miss
summing: P(A survives)= 2/3 = 6/9

scenario 2: Suppsoe A aims C

(1): P(A misses)=2/3 and A will survive with argument from (A)
(2): P(A hits)=1/3 and B will shoot A with 2/3 chance of killing him/her
summing: P(A surviving)= 2/3 + (1/3)(1/3) = 7/9

therefore A should aim for C

Regards
ALvin

Never knew i could contribute to a American TV show like that haha