A Truel

Date: 6 Mar 1995 22:17:31 -0500
From: Anonymous
Subject: Truel

  Here's a puzzler for you all:  You and two of your friends 
get into a dispute and decide to solve it with a "truel", a 
three way duel.  Friend #1 is a crack shot, never missing 
his target.  Friend #2 hits his target 2/3 of the time.  You 
hit your target 1/3 of the time.  It is decided that you will 
take the first shot, the 2/3 marksman will take the second  
shot (if still alive) and the 100% marksman will go last.  
This will continue until there is only one left alive.  On
your turn you get to fire one bullet.
 
  You get to go first.  In order to maximize your chances of 
living thru this, where should you take your opening shot?  
And what are your chances of winning the truel if you 
follow this strategy?
 
  Thanks for taking the time with this,
 
  Dave Zechiel

Date: 8 Mar 1995 15:40:20 -0500
From: Dr. Ken
Subject: Re: Truel

Hello there!

I'll give you a couple of hints about how to get started on 
this problem, and then see if you can take it from there.  
Basically, you want to build a tree of all the various 
possibilities.  Let's call you by the variable A, the 2/3 
marksman by the variable B, and the 100% marksman by 
the variable C.

In the first move, you can either aim at B or C, and let's 
denote that by A>B and A>C.  Let's denote a hit by a @ 
(bull's eye) and a miss by a ~ (a tilde, which is used to 
spell the word pinata, which you try to hit when you're 
blindfolded, and so you usually miss).  So the beginning 
of the tree would look like this:

Move 1:                A>B                    A>C
                  @=1/3/ \~=2/3          @=1/3/ \~=2/3
                B dies/   \ B lives    C dies/   \C lives
          C>A, A dies/    /\             B>A/ B>A/\B>C
              C wins  B>C/  \B>A           /\
                                     @=2/3/  \~=1/3
                                   A dies/    \A lives
                                    B wins     \A>B
                                               /\
                                         @=1/3/  \~=2/3
                                       B dies/    \B lives
                                       A wins

You can tell when any branch has terminated, because you 
can follow it up to the top and you'll pass exactly 2 "dies."  
Can you complete the chart now?  One more thing to keep in 
mind: the other two players are trying to win, too.  So for 
instance, in the fork in the first tree, there aren't 50/50 odds 
that B will aim at C.  Rather, B will look at the situation and 
figure out what strategy is best for himself.  So unless it 
coincidentally ends up being even 50/50 odds, that fork will 
be eliminated, and we'll have only B>C or B>A.  For 
instance, if C is shooting and there are still 3 shooters left, C 
will certainly shoot B, because then A will only have a 1/3 
chance of hitting C on the next shot, compared with B's 2/3 
chance.

What will save this puzzle from being hopelessly difficult is 
the fact that whenever C shoots, a player gets eliminated.  
That's a big help, because that means that the only time the 
shooting can conceivably go on forever is when C is the first 
to die and A & B duke it out.

Once you've completed the tree, you can find out the total 
probability each player has of survival by finding all the 
branches that win for each player, and then multiplying 
together all the different probabilities down the path that led 
to that termination.  This will be the total probability of that
branch's outcome.

By the way, you may want to investigate what happens if 
you shoot your first shot in the air, as opposed to aiming at 
one of the other two folks.  Just for fun.

Incidentally, I thought the line "you and two of your 
_friends_" was a funny set-up for this puzzle.

-Ken "Dr." Math