From: karr@cs.cornell.edu (David Karr)
Subject: Re: Monte Hall II <SPOILER>
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Fri, 18 Dec 1992 16:46:16 GMT

In article <Dec.18.06.50.20.1992.23923@romulus.rutgers.edu> clong@romulus.rutgers.edu (Chris Long) writes:
>
>There are three doors, behind one of which is a good prize and behind
>the other two are booby prizes (each door equally likely to have the
>good prize).  You pick a door, and then of the two remaining doors
>Monte [sic] picks one containing a booby prize and shows it to you, and
>offers you the opportunity to switch.  We all know by now that if Monte [sic]
>choose uniformly at random if there is more than one door with a booby
>prize to choose from you should always switch and your probability of
>winning is 2/3.

... And of course we all know that Monty *must* open a door in this
way, according to the rules. ...  :-)

By the way, "monte" is the spelling for the three-card game, "Monty"
is the name of the game show host.  ;-)

<SPOILER> ahead...


>1.  If doors A, B, and C (door A leftmost, door C rightmost) have the
>probabilities a, b, c with a+b+c=1 of having the good prize behind
>them, what is your best strategy?

Suppose you pick door A first.  Then with probability b+c, the prize
is actually behind door B or C.  If so, Monty's move tells you which
one, so if you switch, you win with probability b+c.  So you should
pick the door with *least* probability and always switch.  This will
make your odds of winning equal the sum of the two largest
probabilities, which beats choosing a high-probability door and
sticking with it.

>2.  If when Monte [sic] is faced with two empty doors instead of choosing
>uniformly he chooses the leftmost door with probability p, the
>rightmost with probability q and p+q=1, what is your best strategy
>now?

The obvious intention here is that Monty's choice now gives you
additional probabilistic information.  I assume for this question, a =
b = c = 1/3.  For sake of argument, assume p = 1 (an extreme case).

Suppose you first select door A.  If the prize is behind door A or C,
Monty opens door B.  If the prize is behind door B, Monty opens C.  So
if Monty opens B, it is even money as to whether to stay or switch,
and if he opens C, switching always wins.  So switching is always
optimal (at worst it's a wash).  By symmetry this is true for p = 0 as
well, and by "interpolation" with the standard case p = 1/2, it is
clear that for 0 < p < 1 it is true also (and if you don't believe it
you're welcome to do the math).

So the answer is:  pick any door and switch, same as in the original
puzzle.

>3.  What if both conditions hold?

Consider the example from part 2, i.e. p = 1, but instead of a = c =
1/3 assume that a = 1/2 and c = 1/6.  Then if you pick A and Monty
opens B, it is three times as likely to be by "choice" (i.e. A is
correct) than by "force" (i.e. C is correct).  So you should stay.
I.e. you have gotten additional information due to Monty's choice
of right or left door.

But the above strategy has you winning only 5/6 of all games, which
is the same as the part 1 strategy, which is to choose C initially
and always switch.  In fact I haven't yet seen a way to improve on
the part 1 strategy in any case, but I'm still thinking about it.

-- David Karr (karr@cs.cornell.edu)

Newsgroups: rec.puzzles