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A game strategy.



The puzzle below was described on page B9 of The New
York Times on January 19, 2002. Like many others I
originally thought that switching the doors should have
no effect on the chance of winning the car. But I was
wrong and I now understand why.

************************************
THE ARTICLE:
Suppose you were a contestant on "Let's Make a Deal," the
old television show. The host, Monty Hall, presents three
doors. One hides a luxury car; the others conceal scrawny
goats. You must choose.

The odds of winning the car are, of course, one out of three,
so without much thought, you pick any door. But then Monty
mischievously opens one of the other doors, revealing a goat.
Are the odds; better if you switch your choice to the other
closed door or if you stay with your original selection? It
seems that it doesn't matter. After all, there are two doors
left, so the chance of your being right is now one out of two.

But actually, if you switch doors, you are twice as likely to
win. This bizarre result led to such controversy—even a mist
aken objection by the great mathematician Paul Erdas -- that
it became a frontpage article by John Tierney in The New
York Times in 1991.

Here's the explanation. Of course a goat was always behind
one of the two unchosen doors; Monty's act doesn't reveal any
new information. So the initial odds of one out of three remain
unchanged. But now for the weird twist. There must be a car
behind one of the two closed doors; if the car isn't behind one
door, it's behind the other. So if one out of every three times
the car is behind the door you've chosen, the other two times
it will be behindthe other door. This means that the odds of
finding the car behind that other door are two out of three.
As the audience might have correctly screamed: "Switch! "

If this counterintuitive analysis holds no interest for you,
stay away from this book. "The Colossal Book of Mathematics:
Classic Puzzles, Paradoxes and Problems" is a grand compliation
of the best "Mathematical Games" columns by Martin Gardner
for Scientific American from 1956 to his retirement in 1991. ...

************************************************
MY COMMENTS:
Why is this problem counterintuitive? Because it is a puzzle;
the designer wanted it to be intriguing. One rule of the game
(see below) was implied rather than explicitly stated. The
author of the puzzle wanted us to assume that the host
chooses among the remaining two doors randomly. But
this is not always the case. To discover the "unstated rule"
play this game with a friend (for example, using paper caps as
doors and scraps of paper representing the goats and the car).

Acting as the host you will quickly discover that the
host is not allowed to open a door hiding the car before
the second selection is made by the guest. That rule is
essential. Suppose the first selection was a goat. Then
the host must open the door hiding another goat. Thus
the guest who switches wins while the guest who
does not switch looses. By the way, the applet for
playing the game on your screen is available at:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

This URL was discoverd by friend, Jeff Giacobbe.
As a pedagogical exercise let me show how I would
explain the correct strategy. Instead of the fourth
paragrap (quoted above) I would say:

Here's the explanation. Of course a goat was always
behind one of the two unchosen doors. The other unchosen
door was hiding another goat (when the car was chosen
in the first selection) or the car (when a goat was chosen
in the first selection). Monty knows the outcome of the
first selection. Can he choose among the remaining doors
randomly? Not always. He can do this only when there is
a gaot behing each door. But he is not allowed to do this
when one of the unchsen doors hides the car. In that case
his choice is predetermined; he must open the door hiding
a goat. Consequently, a player who switches always wins
while a player who does not switch always looses.
Ludwik Kowalski