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Re: A game strategy.



John Mallinckrodt wrote (see his full message below):

In general, Monty isn't likely to follow *any* predictable
pattern of behavior, and the odds of winning the car will
probably remain about 1/3 no matter what you do as well.

Play the game 50 times using paper cups as doors, a scrap of
red paper as a car and two scraps of white paper as goats. Being
the host (Monty) do not "follow any predictable patter", except
for one rule. That implied rule is that "the door hiding a car can
not be open before the second selection is made by the guest."
You will discover a significant difference between the outcomes
of two strategies: (a) switching the original selection and (b)
not switching the original selection. The first strategy, for a
large number of games, will yield twice as many winnings
as the second one. Doesn't this experimental fact contradict
the above expectation?

I suspect the game you are referring to (see above) has rules
different from the one described in The New York Times.
You can also play the applet game but that would not be
convincing because we do not know if the underlying code
is a true simulation designed to validate (or invalidate) a
theory or a fake simulation designed to illustrate what is
"expected according to that theory." I did write codes for
true simulations of (a) and (b) strategies; they confirm the
theoretically expected 1/3 and 2/3 limits. It takes a second
to play more than 10,000 games on my six years old
computer. I will be happy to e-mail (or post) these easy
to follow True Basic codes, if somebody is interested.
Ludwik Kowalski

John Mallinckrodt wrote:

Ludwik reprises the classic "Monty Hall" problem. It's
interesting to examine a few unspoken assumptions in "the
solution."

The usual assumption is that Monty *always* shows you a goat
(because Monty *always* knows where there is a goat) and *always*
offers you a chance to switch. In this case, you win 1/3 of the
time if you never switch and 2/3 of the time if you always switch.
The reason is that Monty's actions provide you with *extremely*
valuable information: After all, *every* time you initially pick
a goat (i.e., 2/3 of the time), Monty then tells you precisely
where the car is.

But what if Monty *only* shows you a goat and offers you a chance
to switch when you have chosen the car? Then the best strategy
is *never* to switch and you end up winning 1/3 of the time--i.e.,
*every* time you are given the chance to switch.

And what if Monty *only* shows you a goat and offers you a chance
to switch when you have chosen a goat? Then the best strategy is
*always* to switch when offered the chance and the result is that
you win 100% of the time.

And what if Monty *always* opens a door and offers you the chance
to switch but does so without knowing what is behind the doors?
Then, 1/3 of the time he shows you a car and you win a goat no
matter what you do. The other 2/3 of the time you win a car with
50% probability whether you do or don't switch. In other words,
you win 1/3 of the time. This is because Monty gives you no
useful information.

In general, Monty isn't likely to follow *any* predictable pattern
of behavior, and the odds of winning the car will probably remain
about 1/3 no matter what you do as well.