🚀 go-pugleaf

In article <92353.100332SHERE@SLACVM.SLAC.STANFORD.EDU> <SHERE@SLACVM.SLAC.STANFORD.EDU> writes:
>Alright, I've had just about all bickering I can stand on this
>Monty Hall drivel [...]
>
>      Permutation:  A B C D E F G H I J K L M N O P Q R S T U V W X
>
>Prize behind door:  1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3
>Originally choose:  1 1 1 1 2 2 3 3 1 1 2 2 2 2 3 3 1 1 2 2 3 3 3 3
[... plus 4 additional rows distinguishing cases ...]

Lee then goes on to count winning and losing cases from the above
table.  But the table counts 4 different ways for the prize to be
behind door 1 if you choose door 1, but only 2 ways for it to be
behind door 2 if you choose door 1!  In fact those two events should
be equally likely.  Since the probability calculations are all based
on just counting cases, of course they are all biased by this
underrepresentation.

I thought the original Monty Hall puzzle died long ago (barring
residual quibbling over what constitutes correct wording of the
question), but I guess I was wrong--here's someone who agrees Monty
has to open the door but still doesn't get 2/3 chance of winning.

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