🚀 go-pugleaf

Lets formulate the problem slightly differently.  Your are show two doors, D1
and D2. One of which contains a prize and the other a goat. 50-50 right.
Ah, but now I tell you that it is twice as likely that the prize is behind D1
than it is behind D2.  Do you still insist that because there are two doors it 
still a 50-05 pick?  The extra rig-am-roll that Monty goes through (letting
you pick one of three dorrs, then showing you that one of the other 2 doors
has a goat) gives you exactly the same new information that the extra
statement above gives you (i.e that it is twice as as likely that the 
prize is behind the door you didn't pick).

Another way to look at it is like so: you decide to never switch and always
stay with the door you picked first.  As you have only a 1/3 chance of picking
right door, you can win only 1/3 of the time.  but that means that the
alternitive (switching) must win 2/3 of the time.

--
Mark Biggar
mab@wdl1.wdl.loral.com