So...if the contestant doesn't switch doors after the host reveals Goat A, 
then the host exposes Goat B behind the door the contestant didn't switch 
to, the contestant still has a 1-in-3 chance of there being a car behind the 
remaining door?  Monty's studio audiences must've felt pretty silly with all 
their premature applause (at least two times out of three).

But why argue about it?  Mathematicians can prove that 1=0.  We're techies! 
Simulate, observe, record, _then_ argue about one another's methods. ;)

RR


----- Original Message ----- 
From: "Michael Rigby-Jones" <Michael.Rigby-Jones@bookham.com>
To: "Microcontroller discussion list - Public." <piclist@mit.edu>
Sent: Wednesday, December 05, 2007 5:51 AM
Subject: RE: [OT] The Monty Hall problem


>
>
>> -----Original Message-----
>> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On
> Behalf
>> Of Alan B. Pearce
>> Sent: 05 December 2007 13:23
>> To: Microcontroller discussion list - Public.
>> Subject: Re: [OT] The Monty Hall problem
>>
>> >The host does not act the same. That's where the probability shift
>> occurs.
>> >
>> >The host knows where the car is. So he will never reveal the car.
> That
>> >means that his behavior changes depending on if the car is selected
>> >initially or not.
>>
>> Is that really correct? Surely the host always picks a known non-prize
>> door,
>> and gives the contestant a second choice, whether or not they picked
> the
>> correct one?
>
> That's my understanding as well, the hosts behaviour is always the same.
> After the contestant selects a door, the host will always open one of
> the other doors with a goat behind it.  The behaviour is the same
> whether or not the contestant picked the car on his first choice.
>
> Regards
>
> Mike
>
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