🚀 go-pugleaf

In article <BzqJx8.3Fv@unx.sas.com>, sasjec@asimov.unx.sas.com (Jerry M. Cox) writes:
>Odds are that I could stand a review course in probability. But hear me 
>out anyway :-) If the contestant selects one door of three and that was 
>all there was to it then the odds of initially selecting the winning door 
>are one out of three. I have no problem with that, but I don't think 
>this describes the situation. If Monty is *always* going to reveal a 
>goat behind one of the doors, that door is removed from consideration. 
>The final choice then becomes a choice between one of two doors. If 
>Monty *always* removes one of the doors from consideration, then the 
>final choice (as well as the initial choice) is between a door with a 
>car behind it, and a door with the "other" (the one Monty doesn't show) 
>goat behind it. The odds of picking the car are 50/50 whether you switch 
>or not because Monty is *always* going to take away one of the losing 
>choices.

But YOUR choice of a door will, two times out of three, limity Monty's choice
of a door:  If you picked a door with a goat behind it, then Monty CANNOT open
that door.  So the underlying symmetry opon which your argument rests is
broken, two times out of three (i.e., when you pick a door with a goat).  Hence
your analysis doesn't apply.

>If he said "You can have the door you picked or the TWO doors that you 
>didn't pick", and didn't reveal anything about the two doors, then I 
>would agree that the best thing to do would be to switch. But, Monty 
>changes the odds when he shows you one of the goats behind one of the 
>two doors. He removes that door from consideration completely.

Then consider the following hypothetical offer from Monty:
	You can either keep the door that you picked, OR you can pay me a goat
	and have both of the other two doors.
The payoffs for that offer are identical to the payoffs from the original
problem.

>Unless 
>you would like to take the goat Monty just showed you, you are left with 
>a one out of two choice. He equivalently says to you: "You can have the 
>car or the goat that I didn't show you", or "You can have the door you 
>initially chose or the door I didn't show you". That sounds like 50/50 
>odds to me.

>To sum up the initial choice is not a one of three choice between 
>a car or goat #1 or goat #2. The initial (and final) choice is between a 
>car and one of the 2 goats. The other goat will be removed as a choice 
>by Monty. 

That's right.  Which is what causes you to win 2 times out of three by
switching.  If Monty randomly opened a door, it wouldn't alter the odds of your
winning by switching.  You'd still have a 1/3 chance of winning.  But Monty
CAN'T randomly open a door:  He's GOT to pick a door that does not have a goat
and that you did  not pick.  2/3 of the time, he's got only one choice, and
his pick tells you which door has the prize.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.