🚀 go-pugleaf

don@perseus.spk.wa.us (Donald Byrne) writes:

>So unless I'm not seeing the game properly it seems to me the 
>final results should be 50%.
>     Now explain why all of you say 66% of a chanch if he changes to the 
>other one of the remaining two??
>     I remain confused??

The initial odds that you guessed right are 33%, as your first guess is
random.  That means there's a 67% chance the prize is between one of the
other 2 doors.  Now Monty always reveals a "goat" door, so this gives you
*new* information, the prize is not behind the "goat" door he reveals.  Since
your initial guess is still 33% at being correct (revealing the goat does
NOT tell you anything about your door as Monty can *always* reveal a goat)
this means the remaining door is 67% at being correct.

Or change the rules a little.  After you pick a door, Monty, instead of
revealing a "goat", offers a choice of switching your door for BOTH the other
doors, but you then have to give Monty a "goat".  Do you switch?

This actually works out to be the same game, just described differently.

-Mike

I can't believe we are going through this again, but I suppose it is
somewhat educational.

In <X382VB1w164w@perseus.spk.wa.us> don@perseus.spk.wa.us (Donald Byrne) writes:
>     He was asked to change only after the elimation of one of the other 
>doors. Now we know that there are two non-win doors in the game, and the 
>object of Monty Hall is to elimanite one by the rules. If our player has 
>any smarts he knows that he will either have a winner or a looser, and 
>that Monty Hall will asure him of that. The reason is that of the three 
>doors he will only really deal with two. Monty Hall will elimaniate the 
>other. So unless I'm not seeing the game properly it seems to me the 
>final results should be 50%.
>     Now explain why all of you say 66% of a chanch if he changes to the 
>other one of the remaining two??

Think about it this way:  If the person guessed correctly initially,
then switching would not benefit him.  Since he selected one door from
three, the odds of having a correct initial guess are 1/3.  There should be
no dispute on this point or a review course in probability is necessary.

Now, this isn't quite as easy to see, but if the person guessed
INCORRECTLY initially, then he automatically wins by switching.  Why?
Because Monty always reveals a non-winning door.  He equivalently says to
you: you can have the door that you picked, or the TWO doors that you
didn't pick.  Since He always reveals a goat, if the initial guess was
incorrect, then the remaining door contains the prize.  The odds of the
initial guess being incorrect are 2/3 by the same reasoning as above.

If you want an empirical demonstration, we can set up an experiment and I
will win all of your money without cheating.
-- 
    / \      Reverend fnord                 | "King Kong died for your sins!"
   /   \     fnord@panix.com                | 
  / <0> \                                   | "Don't just eat a hamburger, 
 /_______\   Church of Obfuscatology, Inc.  |  eat the HELL out of it!"

In article <X382VB1w164w@perseus.spk.wa.us> don@perseus.spk.wa.us (Donald Byrne) writes:
 [Argues that there are only two possibilities, a goat or a car.]
>So unless I'm not seeing the game properly it seems to me the 
>final results should be 50%.
 
 What are the odds that the sun will rise tomorrow?

 There are only two possibilities. Either the sun will rise tomorrow, or it
 won't. Therefore, the probability that the sun will rise tomorrow must be 50%,
 right?

>     Now explain why all of you say 66% of a chanch if he changes to the 
>other one of the remaining two??

There is a 1/3 chance of choosing the car from the original three doors.
Whenever this happens, you lose by switching.

There is a 2/3 chance of choosing a goat from the original three doors.
Whenever this happens, you win by switching.

--
Dave Seaman
ags@seaman.cc.purdue.edu

In article <1992Dec20.180350.7276@panix.com>, fnord@panix.com (Cliff Heller) writes:
|>
|> I can't believe we are going through this again, but I suppose it is
|> somewhat educational.
|>
|> In <X382VB1w164w@perseus.spk.wa.us> don@perseus.spk.wa.us (Donald Byrne) writes:
|> >     He was asked to change only after the elimation of one of the other
|> >doors. Now we know that there are two non-win doors in the game, and the
|> >object of Monty Hall is to elimanite one by the rules. If our player has
|> >any smarts he knows that he will either have a winner or a looser, and
|> >that Monty Hall will asure him of that. The reason is that of the three
|> >doors he will only really deal with two. Monty Hall will elimaniate the
|> >other. So unless I'm not seeing the game properly it seems to me the
|> >final results should be 50%.
|> >     Now explain why all of you say 66% of a chanch if he changes to the
|> >other one of the remaining two??
|>
|> Think about it this way:  If the person guessed correctly initially,
|> then switching would not benefit him.  Since he selected one door from
|> three, the odds of having a correct initial guess are 1/3.  There should be
|> no dispute on this point or a review course in probability is necessary.

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.

|> Now, this isn't quite as easy to see, but if the person guessed
|> INCORRECTLY initially, then he automatically wins by switching.  Why?
|> Because Monty always reveals a non-winning door.  He equivalently says to
|> you: you can have the door that you picked, or the TWO doors that you
|> didn't pick.  Since He always reveals a goat, if the initial guess was
|> incorrect, then the remaining door contains the prize.  The odds of the
|> initial guess being incorrect are 2/3 by the same reasoning as above.

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. 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. 

In <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. 

It is precisely because Monty ALWAYS reveals a losing door that makes it
significant.  

All you have to see is that if the initial guess was incorrect, the person
will always win when they switch.  Think about that very carefully.  The
initial guess is incorrect - of the two door, one has the prize, and the
other has the goat - Monty will reveal the goat, so the door switched to
will win.  This is only in the case when the initial guess was a goat.


Now, the probability that the initial guess was incorrect is 2/3.  That
cannot change, because a selection was made from 3 doors, only one of which
is correct.  Monty doesn't remove the door from consideration, because it
was there when the selection was made.    

Sure if he took away one of the doors, and scrambled the prizes and then
let you select randomly from the two, it would be 50/50, but the guess was
made from three doors.  Revealing a door does not change the probability
of the original guess, it can't.  It does reveal a lot of information about
the remaining door.

>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.

Just because it is a choice between two doors does not make it 50/50
probability. 

>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. 

Ah, but he does equivalently say this.  Look at the cases very carefully.
Any time the prize is in one of the two doors you did not pick, you win,
and any time the prize is the one that you initially chose, you lose.


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. 

No one, not even Monty can change the probability of something that you
already selected from three choices.  That remains fixed at 1/3.  It cannot
change. 



-- 
    / \      Reverend fnord                 | "King Kong died for your sins!"
   /   \     fnord@panix.com                | 
  / <0> \                                   | "Don't just eat a hamburger, 
 /_______\   Church of Obfuscatology, Inc.  |  eat the HELL out of it!"

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.