Monty Hall Strikes Again

Subject: Probability Question from OZ!
Date: Wed, 2 Nov 1994 17:13:05 +1100 (EST)
From: "Sean Pryor"

Hi...

Basically the problem goes like this.  There are three cups, one of which 
is covering a coin.  I know the whereabouts of the coin, but you don't.  
You pick a cup, and I take one of the remaining cups, one which DOESN'T 
contain a coin.  Both you and I know the cup I pick doesn't contain a 
coin.  You then have the option to swap your cup with the third, 
remaining cup, or keep your first choice.  What is the probability of the 
coin being in the cup if you keep your first choice, or if you decide to 
swap them?

To summarise:

Three cups with a coin under one.
You pick one, I pick one that DOESN'T have the coin.
You then either stay with your choice, or swap it with the remaining cup.
What is the probability of getting the coin, either way?

BTW the answer I get is 50% either way - though it has been suggested 
that you have a 2/3 chance if you swap cups....  I disagree with this, 
but I don't think my maths is capable of giving a definitive answer 
(which is why you're reading this!)

(This problem has placed a cool Australian $50 on the line here in a bet - 
so I need an answer proving me right! :-)  My mate would never let me 
forget it if he was right...)

Thanxx

Sean Pryor

Date: Thu, 3 Nov 1994 08:20:34 -0500
From: Phil Spector

Sean-

This question is a famous brain teaser that is usually described in terms
of the game show Let's Make a Deal.  Indeed, you do have a 2/3 chance 
of winning if you swap cups.  The answer is anti-intuitive, and my efforts 
at explaining it usually fall a little flatter than others' explanations, so
I'm going to let other Dr. Math's try to provide an answer, and I'm going
to work on a productive explanation.

Basically, the solution is based as follows:
Let's say it was under cup 1.
Option 1: You originally choose cup 1.  Then, you get shown one of the
empty cups (let's say cup 2 or 3) at which point if you CHANGE (to the
remaining empty cup, either cup 2 or 3), you LOSE, but if you REMAIN 
with cup one, you WIN.

Option 2: You originally choose cup 2.  Then, you get shown the remaining
empty cup (cup 3), at which point if you CHANGE (to cup 1), you WIN, 
but if you REMAIN with cup 2, you LOSE

Option 3: You originally choose cup 3.  Then, you get shown the remaining
empty cup (cup 2), at which point if you CHANGE (to cup 1), you WIN, 
but if you REMAIN with cup 3, you LOSE.

Each of the three options has an equal chance of occuring, based upon your
original random pick.  If you change 1/3 of the time, you lose (in the case
of option 1, which has a 1/3 probability of occuring), while if you change
2/3 of the time (with options 2 or 3, which have a 2/3 probability of
occuring), you win.

Therefore, the proper choice is always to change!  Sorry about that 50
dollars. :(

I think the place where things skew to the anti-intuitive is the fact that
you will -always- get shown an EMPTY cup by the game-show host/tricker, 
not necessarily a random cup.

Hope this helped.

Phil, a Dr. Math who knows only statistics revolving around game shows...

Subject: Probability Question from OZ!
From: "Sean Pryor"

Hey, it wasn't my $50!  It was a mate you was silly enough to bet on a
maths problem!! :-)

I think I understand.  My query is that perhaps you have a 2/3 of getting
the coin when changing, taking into account the whole process.  If you
isolate the last choice and say - now it's either in this cup or that and
I can choose this or that, its 1/2.  Is that a reasonable method of
explaining the anti-intuitive bit?  Or have I missed the point here...

X-Sender: pspecto1@cc.swarthmore.edu
Date: Fri, 4 Nov 1994 18:46:18 -0500

Exactly.  Nope, you explained it better than I did...

 Phil, a Dr. Math who knows only statistics revolving around game shows...

>What better place to start - my favourite was always the Wheel of Fortune
>- spinning wheels, angular momentum, friction, force, acceleration, sectors,
>degrees, probability - that show had it all! :-)

True, but I always thought Pat Sajak was a dweeb. :)  Glad Dr. Math could
help, and of course, write back with any other problems you have...

Phil