Probability: Let's Make a Deal

Date: 4/29/96 at 21:23:4
From: Bob Gordon
Subject: Probability: Let's Make a Deal

Dear Dr. Math,

I have been having difficulty understanding certain aspects of the 
"Let's Make a Deal" problem, and I was hoping you could help me.  I 
teach middle and high school mathematics classes.

I will state the problem, indicate my understanding of the solution 
and pose the related question that perplexes me.

Here is the problem:  There are three closed doors at the conclusion 
of a game show. A contestant chooses one of the doors, behind which he 
or she hopes lies the GRAND PRIZE.  The host of the show opens one of 
the remaining two doors to reveal a wimpy prize.  The contestant is 
then given the choice to stay with the original choice of a door or to 
switch and choose the remaining door that the host did not open.  The 
problem is: Should the contestant stick with the original choice of 
doors or switch and choose the other door?

The solution is that the contestant should switch and choose the other
door.  The reason for this is as follows.  Let's say that the 
contestant chose door #1.  This means that the probability that the 
grand prize is behind door #1 is 1/3 while the probability that the 
grand prize is behind one of the other two doors is 2/3.  Once door #3 
is opened, the probabilities do not change.

This would be analagous to a situation where I had a choice of picking 
one lottery ticket among 100 tickets where the winning ticket was 
among the 100.  Clearly, the probability that the winning ticket is 
among the other 99 is 99/100, so that if all but one of those 99 
tickets were revealed to be losing tickets, it would be in my interest 
to switch, as the winning ticket was (and still is) among the other 
99.

I see two different questions that need some clarification, and this 
is where I begin to need help.

Are the following statements true?

1) Once door #3 is shown to hide a poor prize, then the probability 
that I will choose the correct door is 1/2.

2) Once door #3 is shown to hide a poor prize, then the probability 
that the prize is behind door #2 is 2/3.

Here is my question:

Suppose that there is a second contestant from the beginning of the 
door choices.  Contestant A chooses door #1.  Contestant B chooses 
door #2. Door #3 is opened and a silly prize is behind it.

According to the above logic:

1) both contestants should switch choices of doors; and

2) the probability that the grand prize is behind door #2 is 2/3 
AND the probability that the grand prize is behind door #1 is 2/3.

Clearly, something is amiss. I suspect that the error lies with the
language I have used, the referrant for the probabilities or a misuse 
of definitions. I would appreciate your help, please.  Thank you.

Date: 4/30/96 at 11:18:15
From: Doctor Aaron
Subject: Re: Probability: Let's Make a Deal

Hello,

Interesting problem.

The problem is that you can't use the logic from the one-player game 
to explain the two-player game.  In the one player game, we have only 
2 distinct sets: the door I have chosen (door 1), and the doors I have 
not chosen. Because the probability that the Grand Prize is behind a 
given door is 1/3 for each door, the probability that it is behind my 
door is 1/3, while the probability that it is behind one of the two 
remaining doors is 2/3.  

When I get information that of the doors I have not chosen, the prize 
cannot be behind door 3, the probability is still 2/3 for the doors I 
have not chosen, but the only unknown of these is door 2 so the 
probability the prize is behind door 2 is 2/3, therefore I should 
switch.  This logic is sound, but it does not hold in the two player 
game.  

In the two player game, we have 3 distinct sets of doors: 
The door I chose (#1), the door player B chose (#2) and the door that 
neither of us chose (#3), each of which has a 1/3 probability of 
containing the prize.  

Now Monty has less freedom in choosing which door to reveal - he must 
reveal the door that neither of us has chosen.  This gives us 
information about an entire set of doors instead of about a single 
door within a set.  In the 2-player case, the information we gain is 
sufficient to update the probabilities, so each of the remaining doors 
has a 50/50 chance of containing the prize.  

Another interesting case would be to suppose that we had 2 players 
and n doors.  In this case each of the chosen doors has a probability 
of 1/n of containing the prize, and the set of unchosen doors has a 
probability of (n-2)/n of containing the prize.  If one of the 
unchosen doors is revealed, then the probability that the prize is 
behind the set of unchosen doors is still (n-2)/n, but now we only 
have n-3 relevant elements in this set, each of which has a 
probability of ((n-2)/(n-3))/n, which is greater than 1/n for n>3.  
This tells us that each player should choose from the set of 
previously unchosen doors.  

Other interesting extrapolations concern what happens when Monty 
reveals m doors or when we have l player.

I hope that this has been helpful

-Doctor Aaron,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 5/1/96 at 16:59:48
From: Stephen Weimar
Subject: Re: Probability: Let's Make a Deal

Hi Bob,

It's fun to see your questions in the Dr. Math area. I've been 
monitoring this exchange, making sure you get a good answer but 
holding myself back.

You had a question that went unanswered in your first post:

>Are the following statements true?
>
>1) Once door #3 is shown to hide a poor prize, then the probability 
>that I will choose the correct door is 1/2.

Nope. In fact, this way of asking the question reveals a critical 
piece of the problem. It also addresses your comment made elsewhere, 
which I share, that what may be most interesting about this problem
is what it reveals about the way we think and problem-solve and the 
role communication plays.

The probability that you will pick the correct door is difficult to 
calculate but it has to do with whether and how well you use the 
information you've gained from Monty's showing you door #3.

People have an "agent" they tend to use in their thinking that makes a 
quick calculation of probability based on the number of choices. This 
agent serves us in many situations but not all, and certainly not 
here. The job of the math learning process is to develop other 
powerful agents that can become as trusted and easily used. When our 
thinking runs into contradictory conclusions, this is the prime 
opportunity for the development of new agents or a switch in 
allegiance between agents.

One new agent that has been discussed here is the "new information 
from choices made" agent.

I tend to like rephrasing the problem along the following lines:

You pick a door.

Monty offers to let you stay with your pick or take both of the 
others.

It's clearly a good deal to get the other two doors instead of the one 
you first picked.

When Monty shows you door #3, in essence he is merely showing you 
which of those two doors you should look behind if you're going to 
find a prize.

We get new information from Monty's choice because we know he is going 
to pick a door with a poor choice. This means that 2 out 3 times he 
will have to avoid showing you the good prize behind the other door. 
He has a choice between two doors and he uses information about what 
is behind them to make that choice. His choice passes along some of 
the information to us if we are clever enough to figure it out. Your 
question above asks how likely it is we will figure it out. I guess it 
depends on whether Bob Gordon was your teacher. :-)

So, let's use the "new information from choices made" agent in the 
case of the two-participant version. As was pointed out, Monty's 
choice is different and thus we are getting different information.

In this case he is not picking which door to show us. He is choosing
whether or not to show us the door. If he doesn't show it to us then 
that tells us that the good prize is behind it. If he does show it 
that tells us that the prize is behind one of our doors. That's all it 
tells us.

I agree with you that what this reveals about our thinking process is
perhaps the most fascinating part of the problem. How do we develop 
new agents? What is the environment that makes it easy for people to 
shift their thinking? How do we learn to coordinate all of the 
different agents we develop over time? And so on.

Lots of fun.

- Steve Weimar  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/