Probability: Let's Make a DealDate: 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/ |
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