Gary Fowler, revision date: January 22, 1996.
Monty Hall was the host of a game show called "Let's Make a Deal." This was a very popular show due in part to the finale. The stage was set with three doors. Behind each door was a prize. One prize was very desirable and valuable, e.g., two week, all expense paid trip for two to Hawaii. There was a much less desirable prize, e.g., living room furniture. The remaining prize was undesirable. The undesirable prize is traditionally called a "goat," but since this is the Naval Academy we will call it a "mule." After the contestant selected a door, another door was opened to show the prize and the contestant was given the choice between the already selected door or the other door that had not been opened. A few years ago, the popular press contained several articles debating whether the contestant should switch doors. This debate was sparked by an analysis of the given by Marilyn vos Savant in Parade Magazine in which she concluded that the contestant should switch. She received many letters objecting to her analysis and conclusion. Several of these letter were from college professors who teach statistics. The debate spread to professional journals including The American Statistician.
Many of the disagreements resulted from imprecise statements of the problem. If you choose to debate this analysis with aunt Sue or cousin Jose begin with a careful statement of the problem - as we will.
Should the contestant switch doors? For example, if the contestant selects Door #3 and the host shows a mule behind Door #1, should the contestant switch to Door #2?
My mother watched this show a lot. She is absolutely sure that the contestant should switch. My brother, a CPA, is sure it does not matter, and I have a sister, another CPA, that is sure it is best to stay with the original choice.
It does not make a difference which door the contestant selects. It will make this explanation easier, if we assume the contestant selects Door #3. If you do not like this assumption, first understand the analysis with this assumption and the extend the analysis. Assumption #1 implies
P[Hawaii behind #1]=1/3 P[Hawaii behind #2]=1/3 P[Hawaii behind #3]=1/3
Assumption #3 implies
P[Host opens #1 | Hawaii behind #1]=0 P[Host opens #1 | Hawaii behind #2]=1 P[Host opens #1 | Hawaii behind #3]=1/2 P[Host opens #2 | Hawaii behind #1]=1 P[Host opens #2 | Hawaii behind #2]=0 P[Host opens #2 | Hawaii behind #3]=1/2 P[Host opens #3 | anything]=0
Using the fact that the probability of an intersection is the product of a conditional and the probability of the condition, the following table of intersections and margins can be completed.
Prize behind Door #1 Door #2 Door #3 Host opens #1 (0)(1/3) (1)(1/3) (1/2)(1/3) Host opens #2 (1)(1/3) (0)(1/3) (1/2)(1/3) Total 1/3 1/3 1/3
Note that this table's marginal (totals) are consistent with assumption #1. We want to compute P[prize is behind #3 given the host opens #1]. The calculation is
P[prize is behind #3 and host opens #1]/P[host opens #1]=(1/6)/(0+1/3+1/6)=1/3.
Also P[prize is behind #2 given the host opens #1] = P[prize is behind #2 and host opens #1]/P[host opens #1]=(1/3)/(0+1/3+1/6)=2/3. Therefore, the contestant should switch. If you don't switch, you get your original door which had and still has a 1/3 chance of winning. If you switch, you get the benefit of both of the other doors. The host will tell you which of these two is the wrong one.
To perform a simple simulation of this strategy, choose one student to be Monty and one to be the contestant. From a deck of playing cards, select an ace and two lower cards (to represent the prize and the mules). Monty holds up the three cards so only he can see their value. The contestant picks a card which is placed aside. Monty then reveals one of the remaining cards which represents a mule. (He chooses if there are two mules possible.) If the contestant is a "switcher", then he will only lose in those cases where his original guess was the prize. Therefore, he will win in the remaining 2/3 of the cases.
An acquaintance of mine Harry enjoys a good bet. He always keeps the bet small enough to be affordable, but big enough to keep him in pocket money. One of his favorites uses three cards and a brown bag:
Harry: I have an empty brown bag and three cards. One card is black on both sides, one card is red on both sides and the other is black on one side and red on the other. I'll put all three cards in the bag and mix them well. Now you carefully pull one card out so that we see only one side. Look, its red. That means it can't be the card that is black on both sides. So its one of the other two cards and an even bet that its black on the other side. What do you say we each bet $1. You get black and I get red.
Harry likes this game so much he wants to continue playing, always giving you the color not showing. Since the color showing has already been used once making it less likely and he just plays for the entertainment.
One of the following statements about Harry is true. Select the true statement and show that it is true.