The Monty Hall problem is derived from a similar dilemma that was frequently brought up on the television game show “Let’s Make A Deal,” hosted by Monty Hall. Though the game show’s version usually involved some grand prize and then other smaller prizes to compare it to, the concepts of the two versions are the same. Here is the scenario: There are three doors. One of the three doors is hiding a prize, while the other two are empty or contain something undesirable, such as a goat. The contestant is asked to try and guess which door is the prize door.
If the contestant guesses correctly, he wins the prize. ) Once the contestant makes his choice, the host says something like, “Are you sure that is the right door? How about this; I’ll up your chances and open one of the other two doors. ” The door is opened to reveal a goat. “Now, do you still think the prize is behind your original choice, or would you like to switch to the other closed door? ” The contestant must make a decision between the two doors remaining closed. Here is the big question: Should the contestant stay with his initial selection or should he switch, and does it really matter?
This question was brought to the attention of Marilyn vos Savant and plublished in “Ask Marilyn,” her column in PARADE Magazine. Her reply caused an uproar and she received several letters, many from mathematics professors and probability experts, claiming that her answer was totally false. In her reply, she had said that the contestant was more likely to win if he switched. (I agree with this theory. ) The controversy from Marilyn’s column spread to other publications and spurred on a huge debate over the probability of winning if the contestant stays with or switches his choice.
At first glance, the probability of winning seems obvious. After the host opens one of the doors, two are left closed, each having an equal likelihood of containing the prize. So there is a fifty-fifty chance of choosing the correct door. It does not matter whether the contestant switches or stays, because the probabilities are the same. But if the problem is considered a little more closely, you might change your mind. Here is a different approach to the puzzle: First of all, there are three doors, and the prize can only be behind one of them.
So there is a one-third chance that the contestant will get the prize, and a two-thirds chance that he will not get the prize. If the contestant decides to stay with one door, then his chances of hitting the prize will be one-third. Let’s say the contestant has chosen a door and it is not the prize door. This has a probability of two-thirds. Then the prize must be behind one of the other two doors. Furthermore, the host will open one of these two doors, revealing the one that is empty. So the prize has to be behind the other door remaining closed.
Therefore, if the contestant chooses the wrong door at first (this having a two-thirds probability,) and then switches, he is certain to get the prize. Summing up, if the contestant switches, his odds of winning is two-thirds, and if the contestant does not switch, his odds of winning is one-third. A simple theory, though, does not hold very much importance until it is proven. With this in mind, I ran several trials of different experiments to see if Marilyn vos Savant’s assumption was actually true.
The first investigation I ran was several trials on a simulation of the game I found on the internet (at http://www. intergalact. com/threedoor/threedoor. cgi). The simulation asked me to pick a door, then it opened another door, and finally asked me to either stay with my previous selection or switch. I played twenty times; ten times staying and ten times switching. The results are shown in Table A. When I switched, I won eighty percent of the time, whereas when I stayed, I only won forty percent of the time. In a separate simulation of the problem on a different web site (at http://cartalk. rs. com/Tools/monty. pl), after completing one trial, it gave the cumulative results of all the trials ever run on that site (shown in Table B). Almost exactly one-third of the games were won by those who stayed, and the other two-thirds were won by those who switched. The second test I performed was with a calculator program. (With the help of Mark Lewis. ) I wrote two separate programs. One played the game fifty times, switching each time, and the other played fifty times staying with the first choice each time. The programs were written to choose the “doors” totally randomly.
After the fifty trials were finished, the screen displayed how many times the game was won and how many times it was lost, along with the percentage of wins and losses. I ran each program three times, and the results are shown in Table C. According to this data, the “switching” program usually won over half of the time, and the “staying” program usually won less than half the time. Because you can never trust a machine, my final experiment dealt with actual human beings. I created a small imitation of the game using three plastic bathroom cups and a quarter.
I played the host, and I let nineteen of my relatives play the contestant, one time each. Only seven people decided to switch their choice, four of which who won. That means 57. 1 percent won the game when they switched. On the other hand, twelve of my relatives stuck with their original selection, and four of those people won. So only 33. 3 percent won the game when they did not switch. In conclusion, all of the experiments I conducted seemed to prove that the contestant in this three-door situation will have a much better chance to win the prize if he switches his choice.
According to logical reasoning and probability, that chance is about a two-thirds possibility, as opposed to only a one-third possibility of winning if the contestant stays with his original selection. It is odd to think that there is such a controversy over this problem when the correct way of thinking can very easily be proven. The moral of the story? If you are ever a contestant on any game show with three mystery doors and the host opens one of them, just remember to stop and think about what probability has to say, and then make your decision.