Wednesday, June 18, 2008

A Question About the Monty Hall Problem

A reader forwarded me a link to an excerpt from a new book—The Drunkard’s Walk: How Randomness Rules Our Lives by Leonard Mlodinow. The excerpt discusses a widely misunderstood issue in probability theory that has become known as the “Monty Hall” problem. The reader asked me: “What is going on in this one? I looked through your book [The Two Headed Quarter] for a possible answer, but no luck.”

Coincidentally I just started reading Mlodinow’s book a few days ago and at least through the first two chapters I’m enjoying it. The book’s subject has always fascinated me and the author is a fellow physicist.

My book does not discuss the Monty Hall problem explicitly, but I do in the last chapter discuss how people become deceived by events that are conditionally probable. The Monty Hall problem is an extremely subtle example of how people are fooled by conditional probabilities. It is so subtle it has fooled professional mathematicians including Paul Erdos, one of the 20th century’s most prolific mathematicians.

Here is the problem. You are a contestant on the game show Let's Make a Deal. The host, Monty Hall, asks you to pick from one of three doors. Behind one is the grand prize and behind the other two are worthless consolation prizes. You choose door number one. Monty Hall, who knows what is behind all three doors and will not reveal the grand prize, opens door number two to reveal a consolation price and asks if you want to switch your choice to door number three. Should you switch or stay with door number one?

Many people argue that at this point your chances of winning are 50-50 and switching your choice to the other door will not improve your chances. But the reality is you should switch because you will win two-thirds of the time if you do. This fact is hard to believe. In his biography of Paul Erdos, The Man Who Loved Only Numbers, Paul Hoffman describes how Erdos, did not understand why switching improved the contestant’s chances. A friend wrote a computer program that simulated the Monty Hall problem and showed that switching does win two-thirds of the time, but Erdos still did not understand the reason why.

The reason you should switch is that once the all-knowing Monty Hall opened a door, he gave away information. His choice is not random; which means that your choice now has conditional probabilities associated with it. Had Hall acted first and then presented you with a choice of two doors, your chances would be 50-50. But he acted after you did and now you have a chance to respond. These are the three possible scenarios:

Prize behind door 1 – Hall opens either door 2 or 3, you switch to the one he doesn’t open and loose.
Prize behind door 2 – Hall must open door 3, you switch to door 2 and win.
Prize behind door 3 – Hall must open door 2, you switch to door 3 and win.

Switching wins two-thirds of the time and looses one-third of the time. But if you do not switch you have ignored the critical information that Hall provided. You win only the one-third of the time your original choice was correct.

That you do not even win half the time when you do not switch is another surprise. But if you ignore the information provided, the original odds cannot change. The mere act of opening another door has no effect on your original one out of three chance. You have to change your choice based on the new information. Notice the effect of the word “must” in the last two scenarios. Hall does not have a choice in these two scenarios but you do.

Note that you only lose by switching the one-third of the time your first choice was correct. The two-thirds of the time your first choice is wrong, switching guarantees a win. To understand this it helps to imagine an extreme example. Suppose there are 100 doors and you are asked to pick one. Your chances of being correct are 1%. Monty Hall then opens 98 doors revealing consolation prizes behind each. You are faced with a choice of two un-opened doors. I’d switch immediately to the one door he avoided opening. It will win 99% of the time. Only on the 1% chance that my first choice is correct will it be possible to lose.

The Monty Hall problem is subtle but people who play poker should recognize it as the entire premise of the game. In poker, players are dealt random cards, but they do not play random cards. Once the players act and additional cards are exposed it is not correct to say that someone vying for a pot could be holding any of the possible hands. In Texas Hold’em the odds against being dealt two pocket Aces are 220-1. But, if someone is betting and acting as if he or she has a great hand, the odds are much better than 220-1 that player has pocket Aces. The deal might be random but, like Monty Hall, a player’s actions are usually deliberate.

Joseph Ganem is a physicist and author of The Two Headed Quarter: How to See Through Deceptive Numbers and Save Money on Everything You Buy


C.Leach said...

The Monty Hall solution is always stated as if you must change in order to win the car. The solution indicates that you have one chance in three of losing. In the long run, your choice should always be to change, but in fact, in the Monty Hall game, you have only one chance in a lifetime of winning. The odds of winning therefore are only of interest to the supplier of the car to the game, and they would likely to be happier if more people knew the solution was as proven--more winners, more interest in the audience.

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