The problem gained attention when it became the object of a Marilyn Vos Savant column in Parade magazine where it was stated in the following form:
- Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what is behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Vos Savant says that the probability of obtaining the car improves if you switch. Many people find this answer runs counter to intuition. This answer like all answers depends on the assumptions that are made especially in regard to the statement
- the host, who knows what is behind the doors,...
Let us assume that the host does show a goat to the contestant no matter which door is originally chosen. The probability that the car was originally chosen is 1/3rd. The probability that the contestant did not chose the car is 2/3rds, so it follows that the contestant will obtain the car 2/3rds of the time by switching after the host shows a goat.
It is easier to see this relation by considering the question when the number of doors is large. For instance, suppose there are 100 doors. We further assume you pick one door at random and then the host opens 98 doors with goats in them. Obviously, the probability is 1/100 that you originally picked the right door and the probability that the car is behind the remaining door is 99/100.
But suppose the host wishes to minimize the likelihood that the contestant will get the car. What strategy ought he to pursue ?
The question you asked at the end is too open-ended to answer. It depends on what freedoms you allow the host to take, and what information you assume the contestant knows about his strategy. His overall best approach would be to end the game whenever she chose a goat originally, and open a different door and offer a switch whenever she chose the car. This way, he will give away a car at most 1/3 of the time. If the contestant knows this, the host cannot better that 1/3 chance if she can implement an optimum strategy.
ReplyDeleteBut I'm more interested in your explanation of the classic problem. Your solution fails to satisfy most people who answer that the chances are 1/2, because they feel they provided a mathematical solution (the car could have been placed behind any of three doors with equal probability, but one is eliminated; the remaining two are equally likely) while you just made an analogy, a process that can be subtly flawed.
The correct solution is that there is a hidden random variable (as is the case in most probability controversies). If the contestant chose the car originally, the host had a choice of two doors to open. So, where the supposed "mathematical solution" considered three cases, four are needed. The case where the contestant originally chose the car needs to be split into two cases: one case where the host opened the door you saw him open, and one where he would have opened the other door.
This way you "eliminate" not only the case where the car was behind the door that was opened, but also the case where it had a goat but the host would have opened the other door. As a result, the two remaining cases are no longer equally likely. The one where the contestant would win by *not* switching gets split in half. So she wins twice as often if she switches.
And I bring all this up, because the supposedly "correct" answer to a similar problem is wrong. "I have two children, and at least one of them is a boy. But maybe both are; what is the probability of that?" If you answer 1/3 - because from the four combinations BB, BG, GB, and GG, only one is eliminated - you are making the same mistake made by those who answered 1/2 to the Game Show Problem. In the BB case I had no choice about what to tell you. But in the BG and GB cases I could have told you one offspring was a girl. Those cases need to be split in half the same way, making the answer 1/2.
And you can test the intuitiveness of this, if I change my problem to "I have two children, and at least one of them is a boy, and he has red hair. But maybe both are boys; what is the probability of that?" If you think that adding information about red hair cannot change the answer, it means you intuitively feel the facts I gave you are an observation made after I choose an offspring of mine to describe.
(And to resolve any questions you might have, red hair can darken with age. My first son's, like mine, would no longer be called "red" by most people. My younger son will forever be called a copperhead.)
You are absolutely correct in saying that the question is too open ended.
ReplyDeleteI will eventually write an analysis of the strategy problem which will include
a variety of strategies. But I will say that you are basically correct
to say that the strategy of the host must be based on whether the contestant has picked a car.
In fact, the host must apply a mixed strategy to minimize cost. Monte Hall employed a mixed strategy in the game show. If he had not, the contestants would have been able to read the appropriate choice of door by observing his behavior.
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