1 Transitional//EN"> Car and Goats Problem

# Car and Goats Problem

NOTE: On 2003 August 23 I read an article by Keith Devlin of Devlin's Angle, insisting that the probability of the contestant winning if he switches is 2/3. I will stick with what I say in this article, which says that there is not enough information in the statement of the problem to solve it. But in reviewing this article I did find an error: if Monty Hall uses a 50%-50% split for the Door 3 only and the goat only methods, the probability that the contestant wins if he switches is 8/13, not 7/12. I explain this later in the article. I also revised the rest of this article slightly.

A number of years ago, a probability problem based on the show "Let's Make a Deal", with its host Monty Hall, seemed simple enough, but it has created a large number of papers due to the controversy it aroused. It apparently came from one episode of the show. In this episode, a contestant was confronted with three doors:

 1 2 3

and was asked to pick one. The contestant picked number 1. The master of ceremonies opened door 3, to reveal a goat:

 1 2 3

He then offered the contestant this deal: the contestant can, if she wants to, switch doors and choose number 2 instead of number 1. The question now is: should the contestant switch? Further, what is the probability of winning if the customer does not switch, and what is the probability of winning if the customer does switch?

It looked fairly easy. There were three doors. The goat could be behind each, so the probably was 1/3 for the goat being behind each door. The opening of a door eliminates a door and a goat, meaning there is only a car and a goat to consider. Since there are now two possibilities, and the odds were initially the same for each goat, it would appear they are still the same for the remaining doors, so the probability that the car is behind door 1 is now 1/2, and that behind door 2 is also 1/2, so it appears not to make any difference which door you choose. The mathematicians attempted to demonstrate this using a complex formulation known as Bayes' formula:

p(c1|g3) = p(g3|c1)p(c1)/[ (p(g3|c1)p(c1) + p(g3|c2)p(c2) + p(g3|c3)p(c3) ]

= 1* (1/3) / [ 1 * (1/3) + 1 * (1/3) + 0 * (1/3) ] = 1/2

where gi is the probability that there is a goat behind door i, ci is the probability that there is a car behind door i, p(x) is the probability of x, and p(x|y) is the probability of y.

Someone posed the question to Marilyn Vos Savant, the high-IQ columnist, and in a column dated 1990 December 2, she said that this was not correct. Her argument, with hardly a mathematical symbol, is that the probability of the car being behind door 1 was 1/3 when the contestant was shown the doors, so therefore it is still 1/3, since nothing has happened that has affected the odds one way or the other. The master of ceremonies' picking a door that had a goat behind it did not affect the situation, since we knew in the original situation that at least one door that the contestant did not select had to contain a goat. Since the probability for door 1 is 1/3, the probability for door 2 is 1 - 1/3 or 2/3. Therefore, the contestant should switch.

This caused a huge number of papers in the mathematical, operations research, and statistical fields, and a lot of argument among mathematicians. Many sided with Marilyn, and others insisted just as much that the above Bayesian analysis is the one that holds. It still has not been completely settled. However, I will make a stab at this problem. It is my belief that there is too little information to be able to solve the problem.

First of all I will bring up some of Marilyn's reasoning:

Let's say we play a shell game. You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger.

She is saying that you learn nothing from her lifting one of the shells and revealing nothing, because she can always do this. There is always an empty shell that is not selected by you, since there are two empty shells and you select at most one. So she can always lift a shell and reveal nothing. She accentuates her argument by saying what if there were a million doors, and after you select one, the master of ceremonies opens all of them except yours and number 777,777, to reveal 999,998 goats. She said you'd switch pretty fast. My way of putting it would be like this: You buy a lottery ticket. The odds of winning are one in 7,059,052. A friend of yours who works for the lottery sees you buy the ticket, and she has just seen a ceremony where the winning ticket was selected; it has not been announced publicly yet. She takes you aside in the building, pulls out a lottery ticket, puts it by you, and says, "I will swap tickets with you if you wish. I will tell you that I know what the winning ticket is, and I know that either your ticket or mine is the winner." Most of us would grab for her ticket in less than a second. Why? Because yours was selected randomly without knowledge of the winning number, and hers was selected with the knowledge of the winning number.

In favor of the probability being 1/2, one can say that information was learned; namely, that the car is not behind Door 3. This means there are only two possibilities left, and they had equal odds originally so they should still have equal odds.

So which is it, 1/2 or 2/3? The answer depends on whether a chance was taken. Suppose you deal a card from an ordinary 52-card deck onto the table face down. As of now the chances of that card being red or black is 1/2. You draw a card. It is black. The odds have now changed, to 0 for red and 1 for black. You took a chance and obtained information on drawing that card. It could have been red or black and by drawing you eliminated one of the possibilities. Now suppose that every card in the deck was black. The chances for drawing a black card are certainty, 1. You draw the card and it is black. You did not take a chance and obtained no new information. The odds are still certain you have a black card. Suppose now that you draw a card from an all-black deck, and one from an all-red deck. Then you draw randomly from these. You took a chance. It was 1/2 red, 1/2 black before you drew from them, and it is now 0 for one of these and 1 for the other, depending on which you took. This seems simple enough.

But now take this situation. There are three cards on the table, and someone across the table knows what they are. You know that there are two black cards and one red one among them. You draw one card from these three. The probability of a red card is 1/3. You ask him to draw a black card for you. You say that you know for certain that one of the two remaining cards is black, so he is giving nothing away to you by showing you a black card, since certainly one of these cards is black. He draws one. It is black. Isn't this situation the same as that of the previous paragraph? You have not learned anything. Therefore, the odds that the card you selected is red is still 1/3. This means the other card on the table has a 2/3 chance of being red.

Translate this into the car and goats problem. The contestant selects a door. The MC opens another door and shows a goat. The reasoning of the previous paragraph holds provided the MC knew what was behind the doors and deliberately selected to show a goat. The problem statement does not say this. If the MC purposely drew to draw a goat, then yes, the odds on the second door are 2/3 and the contestant should switch.

Maybe that was not the MC's objective. Maybe he wanted to select a random door, or Door 3, no matter what. With these strategies there is a non-zero chance that the MC will pick the door with the car behind it. The problem does not state what happens in this case. I presume that he would say, "Congratuations! You just won a Lexus Aleph One Convertible!" or "Sorry! but your consolation prize..." as the case may be. We presume that this possibility was not mentioned because it did not occur. The MC was going to select Door 3, no matter what, and it just happened in this case that it contained a goat. Clearly the MC took a chance on Door 3 having the car. The contestant found that it didn't, so he learned information, and this will change the probability. In this case, indeed the probability of the car behind Door 1 is indeed 1/2.

This means the problem can't be solved due to insufficient information. If the MC wanted to deliberately pick a goat, then the probability of winning if the contestant switches is 2/3. If the MC was going to pick Door 3 no matter what (or if he picks randomly, or if he picks the door to the right of the chosen door), then the probability of winning if the contestant switches is 1/2. The probability could even be 8/13. For example, Monty is the main MC, and Nancy is the backup MC. Suppose Monty has been sick, and the doctor gave him a 50% chance that he could MC the "Let's make a deal" show. Suppose Monty chooses deliberately a goat, and Nancy chooses at random. Then the chances of Monty being the MC are 50%, and in that case the contestant wins with probability 2/3. It is 50% for Nancy, in which case the odds for the contestant winning is 1/2. Then one could try to compute the odds overall for the contestant as

1/2 x 1/2 + 1/2 x 2/3 = 7/12.

This is not correct. It is 8/13 instead. This is because, by opening a door with the goat, the MC makes himself more likely to be Monty than Nancy. This table examines the situation completely:

MC Prize Cont MC Switch Outcome
Nancy 1 1 3 Loss
Nancy 1 2 3 Win
Nancy 1 3 3 SORRY!!
Nancy 2 1 3 Win
Nancy 2 2 3 Loss
Nancy 2 3 3 SORRY!!
Nancy 3 1 3 SORRY!!
Nancy 3 2 3 SORRY!!
Nancy 3 3 3 CONGRATULATIONS!!
Monty 1 1 2 or 3 Loss
Monty 1 2 3 Win
Monty 1 3 2 Win
Monty 2 1 3 Win
Monty 2 2 1 or 3 Loss
Monty 2 3 1 Win
Monty 3 1 2 Win
Monty 3 2 1 Win
Monty 3 3 1 or 2 Loss

For example, if Nancy is the MC, the prize is in Door 1, the contestant selects Door 1, Nancy will pick Door 3 since she always picks Door 3. That reveals a goat. If then the contestant switches, he loses. And so forth for the other cases. In five cases, Nancy reveals the car, and in those cases, the result is immediate: "SORRY!!" or "CONGRATULATIONS!", as the case may be. However, the problem gives us that this did NOT occur. If you count the number of "Win"s and the number of "Loss"s, you get 8 wins and 5 losses. Since the probability of Monty and Nancy was equal, and all the other initial probabilities were equal, the odds in favor of switching are now 8 in 13, not 7 in 12. Basically, it is not 7/12 because when the MC reveals a goat, we learn something about which game he or she may be playing. If there were a million doors and the MC reveals 999,998 goats, then almost certainly Monty is the MC, right?

Note that the Nancy or top half of the table shows what happens if Nancy is known to be the MC. There are two Wins and two Losses, so the probability upon switching is 1/2. In the Monty or second half, there are six Wins and three Losses, so the probability upon switching is 2/3. This confirms my earlier statements.

And are these the only games that Monty plays? Let the original game be Game 1, and suppose that he plays Game 2 as well. In that game, if the contestant chooses the correct door, it is opened, and the contestant is congratulated for having won the car. If the contestant chooses a wrong door, then he says, "No, that's not it. But I will give you a second chance. Choose one of the other doors.". Maybe Monty plays games with the games. This is quite likely on that gamey show. Maybe he plays Game 1 if the contestant chooses the correct door and Game 2 if he chooses a wrong door. In that case, if Monty shows a goat, then it is certain that Door 1 contains the car, and so switching will mean losing for sure.

With the probability of winning given switching being one of 0, 1/2, 8/13, 2/3, or an uncountable infinity of other numbers, this problem has insufficient information for sure.

When I first heard about the problem in 1990, I sent Ms. Vos Savant a letter, the text of which follows:

Dear Marilyn:

I saw your column in Parade Magazine of December 2, 1990, about the game show puzzle with the car and the goats. A contestant has a choice of three doors. He picks door No. 1. The host, who knows what is behind the doors, opens No. 3 and reveals a goat. He offers the contestant an opportunity to switch to Door No. 2. You say he should, since he would then have a 2/3 chance of winning, while with door No. 1 he would have a 1/3 chance. You then published three letters from Mathematics PhDs who said that you are wrong; the probability for either door 1 or door 2 is 1/2. You then argued that your original choice was correct.

The key phrase is "who knows what is behind them", and in fact the whole problem is an example of what I call a trick problem. High school and college teachers and textbooks all the time give their students trick questions. This turns many students off of mathematics and in fact may be one source for the phenomenon known as "math anxiety". For who likes to be tricked? In your case, you trick Mathematics PhDs.

In fact, the answer to the problem cannot be determined from the information given. Your answer is correct provided that the host uses his knowledge to ensure that the door he opens has a goat. Perhaps the show is operated like this, or perhaps not. Nothing in the problem excludes the case of the game show host knowing what's behind the doors, but deliberately ignoring that information and either (1) picking Door No. 3 in any case, or (2) throwing a die and choosing door 1 if the die comes up 1 or 2, door 2 if it comes up 3 or 4, and door no. 3 if it comes up 5 or 6. If the host reveals the car, he would perhaps either (1) sound a buzzer and say "Awwwwwhhh... you did not win. But for a consolation prize, you have won a free trip to the Bahamas...", or (2) "Congratulations!! The car is yours!", as the case may be.

If this is the case, the probability that No. 2 contains the car is indeed 1/2, for we have excluded door No. 3 as a possibility and have excluded the possibility that the host reveals the car.

It is even more involved than this. The probability for door No. 2 could be 7/12. It would be this if both my game show and your game show were held in two different cities, and the contestant does not know which one he is in. It is even possible the host may not know which door contains the car! The phrase "who knows what's behind them" could mean simply that he knows there are cars and goats behind the door.

The problem would have been better phrased, without the trick, if it had said, "who knows which door has the car and uses that information to deliberately select a door with a goat...". Presenting math questions in such a more honest manner would do a lot to help people overcome "math anxiety" and would be good public relations for mathematics.

Yours truly,

James V. Blowers
PhD, Mathematics

Unfortunately, I got no reply from Ms. Vos Savant. Is it because I had 7/12 instead of 8/13? Or more likely she had so many replies that she could not answer them all? But you can still try it for yourself. Create some goat and car cards, and try playing the game a few times, under first the "purposely select a goat" hypothesis and then under one of the other ones, or try one of the many Java or other program examples of the game on the Web. You will find that switching gives you twice as many wins as losses, if the MC purposely select a goat.

Jim Blowers