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© Copyright 1998, Jim Loy
This puzzle (also called the Monty Hall problem) has become fairly famous. If you've never seen it, you won't believe the answer that I will give below.
Suppose that Monty Hall (on TV's Let's Make a Deal) asks you to choose between three doors: #1, #2, and #3. Behind a random door is a new Rolls Royce. Behind each of the other two doors is a goat. Let's assume that you would prefer a Rolls Royce to a goat. You choose a door. Now, Monty, who knows which door hides the Rolls Royce, shows you a goat behind one of the two doors that you did not choose. He then gives you the opportunity to change your choice. Assume that Monty always does this, regardless of your guess. Should you change your choice?
There is a fairly simple solution to this. But, people don't believe it. There has been much heated debate over it. I'll show my reasoning below. And then, see if you agree. Originally, your chances of choosing the correct door are one in three. Monty, who knows where the Rolls Royce is, shows you a goat behind another door. It will turn out that he did not change your odds. You still have one chance in three of being right. So you should switch to the remaining door, which now has odds of 2/3.
You probably don't believe that Monty did not change the odds of your first guess being right. He showed you a door, seemingly at random. So, aren't the odds now 50-50 for the two unseen doors? Of course, Monty's choice of doors was not at random. But, what difference does that make?
I guess I will have to go into some detail. We are going to test my recommended strategy of always switching doors. There are two cases:
Case I - You guessed right the first time, chances 1/3.
Case II - You guessed wrong the first time, chances 2/3.
You do not know which case (I or II) has just happened. So, you either switch your choice or not. Let's say that you do change your choice, as I recommended:
Case I - You guessed right, then you switch. You lose, chances 1/3.
Case II - You guessed wrong, then you switch to the only remaining door. You win, chances 2/3.
Hey, every time you guessed wrong the first time, you switch to the correct door, and win. Convinced?
There's another way to prove this. I'll have my computer simulate this process a few thousand times, and we will see how often I win if I switch every time.
OK, I played Monty's game 10,000 times. I made my first guess at random, and I switched doors every time. I won 6657 times, and lost 3343 times. That's pretty close to 2/3. That's not a pure mathematical proof, but it's pretty good. Convinced?
Here is a description of the program:
Do the following five steps, 10000 times
- Place the Rolls at random.
- Make the first guess at random.
- Show a goat (at random if there is more than one available goat).
- Switch to the other unseen door.
- Record whether I won or lost.
This method is an example of the Monte Carlo method.
Marilyn Vos Savant, reportedly the person with the highest IQ in the world, published this puzzle in her newspaper column. And she got a lot of criticism for giving the correct answer, making the front page of the New York Times. We all like to show that we are smarter than the worlds smartest human. But she was right, all along.
Apparently, Martin Gardner first published this puzzle, in Scientific American, many years ago. His version did not involve Monty Hall.
Above, I showed you how to avoid the trap. Here is the actual trap. You have made your guess, and Monty has shown you a goat. You can now choose between two doors, and you cannot know which is the correct door. Doesn't that imply a probability of 1/2 (50-50 odds)? Well, the above article actually shows that it does not imply 1/2.
We need to have an example. Let's say we number a die (half a pair of dice) 0-2-3-4-5-6. You can choose even or odd, then I will roll the die. You can choose between two choices (just like the two unseen doors that Monty provides), and you cannot know which is the correct choice (just like your situation above). But that does not imply 1/2 probability. You should choose even, which has a probability of 2/3.
Perhaps the following example will show why 1/2 seems so logical. I have bent a coin so that one side has a probability of 1/3, and the other side has a probability of 2/3. I will not tell you which side has the higher probability. Which do you choose, heads or tails? The odds are not 50-50, but you have to pretend that they are. Let's say you choose heads. I may now know that you are likely to win (depending on how I bent the coin), but you still do not know. In the Monty Hall Trap, we guess 1/2 because we have not reasoned out that we should change our guess. Not knowing the real probabilities, we have to pretend that they are 1/2. But, because we have not reasoned out the probabilities does not mean that we could not reason them out.
First of all, if you don't believe the above arguments, please notice that Monty does not have any choice at all if your initial choice was wrong (2/3). Two thirds of the time he knows you picked a goat, and he has to show you the other goat, and that gives the game away.
Second, if you don't believe the above arguments, please reread the above section labeled "Reasoning." It should clear this all up, if you actually read it.
OK. I received email saying that the two choices are obviously independent, and therefore my arguments are wrong. What is meant by "independent?" I flip a coin (50-50), and it comes up heads. What is the probability of a head on the second flip? It is .5 (50-50), because the second flip is independent of the first. In other words, the first flip does not influence the second flip in any way. We say that the coin has no memory. Many people don't believe this, and that is the famous Gambler's Fallacy. How about my bent coin, mentioned above? Let's say the probability of a head is 2/3, and tails is 1/3. I flip a head. What is the probability of a head on the second flip? It is still 2/3 because this flip is independent of the first. Let's pick a card out of the deck. What is the probability of choosing an ace? It is 1/13 (4/52). Well, I pick a card, and it is the five of spades. What is the probability of an ace on my second pick (assuming I don't put the five back into the deck)? It is 4/51, slightly greater than it was for my first pick because there is a card missing from the deck. Those two picks were not independent. The first pick affected the second pick. We say that the cards have a memory.
In the Monty Hall trap, are the two choice situations independent? No. Monty's clue depended heavily on our first choice. And if we are smart, and change our choice, our strategy depends heavily on Monty's clue. Our choices are not independent, unless we ignore Monty's clue and flip a coin to make our choice.
I've never heard if Monty Hall has actually played this game. He does something similar. But he does not have to show you what is behind some other door. If he has a choice, in every situation, then he is not conforming to the Monty Hall trap, and the above arguments do not apply.
If we expand the Monty Hall trap to include four doors, then it becomes more difficult, in my opinion. Then Monty always has a choice of doors. And I am sure that even fewer people would believe my arguments. You should still switch. But, I think the whole thing is much less clear.
The following may be Marilyn Vos Savant's argument, which I didn't read at the time. A glance at it gave me the impression that it was too complicated. But it does work well. We modify the game to include a hundred doors. We choose a door. The chance that we chose a car is 1/100. The chance that we chose a goat is 99/100. Monty now shows us 98 goats, which were behind 98 of the doors. Only two doors remain, our first choice, and the 100th door. Which one contains the car?
Almost certainly (99/100) we guessed wrong the first time. In that case, Monty is going to show us 98 goats. We chose a goat; he will show us the other 98 goats. One other door will be suspiciously unopened, the one with the car behind it. We change our choice, and choose the car. If we were incredibly unlucky, we chose the car first guess, changed our choice, and lost. And the crowd groaned.
After Monty showed us 98 goats, we don't know where the car is. Does that make it 50-50? Was our original guess actually 50-50, with 100 doors and one car? Certainly not. It was 1/100 then, and it is 1/100 now that we have seen 98 goats.
Instead of Monty showing us 98 goats, I originally assumed that Marilyn specified that Monty would show us only one goat. That variation would not seem to make the situation any clearer.
I received email saying that no statistician had come forward to support Marilyn's claim that you should change doors. The reason for that is that people don't come forward to say that other people are right. They only come forward to say that other people are wrong. You always get a biased sample, when reading letters to the editor. In this case, the mathematicians who came forward looked foolish. I also received several emails saying that the above is the best explanation of this famous trap, that these people had seen.
A pool player has apparently made up the amusing Buddy Hall paradox:
You have three slips of paper to pick for your next pool opponent, and one of them is Buddy Hall (famous pro). You pick your choice then the tournament director removes the third choice after your pick (definitely not Buddy Hall). Do your chances of playing Buddy go from 33% to 50%?
The answer is that no, your chances of playing Buddy are still 1/3. If you are offered the chance of switching your choice, then the puzzle becomes the Monty Hall trap. And your choice depends on whether or not you want to play against Buddy.
I received email suggesting that we actually test the "always switch" strategy by playing the game by hand, with playing cards (ace for the car, whatever for the goats). You can be both Monty and the contestant who always switches, and in that way you will see how restricted Monty often is in his actions. Shuffle the three cards, and place them on the table, face down. Choose a card. Now, Monty shows a goat (after refreshing his memory by looking at the cards), and then you switch your choice. And you will see that there is no fifty-fifty about it.
Another email objected that, since I always changed my choice, I never did choose the door of my first "choice." I was instead choosing the other pair of doors. This is just a little semantics, it would seem. It is absolutely right, that I am not really choosing my first door hoping to find a car there, I am choosing it hoping to find a goat there. But, if we say that I am choosing a pair of doors, as this person suggests, then already my chances of winning are 2/3, and Monty will clear it all up by showing a goat behind one of the two doors that I chose. So the Monty Hall trap may actually be clearer from this perspective.
Martin Gardner's version, published in October 1959, involved three condemned prisoners, one of whom will be pardoned at random. One prisoner cons the warden into naming one of the other prisoners (other than the prisoner who is asking this of the warden) who will not be pardoned. Do this prisoner's (the one talking to the warden) chances of being pardoned then go up to 50%? This is identical to the Monty Hall trap, and this prisoner's chances are still 1/3, but the probability that the third prisoner will be pardoned have gone up to 2/3. Mr. Gardner got a flood of mail about this, much smaller than Ms. Savant's flood of mail.
Marilyn Vos Savant's column was published in Parade magazine, on September 9, 1990. Subsequent readers' comment appeared on Dec. 2, 1990, Feb. 17, 1991, Jul. 7, 1991, Sep. 8, 1991, Oct. 13, 1991, Jan. 5, 1992, and Jan. 26, 1992. Also see The New York Times of July 21, 1991 (front page) and August 11, 1991 about the furor. Several articles in mathematical journals were also devoted to this.
See The Monty Hall Problem for a clear picture showing that you should switch doors.
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