Classic Monty Hall
You are on a gameshow hosted by a man named Monty Hall, where you are faced with three doors: behind two of them are goats and behind one is a car. You choose a door; when it is opened, you win the prize behind it. Before it’s opened, Monty (who knows what is where) opens a door. You know he will open a door with a goat behind it; if there are two goats, Monty picks one at random. You have the option to stick with your door or to switch. What should you do?
You say it doesn’t matter? Is that your final answer?
The answer that you’re equally likely if you stick or switch fits with most people’s intuition, but it’s wrong. The correct answer is that you should switch, because then you have probability ⅔ of getting a car, and ⅓ if you stick. This is not obvious! There are a few ways to see it, but one that will let us easily tackle variants on the problem uses Bayes’ Theorem.
Let’s first look at a case where we do get an intuitive answer: suppose Monty doesn’t know what’s behind the other doors either, and opens one of them at random. Of course, now there’s a chance that Monty will reveal a car, and it’s game over. Let’s suppose you picked the door on the left, and Monty picked the door in the middle. (Everything in this puzzle so far is symmetrical, with no distinction between left, middle and right, so we will get the same answer regardless of which door we picked and which door Monty opens.) What we’re interested in is the probability that you picked a goat, given that you know Monty has revealed a goat.
The probability that you picked a goat at the start is ⅔.
Suppose you picked a goat, Monty now has two doors, one of which contains a goat. Monty picks at random, so the probability of them opening a goat is ½. If, however, you picked a car, Monty only has goats left, so will certainly reveal one.
The last thing we need to use Bayes’ Theorem is the overall probability of being shown a goat, which isn’t immediately obvious—but we can work it out. The probability that you choose a goat and Monty shows you one is ⅔ times ½, or ⅓. The probability that you don’t choose a goat, but Monty shows you a goat, is ⅓ times 1, or ⅓ again. So the probability that Monty shows you a goat can be found by adding up the two cases, which each happen with probability ⅓.
Now we can use Bayes’ Theorem:
This is what we expected.
What about the Monty Hall case (i.e. Monty does know where everything is)? Well, our first probability is the same: there is still a two-in-three chance of picking a goat.
But now the probability that Monty shows us a goat given that we chose one is 1, because Monty knows where the other goat is. And if we picked the car, Monty still only has goats to reveal, so we get 1 as well. Indeed, whatever was behind our door, Monty will show us a goat.
So now we can apply Bayes’ Theorem again:
That’s our probability of having a goat, updated with the information that we’ve been shown a goat. So we’re more likely than not to have a goat still, and we’d better switch.
Monty Hall Revisited
As above, but now Monty, instead of picking at random, always opens the door on the left, if they have a choice. What should you do if Monty picks the left door? What about the right?
If you’ve read the previous answer, you might think that surely this doesn’t make any difference, and it’s still better to switch. Alas, probability has come on down to fool you once more.
In the answer above I made a comment that it didn’t matter which door we assumed you and Monty had picked, because everything was symmetrical. This is no longer true! However, it still doesn’t matter which door you pick; what matters are the relative positions of the other two doors. (Still, if you want, you can assume that you pick the door on the left, or in the middle, or on the right; you will get the same answers.) From now on, when we say “the left door” or “the right door”, we’ll mean “of the two that you didn’t pick”.
We can use Bayes’ Theorem again. As ever, the probability that you pick a goat at the start is ⅔.
Let’s suppose Monty opens the door on the left. If you chose a goat, Monty doesn’t have a choice. So whether they open the left door or the right door is determined by the initial random arrangement of the goats and the car. The two options are equally likely, so the conditional probability is ½.
Now suppose you picked the car at the start. This means Monty has a pair of goats to choose from, and will choose the one on the left. So we have that
In exactly the same way as before, to get the last thing we need—the probability that Monty opens the left door—we can add up over the cases. The probability that you chose a goat and they opened the left door works out at ⅓, as does the probability that you chose a car and they opened the left door.
You might notice that these are the exact same numbers as in the case where Monty knew nothing about where the goats were. So it’s not surprising that when we plug everything into Bayes’ Theorem, we get:
So if Monty chooses the left whenever they have a choice, and that’s what they do with us, that increases the chance that our door has a car behind it. Weird, right?
The intuition here is that both the knowledge-free host choosing at random, and the knowledgeable host going to the left, have given us some information about what’s behind the other doors, that points towards the two goat case. In the first case, it’s because they were more likely to open a goat (and not a car) if there were two goats for them to pick from; in the second, the presence of the second goat makes them more likely to go to the left. (In either case, Bayes’ Theorem lets us quantify how much information that gives us.) In the case where Monty doesn’t do anything we didn’t know they would do, we can’t update our probability of what is behind our door: it stays the same as at the start.
What if we know Monty goes to the left when they have a choice, but they pick the right door in our game? We don’t actually need Bayes’ Theorem for this: we can just logic it out. Suppose we had picked a car. Then Monty would have had two doors to choose from, and would have chosen the left one. But that’s not what they did, so we must have picked a goat. Monty’s choice has now given us all the information we need to determine where everything is—and if you go through the Bayes’ calculations, you’ll see that you get an answer of
So you’ll definitely get a car if you switch. Didn’t you do well?
Here’s a final summary of all the cases:
|Monty opens a goat unknowingly||Stick and switch win equally often|
|Monty opens a goat at random||Switching wins twice as often as sticking|
|Monty opens the left if possible, and he opens the left||Stick and switch win equally often|
|Monty opens the left if possible, but he opens the right||Switching always wins|
To learn more about the history of the problem click here.
For some more fun extensions to the classic problem click here.
[…] Monty Hall revisited—Tom Rocks Maths […]
[…] the first article ‘Monty Hall Revisited‘ for the answer to the classic problem (and a bonus extension problem where Monty always […]
[…] For an explanation of the original problem click here. […]
[…] (such as the Unexpected hanging paradox), and sometimes it is perfectly valid (such as the the Monty Hall Problem). Paradoxes are not only an interesting weirdness, they are a great way to test your mathematical […]