Monty Hall Problem and Variations: Intuitive Solutions

Estimated read time (minus contemplative pauses): 26 min.

Here are six variations on the Monty Hall problem, starting with the classic version.

Each variation is solved in multiple ways with the aim of making the correct answer intuitive. These solutions can easily be generalized by adjusting the relevant probabilities or frequencies, etc. I’ll give fewer details as steps become repetitive and obvious.

1. Classic Monty Hall (Three Doors)

You stand before three closed doors. The doors are evenly spaced and appear identical, aside from being numbered from 1 to 3. One of the doors conceals a car, while each of the other two doors conceals a goat. The host of this game, Monty Hall, asks you to select a door. If you select the car door, you get to keep the car. You select Door 1, hoping to win the car. But wait. Hall opens Door 2 to reveal a goat. Hall, who knows where the car is, always reveals a goat. (Also assume that, when a contestant chooses the car door, Hall chooses which of the two goat doors to open with equal probability.) You chose Door 1, but are now given the option to switch your guess to Door 3. What’s the probability of winning if you switch?

The quick intuitive answer here is 1/2. The correct answer is 2/3. This makes sense when you think of it this way, given the conditions described above: 2/3 of the time, you’ll choose a goat door and 1/3 of the time you’ll choose a car door. In other words, you’ve got twice the chance of choosing a goat door as you do a car door. Which is to say that switching will win twice as often as it loses; i.e., 2/3 of the time.

For example, suppose you play the game 60 times. You’ll choose the car door 20 times, in which case you’ll lose 20 times with an “always switch” strategy. But you’ll choose a goat 40 times, so that strategy gets you 40 wins. In other words, you’ll lose 20 of the 60 times you play, and will win 40 of the 60 times, and 40/60 = 2/3.

Here’s a tree diagram illustrating the 60 games.

monty-hall-problem_01

Those are hypothetical frequencies—idealized events whose proportions simplify to the probabilities we’re looking for. That in mind, here’s a tree just using probabilities (though you can think of it as playing a single game rather than 60):

monty-hall-problem_02

Each branch shows the probability of ending up on that branch from the preceding point along the path. Find the probability of taking a particular path by multiplying the probabilities of each branch on that path. For example, you choose the car door with a probability of 1/3, and then have a probability of 1 of losing by switching. Multiply those together to get the probability of losing by switching: (1/3)(1) = 1/3. Do the same to find the probability of winning by switching: (2/3)(1) = 2/3.

Finally, let’s use Bayes’ theorem to solve this. It’s overkill here, but it’ll be good practice for some of the trickier variations. The idea is that, when you started, the probability was 1/3 that you chose the car door. Actually, let me say this in more Bayesian-like terms: when you started, your degree of belief in the proposition (or hypothesis, etc.) that you chose the car door was 1/3. But what about after Monty Hall reveals a goat? Should you increase or decrease your degree of belief in that proposition?

Recall that Monty Hall always reveals a goat and chooses either goat door with equal probability (see below what happens if either of these assumptions are false). In case you’re not familiar with Bayes’ theorem, I’ll try to make its parts intuitive as I go along. After a few variations, it should make more sense. Here it is:

\displaystyle P(A|B) = \frac{P(B|A)\, P(A)}{P(B)} 

Let’s relate this to the problem at hand.

Where:

G represents the event “chose a goat door.”
H represents the event “Hall reveals a goat.”
P(A) is read as “the probability that event A happens (or has happened, etc.).”
| is a vertical bar read as “given,” like so: P(A|B) is something like, “the probability that event A happens given that event B happens (or has happened, etc.).”

Now for the probabilities (I’ll use the technical part names for convenience, but I don’t think the names matter much here):

Prior probability for G: This is just the probability that you choose a goat door. It’s P(G) = 2/3.

Prior probability for H: This is just the probability that Hall reveals a goat. He always reveals a goat in this variation. So it’s P(H) = 1.

Posterior probability: P(you chose a goat door given that Hall reveals a goat) = P(C|G). This is what we’re trying to find.

Likelihood: P(G|C). This is the probability that Hall reveals a goat given that you chose a goat door. Again, he always reveals a goat, so that’s just 1.

Now we plug all that in:

\displaystyle P(G| H) =\frac{P(H| G)\, P(G)}{P(H)} =\frac{(1)\times (\frac{2}{3})}{1} = \frac{2}{3} \

Again, we see a 2/3 chance you chose the goat and a 1/3 chance you choose the car. (This is clear from the fact that those two probabilities must add to 1, but you can also re-run the above process replacing G with C = “choose the car door.”) And so there’s a 2/3 chance of winning by switching.

I’ll say a little more below about what this formula is doing—namely, why we divide by P(H). It’s not so helpful to get into that now, since P(H) is just 1 here. I will now say about the numerator, however, that we can interpret it as asking for the probability that you choose a goat door and then Hall reveals a goat. The order in which events occur isn’t always important with the application of this formula (and it certainly isn’t important for the math, as multiplication is commutative), but it often makes its application more intuitive to think of a sequential story. In which case we might switch the terms: P(G)×P(H|G) = P(G ∩ H) = probability you choose a goat door AND then Hall reveals a goat. (We can read “∩” as “and,” but it technically denotes the intersection of G and H, which, as events, are subsets of the sample space, as is their intersection. I don’t plan on getting into such details here—”and” is good enough.)

Before moving on, it’s worth asking: Why do people get stuck on the answer of 1/2? (For a brief history of the problem, see this footnote.1) I think the simplest explanation is that we see two doors that look the same and we don’t feel we know any more about one than the other, so we get a strong impression of symmetry—the doors appear as two sides of a coin. Merge this with our basic, default intuition that if you have two potential events—e.g., “Will I or won’t I guess the correct door?”—and you don’t know which will occur, you figure, “eh, it’s a coin flip, 50-50.”

Here’s a bad example of how this reasoning can go wrong—bad, but also useful, because the mistake is so obvious. When you roll a fair die, you either will or won’t land a 6. Are the odds here 50-50? Of course not, they’re (1/6)-(5/6), or roughly 16.7-83.3, because one side has a 6 and the other five sides each has some non-6 number. The Monty Hall problem is a little more complicated than that because it’s easy to not see all the sides, as it were—that is, it’s easy to not realize that we must work with, or condition on, the background knowledge that Hall reveals a goat with probability 1. (Notice that, after Hall reveals the goat, were someone to then randomly shuffle the goat and car around behind the two closed doors, this undoes that knowledge and really would yield a kind of symmetry that results in 50-50 odds.)

The Monty Hall problem is simple compared to the messiness of the real world. Imagine how often we must fall into this sort of fallacy in daily life. Yikes.

Moving on, we’ll avoid this and similar mistakes by properly conditioning on what we know at any given stage of the story.

2. Classic Monty Hall (One Million Doors)

Here, the correct intuition is more easily stoked. There’s a one-in-a-million chance you chose the car door to begin with. So once Monty Hall has opened 999,998 goat doors, the probability is overwhelmingly in your favor to switch to the one door he avoided opening (“avoided,” that is, unless you actually did choose the car). In fact, you have a greater than 99% chance of winning, because more than 99% of the time, you chose a goat to begin with.

I won’t go to the hassle of making diagrams for this one, as all it requires is replacing all the stuff I did above with the adjusted numbers of P(G) = 999999/1000000 and P(C) = 1/1000000. In other words, hypothetically, if you ran this game a million times, you’d choose the car door once, and would choose a goat door 999,999 times, so you’d lose once by switching, but would win 999,999 times by switching.

This version of the problem is really just to drive home the point that switching is better in the classic Monty Hall case!

Bonus Observation: The above approach to Classical Monty Hall scenarios doesn’t depend on Hall leaving all but one door closed. He could leave five doors closed, for example. In which case the probability for success by switching generalizes as:

\displaystyle \frac{n-1}{n(n-1-g)}

Where n is the number of doors overall, from which we subtract 1 for the numerator. For the denominator, we multiply the n by the following quantity: n minus the door you chose minus the number of doors Hall opens (represented by g). Trying this formula with the last two examples is illuminating, but using it to tinker with an example where there are several doors and Hall only opens only a few of them is even more illuminating.

3. Forgetful Monty Hall (Three Doors)

This example has the same rules as in the classic case. You expect Monty Hall to reveal a goat and thus expect a 2/3 probability of winning by switching. You’ve chosen Door 1, and are eager to say, “Switch.” But this time, just as Hall reveals the goat, he says, “Phew, that’s lucky. I’d forgotten where the car was, so I just hoped for the best opened a door at random.”

What to do with this new knowledge?

Intuitively, the probability of winning by switching must change from 2/3 to something smaller. That is, there was a 1/3 probability you chose the car door. But if you did choose the car door, this would make it easier for Hall to reveal a goat if he’s randomly choosing which door to open. So, the fact that Hall randomly revealed a goat provides at least a little evidence that you chose the car, so the 1/3 must be updated to a higher number. But how much higher?

There’s a 1/3 chance you picked the car door to begin with. In which case Hall will always reveal a goat. You’ll be in that situation (1/3)×(1) = 1/3 of the time.

But 2/3 of the time, you’ll pick a goat door, and in those worlds Hall reveals a goat with only 1/2 probability. That’s a critical difference from when he reveals a goat with a probability of 1, and we must condition accordingly. This means that, (2/3)×(1/2) = 2/6 = 1/3 of the time, you’ll be in a world in which you chose the goat door and Hall reveals a goat.

The probabilities of the car-door-then-reveals-goat and the goat-door-then-reveals-goat worlds are each 1/3, making the odds (1/3)-(1/3), or 50-50, for winning by switching (or by staying; it makes no difference). In terms of probability, we show this by finding all the worlds in which Hall reveals a goat and then finding the proportion of those worlds in which you chose the car door:

\frac{P(car-door-then-reveals-goat)}{P(car-door-then-reveals-goat) + P(goat-door-then-reveals-goat)} =\

\displaystyle \frac{(\frac{1}{3})}{(\frac{1}{3})+(\frac{1}{3})} = \frac{1}{2} \

We can also confirm this with hypothetical frequencies. If you play this game 60 times, 20 of those times you’ll choose the car door and switching loses. But 40 of those times you’ll choose a goat door, and 20 of those times Hall will reveal a goat and switching will win (the other 20 times, he ruins the game by accidentally revealing the car). That’s 20 wins and 20 loses by switching, which comes out to 50-50 odds and a probability of 1/2 for winning (or losing) by switching.

Here are trees for the frequency and probability approaches:

monty-hall-problem_03

 

monty-hall-problem_04
After crossing out the situation we know we’re not in, we “renormalize” the unconditional probabilities of the outcomes that remain in the sample space so they add to 1. Here we divide 1/3 by (1/3 + 1/3) = 2/3. This update yields the relevant conditional probabilities.

Finally, we can plug stuff into Bayes’ theorem. I’ll just find the probability you chose the car because once we’ve worked that out, it’s easy to work out probabilities for winning by switching or staying. Here are the parts we need:

Prior probability for event C = P(C) = probability of choosing the car (without conditioning; i.e., at the offset) = 1/3.

Prior probability for event H = P(H) = probability that Hall randomly reveals a goat, whether or not you chose the car door. This is a bit more complicated than in the classic case, where Hall always reveals a goat. What we need to find here is the probability we’d assign to Hall revealing a goat prior to his actually doing so; that is, when all we know is that there’s a 1/3 probability you’ll choose the car door, a 2/3 probability you’ll choose a goat-door, and Hall has forgotten where the car is. Another way to put this is that we need the probability of seeing a goat revealed, period.

There are a few ways to think this through. The goal is to break things down into manageable events. It should be clear that this is just a repeat of the work we did a moment ago. Hall reveals a goat whenever the car door was chosen, which is 1/3 of the time. He reveals a goat 1/2 the time a goat door was chosen, which is 2/3 of the time, and 1/2 of 2/3 is 1/3. So, P(H) = (1/3) + (1/3) = 2/3. Again, we can bear this out by referring back to the above trees. For example, if you play the game 60 times, you’ll see a goat revealed in 40 out of 60 games you play, which is 40/60 = 2/3. So, 2/3 of the time, Hall will reveal a goat. That’s P(H).2

Posterior probability for event C = P(C|H) = probability you chose the car given that Hall randomly reveals a goat = this is what we’re trying to find.

Likelihood = P(H|C) = probability that Hall randomly reveals a goat given that the car door was chosen = obviously 1.

We now plug all this in:

\displaystyle P(C|H) =\frac{P(H|C)\, P(C)}{P(H)} =\frac{(1)\times (\frac{1}{3})}{(\frac{2}{3})} = \frac{1}{2} \

Do you find it as strange as I do that, even if this game is only ever played once in the history of all possible worlds, if Monty Hall forgets where the door is but pretends like he knows where it is to save face, and he gets lucky and reveals a goat, and you switch and win… it will, in all outward respects, look exactly like a game in which he remembers where the car is, but the “real” probabilities will differ, simply due to a change in Hall’s cognitive state.

4. Forgetful Monty Hall (One Million Doors)

Here again Monty Hall forgets where the car is, but must open 999,998 doors without accidentally revealing the car. Assuming he succeeds, my initial intuition is that you must have made it easier for him to reveal only goats by choosing the car door. After all, while the probability of picking the car door is extremely low, the probability that Hall, after you’ve chosen a goat door, will avoid the car door in 999,998 random tries must be even lower—much lower. So, I overwhelmingly feel that you must have chosen the car and so should stay, not switch!

But this is wrong! Although it’s true that, when you choose a goat door, Hall has a very low probability of not revealing the car, it’s not quite as low as your initial probability of choosing the car door. But it doesn’t start out low. When he opens the first door, he has a 999998/999999 probability of revealing a goat. For the next door, it’s 999997/999998, and so on. Once he has ten doors left closed, he still has a 9/10 chance of revealing a goat. Notice that, when you multiply those first two fractions together, the 9999998’s cancel out, leaving a denominator of 999999; this will continue successively, until we’re left with 1/999999, which is the probability of revealing only goats. For example, if we had ten doors, nine of which hide goats and one hides a car, the probability of selecting one door at a time and revealing only goats will be:

\displaystyle \Big(\frac{9}{10}\Big)\Big(\frac{8}{9}\Big)\Big(\frac{7}{8}\Big)\Big(\frac{6}{7}\Big)\Big(\frac{5}{6}\Big)\Big(\frac{4}{5}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{2}{3}\Big)\Big(\frac{1}{2}\Big) = \frac{1}{10}

Now imagine doing that starting with 999998/999999. This is a bulky way to work, but thinking about it helps get the intuition right.

So, the probability that he avoids the car is higher than the probability that you chose the car to begin with. But only slightly. After all, Hall has 999,998 opportunities to accidentally reveal the car, so the probability that he does so is 999998/999999. We can see this quickly by simply recognizing that it’s the complement to choosing all goats: 1 – 1/999999 = 999998/999999. We can also use the same sort of bulky math as above to see that lots of cancellations happen. In other words: Hall opens the car door on first guess (with probability 1/999999); OR Hall opens a goat door on his first guess then a car door on his second guess (with probability 999998/999999 × 1/999998 = 1/999999)… and so on. You take the sum of those, results to get 1/999999 + 1/999999 + …. all 999,998 times he opens a door, yielding 999998/999999.

We can get to these numbers even faster by imagining that right at the start Hall chooses one of the 999,999 doors with the hope that it’s the car, with probability 1/999999, and says, “Open the other 999,998 doors.” This would be like randomly pulling one card from a well-shuffled deck while hoping for the A♥, versus pulling all but one card from that deck while hoping that the A♥ is the card left behind; both procedures must have the same probability of yielding the A♥: 1/52.

Now, what matters here isn’t that the probability of his avoiding the car (given that you chose the goat) is so very low (if not as low as I first intuited). What matters is that, in the world we’re in, Hall did avoid the car, and that is the knowledge we must condition on. When we do that, we once again have a probability of just 1/2 of winning by switching!

I’ll jump now to the frequency and probability trees to show that:

monty-hall-problem_05

 

monty-hall-problem_06

The key insight here is that, just in the case of three-door Forgetful Monty Hall game, the numerator in the “choose goat door” fraction cancels against the denominator in the “Hall reveals only goats” fraction, leaving behind a fraction equal to the “choose car door” fraction. This makes the odds 1-1 and thus the probability for either outcome 1/2. We’ll see this again when solving with Bayes’ theorem. I’ll use the same parts as in the three-door version, but with updated numbers.

Prior probability for event C = P(C) = probability of choosing the car (without conditioning; i.e., at the offset) = 1/1000000.

Prior probability for event H = P(H) = probability that Hall randomly reveals only goats, whether or not you chose the car door. Again, this is the probability we’d assign to Hall revealing a goat, prior to starting the game. In other words, it’s the probability of the evidence (i.e., seeing a goat revealed) arising in this game, period.

We can think this through just as we did above. Hall reveals a goat whenever the car door was chosen, which is 1/1000000 of the time. And he reveals only goats 1/999999 of the time you chose goat door, which you do 999999/1000000 of the time; multiply those two probabilities together to get 1/1000000 (the 999,999’s cancel each other out because 999999/999999 = 1). We now add together the probability that we see only goats when you chose the car door and the probability that we see only goats when you chose a goat door, to get P(H). Like so (where G is the event that you choose a goat door):

\displaystyle P(C)\, P(H|C) + P(G)\, P(H|G) = \

\displaystyle  \Big(\frac{1}{1000000}\Big)(1) +\Big(\frac{999999}{1000000}\Big)\Big(\frac{1}{999999}\Big) = \

\displaystyle \frac{1}{1000000} + \frac{1}{1000000} = \frac{2}{1000000} \

So, 2/1000000 of the time, Hall will reveal a goat. That’s P(H). Notice that as in the three-door case, this is twice the probability of choosing the car.

Posterior probability for event C = P(C|H) = probability you chose the car given that Hall randomly reveals only goats = this is what we’re trying to find.

Likelihood = P(H|C) = probability that Hall randomly reveals only goats given that the car door was chosen = 1.

Now just plug in to get 1/2:

\displaystyle P(C|H) =\frac{P(H|C)\, P(C)}{P(H)} =\frac{(1)\times (\frac{1}{1000000})}{(\frac{2}{1000000})} = \frac{1}{2} \

It occurs to me that these Forgetful Monty Hall problems are similar another classic probability problem. One hundred passengers are waiting to board a 100-seat airplane. Each passenger has an assigned seat, but the first passenger who boards doesn’t even look at his boarding pass. He just randomly chooses a seat to sit in. The next 98 passengers to board each sits in their assigned seat if possibly; otherwise, they choose a seat at random. What’s the probability the last passenger to board gets to sit in their assigned seat?

5. Lazy Monty Hall (Three Doors)

This variation is a slightly altered version of a question from the wonderful textbook Introduction to Probability, by Joseph Blitzstein and Jessica Hwang (page 81). It’s also available in the 3rd homework assignment of Blitzstein’s Stats 110 course website, where you’ll also find video lectures. Highly recommended!

This is just like the classic Monty Hall problem, but with a slight twist. When Hall has a choice between opening Doors 2 or 3 to reveal a goat, he more often opens Door 2 because he’s usually standing closer to it. This happens 3/4 of the time, in fact. Maybe this is more efficient than lazy—but maybe it’s not best practice, as it gives the contestant a clue. What I like most about this variation is that it demonstrates the importance of whether Hall favors a particular door, even though its importance is rarely brought up in the classic version of the problem.

Suppose you’ve chosen Door 1 and Hall reveals a goat behind Door 2. What’s the probability that switching to Door 3 wins, given that, when you’ve chosen Door 1, Hall opens Door 2 with probability 3/4?

We are concerned here strictly with those worlds in which you choose Door 1 and then Hall opens Door 2 to reveal a goat. Once we isolate those worlds, we then just need to calculate the proportion of those worlds in which switching wins. When choosing Door 1, you choose the car door 1/3 of the time, after which Hall opens Door 2 to reveal a goat 3/4 of the time. Switching always loses in this scenario. So, you lose by switching (1/3)×(3/4) = 1/4 of the time. This means if you run this game 60 times, choosing Door 1 each time, 20 times the car will be behind that door, and 15 of those times you’ll see a goat behind Door 2 and switching will lose.

But 2/3 of the time, Door 1 will conceal a goat. In that scenario, Hall is forced to reveal the other goat. We assume that Door 2 and Door 3 are equally likely to hide the other goat. So, in 1/2 of these worlds, Hall opens Door 2 to reveal a goat, and switching wins. So, switching wins (2/3)×(1/2) = 1/3 of the time Door 2 reveals a goat. This means if you run the game 60 times, choosing Door 1 each time, 40 times a goat will be behind Door 1, and 20 of those times you’ll see a goat behind Door 2 and switching wins.

Now, you don’t know whether there’s a car or goat behind Door 1. But you do know you’re in a world in which you choose Door 1 and then Hall opened Door 2 to reveal a goat. You’ll find yourself in such a world (1/4) + (1/3) = 7/12 of the time. And the proportion of time you switch and win is (1/3)/(7/12) = 4/7. That is, a policy of switching will win 4/7 of the time in this scenario. We see this with the frequencies as well. That is, of the 20 + 15 = 35 games in which you’re in this scenario, 20 out of 35 of them, you win by switching; and 20/35 = 4/7.

Here are the trees:

monty-hall-problem_07

 

monty-hall-problem_08

I’ll also pop it into Bayes’ theorem, in order to find the probability of event G (i.e., the event that you chose a goat door) given event H (i.e., the event that Hall opens Door 2 to reveal a goat). This is the same as finding the posterior probability for winning by switching conditional on Hall opening Door 2 to reveal a goat. This time I won’t list out the parts, as what I’m doing should be clear by now:

\displaystyle P(G|H) =\frac{P(H|G)\, P(G)}{P(H)} = \

\displaystyle \frac{P(H|G)\, P(G)}{P(H|G)\, P(G)  +  P(H|C)\, P(C)} = \

\displaystyle \frac{(\frac{1}{2})(\frac{2}{3})}{(\frac{1}{2})(\frac{2}{3}) + (\frac{3}{4}) (\frac{1}{3})} = \

\displaystyle \frac{(\frac{1}{3})}{(\frac{1}{3}) + (\frac{1}{4})} = \

\displaystyle \frac{(\frac{1}{3})}{(\frac{7}{12})} = \frac{4}{7} \

We see, then, that, given this knowledge about Hall’s door-choosing tendencies, your confidence in having chosen the car should increase from about 33% to about 43% when you see a goat emerge from Door 2, since it’s now about a 57% probability that you chose a goat door. This may at first seem strange. It did to me, anyway. Why should it matter which door Hall opens when both available doors conceal a goat? But it makes intuitive sense on reflection. Hall favors opening Door 2 over Door 3 when he can, and has the greatest opportunity to express that preference when both doors are available.

I find that turning the knobs here helps solidify this intuition. Suppose Hall chooses Door 2 every time he’s given the option. This means all 20 times Door 1 hides the car and you choose Door 1, you’ll see Door 2 opened. That’s half of the times you’ll see Door 2 opened in this variation overall. So, seeing a Door 2 goat should take your confidence in having chosen the car door from 1/3 to 1/2. That’s as high as seeing a Door 2 goat can take you, in fact. But notice what this means for when Hall opens Door 3. It’s then 100% certain that you chose a goat door and that the car is behind Door 2, because Hall must have been forced to open Door 3!

We can test these specific solutions against the general solutions given in the book and Homework PDF, which conclude in generalized formulas rather than specific numbers. All we have to do to get there is to assign p (where p is some number from 1/2 to 1, inclusive) as the probability that Hall selects Door 2 when given the opportunity, and plug that, rather than 3/4, into Bayes’ theorem above. The resulting formula outputs the probability function that switching wins.

I modeled the problem a little differently than in the book, but it works out the same:

\displaystyle \frac{(\frac{1}{2})(\frac{2}{3})}{(\frac{1}{2})(\frac{2}{3}) + (p) (\frac{1}{3})} = \frac{(\frac{1}{3})}{(\frac{1}{3}) + (p) (\frac{1}{3})} = \frac{(\frac{1}{3})}{(\frac{1}{3})(1+p)} = \frac{1}{1+p} \

If we plug in 1 for p, we get, just as above:

\displaystyle \frac{1}{1+p} = \frac{1}{1+1} = \frac{1}{2}

What if Hall opens Door 3? What’s the probability there of winning by switching? All we have to do to find that is replace the p in our new generalized formula with the probability of Hall not choosing Door 2 when given the option, which is 1–p, and then substitute the specific probability we wish to test of Hall’s opening Door 3 when Door 2 is an option; in this case, that specific probability is 0:

\displaystyle \frac{1}{1+(1-p)} = \frac{1}{1-0} = \frac{1}{1} = 1

Again, just as above, if Hall ever chooses Door 3 when his total preference is for Door 2, there’s a 100% chance of winning by switching.

6. Monty Hall Has a Tell (Three Doors)

This time, Monty Hall gives off a tell. You know this because you’ve watched him host this game thousands of times. To be clear, you haven’t watched Let’s Make a Deal thousands of times, the show the real Monty Hall hosted for many years. The real Monty Hall has said he never gave contestants the chance to switch.3 But if he did offer the chance to switch, you would be in the world Marlyn vos Savant described, and which I describe at the beginning of this post, and an “always switch when possible” strategy will win 2/3 of the time (and, of the times he didn’t offer you any alternatives, you’d win 1/3 of the time and lose 2/3 of the time).

At any rate, you’ve watched the tidily idealized, “classic Monty Hall problem” version of this game thousands of times, and have observed that, when the contestant has initially chosen the car door, Hall often subtly arches his left eyebrow at least three times between revealing the goat and offering the option to switch. Hall gives this tell 95% of the time the car door is chosen, in fact. On the other hand, he sometimes gives false tells—i.e., arches his left eyebrow in this way when the contestant has chosen a goat door. He does that 10% of the time a goat door is chosen.

Suppose you spot the tell. What’s the probability of winning by staying? In other words, what’s the probability you’ve chosen the car?

I’ll go right to frequencies for this one, but will also show a probability tree below. If you play 60 times, you’ll select the car door 20 times. You’ll see the tell 19 of those time, and will get no tell 1 of those times. But if you choose a goat door, which will happen 40 times, you’ll see the tell 4 times, and get no tell 36 times. We are concerned here with the worlds in which a tell appears. There are 19 + 4 = 23 such worlds. In 19 out of those 23 worlds, staying wins; that’s 19/23 ≈ 83%. In 4 out of those 23 worlds, staying loses; that’s 4/23 ≈ 17%. So, if you see the tell, you should update your confidence in having chosen the car from about 33% to about 83%. Which is to say you have about an 83% probability of winning by staying.

The tree diagrams:

monty-hall-problem_09

 

monty-hall-problem_10

And here I use Bayes’ theorem to work out the probability of winning by staying, given that Hall gives off a tell. I trust the notation I choose is clear by now.

\displaystyle P(C|T) =\frac{P(T|C)\, P(C)}{P(H)} = \

\displaystyle \frac{P(T|C)\, P(C)}{P(T|C)\times P(C)  +  P(T|G)\, P(G)} = \

\displaystyle \frac{(\frac{95}{100})(\frac{1}{3})}{(\frac{95}{100})(\frac{1}{3}) + (\frac{5}{100}) (\frac{2}{3})} = \

\displaystyle \frac{(\frac{19}{60})}{(\frac{19}{60}) + (\frac{4}{60})} = \

\displaystyle \frac{(\frac{19}{60})}{(\frac{23}{60})} = \frac{19}{23} \

This only applies, of course, if you’re lucky enough to get a tell. How often does that happen? Not as much as we’d like. Above, we saw the tell 23 out of 60 games; that’s 23/60 ≈ 38% of games. Maybe with more research we could find other tells. Maybe Hall is more likely to give a tell if you wear a red shirt or look depressed or give him a long hug or laugh at his jokes or if he’s low on time (because the long-hugging contestant wouldn’t stop laughing at his jokes). Maybe he has other tells. This is starting to sound more like real life.

7. Monty Hall: So on and So Forth

There are, of course, as many Monty Hall variations as we care to imagine. Some more complicated than others—involving multiple cars; multiple chances to switch; and, I don’t know, how about doors that open into numbered chambers that permute in cycles with some regularity (ugh, not today); and whatever else you care to delight (or torture) yourself (or your students) with. A quick Google search brings up examples such as this article I just found and skimmed: “Higher Variations of the Monty Hall Problem.”

8. More Counterintuitive Conditional Probability Problems

Here are some other conditional probability problems I’ve written about that have intuition pitfalls similar to those above. Listed in reverse chronological order:

(Strange?) Probability with Playing Cards
A fun little problem showing that some seemingly insignificant knowledge can matter (if only a wee bit).

Bertrand’s Box Paradox (with and Without Bayes’ Theorem)
Some classic conditional probability from 1889.

Two-Child Problem: A Brief Explanation
There’s a longer explanation of this problem linked below.

A Secret Intention’s Strange Effects on Probability
This one weirds me out. Maybe you won’t find it strange.

Cancer Screening: An Easier Counterintuitive Conditional Probability Problem (with and Without Bayes’ Theorem)
A commonly encountered problem—and rightly so—in discussions of just how wrong things can go when we don’t condition properly. The results here are quite surprising until closely considered.

Two-Child Problem (when one is a girl named Florida born on a Tuesday)
This is a longer exploration of the Two-Child problem.

Counterintuitive Dice Probability: How many rolls expected to get a 6, given only even outcomes?
This is by far my most consistently popular post on probability and—since publishing it nearly a year ago—on this website overall.


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Further Reading

Footnotes:

  1. The problem first came to public attention in 1990 when it was sent in to Marilyn vos Savant’s “Ask Marilyn” Parade magazine column. She printed the letter and responded with the correct answer. Around 10,000 readers, supposedly including nearly 1,000 PhD’s (some of whom were math professors), wrote in to say she got it wrong. And they weren’t always nice about it. These days, people “in the know” seem to take for granted—as super obvious—that the answer is 2/3. I’m one of those people. But I think it’s very much worth remembering and ruminating on the fact that, not so long ago, even Paul Erdős, one of the brightest probability theory mathematicians around, was reportedly unconvinced of the 2/3 answer until he saw a computer simulation. We shouldn’t take probability for granted, especially when it’s counterintuitive. And even more especially when it’s not! No one’s intuition is safe! For a deeper look into the Monty Hall problem’s history, see its Wikipedia entry.
  2. This is technically an application of a rule known as the law of total probability. I’m not going to delve into that topic, but will say that this can be formally represented here as P(H|C)P(C) + P(H|not-C)P(not-C), where “not-C” is the event that you didn’t choose the car door, which is to say the event that you chose a goat door. This is, just as we’ve already thought through: (1)(1/3) + (1/2)(2/3) = (1/3) + (1/3) = 2/3. You’ll often see Bayes’ theorem represented with the denominator already broken down in this way.
  3. In the linked video, he first says he doesn’t recall ever offering the chance to switch, and then says he never did. According to obituaries I’ve seen, he hosted the show more than 4,700 times!

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