The Monty Hall problem, but different

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 №1, and the host, who knows what’s behind the doors, opens another door, say №3, which has a goat. He then says to you, “Do you want to pick door №2?” Is it to your advantage to switch your choice?

It is mostly clear that:

  • you prefer to win the car (you will sell it and buy a lot of goats)
  • you need to choose which strategy yields a better conditional probability of winning the car
  • the host does not want to let you win the car

However, why does the host open another door? This is not clear.

If I was the host, and I did not want you to win the car, I would do the following (0): just force you to keep your first choice (if it was wrong), or show another door and persuade you to switch (if your first choice was correct). In that situation, the answer to the question is clearly NO, don’t switch. And that answer seems consistent with the problem statement.

But let’s assume (*) that he has to help you, by opening one of the three doors, and he has to give you another choice (out of the three doors) after revealing. Of course the host will never show the door with the car, and you will never pick the door with a revealed goat.

Even after this assumption, there are many possible interpretations, which I have asked about in a Twitter quiz. In the following situations, is it better to switch, better not to switch? Or maybe both are equally good, or maybe none of the above (it depends, etc.)? Most people on Twitter did not get the answer right (in most cases), although the statements has to fit in two tweets and thus they were a bit less precise (which kind of proves the point that the problem is ambiguous).

(1) The host randomly opens of of the two doors which do not contain the car (and you know it).

In this case, assuming you have picked door #1, we have the following possibilities:

a) the car is #1, the host opens #2 (does not meet the condition)

b) the car is #1, the host opens #3 (should not switch)

c) the car is #2, the host opens #1 (does not meet the condition)

d) the car is #2, the host opens #3 (should switch)

e) the car is #3, the host opens #1 (does not meet the condition)

f) the car is #3, the host opens #2 (does not meet the condition)

By the definition of the conditional probability, you win in 50% cases if you switch and in 50% cases if you do not switch. (Above we have assumed that the condition is that you have picked #1 and the host has opened #3. You can also relax the assumption that you have picked door #1, or that the host opened #2 — he just did no open the door that you have picked; the result is the same.) So both are equally good.

(2) The host is lazy: he usually opens the door you have picked, unless you picked the correct door — then he chooses randomly one of the other doors. And you know it.

In this case, it is quite obvious that you should not switch.

(3) You believe that the host believes that you believe Marilyn vos Savant. Marilyn vos Savant says that you should switch.

In the case when you have picked a wrong door (#1), the host has a choice of showing you the other wrong door (say #2) or your door (#1). Knowing that you are going to switch, he sees that you will win if he opens #2, and win in 50% cases if he opens #1. So he will open your door if you were wrong. If you were correct, he will obviously open one of the wrong doors (and if you believed Marilyn vos Savant, you will switch and lose).

So: if you switch, you always lose. If you don’t switch, you always win. Better not to switch.

(4) The host acts according to the Nash equilibrium strategy (and you know it).

The Nash equilibrium is a part of (randomized) strategies where neither side can achieve a better outcome (higher probability of winning) by changing their strategy.

In our case, you can clearly achieve 50% probability of winning if you randomly decide to switch or not. The host also can clearly achieve 50% probability of winning by using strategy (1). This pair of strategies is a Nash equilibrium (the Nash equilibrium need not be unique).

However, let p be the probability of you winning if you switch (under the condition that the host has opened a different door), and q is the probability of you winning if you don’t switch (under the same condition). Clearly, p+q=1, so if pq, one of them is greater than 1/2. If the host opened your door, you also win with probability 1/2, so you have a strategy winning with probability greater than 1/2. Which means that the host’s strategy was not a part of the Nash equilibrium.

Thus, if the host acts according to the Nash equilibrium, both are equally good.

(5) The host does what he thinks is their best to not let you win the car (you know it, but you do not know what he thinks is the best).

The Nash equilibrium strategy is the most safe, but not necessarily the best. For example, in Rock-Paper-Scissors, the Nash equilibrium strategy says that you should pick every choice with probability 1/3. However, there are tournaments of computer programs playing Rock-Paper-Scissors, and apparently the best programs win about 80% games! If you are good at predicting what your opponent will do, you will do better than the Nash equilibrium.

Since, according to the Twitter quiz, the overwhelming majority of people believe that it is better to switch strategy (3) appears to be better than Nash equilibrium (and then you should not switch). However, we would need more testing to be sure of the answer. (Most sources seem to claim that most people believe that both are equally good, so Twitter might be a strange sample.) The intended answer was “none of the above”.

(6) The host always open the door which is to the right from the car (or the leftmost door if the car is in the rightmost door), and wonders when the guests will notice. Well, you have noticed this.

This has been placed on the Twitter quiz as an another case where “none of the above” is correct. Whether you should switch or not, it depends on the placement of the doors.


The Monty Hall problem is famous for confusing people. According to Wikipedia: “Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant’s predicted result.”.

However, it seems the problem is confusing because the statement is actually ambiguous! In my opinion, the most natural interpretation is (1), where switching is not beneficial. The original discussion can be found here. Marilyn vos Savant assumes (*), and she also assumes (7) that the host never opens the door which has been picked by you. Her answer is correct under this assumption, however, neither the original problem statement, nor her answer make this clear. The three PhDs seem to have assumed (1): they are rude, but are also correct under this assumption. Marilyn vos Savant writes another explanation, but she still does not clarify her assumptions, so people are still arguing with here (even more rudely). In the next explanation, her interpretation can be seen clearly form the experimental setup she suggests — and people start to agree with her. So it’s like the dress, people are interpreting the same thing differently, but everyone seems to be correct according to their interpretation. The only weird thing to me is that nobody seems to have understood the source of the confusion (even people who have changed their own opinion, or people who had easy access to people who think otherwise).

Likewise, in the Erdős case, I think it is more likely that Andrew Vazsoniy did not state the problem clearly, than that Paul Erdös understood the problem and was wrong. Especially that when Ron Graham later explained the solution to Paul Erdös, Andrew Vazsoniy could not understand this explanation.

Vos Savant and Vazsoniy mention that computer simulations prove that they were right. When I have heard about the Monty Hall problem for the first time, I believed that the answer “both are equally good” was correct. And I have created a computer simulation —I do not remember the results and I do not think I could find that simulation, but I think that the problem was not stated clearly, I have assumed interpretation (1), written a program according to that interpretation, which has proven me right. Here is an example program which works according to interpretation (1). (The next time it was stated more clearly and I agreed it was better to switch.)

Mathematics, game development, art, roguelikes, hyperbolic geometry. Sometimes all at once.