Solutions to probability puzzles can be very counter-intuitive. This is often related to conditional probabilities and Bayesian reasoning, as in the famous Monty Hall Problem.

Monty Hall Problem

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 says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors? (vos Savant 1990)

 Intuitive Solution

“If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2”, according to Robert Sachs, Ph.D. of George Mason University in (vos Savant 1990).

Other Solution

“Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?”, according to Marilyn vos Savant in (vos Savant 1990).

Who is right?

We can solve this problem with Dione by loading a set of data to represent the probability space of all possibilities and learning a Bayesian network for these data. The data look like this:

Assuming the doors are numbered 1, 2 and 3, variable CarDoor represents the door that hides the car, GuessedDoor represents the door that is guessed by the player, and OpenedDoor represents the door that is opened by the host after the initial guess. Variables Stay and Switch indicate whether the player wins the car after staying with their initial guess and after switching respectively.

After learning the network structure, we edit the network to reflect the given dependency structure of the problem, as shown at the right hand of this diagram. At the start of the show, there is a car behind one of the doors, represented by variable CarDoor, and the player makes an initial guess, represented by variable GuessedDoor. The door opened by the host, represented by variable OpenedDoor, depends on the door hiding the car and the player’s initial guess. Depending on the door opened by the host, the player then decides to stay with the initial guess or to switch to the other still closed door, with variables Stay and Switch representing win or loss for each case. We calculate conditional probabilities of each node, as shown here for the nodes that represent the door hiding the car (CarDoor) and the door guessed by the player (GuessedDoor):

At the start of the show, we do not know behind which door the car is, so each door has equal probability of 1/3 to hide the car. We also do not know the initial guess of the player, so each guess has probability of 1/3.

If the initial guess was the door with the car, the game host opens  a random other door to show a goat. If the initial guess was a door with a goat, the host opens the other door with a goat, as shown here for the conditional probabilities of OpenedDoor after an initial guess by the player of Door 1:

The first line in the conditional probabilities table shows that, if the car is behind Door 1 and the player correctly guesses Door 1, the host opens Door 2 or Door 3 with equal probability 1/2. If the player guesses Door 1 and the car is behind Door 2, the host opens Door 3 with probability 1, and if the player guesses Door 1 and the car is behind Door 3, the host opens Door 2 with probability 1.

The probabilities of nodes Stay and Switch now show that the probability of winning the car when staying with the original guess is only 1/3 and the probability of winning when switching is 2/3:



Selvin, S. (1990). A Problem in Probability.

vos Savant, M. (1990). Game Show Problem.

Wikipedia. Monty Hall problem.

1 Comment

Jeff Jo · 17 July 2021 at 5:32 pm

Marilyn vos Savant gave the right result, but not the right solution. In fact, she didn’t offer a solution at all. Just a different way of applying intuition. And the problem with that, is you have to identify what is wrong with the other approach before anyone will change their mind. (And imho, her response has done more to confuse people about probability than any other.)

Sachs is wrong when he says “neither of which has any reason to be more likely.” There is a reason. There was a 1/3 probability that the car was behind door #2, in which case (by the usual assumptions that the host had to open a goat door that the contestant didn’t pick, and that are left out of the problem statement) the host had to open door #3. There also was a 1/3 chance it was behind door #1, but then the host had a choice to open #2 or #3. Assuming he is unbiased, this makes #2 twice as likely as #1.

I don’t blame Sachs for thinking Savant made a mistake, since she did by not including a reason. But he made the same mistake by not looking for one.

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