Today’s entry on Jeff Atwood’s weblog, “Coding Horror” talks about the time-honored probability puzzler known as the Monty Hall problem. If you are unfamiliar with the problem, it deals with devising the optimal strategy for playing a game that was common on the game show “Let’s Make a Deal“, starring Monty Hall, and has a solution that is commonly perceived as unintuitive.

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 No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? (Whitaker 1990)

This, much like airplane on a treadmill, and 0.999… == 1 are common topics of long, drawn-out arguments on the Internet. But what I want to talk about is not the problem itself (which has been done to death), or Jeff’s post. Rather, I want to discuss a variant problem humorously called the “Monty Fall Problem” proposed by Professor Jeffrey S. Rosenthal of the University of Toronto, which is included in an article titled “Monty Hall, Monty Fall, Monty Crawl”, which was linked from the Coding Horror article.

The Monty Hall problem in its most common form has never really seemed all that unintuitive to me. Admittedly, yes, when I first heard the problem my immediate reaction was to claim that the probability of winning was 50-50, but the logic behind why that is not correct was clear when it was explained.

When I explain the solution to this problem to others, I use a variant of what Rosenthal refers to as “The Shaky Solution”. He gives it that derogatory name because the logic employed in that explanation is not readily generalizable to variants of the Monty Hall problem. Indeed, the main subject of his article is to present a method of reasoning that does generalize to variants. But the explanation I use (which is perfectly valid for the vanilla problem) is as follows:

In the Monty Hall problem, the option to switch is equivalent to the option of trading the contents of your original door for the contents of

bothother doors. From this it is obvious that picking two doors (switching) improves your chances of winning over picking only one.

Put another way, the choice to switch asks you to wager on whether you were right or wrong, and when you made your original selection, you had a 1/3 chance of being right.

So that is straightforward enough. I read Rosenthal’s article, and most of the other examples seemed similarly straightforward when they were explained. But the solution given to the “Monty Fall” problem made me pause and briefly doubt that the argument he was using was necessarily correct. The Monty Fall problem is stated as:

In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?

And his solution, which claims that there is an equal probability of winning whether you switch or not, is given as:

In the Monty Fall problem, suppose you select Door #1, and the host then falls against Door #3. The probabilities that Door #3 happens not to contain a car, if the car is behind Door #1, #2, and #3, are respectively 1, 1, and 0. Hence, the probabilities that the car is actually behind each of these three doors are respectively 1/2, 1/2, and 0. So, your probability of winning is the same whether you stick or switch.

The question that came to mind is this: The Monty Fall problem seems to give no information beyond that which is available in the traditional Monty Hall problem. In particular, both problems seem to say only that a door other than the selected door was opened, and a goat was revealed. If the game is the same and the information is the same, then it would follow that the probabilities involved cannot possibly differ.

The key is, of course, that the information is *not* the same, and I will now go through the reasoning I used to convince myself that Rosenthal’s solution to the Monty Fall problem is indeed correct.

First, the problem statement as presented by Rosenthal elides what is actually a key piece of information. The problem says that we, the player, are operating under the assumption that Monty fell, and that the door he opened as a result of his spill was not the door that we selected, and also did not contain a car. But wait – Monty’s fall itself is implicitly described as a probabilistic event! Why are we allowed to ignore the other possibilities?

Let’s work this through, using Rosenthal’s own method of applying the Proportionality Principle. Rosenthal’s generalizable method for the Monty Hall problem is, in essence: **For each door, calculate the probability that the described situation occurred given that the car is behind the door in question.** Then, if you normalize the probabilities, the probability of winning by *not* switching is the probability that the described situation occurred given that the car was behind the door you originally selected.

But here, there are actually four possible “situations”. There are two binary choices associated with Monty’s fall. Either the door he opens is the player-selected door, or it is not. And either the door he opens contains the car, or it does not. In lieu of any other stipulations, I’ll make the reasonable assumption that the door that Monty falls into is chosen uniformly at random from all three available doors.

Then, the probability that he opens the player selected door is 1/3 (since the player selects only of three doors), and the probability that he opens the car door is also 1/3 (since the car is placed behind only one door). From this, we can draw up a table of the probability of being in any given situation after the fall:

Reveals Car | Reveals Goat | |
---|---|---|

Opens Selected | 1/9 | 2/9 |

Opens Other | 2/9 | 4/9 |

The values given are simply the products of the probabilities of the two events in the row and column headers. Rosenthal’s explanation is limited to the case of Opens Other / Reveals Goat, which I will abbreviate as OO/RG.

Now, to use Rosenthal’s method, we must determine, for each situation, the door-by-door probability that each situation occurs, given that the car is behind the door. Without loss of generality, let’s denote the door we the player choose as Door #1. If Monty falls into a door other than yours, we call that Door #3.

We can compute this in two steps. First, we list the situation/door combinations, and mark a 1 when the combination is *possible* and a zero when it is not.

Door 1 | Door 2 | Door 3 | |
---|---|---|---|

OS/RC | 1 | 0 | 0 |

OS/RG | 0 | 1 | 1 |

OO/RC | 0 | 0 | 1 |

OO/RG | 1 | 1 | 0 |

This gives us a description of all possible car-placement / situation pairs. For example, the intersection of the Opens Selected / Reveals Goat situation and the Door 1 car placement has a zero because it is impossible that the selected door (Door #1 by assumption) has the car and is also opened to reveal a goat.

Now, if we multiply the contents of each row by the probability that the situation occurs, we get this:

Door 1 | Door 2 | Door 3 | |
---|---|---|---|

OS/RC | 1/9 | 0 | 0 |

OS/RG | 0 | 2/9 | 2/9 |

OO/RC | 0 | 0 | 2/9 |

OO/RG | 4/9 | 4/9 | 0 |

And if we normalize these probabilities (so that they add to one) we get

Door 1 | Door 2 | Door 3 | |
---|---|---|---|

OS/RC | 1/15 | 0 | 0 |

OS/RG | 0 | 2/15 | 2/15 |

OO/RC | 0 | 0 | 2/15 |

OO/RG | 4/15 | 4/15 | 0 |

What this table describes is the complete, original state of the Monty Fall problem *before* Monty falls. For example, there is a 2/15 chance that the car has been placed behind Door #2 and also that Monty will fall into Door #1 (the player-selected door) to reveal a goat.

The important point is that this table is based on **zero information** beyond what is known in the vanilla Monty Hall problem. It is just as if we are speculating on what might happen if Monty were to fall. Therefore, we should expect that the probabilities implied by this table should be identical to the probabilities of the orignal problem. In particular, this table should tell us that if we must commit to a strategy before the game begins, the strategy of “pick a door and stick with it” works 1/3 of the time, and “pick a door and switch” works 2/3 of the time.

Does it? Add up the columns to find out the probabilities of the car being behind each door.

Door 1 | Door 2 | Door 3 | |
---|---|---|---|

Total | 5/15 | 6/15 | 4/15 |

Now, recall that we “Door #1” is just a label for “the door the player selected”. According to my version of the so-called “Shaky Solution” logic, the choice of whether to switch or not amounts to the choice between door 1 or both of doors 2 and 3. If we add the probabilities (and reduce the fractions), we get:

Door 1 | Doors 2 & 3 | |
---|---|---|

Total | 1/3 | 2/3 |

Which is exactly what one would expect. We have added no information to the problem, and so we have come up with precisely the same solution, even with all that mumbo-jumbo about what might happen if Monty were to fall.

But the key realization is this: While the knowledge that Monty *will* fall gives you no additional information, the knowledge of exactly *how* he fell (once he falls), does indeed give you new information. In particular, **knowledge of how exactly Monty fell eliminates 3 of 4 situations from the speculative probability table above**.

From a numerical standpoint, when Monty actually does fall into Door #3 and reveals a goat, the new information that you have been given is that all car-placement / situation pairs other than Opened Other / Revealed Goat are impossible, and should have their probabilities in the full table updated to zero. When you recompute from that point, you end up with the 1/2 probability of winning by staying that Rosenthal gives.

From a logical standpoint, when Monty actually does fall into Door #3 and reveals a goat, the player can infer that certain placements of the car were more likely to cause the random event of Monty’s tumble to reveal exactly what it did. That constitutes new information through indirect observation. A non-rigorous analogy would be that hearing your doorbell ring constitutes new information about whether or not someone is standing outside your door. It is *possible* that it rang of its own volition, or that a child many yards away just struck your ringer with a baseball, but it is *more likely* that someone is standing there ringing it.

In summary, the difference between the vanilla Monty Hall problem and the Monty Fall variant is that in both problems it is known *a priori* that Monty will attempt to open a door other than your selection which contains a goat. But in the Monty Fall problem, the extra information is that Monty failed to do this in a very specific way, and that **his exact method of failure is something that is strongly influenced by the position of the car that you are trying to find.**

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Whoa there:

“And if we normalize these probabilities (so that they add to one) we get”

In what world is it acceptable to “normalize” probabilities so they add up to 1 simply to get out if a situation where you must have made some invalid assumptions?

Care to share what you think the invalid assumptions are?

Since probability normalization is valid when you have a set of numbers describing all possible situations where the magnitude of the numbers is proportional to the probability of occurrence, I doubt you’re objecting to the normalization step itself. You must either think that the four situations I enumerated do not describe all possibilities, or you think that the probabilities that I assigned to those situations are not correct, so, please elaborate on what you think is wrong.

Hi,

First of all, thank you for your very interesting website.

I had not heard about the Monty “Fall” problem and after reading Prof. Rosenthal’s explanations I am surprised.

Both problems seemed different to me, though:

In the “Hall” problem there is one possibility for game-over, while in the “Fall” problem there are two steps.

Hall:

1.You choose

2.Monty opens knowingly one wrong door

3.You decide whether or not to change

4.Game over

Fall:

1. You choose

2. Monty falls and opens accidentally one of the two remaining doors.

3.If he opned the car door, then Game over

4.If you are still alive, then you decide whether or not to change

5.Game over

So I wouldn’t address it from the point of information ammount. It’s easier I think.

Suppose the game behaves exactly according to probability:

We have in the summer 18 tv-shows.

in 1/3 of the shows (6) the candidate has chosen the prize door from the very beginning.

Monty selects one of the two other doors randomly(could be coin tossing). In 1/3 of the games he will open an empty door, regardless which one he chooses (falls into).

In the 12 (2/3) remainig shows he has 50% probability to choose the empty door, so 6 times (1/2 *12) he will send you home. Here is the difference.

So you happen to be alive in one of the remaining 12 shows; chances are that 6 times you selected correctly and loose by changing and six times you selected wrong and win by changing.

Funny, it’s really 50:50, though initially I thought differently.

When Monty knowingly opens an empty door he is eliminating those 6 possibilities of sending you home, he is therefore increasing you odds of winning to 2/3 if you change. He is joining a two step issue into a one step game with this elimination.

Gary is correct. While you can normalize when appropriate, it is not appropriate in this case, because, as Gary said, it is erasing an error you made.

Your binary table of whether things can or cannot occur has probabilities so simply using 1 or 0 skews the results. Your final result should show that the prize is equally as likely to be behind any of the three doors, however, it is indicating that the prize is most likely to be behind door #2 which we know cannot be true.

The correct table looks like this:

Prize Behind Door #

1 2 3

osrc 1/9 0 0 1/9 chance host opens selected car

osrg 0 1/9 1/9 2/9 chance host opens selected goat

oorc 0 1/9 1/9 2/9 chance host opens non-selected car

oorg 2/9 1/9 1/9 4/9 chance host opens non-selected goat (monty fall)

1/3 1/3 1/3 1/3 chance car is behind each door

From there you can see that you have an equal chance of selecting the car or not selecting the car (2/9 vs 1/9 + 1/9).

This does not show the monty hall scenario. The Monty hall scenario always reaches oorg with 100% probability.

That is

Prize Behind Door #

1 2 3

osrc 0 0 0 0 chance host opens selected car

osrg 0 0 0 0 chance host opens selected goat

oorc 0 0 0 0 chance host opens non-selected car

oorg 1/3 1/3 1/3 100% chance host opens non-selected goat (monty hall)

1/3 1/3 1/3 1/3 chance car is behind each door

In the Monty Fall, Monty’s door is independent of the prize door and oorg is dependent on the prize door.

In the Monty Hall, Monty’s door is dependent on the prize door and oorg is independent of the prize door.

That is:

Monty Fall

P(C and A) = P(C | A) * P(A)

Monty Hall

P(C and A) = P(A) * P(C)

If oorg occurs (which is 4/9 of the time in the Monty Fall) then you can normalize the probabilities:

Prize Behind Door #

1 2 3

OORG (Monty Fall) 1/2 1/4 1/4

OORG (Monty Hall) 1/3 1/3 1/3

Now you can see that half the time during the Monty Fall scenario you will choose the prize and half the time you won’t so it doesn’t matter if you switch. But in the Monty Hall scenario you only have a 1/3 chance of choosing the prize so switching gives you a 2/3 chance of winning.

The only difference is that in one Event A and Event B are independent and in the other they are dependent.