June 06, 2011
Probability can be enormously puzzling. We have already written plenty on probability confusions, but until now we’ve neglected the most famous of such puzzles, the Monty Hall Problem. It’s time to make amends.
Many people have written many words on the Monty Hall Problem. The mathematician Jason Rosenhouse has even written an excellent book on the problem. Fair enough: when apparently pigeons are better at the Problem than some people, even some mathematicians, it is definitely worthy of attention.
The problem is named after Monty Hall, host of the American TV show Let’s Make a Deal. The game consists of the contestant, let’s call him Joe, being presented with three doors. Behind one door is a terrific prize, a kitten (we really like kittens!), and behind each of the other two doors is a booby prize, a calculator. Joe then chooses a door for Monty to open, winning the prize behind.
The probability that Joe has chosen the correct door is 1 in 3 (assuming that, like all sensible people, Joe prefers kittens to calculators). That’s easy. The confusion comes when Monty delays opening Joe’s chosen door: instead, Monty opens a different door, revealing a calculator. He now offers Joe the opportunity to switch, to instead take the prize behind the other unopened door.
The question is, what now is the probability that Joe has initially chosen the correct door? Does the probability remain at 1/3? Or, given that the kitten can now only be behind one of two doors, has the probability increased to 1/2?
To make it clearer, let’s suppose Joe initially chooses door 1 and then Monty opens door 3. So, the kitten is either behind door 1 or door 2, and we want to know the probability for each.
It is tricky. In fact, using only the information that we’ve so far provided, there is no proper answer.
Probability is based upon information, depending fundamentally upon what you do and do not know. For example, suppose the Maths Masters flip a coin, and Burkard sees the coin land but Marty does not. What then is the probability that the coin landed heads? Clearly, there is no one answer.
For Burkard, it barely makes sense to talk of probabilities, since he simply knows whether the coin came up heads: the probability is either 0 or 1, depending upon what actually happened. However, unobservant Marty has no such information, and so from Marty’s perspective the probability of heads is the usual 1/2. Of course, if Burkard somehow indicates to Marty how the coin landed, the probability for Marty will then change.
The Monty Hall problem is similar. Joe wants to know if the kitten is behind door 1. Initially he knows nothing and the probability is simply 1/3. The question is, what information does Monty Hall impart to Joe about door 1 by opening door 2? The answer depends critically on how Monty decides to open a door and which door to open.
For example, let’s call it Scenario A, suppose Joe knows that Monty is trying to avoid giving away the valuable kitten: Monty will only open a door, and so tempting Joe to switch, if Joe’s original guess is correct. (According to Jason Rosenhouse, the actual instances on Let’s Make a Deal were commonly of this nature). In this scenario there is no question that Joe has chosen correctly, and the “probability” is simply 1.
On the other hand, let’s call it Scenario B, perhaps Monty Hall never attends dress rehearsal, has no idea where the kitten is, and always simply opens one of the doors at random. Then, 1/3 of the time Monty will reveal the kitten. That did not happen to Joe, leaving a 50-50 chance for both door 1 and door 2.
However, the plausible game show context, we’ll call it Scenario C, is that Monty knows exactly where the kitten is, so that he can always safely open a door to reveal a calculator, and he always does so. Furthermore, if Monty has a choice of two doors to open (that is, if Joe happens to have chosen correctly), then Monty chooses one of the two doors randomly.
Scenario C encapsulates the assumptions implicitly or explicitly assumed in the usual version of the Monty Hall Problem. We shall try to convince you that in this scenario the probability that Joe has chosen correctly has not changed: there remains a probability of 1/3 that the kitten is behind door 1.
All that Joe cares about is whether the kitten is behind door 1, the door he chose. The probability of that will only change from 1/3 if Monty has given Joe new information in regard to what’s behind door 1.
It’s definitely possible to give information about door 1 without opening it. For example, if Monty opens both doors 2 and 3, revealing two calculators, then there is no doubt that the kitten is behind door 1. Scenario A also exactly locates the kitten as being behind door 1. Scenario B is a step along this line, giving some but not all information about what might be behind door 1, leading the probability to jump to 1/2.
However, in Scenario C, the usual Monty Hall Problem, absolutely no information has been given as to what’s behind door 1. Wherever the kitten actually is, Monty would have opened one of the other doors at random to reveal a calculator. Since that was always going to happen, it cannot tell Joe anything new about what’s behind door 1.
That’s all there is to it. Since Joe has no new information on what is behind door 1, the probability must remain at 1/3.
Of course, and this is the really puzzling aspect, it means the probability of the kitten being behind door 2 must have jumped to 2/3, indicating Joe should switch to that door. How did that happen? We’ll happily discuss that in the comments. Or, we could all just consult some expert pigeons.
Puzzle to Ponder: Consider Scenario D, where Monty always knows where the kitten is and always opens a door to reveal a calculator, using the following method to choose the door to open (given Joe has initially chosen door 1): if door 3 leads to the kitten then Monty opens door 2, and otherwise he opens door 3. So, given Joe chose door 1 and Monty opened door 3, what is the probability that the kitten is behind door 1? What if Monty had instead opened door 2?
Free Public Lecture, Sunday June 12
Burkard will give a talk entitled The Problem with Probability. See Burkard at the Melbourne Museum, 11:30-12:30. For more information and to register, please visit the MAV website.
Burkard Polster teaches mathematics at Monash and is the university's resident mathemagician, mathematical juggler, origami expert, bubble-master, shoelace charmer, and Count von Count impersonator.
Marty Ross is a mathematical nomad. His hobby is smashing calculators with a hammer.