Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."
The classical solution (assuming one contestant) is as follows. The contestant picks door A. There is a 2/3 probability before the doors are opened that the prize is not behind door A. Once a different door, door B, is opened and reveals a goat, the 2/3 probability is compressed into the remaining door, door C. Hence the contestant should switch. All that is well and good. Now consider the following, equally valid argument.
The contestant selects door A. Before any of the doors are opened, there is a 2/3 probability the prize is not behind door C. Door B is opened and contains no prize. Since there's a 2/3 probability the prize is not behind door C, there is a 2/3 probability that it is behind door A. So the contestant should not switch. The contestant's chances, it seems, depend on how exactly he or she looks at the problem -- clear nonsense. It's as if the point in space which is the solution to an equation depends on which coordinate system you use! A similar argument holds for the 2/3 probability that the prize is not behind door B (in which case the contestant has not learned any more information about the remaining two doors). In that case, since the two remaining doors are more or less equal in the contestant's eyes, switching does not matter.
On the classic problem, I explain it this way: When you choose door A, you have only and 1/3 chance to win. Clearly the better odds of 2/3 suggest you should switch, but since you don't know what door to switch to, you would only win half the time you switch. This cuts the 2/3 down to 1/3. You might as well stay with door A. Now Monty shows that door B has a goat. Assuming Monty knew this (and was going to show you a losing door, which he could always do), you now know that, if you should have switched (2/3 of the time) because door A was a loser, you can now switch to the winning door. So it is better (2/3) to switch.
On the other hand, what if Carol Merrill accidentally tripped and pulled down the curtain in front of door B, revealing the goat? Then you might as well stay. If door A is a loser, then only half the time would door B also be loser. If door A is a winner, then door B is certainly a loser. The probability (thanks to the elimination of cases where door B is a winner) changes to 1/2. This is a good time to mention Bayesian statistics for those further interested. I might also mention to some people, that, while PhD's are still awarded in "mathematics," the field is so broad that it doesn't necessarily mean you know anything about a branch that isn't your specialty.
Why? Simply because Monty cannot touch door A, no matter what he chooses to do. So he is unable to select it. Why is that relevant? Because by not selecting a door he is giving you information about that door. Assuming that he always opens a door with a goat in it there is a chance of 1/3 that you did choose the correct door, in this case monty will "bluff" and choose one of the doors, and no matter which he selects you will lose by moving. On the other hand there is a chance of 2/3 that you didn't choose the correct door. If this does happen then you will always win by switching doors - he shows you the correct door by opening the incorrect one.
The conclusion is that by not opening door C he is bluffing 1/3 of the time, and indirectly giving you information 2/3 of the time.
He first argues that, after the contestant has chosen door A, "once a different door, door B, is opened and reveals a goat, the 2/3 probability is compressed into the remaining door, door C." The probability of the prize being behind door B has not moved, or "been compressed" to door C. Instead, it has been eliminated, along with half the chance that the prize is behind door A, and the remaining half of the probability has been scaled up to sum to a total of one.
As an example of "probability scaling," imagine a standard six-sided die. There are six faces, each equally likely to turn up. You might imagine the probability of each side turning up to be one unit. However, by convention the total probability for all non-intersecting result sets must be one. Therefore, each unit of probability in this case must be scaled to 1/6. This "probability scaling" process is called normalization.
If you decide that each time the die turns up a six, you will roll it again until it turns up something other than a six, you might say that you have "split" the 1/6 probability of the six into five parts, and equally distributed them to the other faces, resulting in a 1/5 probability of each face turning up.
What if, equivalently, you simply discounted all rolls of six as if they never happened? You never bothered to include them on your results tally. The way to see this is that there are now only five units, each equally likely, as before. To get the sum of all probabilities to equal one, you must "scale" the units from 1/6 up to 1/5.
Recall one other fact about (independent) probabilities - they compound through multiplication. For example, if I roll a six-sided die, each face has a 1/6 probability of turning up. If I flip a coin, each face has a probability of 1/2 of turning up. If I do both, then each compound result - for example, heads and a five, or tails and a two - has a 1/12 probability of occurring.
What if I have two dice - a five-sided and a six-sided die. I flip a coin. If it turns up heads, I roll the five-sided die, but if it turns up tails, I roll the six-sided die. Then, the probability of the combination "heads-2" is 1/10, but the probability of "tails-5" is 1/12. Each die-roll can only "divide up" the portion (half) of the total probability space handed down to it by the coin flip result.
Now consider the Monty Hall Problem. We'll call the door I pick 'A.' This naming does not affect the analysis. Initially, the prize has an equal likelihood of being hidden behind any door, so the probability that any given door hides the prize is 1/3.
Now comes the second, compounding probability. If the prize is behind door A (1/3 chance), Monty has an equal chance of opening either of the other doors. Thus the combination "prize-behind-A, Monty-opens-B" has a 1/6 probability, as does "prize-behind-A, Monty-opens-C." However, if the prize is behind door B (1/3 chance), Monty will always open door C. Therefore, the combination "prize-behind-B, Monty-opens-B" has a zero probability, and "prize-behind-B, Monty-opens-C" has a 1/3 probability. Likewise, "prize-behind-C, Monty-opens-B" has a 1/3 probability and "prize-behind-C, Monty-opens-C" has a zero probability.
In the outline below, the left-aligned items represent the first division of probability into three equally likely prize locations. The second, indented level represents, for each prize location, the probability further split by Monty's door-opening choice. The total probability for each level sums to one. You can imagine this outline as a tree with branches.
Prize-behind-A (1/3) Monty-opens-B (1/6) Monty-opens-C (1/6) Prize-behind-B (1/3) Monty-opens-B (0) Monty-opens-C (1/3) Prize-behind-C (1/3) Monty-opens-B (1/3) Monty-opens-C (0)
Now you can see what happens when Monty opens door B. The "Monty-opens-C" possibilities are crossed out. This includes half the chance of "Prize-behind-A," and all the chance of "Prize-behind-B." The total remaining chance of "Prize-behind-A" is 1/6, while the total remaining chance of "Prize-behind-C" is still 1/3. Since this is all the remaining chance, it must be scaled up so that the total probability equals one. This is done by multiplying each value by two, yielding the normalized result that "Prize-behind-A" has probability 1/3, and "Prize-behind-C" has probability 2/3.
So, the faulty assumption in Andrew Goldish's second analysis is that the probability that C is not the door hiding the prize must remain fixed at 2/3. In the first argument, the fact that the probability of the prize not behing behind door A remained invariant at 2/3 is a merely a coincidence of the problem structure.
For example, imagine if the prizes were distributed as follows: before the show, Monty flips a coin. If the coin turns up heads, he puts the prize behind door C. If it turns up tails, he flips the coin again. If the coin turns up heads the second time, he puts the prize behind door A, but on tail he puts the prize behind door B. Then, there is a 1/4 chance each for doors A and B to hide the prize, but a 1/2 chance for door C.
In this scenario, if you choose door A and Monty opens door B, there is a 1/5 chance the prize will be behind door A but 4/5 chance it will be behind door B. If instead he opens door C, that tells you there is a 1/3 chance door A hides the prize, and a 2/3 chance door B hides the prize. Do the math and verify this yourself. You can see that there is no hard and fast rule about which pairs of doors must conserve their total probability.
When the contestant selects door A, before any of the doors is opened, there is a 2/3 probability that any of the three doors, such as door C, does not hide the prize. The reason to switch doors, however, is that there is a 2/3 probability that one of the doors not chosen hides the prize after the other door has been opened.
It may look as though there is a 2/3 probability that the prize is behind a different door once a door has been opened; that is only true because we assume that a) the host will not open the door the contestant has selected, and b) the host will not open the door hiding the prize. If there is a 2/3 probability that door C does not hide the prize, and door B is opened, door A would only be a 2/3 probability if either door A or door B could have been opened. The host could only open door B, so there would be a 1/3 probability that door A hides the prize (which is part of our assumption).
Here we have a similar confusion, but twisted around. One must remember that in the standard version the host knows where the prize is, and his actions leak this information to the contestant. When A is picked and the prize is behind C, the host must reveal B; were the prize behind B, C would have to be opened. Only when it is behind A can the host pick either. So in Andrew's scenario: A is picked. There is a 2/3 probability it is not behind C, but this is unimportant. He asserts, in effect, that "nothing changes" when the host opens B. This is wrong. What is invariant to the host's action is that there is a 1/3 chance it is not behind C OR B. When B is opened showing no prize, the probability that is is not behind B or C revealed to the contestant to be equal to the probability that it is not behind C. Thus, there is a 1/3 chance that it is behind A, and, yes once again, the contestant should switch.
A related objection voiced by many students is the "switch" advice seems oddly symmetric: we always say the first choice is A, B is opened and the best move is to switch to C. But if C were chosen and B revealed, we'd be saying switch to A. Since in one case you end up at C and the other A, surely they must really be equivalent, and so it doesn't matter which you choose. False. The problem is that the host reveals information that makes choosing the right door easier, but since that information depends on which door you choose first, it will tell you to do different things depending on that choice.