Steven Dutch, Natural and Applied Sciences, Universityof Wisconsin - Green Bay
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There are some puzzles where the "correct" answer becomes so ingrained in popular culture that it becomes almost impossible to convince people the answer is wrong.
For example, my father used to be fond of a riddle that went "Brothers and sisters I have none, but this man's father is my father's son." He was absolutely convinced the answer was that the man was looking at a picture of himself. So I even drew him diagrams like the one below:
| My Father | |
| This Man's Father is | My Father's Son (Me) |
| This Man | My Son |
The popular answer is wrong. The riddle actually refers to the man'sson. Do you think I could make my father see it? His response was to repeat the riddle, emphasizing different parts of it, as if saying it louder somehow overcame logic.
Pretty much the same thing has happened with the infamous "goat problem" or "Monty Hall problem." You are confronted with three doors. Behind two of the doors is a goat and behind the other is a car. You pick a door. Monty opens one of the other doors and shows you a goat. Now, should you switch doors or not? Many people intuitively say it doesn't make any difference but the answer, most everyone who writes about this subject assures is, is yes - you have a 2/3 probability of winning if you switch. We even have the word of Marilyn vos Savant, world renowned filler-in of bubbles on IQ test sheets.
This is a serious challenge. I'm not merely being contrarian.I have never seen a rigorous proof of the "Goat Problem". People parrot the conventional solution in broad, arm-waving terms but never get rigorous.
This is a totally discrete problem. There are possible configurations of doors, cars, and goats, possible choices of doors by the contestant and Monty, and possible changes of mind. None of them overlap, none grade into one another, and all the initial ways of arranging cars, doors and goats are equally likely. All the subsequent choices are yes-no choices.
In a discrete problem like this, it should be possible to list every single possible state of the system, all of which should be equally probable. If two states are not equally probable it's because you missed something and have not broken the problem down into its most fundamental terms. For example, since a roll of a pair of dice can yield any total from 2 to 11, you might naively say that there are ten possible totals so your chance of rolling a seven is 1/10. You missed something. In reality, there are 36 possible rolls of the dice, six of which add up to seven, so the probability is actually 1/6.
In the case of the goat problem, if you claim something has a probability of 2/3, there must be three possible configurations, two of which lead to the outcome you claim and one of which leads to something else. Fine.List them explicitly and make sure they are fundamental. That is, you start off with equally probable outcomes, apply choices that limit some possible states, and end up with a 2-1 ratio of final states.
I am not interested at all in arm-waving generalities. If you're going to argue that the probabilities change because "Monty can always choose which door to open," you don't understand what a rigorous proof is, so don't waste your time or mine.
Most people frame the problem by having the contestant Pick door 1 and Monty opening Door 2. As far as anyone has shown so far, that simplifies the problem without losing generality. That is, you'd get the same outcomes if you chose Door 2 and Monty opened Door 3, and only the labels would change. So I see no problem in following that practice. But I want a rigorous, exhaustive listing of fundamental outcomes with 2/3 of them leading to a win by switching doors.
The odds of something happening are given by a simple formula:
Probability = (Number of ways an event can happen)/(Number of all possible outcomes)
For extremely complex cases, calculating those numbers can get very complex. But for simple cases, it's easy to list all the possible outcomes exhaustively. For example, what are the odds of rolling a five with two dice? You can roll 1,2,3,4,5 or 6 with the first die and 1,2,3,4,5 or 6 with the second die, so your possible rolls are 1-1, 1-2, 1-3, .... 6-4, 6-5, 6-6, or 36 rolls in all. To get a five, you can roll 1-4, 2-3, 3-2, or 4-1, four possibilities. So your odds are 4/36 = 1/9.
Note, you have to consider the two dice separately. You can't just consider 1-2 and 2-1 equivalent, because there are two ways one die can be 1 and the other 2, but there is only one way to roll, say, 3-3.
Now you'd think three doors with two goats and a car is a simple enough system that we can settle the issue by enumerating all the cases exhaustively. But you never see this in explanations of the 2/3 answer. Instead you get verbiage like this at <http://www.damninteresting.com/?p=211#more-211> by Alan Bellows. I'm not singling Bellows out for particular abuse here, it's just that he's representative of all the discussions of this problem I've read.
The best way to look at it is to imagine that when the contestant selects a door, she divides the doors into two sets: A) The doors she DID select, and B) the doors she did NOT select. At this point, each individual door has a one-in-three chance of being the winning door, but the two sets have differing odds… Set A has a one-in-three chance of containing the new car, but Set B, having twice as many doors in it, is twice as likely to contain the winner.When Monty opens one of the doors in Set B to show it isn't the winner, Set B still has a two-in-three chance of holding the winner. The only difference is that there is only one door with unknown contents, so the 2/3 odds go to the unopened door in Set B, while Set A still has it's 1/3 odds. So revealing the contents of door #2 didn't make the contestant's odds any worse, but it did make the odds for door #3 improve.
We'll examine the logic later. Let's do what Bellows didn't bother to do - list all the possible outcomes. And let's be thorough and take into account that there are two ways a goat can be behind a door. So let's abbreviate the two goats G and g (one's a baby), and the car with C. There are six possible ways to position them behind three doors:
CGg, CgG, gCG, GCg, GgC and gGC.
So there are 2/6 ways you can have a car behind any given door and 4/6 ways to have a goat. Probability of a car = 1/3, probability of a goat = 2/3.
Now, let's say you choose door number one, Monty opens up door number two and shows you a goat. (You can tally up all the possible sequences of choosing door 2, Monty opens Door 1, etc. if you like - the results will be the same.) Do you switch? Well, Monty has eliminated all the possible outcomes with a car behind door number two:
CGg, CgG, gCG, GCg, GgC gGC.
There are four possible outcomes remaining. The first red flag should be how do we get a probability with 3 in the denominator when there are four outcomes left? Two of the four outcomes have a car behind door 1 - you win by standing pat - and two have the car behind door 3 - you win by switching. So conventional wisdom isright. It makes no difference.
Let's revisit Bellows to see where he went wrong:
The best way to look at it is to imagine that when the contestant selects a door, she divides the doors into two sets: A) The doors she DID select, and B) the doors she did NOT select. At this point, each individual door has a one-in-three chance of being the winning door, but the two sets have differing odds… Set A has a one-in-three chance of containing the new car, but Set B, having twice as many doors in it, is twice as likely to contain the winner. Up to this point, he's right.When Monty opens one of the doors in Set B to show it isn't the winner, Set B still has a two-in-three chance of holding the winner. Wrong! By revealing information about the system, he has changed the odds. You no longer have a 1/3 probability of a car behind door 2, you have a zero probability of a car behind Door 2. If he opens door 1 to reveal a car, the probability of Set B having a car isn't still 2/3, it's zero.
If you have two mutually exclusive events with probabilities p1 and p2, the probability of one of them happening is p1+p2. Without knowing anything about the doors, the probability of having a car behind door 2 is 1/3, and behind door 3 is 1/3. The total probability is 1/3 + 1/3 = 2/3. But by opening door 2 to reveal a goat, he has dropped that probability to zero. It's not the same problem any more.
The only difference is that there is only one door with unknown contents, so the 2/3 odds go to the unopened door in Set B, while Set A still has it's 1/3 odds. So revealing the contents of door #2 didn't make the contestant's odds any worse, but it did make the odds for door #3 improve.
That's not how it works. This is sort of a variation on some gambling fallacies, like if you've had a run of successive heads flipping coins, you're "due" for tails, or if a certain roll of the dice hasn't come up, it's "due." If you don't achieve some outcome during a partial run, the probability doesn't accrue to the outcomes still remaining. The probability changes depending on how many outcomes are still possible and how many ways there still are to win.
Actually, as we'll see below, fixating on whether the contestant's odds get worse is misleading. The odds for both door 1 and door 3 get better.
Whoa, wait a minute. We started with three doors, all with 1/3 probability. By opening door 2 and finding a goat, we drop the probability of that door to zero. Where did that remaining 1/3 go? The answer, as we'll see below, is that by revealing information about the hidden door, we changed the odds. Actually, the missing 1/3 doesn't "go" anywhere because the original odds cease to exist once we open door 2. We now have two doors, one car, and one goat. It's not the same problem any more. If you sit down to a poker hand with three other people, your odds of winning the hand are 1/4. Once you start dealing the cards, the probabilities change. If you draw three aces, your odds go up dramatically. Not only is your chance of winning enhanced, but the odds of everyone else having a strong hand decreases. The original 1/4 odds for each player no longer exist. There is no such thing as conservation of odds, except that all probabilities have to add up to 1. Knowing something about previously hidden information changes the odds.
If you knew for a fact that the car was behind doors 2 or 3, then opening door 2 would change the probability of finding it behind door 3 - to 100%, in fact. But there is also a probability that the car is behind door 1, so by opening door 2, you change that probability as well. The probabilities are not locked in at 1/3 for Set A and 2/3 for Set B, any more than they're locked in at 1/4 for the players at a poker table. They change as new information becomes available.
A related problem, described by Martin Gardner in Scientific American when he discussed the goat problem, is this: You're on an airliner and your seatmate tells you he has two children. A little while later, he mentions his son. What are the odds that the other child is a girl?
Well, normally the odds of having one of two children be a girl is 3/4. There are four possibilities for two children: BB, BG, GB and GG. However, by mentioning his son, the passenger has eliminated GG. That leaves only BB, BG and GB, so that the odds of the other child being a girl drop to 2/3.
However, just like with the goat problem, not only does the probability of the second child being a girl change, so does the probability that the second child is a boy. Before you knew your seatmate had a son, the probability of his having two sons was 1/4. After he tells you he has a son, he eliminates possibility GG and the probability of the second child being a boy rises to 1/3 - it is not locked in permanently at 1/4. There's only one way he can have a second son, but two ways he can have a daughter, so the odds of a daughter are twice as great.
Remember our original formula:
Probability = (Number of ways an event can happen)/(Number of all possible outcomes)
Let's call probability p, the number of ways an event can happen n, and the total number of outcomes N. We have p = n/n. Now, let's say that we obtain information that eliminates x possible outcomes (from both n and N). We now have p = (n-x)/(N-x)
Apply this to the airline passenger: originally the odds of one child being a girl are 3/4. We have n = 3 and N = 4. By mentioning his son, the airline passenger eliminates one possible outcome: x = 1. So we now have p = (3-2)/(4-1) = 2/3.
Back to the goats. There are three possible doors for the car, and two of them are doors you didn't pick. Again, n = 2 and N = 3. Now, by revealing that there's a goat behind one of those doors, you've eliminated one possibility: x = 1. So we have p = (2-1)/(3-2) = 1/2 that the car is behind the remaining door.
Bellows goes on to write:
In explaining the effect, it helps to increase the scale of the question. Imagine that there are 100 doors to choose from instead of three. The contestant chooses a door, and then the host opens 98 other doors to show that they don't contain prizes. Which is more likely to hold the prize… the door she selected initially, or the one door left unopened from the 99 she didn't choose? The answer is much more obvious: the door she chose still has a one-in-a-hundred chance of being the winning door, where the other closed door has a 99/100 chance.
He's not alone here. Michael Shermer wrote pretty much the same thing in the October, 2008 Scientific American, except he used the more manageable example of ten doors. So let's go that route. You have ten possible situations:
Cggggggggg gCgggggggg ggCggggggg gggCgggggg ggggCggggg
gggggCgggg ggggggCggg gggggggCgg ggggggggCg gggggggggC
(I once took a diabolical test in the military to test attention to detail. It consisted of long strings of upper case C's with a few O's interspersed. To spare you that ordeal, I used lower case g's for the goats. We could number them g1, g2, g3... and consider the 9 x 8 x 7... = 362,880 possible sequences for each case. Take my word for it, it doesn't change anything.)
As you successively open doors 2-9, does the likelihood of door 10 having the prize increase? Yes it does, relative to its original probability. At the start, the probability is 1/10. After opening door 2 it's 1/9, then 1/8 and so on. After all the doors are open, the remaining possible states look like this:
Cggggggggg gCgggggggg ggCggggggg gggCgggggg ggggCggggggggggCgggg ggggggCggg gggggggCgg ggggggggCg gggggggggC
We start off with nine possible ways the prize can be behind the last nine doors, out of ten possible states. So p = n/n = 9/10. By opening the middle doors, we successively eliminate eight possibilities: x = 8. So the final probability after all eight doors are open is p = (9-8)/(10-8) = whaddya know! 1/2.
The fact that the probability of the last door having the prize increases as we open doors may account for why so many mathematically literate people who should know better have been taken in. Yes, opening one door raises the odds from 1/10 to 1/9, to 1/8, and so on, ending at 1/2 when the last door is open. And 1/2 is a heck of a lot better odds than 1/10. But there is always the possibility that that first door has the prize. And that probability increases as well.
Let's say that five doors have been opened. The odds of the prize being behind any remaining door is 1/5. The odds of the last door having the prize have doubled. But so have the odds of the first door.
To use Bellows' example of a hundred doors, p = n/n = 99/100 that we picked the wrong door. Opening doors 2-99 eliminates 98 possibilities (x = 98). The odds of the remaining door having the prize are now (99-98)/(100-98) = ta-da! = 1/2. We started with a 1% chance of winning and end at 50%.
Again, if you knew for a certainty that the car was behind doors 2-10, each door you opened would increase the probability of the car being behind door 10. But since the car could also be behind door 1, the probability of finding it there also increases every time you open a door to reveal a goat.
Suppose instead of doors, it's a job interview with 100 candidates. One person is called away for a serious emergency and is told he will be informed by phone if he gets the job. You go to the bathroom and come back to see candidates filing dejectedly out of the room. So what do you think? Every person who leaves means your chances have improved. Or does it mean that you're more likely to become one of them? Finally it's you and the guy who left early. Do you think you have a 99% chance of getting the job?
Here's the most complete analysis I can come up with. We start off with the possible arrangement of goats and car:
CGg, CgG, gCG, GCg, GgC gGC.
There is no reason to assume any of those are inherently less probable, so all have a probability of 1/6. We can assume that the contestant picks Door 1. Monty then can pick only Door 2 or 3, depending. We'll assume he plays by the rules, but if he has a choice, he picks door 2 or 3 randomly. We'll add a numerical suffix to indicate which door he chooses.
If the car is behind Door 1 Monty can pick door 2 or door 3. We have CGg2, CGg3, CgG2, CgG3. But each of those has a probability of 1/12, since CGg and CgG each have a probability of 1/6 and there's a 50% chance he could pick either door.
For the remaining choices, Monty has no choice. He has only gCG3, GCg3, GgC2, gGC2, each with 1/6 probability.
Finally the contestant gets the option of choosing to switch doors. So let's add a second suffix to indicate his final choice of door. For example, CGg21 means Monty picks door 2 and the contestant stands pat with Door 1. Further, we'll assume he chooses randomly just to see what happens.
We have
Now let's color winning configurations blue and losers red:
However, this is not how the problem is often stated. If Monty picks door 2, then we have
We have
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