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Choosing The Prize Behind The Right DoorThere's an old probability question which makes a great discussion and experiment topic for a high school math class. The question centers around a game show, in which the game show host asks you to pick a prize behind one of three doors.
After he asks you to pick, the game show host doesn't open the door you picked. Instead, he opens a different door...one which he already knows is empty. Now, having eliminated one of the choices, he asks you, "Do you want to keep your guess? Or change it?"
Typically, people will answer, "There's a 1 in 2 chance that I'll get it right, so it doesn't make any difference if I change my guess or not, the probability is 1/2 either way!"
And the typical answer is not correct. Surprisingly, your odds are better if you change your guess. Why? Because the odds that you guessed incorrectly in the first place are 2 in 3, or 2/3. The game show's action of opening a door does not change that probability. Which means the odds are better if you switch doors. In fact, 2 times out of 3, you'll be better off switching.
If your students don't believe it, there are two ways you can help them see this.
Experiment
Have the class break up into pairs. In each pair, one person is the game show host. This student will pick a number (one, two, or three), but not tell the other student. The other student makes a guess, and the first student then removes one of the other numbers.
Have the first student always keep his guess the same, and count up how many times he wins. After they have done this many times, switch it around, and have the student always change his guess.
The students can now report back that they do win more frequently by changing their guesses.
Generalize
My favorite way of looking at this problem, which I found here: The Problem Site's Treasure Hunt, is to consider what would happen if there were a million doors. So the player makes his choice, and obviously his odds of having the right door are incredibly small. But now the game show host opens all but one of the remaining doors.
All of a sudden, students see the problem in a whole new light, when they realize that the odds strongly favor the one door the game show host didn't open.
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montyhall says:
Dear Educators,
The mythology around this probability scam is amazing. It must be that the suggested answer is so wrong that no one bothers to question it.
Let me see if I understand the faulty logic: If I pick one door out of three randomly and then I am shown that one of the others is empty, I have DOUBLE the chance to win the prize if I switch. In some articles this is extended to say that if there are 1000 doors and I pick one and then am shown 998 empty doors, I now have a 999 to 1 chance to win if I switch! Both of these bizarre answers are based on the incorrect logic that the probability of my original pick does not change when I am shown an empty door. That is, my chance is still one-third so therefore, the chance to win if I switch must now be two-thirds.
Let's look at it another way: Assume that I don't pick a door at all and then the host shows me an empty door. Which door now has the two-thirds probability? Of course the answer is neither and so my chances are one-half for each door.
Let's try another way. Assume that I pick door #1 and the host shows me that door #3 is empty. Originally, the chance that the prize was behind any one of the doors was one-third. I have now eliminated one door and so that "one-third" is gone. Door #1 had one-third and door #2 had one-third and nothing has changed to favour one door over the other, but now because there are only two doors left, with equal probability, that must add up to a total of 1, each must have a probability of one-half.
Finally, let's use the well known mathematical concept of "conditional probability", which says that the probability of event A happening given that event B has already happened is the probability of the intersection of event A and event B (i.e. both happening) divided by the probability of event B happening.
When we start the game there are only three possible ways that the doors can be set:
case #1: prize, empty, empty
case #2: empty, prize, empty
case #3: empty, empty, prize
I choose door #1. Let's define event A as the probability that door #1 has the prize (clearly it starts out as one-third). The host shows me door #3. Therfore, event B is the host showing me the empty door #3. So, what is the probability that my door #1 has the prize given that the host shows me an empty door #3?
From the three cases above, there is only one case where the "intersection" occurs; that is there is only one case where door #1 has a prize AND door #3 is empty. Therefore, the probability of A intersect B is one-third. From the cases listed above, there are two cases where door #3 is empty; therefore, the probability of event B is two-thirds.
Therefore, using the formula, the probability of door #1 having a prize GIVEN that door #3 is shown to be empty is one-third divided by two-thirds, which EQUALS ONE-HALF.
Monty Hall
Douglas Twitchell says:
Hi Monty,
I assume that, considering you signed yourself Monty Hall, you intended this to be tongue-in-cheek, and not be taken seriously.
In case you were serious, I'd suggest you run a simulation. Either get a friend and go through the process a hundred times, or write a computer program to simulate it.
Either way, you'll come to the inescapable realization that if your mathematics doesn't match reality, there's something wrong with your mathematics, not with reality! 
MrWallace says:
I couldn't resist the temptation to respond to Monty.
His first and third explanations imply the existence of a time traveling machine which allows us to alter the probability of a current event based on future actions. As far as I know, such a machine doesn't exist.
As for his second argument, he's making the rather absurd assumption that he can change the conditions of a problem without altering the outcome, which is TOTALLY bizarre!
And his last argument...I have no idea what he's even talking about - maybe someone else can make sense of it?
montyhall says:
People, people,
Let me start with the response by Mr Wallace to my first thought exercise: I am not postulating any sort of time machine, simply examining the illogic of the original case.
In my second point, I do not suggest changing the conditions, I simply accept the premise of the original question as posed on the website: The game player gets shown an empty door, there are two doors left, there is a 50-50 chance that the prize is behind one of the remaining doors.
Finally, the crux of the stupidity is revealed in Mr Wallace's third comment (no offence meant to Mr Wallace since others have fallen into the same trap). He indicates that he does not understand my third point. My third point is in fact based on the existing, well proven (not by me) actual mathematics of the problem. I did not present it first based on my assumption (proven correct) that people did not wish to understand the actual math involved. If people do not want to believe the actual mathementics involved then there is not any point in pursuing this further.
Permit me to go back to one more attempt to expose the illogic of the original answer. In the answer it is stated that the original 2/3 probability that you are wrong in your choice DOES NOT CHANGE when one empty door is revealed. Let us pursue this to the next step: Assume the host now opens another empty door and only one is left. According to the logic above, the probabilities do not change; therefore, even if there is only one door left and that door MUST have the prize. You people must (according to your own logic) still insist that there is only a 1/3 chance that the prize is actually behind that door. (I wonder where the prize goes two-thirds of the time?)
Oh my!!!
Monty Hall
Douglas Twitchell says:
While MrWallace's comments are a bit tongue-in-cheek, he is essentially correct.
His reference to a time machine comes from here: "Both of these bizarre answers are based on the incorrect logic that the probability of my original pick does not change when I am shown an empty door." Monty has stated that the probability of a past action changes based on a present event.
And MrWallace's observation about changing conditions of the problem is also correct. Monty, you have made the faulty assumption that the game show host's choice is not affected by my choice, so yes, you did change the conditions of the problem.
Regarding your final argument in the first post, to be honest, those of us who have actually done the simulation know your answer is wrong, which I suspect is what makes MrWallace not eager to try to figure out what you're talking about. I also was not interested in trying to decipher your explanation. Using phrases like "well known" and "well proven" does not exempt you from the responsibility of writing clear and accurate mathematical proofs if you want people to believe them.
Regarding your last argument in the second post, your problem is that you have assigned to Monty an action which can only be accomplished 2/3 of the time; if my initial guess is incorrect, he CANNOT open two empty doors.
Now, I have two requests to make. First, on both sides of this discussion, please maintain respectful courtesy and avoid sarcasm.
Second, Monty, I would request that before you post more discussion on this, you actually perform the simulation, and report to us your results.
Thank you,
Douglas Twitchell
AFE Administrator
montyhall says:
Hello again. It has been a while. I took the suggestion of the site moderator and ran 3 tests of 100 samples tries each. The results were as follows:
TEST #1: switch: 57% don't switch: 43%
TEST #2: switch: 50% don't switch: 50% (yes, exactly!!!)
TEST #3: switch: 51% don't switch: 49%
Those certainly do not support a 66.7% suggestion by this website. In fact they call into question the "computer simulation" I was asked to try! Can anyone out there wonder whether it would be possible to program a computer simulation to give a 66.7% result? If you cannot, I guess you cannot imagine that there are any computer scams on the internet!
Second, the major question I raised last time has never been responded to at all. That is, if when you open one door that is empty the odds remain two-thirds to one-third, why, if you opened a second empty door would they not still be two-thirds to one-third? When, as everyone would understand, if Monty opens two empty doors, the probability that the prize is behind the last one MUST now be 1.0 (or a certainty). If it changes for the last door, why does it not change to 50:50 when he shows us one empty door? None of those who have insulted me or my mathematics have answered this.
Finally, no one has actually dealt with the actual mathematics of the "conditional probabilities". The theories are not mine they are well established mathematical principles. The only answer I had was to suggest that they had never heard of it and did not understand it. That shows you how much actual mathemetics is involved in this presumed paradox (scam).
The fundamental error in the logic of the posed probelm remains the sentence that reads: "The game show's action of opening a door does not change that probability." This is simply wrong. It does change! Imagine agin if the game show host opened TWO empty doors; would it still be logical to assume that the probability that the prize was behind the door that you chose was still 1/3? Of course not!
Monty Hall
Douglas Twitchell says:
Just a few comments here:
1. I programmed the simulation myself and I can assure you, it does not "cheat".
2. Since both I and other excellent mathematicians have performed simulations both by hand and on the computer, I know that your results are not correct. I'm tempted to assume that you just made numbers up, but I'll give you the benefit of the doubt, and assume that your procedure was flawed. If you like, you can write up your procedure and mail it to me (address on the site). I'll read it just as I would read a lab report. You need to understand that this is a simulation that high school mathematics students perform every year in schools all over the world, and arrive at different results than you obtained.
3. It's a bit presumptuous for you to conclude that people don't know mathematics just because they don't agree with your arguments. Considering I am hired each year to write math problems for state high school math competitions around the country, I can assure you that I DO know a bit about mathematics.
4. Conditional probability IS an established mathematical principle, and no one denies it. No specific application of this principle, however, is well established. This is why proofs are necessary in mathematics. Otherwise we could all just "name a principle", wave our hands, and claim anything we like. This is why I asked you to explain yourself in more precise mathematical terms.
5. The fact that you came up with the results you did in your simulation means that you are either a) lying or b) have misunderstood the premise of the problem and so did your simulation incorrectly. Because of this, I didn't bother trying to decipher your explanations. Send me your detailed description of how you arrived at your simulation results, and I'll take a look at it.
6. If you send a "lab report", after I review your procedure, I will ask some students to follow the directions you give, and if your directions either make no sense, or produce different results than what you've claimed, I will assume you are lying rather than confused, and that will be the end of your permission to post here.
D. Twitchell
AFE Admin
MrWallace says:
I apologize for my earlier snarkiness. I begin from the assumption that someone signing "Monty Hall" is a troll, which might not be fair.
Here is where I lost interest in trying to decipher Monty's conditional probability: "The host shows me door #3." Monty has made the error of assuming that the host will open door #3, which he cannot assume without loss of generality, since under 1/3 of the conditions, it is IMPOSSIBLE for the host to open that door, as it is not empty.
This is very similar to the error Mr. Twitchell pointed out when he said "assigned to Monty an action which can only be accomplished 2/3 of the time"
It appears that "Monty" has misunderstood the conditions of the problem, since he repeatedly assigns impossible actions to his game show host.
I hope that is helpful without being too snarky.
MrWallace says:
Sorry, I had one more thing.
Monty says: "That is, if when you open one door that is empty the odds remain two-thirds to one-third, why, if you opened a second empty door would they not still be two-thirds to one-third? When, as everyone would understand, if Monty opens two empty doors, the probability that the prize is behind the last one MUST now be 1.0 (or a certainty)."
This was addressed (in a snarky way) by my "time machine" comment. You seem to have forgotten that the contestant picks the door BEFORE the host opens anything.
You are confusing the probability of an event with the conditional probability of an event given other conditions.
The probability that he is correct is NOT the same as the probability that he is correct GIVEN that another choice ISN'T correct. If they were the same, book me a flight to Vegas, because I can win every time!
Douglas Twitchell says:
Thank you, Mr. Wallace - the whole idea of "assigning" actions (and not always possible actions) seems to be at the heart of Monty's explanations. I'm going to guess that this is also the explanation for his incorrect results to his simulation. If I had to guess, I would say that Monty probably tried to do the simulation by himself, instead of asking a second person to take the role of the game show host.
If he did this, then he would have to randomly assign responses to the game show host, which - as you pointed out - would result in impossible actions, and would produce skewed results.
I have not yet seen any mail from Monty; I'm still waiting!
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