Comments
I had the wrong Monty Hall answer at first until I read about the computer simulation and decided to rethink the problem, which then prompted me to develop a new understanding. When confronted with an anomaly like conflicting results from a computer model, we all respond differently (locating the error in self vs. other/other's model) because even people who share a belief in logic as a means of solving problems have widely varying degrees of faith in their own powers of reason. Dissent is healthy but digging in one’s heels in the face of compelling evidence hinders progress, and the line between the two isn’t always clear. (Forget Monty Hall, how do you persuade others that humility is a virtue?)
Rather than continuing a fruitless debate or acknowledging dissent and moving on without resolution, a third option is to try translating the problem into different terms. In this case we can use the language of human behavior, recognizing that Monty Hall is actively responding to the player’s initial choice. To make it even clearer we can imagine that Monty Hall wants the player to lose; he’s greedy and wants to keep the prize.
Consider the only two possibilities: the player’s initial door selection either has or does not have the prize behind it. As a player, you don’t know if you’ve selected the door with the prize. But Monty Hall does “know” and acts accordingly. (Even if this knowledge isn’t explicitly stated, it’s implicit in Monty Hall’s behavior of opening only unchosen doors, coupled with his knowledge of where the prize is.)
Case 1 - If your chosen door doesn’t have the prize behind it (probability of 2/3), then Monty Hall has no choice. He must open the lone door that was neither selected by you nor has the prize, and hope that you don’t switch. Because in this case, switching will always give you the prize.
Case 2 - If your chosen door does have the prize (probability of 1/3), then Monty Hall has a choice; he may open either of the doors you didn’t select. But of course it doesn’t matter which door he opens. He’s just hoping you’ll switch, because if you switch you’ll never get the prize.
Since you don’t know whether the door you chose has the prize behind it (determining whether you’re operating in Case 1 or Case 2), you should switch since it’s twice as likely that you’re in Case 1. Twice as likely that Monty Hall had no choice when he opened the door he did. The “always” and “never” consequences to switching make it mathematically simple - switching gives you the prize 2/3 of the time.
posted by Barbara Blossom
August 15, 2008 @ 3:28 PM
The Monty Hall solution is easily expained by a truth table. Name the doors A,B,C with the prize behind C.
Contestant's Monty's Contestant's
1st choice choice 2nd choice
___________ _______ ___________
A B A or C
B A B or C
C A or B C or B or A
__________ ___ ______
Contestant
wins 1 2
If the contestant stays with his first choice, he has 1 chance in 3 to win. If he changes choices he has 2 chances in three to win.
Bob H
posted by Robert Hatch
August 16, 2008 @ 3:53 PM
This is an effort to figure out the formatting your answer box.
______ ______ _____ ________
A B A or C
B A B or C
C A or B C or A or B
______ ______ _____ ________
Winners 1 2
posted by Robert Hatch
August 16, 2008 @ 4:04 PM
I quess I can't figure out the formatting to get the table to appear as a table. Anyway, it's a short clear explantion.
posted by Robert Hatch
August 16, 2008 @ 4:14 PM
Dear Brian,
I read with sympathy your lament on the difficulties of persuasion, as illustrated by lingering contention over the Monty Hall problem. This problem of persuasion afflicts many areas of science, notably including the creation/evolution debate, where stout fundamentalists insist the world is 6000 years old, as made by godly fiat. Their source is the same ancient text that describes the earth as flat.
But I digress. The Monty Hall problem seems to me to be an instance of a broader debate in statistics, between the classicists and the Bayesians. Classicists approach each statistical question as fresh, discounting history. With that mindset, and Monty having made his choice, they are confronted with two doors and, finding no reason to discriminate, hold that either one is equally likely to conceal the car (my wife, an animal lover, would rather have the goat).
A Bayesian on the other hand would take history into account. Monty's choice would signal--assuming he had advance knowledge of the goods behind the doors-- that switching doors would double the odds of winning (assuming one prefers the car).
This debate among statisticians has consequences. Classical statistics still dominates education, with the Bayesian approach taught, if at all, as an afterthought. This classical dominance explains why Marilyn vos Savant received so many angry responses from PhDs to her original Parade article.
Education and statistics would be better served by a deep and thoughtful exploration of how problem history affects the abstract idea of probability.
Richard LeVitt
posted by Richard LeVitt
August 16, 2008 @ 6:02 PM
Hello Brian;
I've taught Monty hall for ages. It is hard to get it across to everyone. But one trick I have found useful for advanced students is to increase the number of doors. here goes.
Suppose there are 1000000 doors altogether. One has a prize, the rest goats. You pick a door--say number 25. Monty hall then opens door, 1,2,3,4,5,6,...,24,26,27,28,....
up to 324786 which he skips, then he continues opening doors up to one million. So, the choice is now between:
door 25 which you picked
and
door 324786 which Monty picked.
Most peoples view of symetry breaks down here. Some how, the door Monty picked is much more special than the door we picked.
As an aside, I have sometimes taught that 1/2 is the right answer. THis is easy to motivate if you make it into a two player game. Monty's utility is to confuse the contestent. In other words, he wants the croud to be screaming 1/2 for switch and 1/2 for stay. So what strategy should he use? Basically one that only optionally opens doors so that the final conditional probability is .5.
Finally, a fun simulation that I've done in class is on finance. It is about the only example I know of where variance is the one true answer (as apposed to standard deviation). I've written it up for your sister in name journal, "american statistician". I've a on line link of:
http://gosset.wharton.upenn.edu/~foster/research/being_warren_buffett.ps
later,
dean
posted by Dean Foster
August 19, 2008 @ 9:30 AM
I've pondered the Monty Hall problem before, and have now rethought it in what I think is a more concise and hopefully understandable way.
For simplicity, label the door that leads to the prize P, and the doors that lead to empty rooms E1 and E2. If the original guess is P, then switching answers after either E1 or E2 are opened will lose; i.e., this will mean switching from P to E2 or E1. On the other hand, if the original choice is E1, Monty must then open E2, and switching from E1 to P wins. Similarly, if the original choice is E2, then Monty must open E1, and switching from E2 to P wins. Because it is twice as likely for the contestant's original guess to be E1 or E2 than the lone P, the probability of winning by switching doors is 2/3.
Harvey Leff
Claremont, CA
posted by Harvey Leff
August 25, 2008 @ 2:02 PM
Brian states: "The issue is how I can persuade anyone that my answer—or any particular answer—is correct."
Here's another attempt to provide an intuitive insight into why 2/3 is the correct answer...
Suppose there are 100 doors, and all other conditions of the original 3-door puzzle remain intact.
The player chooses one door.
Would everyone agree that the chances of the prize being behind one of the remaining 99 doors is 99/100 ? (And, that the door chosen by the player has only a 1 in a 100 chance of containing the prize?)
"Monty" opens 98 of the remaining 99 doors.
What do you think the chances are of the prize being behind that one door that "Monty" did not open ?
99 to 1, or 1:1 (50/50) ?
I think you'd have a strong feeling that the chances are 99:1, and that you should switch doors. Right ?
Apply that same logic ("feeling") to the 3-door situation and you are left with the choice:
2:1 or 1:1 ?
It seems then, applying the same logic, that the 2:1 odds apply to that remaining unopened door, and that switching doors would be in your best interest.
Hope this analogous problem helps some of you get by the mind-bending nature of the 3-door problem.
Peter
posted by Peter Rauch
August 25, 2008 @ 3:17 PM
Oops, sorry. I didn't see Dean Foster's similar reply before I posted mine.
Peter
posted by Peter Rauch
August 25, 2008 @ 3:20 PM
I introduced this problem into the IT department at my workplace a few years ago and was stunned at the reaction it produced. I thought that I was going to be the victim of physical violence more than once. Even hardware types went off and wrote programs to convince themselves of the solution. I published (in the IT department) several ways of looking at the problem. I think the most direct approach is as follows:
Choose a door and you will be wrong two thirds of the time. Once a door is eliminated (by being opened by Hall) you are still wrong two thirds of the time. So by switching to the remaining door you will be right two thirds of the time.
regards
Barry Allebone
posted by Barry Allebone
August 27, 2008 @ 10:08 PM
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