Deviant argumentation regarding Monty Hall problem
In class number four of the online Artificial Intelligence course from Stanford, Peter Norvig explains “one of the most popular problems in the subject of probability theory, the Monty Hall problem”. His presentation was exactly what I have always been taught, and I got the right answer on the quiz.
Hold; P(goat) = 2⁄3; P(car) = 1⁄3;
Switch; P(goat) = 1⁄3; P(car) = 2⁄3.
Since the last time I looked at that wikipedia article, a number of qualifications and stipulations and elaborations have been added. Apparently this problem has kept its popularity and controversy with undiminished vitality since around 20 years ago, when it was popularized by a national Sunday newspaper magazine columnist Marilyn Vos Savant. Her answer is the one above, which is also what I would call the least complicated wikipedia page answer (as the page sits today.)
Now there is one probable error in Norvig’s lecture. He claimed that Monty Hall didn’t know the answer to his own problem, and quotes a letter he says was written by Monty Hall in 1990. If this is true, I find it impossible to follow the following New York Times story, from 1991 where Monty explains all the subtleties of the problem in much shorter order than all of the complications on wikipedia. To summarize Monty’s take: he is the carnival barker; it’s his game; and he can change the rules in the middle of the game if he feels like making it more complicated than your five minute presentation for your students.
But there is one critical element missing from all of the discussions I have seen to date, except for one which was told to me by a playfully deviant fellow many years ago. His argument is simple: if by this strategy a player can double his chances of winning, the players would have figured this out forthwith; everybody would have switched every time; and the drama and the fun would have been drained from the game. The fact is the show was the most popular game show in it’s hay day. Ergo there cannot possibly be an easy way to game this strategy. My deviant probability theorist pal had confidence in Monty’s game players to find any available edge.
Also the data is in the archives. All you have to do is watch every episode and count the holds and switches and wins and losses and you will have a really high confidence estimation whether the correct answer is closer to 2⁄3 and 1⁄3 or is closer to .5 -- .5. If no statistician is interested in researching this, surely some enterprising historian of science or sociologist of science ought to be up for it.
As a note, there’s no way to do this empirically in the real world, as the problem does not even remotely match a situation that actually occurred on “Let’s Make a Deal”.
I don’t know what monty might have done in the original show to screw with the probability, but the usual analysis is correct and can be verified with as much simulation as you like.
I suspect that players would be unable to make the correct decision given the pressure and psychological manipulations coming from monty. Especially considering that highly intelligent people will argue for days that it’s 50⁄50 up until they do a simulation and actually see the result; it’s not an easy problem for untrained people.
Think of the conformity research. Monty could be saying things like “are you sure you want to switch?”. Also, I think people may have a bias that causes them to think “man I’d feel stupid if I switched and got it wrong, but it would be ok if I didn’t switch and got it wrong”.
This game show is pre-internet so everyone would be on their own with regards to coming up with the solution beforehand. And nearly everyone gets it wrong at first.
I wonder how much selection bias there is in the type of people who actually got on the show.
I don’t understand how the answer could be anything but 50⁄50.
I know the right answer, but if you deleted it from my brain I never would have figured it out.
I guess I’m looking for an explanation that isn’t just following through examples from every scenario.
Does an explanation like that exist?
Sure. Think of things in terms of distributed probability mass. I think we all find in intuitive that each door starts at 1⁄3 of the probability mass. So you chose a door, and the host opens another to reveal no_prize. Once the door was opened, the probability of that door goes to zero. And that probability mass must go somewhere because probability must sum to 1.
So far, so good. Depending on one’s intuitions about probability, the 1⁄3 mass from the wrong door either (a) splits between the remaining closed doors, or (b) stays together so that some door is now has 2⁄3 mass. So which is it?
Imagine a small modification to the procedure. Before the no_prize door is opened, a curtain is closed that covers the unchosen doors. If this curtain was closed before any door was chosen, 2⁄3 chance the prize is behind the curtain. This is still true if the curtain is only closed after you choose a door.
Once the curtain is closed, the no_prize door is opened. But you still don’t know curtain-door is open. From your POV, the curtain is still 2⁄3 likely to include the prize door.
Now imagine a bystander walks in at this point, knowing nothing about the game. She hears the host ask if you want the door or the curtain. If she thinks the game is fair, she things you have a 50-50 chance. But you know more information that the bystander—all that curtain-covers-two-doors stuff. It is this knowledge that justifies the difference between your probability estimate and the bystander’s estimate.
Perhaps this story will help:
Imagine a criminal defendant, represented by a lawyer. The lawyer says, “If we win the motion, your case will be dismissed. Otherwise, you will be convicted.”
Defendant: “So, I have a 50-50 chance.”
Lawyer: “Absolutely not. The motion is totally lacking in legal and factual merit. There’s a 1% chance that the judge will be so confused that the motion will be granted.”
The lawyer’s knowledge of the decision mechanism allows a more accurate prediction than (1 / # of possible outcomes). In the same way, you have seen the whole process (doors, chose, curtain, no_prize opened but unseen), which allows (and requires) you to make a different allocation of probability mass.
These and related issues are addressed in Jason Rosenhouse’s eponymous book on the Monty Hall Problem. I strongly recommend it. (Gratuitous plug: I reviewed it in detail at my blog).
One thing he points out there is that many people even when playing iterated versions of the problem still don’t switch. There’s a fair number of psych studies using Monty Hall to try to understand how humans think about probability and game theory. One major issue is that switching triggers feelings of loss aversion. Many people say they chose not to switch because they’d then feel bad if they had been right after all. So even if Monty had played the game as it is depicted (which he didn’t) it is likely that many people would still not have switched.
Bonus question: suppose that we allow Monty to choose whether or not to open a door and offer a switch. In fact, Monty acts as follows: whenever the guest picks the correct door, Monty always offers a switch; otherwise Monty offers a switch with probability P.
However, the game show has been going on for years, and therefore the guest, who has seen all the past shows, knows what P is. What should the value of P be in order to ensure that, whenever Monty offers a switch, the odds for switching are 1:1?
P=1/2; then you can switch 2⁄3 of the times you play, but you’re still right just 1⁄3 of the time.
Note that for P>1/2, a rational contestant will win more than 1⁄3 of the time by switching, but for P<=1/2, the best strategy is just to never switch.
Correct! I can’t be bothered to create Misha points, so a point to House Orthonormal.
(House Orthonormal’s colors are #010000, #000100, and #000001, which are technically red, green, and blue, but are indistinguishable from black to other houses. The house mascot is the disease-carrying mosquito—a vector.)
What makes you think contestants would have figured it out? People appear regularly on Deal Or No Deal and don’t seem to have figured out that there is no element of skill, and that the ‘deal’ offered is always lower than the expected winnings if the player continues...
It isn’t always lower than the expected winnings if the player continues. It is lower while the big sums are still out there. Once the possibility of winning or losing a lot of money vanishes the deal gets good because producers want the contestant off the stage so they can start a new game. They’re maximizing drama.
The expected winnings are in money though, not utility, so risk aversion may be justified.
Yes.
If someone offered me a sure $100,000 or a coinflip between $250,000 and $1, I’m pretty sure I know which one I’d choose. As far as I can see you can only win the big prize if you are rich enough not to be risk averse about amounts that size.
I don’t think you need to take data.
Don’t you think we would have remembered if Monty had ever screwed the contestant, and showed the Grand Prize as the first door the contestant didn’t win?
“Ok. Let’s look at one of the doors you didn’t select. ….… You didn’t select the Grand Prize! Hurrah! So, do you want to keep your loser door, or take the other loser door?”
Whatever Monty may or may not have know about probability theory, he surely knew that doing that wouldn’t make for an entertaining climax to his show. As long as he was committed and able to avoid that scenario, the usual analysis is fine.
Even if Monty was feeling bitter one day, and decided to screw the contestant, or they just made a mistake and opened the wrong door, that really isn’t the point of the problem.
Of course your best strategy is different for Screw You Monty than for Let’s Have an Exciting Climax to the Show Monty, but that shouldn’t surprise any of us. Any optimal strategy has to assume a prior on the behavior of Monty. No Free Lunch.
My understanding was that Monty Hall knew which door contained the prize. After he revealed a non-prize door, he more strongly encourage players to move when their initial guess was correct.
Unless Monty Hall got a reply to his 1990 letter that explained it to him, and so he knew about it in 1991?
It is trivial to simulate, just do it for yourself.
My results, from back when I did it, follow.
As presented, with Monty always opening a door with a goat that the contestant did not pick:
Switch − 2⁄3 Stick − 1⁄3
If Monty opens a random door that the contestant did not pick, with the contestant losing if the car is revealed:
Switch − 1⁄2 Stick − 1⁄2
Monty should try and predict whether the player is the type to switch or not. If they are not, and a switch would help them, he should offer the switch. In some of the other cases, perhaps randomly, he should refrain from offering the switch.
If he is behaving purely randomly he should not tell people, or it would be like playing rock paper scissors against a random number generator instead of a human—lower drama.