The Monty Hall Problem

This is a very famous puzzle; you may have come across it before. I absolutely love it. I love it because it is counter-intuitive (clue) and controversial to the point of emotional debate** (clue), and yet the problem is easy to describe and the solution is reasonably simple to explain.

Don't get me wrong: complex math does some serious work behind the scenes (so they tell me; I'll take their word for it), but you don't need to be a mathematician to understand the solution...at least on a superficial level.

The Monty Hall Problem gets its name from the host of an old TV game show called Let's Make a Deal. In the show, Monty offers a contestant a choice among three closed doors. Behind one is a new car; behind each of the other two is... I forget. Something less desirable than a new car. Maybe a goat. Yeah, I think that's it: each of the other two doors conceals a goat. (I know I could Google this but I am doing my best to write it down from memory.)

After the contestant makes his or her choice, Monty opens one of the unselected doors to reveal a goat. The car is either behind the contestant's choice or the only other remaining closed door. Then Monty asks whether the contestant wants to switch doors or stick with the original choice.

Question: Should the contestant switch? Does it make a difference to the odds of winning?

What do you think?

** The letters to the editor in the magazine that published the solution to the problem, Parade, got quite heated.


richard said...

It was a goat, and it's a fantastic puzzle-- but I think you're remembering it from a Douglas Adams book, aren't you?

fiona-h said...

I don't recall that, altho it may have been in one. I've come across this puzzle in lots of puzzle collections... I can't remember the very first time I ever read about it :-)

richard said...

Nope, my mistake -- it was in Mark Haddon's Curious Incident of the Dog in the Night-time. I taught that novel a few years ago now! (Excellent read, by the way.)

Mark Hanington said...

I taught this one in math class a few years back. Even after seeing the math I think many of the kids (high school freshmen) still thought it should be otherwise "in the real world."

Mr. Kite said...

Don't get me started! This puzzle is so very counter-intuitive, and hinges on how precisely the scenario is set up and how precisely the question is posed.

I've seen the solution and I think there is some logical flaw in there somewhere. Some inappropriate premise or conclusion, I'm not sure which though.

Forget everything that happens before the final event. The contestant has two doors to choose from. Does it make sense to switch? I don't think so. And I don't think any past events or historical information have any real affect on the final odds of that outcome.

That's my opinion and I'm sticking with it.

fiona-h said...

In it's simple form, the solution is explained this way:
When you make your initial choice, you have a 33.3 percent chance of being correct. That means that there is a 66.7 percent chance that the car is behind one of the other 2 doors. When Monty reveals the goat behind one of the closed doors, the odds don't redistribute to give you a 50% chance of being right. Rather, the whole 66.67% chance applies other closed door, meaning you should switch.

In it's more complex form, the solution is discussed here: http://en.wikipedia.org/wiki/Monty_Hall_problem

fiona-h said...

to be completely unambiguous, these things should be made explicit:
- The host knows what is behind each door
- The host is not trying to trick the contenstant
- The host opens the door in all cases: regardless of whether the contestant initially picks a goat or the car
- The contentest is told what will happen (host will open one of closed doors; contenstant will have chance to switch) before he or she makes the first choice

With these clarifications, do you still disagree?

Mr. Kite said...

Yes, I still disagree because there is still a disconnect between the "perceived" odds based on learing new information and the "actual" odds that disregards the new information.

I assert that after the first door is opened, the odds for the next event do indeed redistribute to the two remaining doors. Why wouldn't they? I think it would be absurd to assume that one of the two remaining doors inherits the entire 66.6%.

There's a bit of slight-of-hand going on with the wording and the assumptions being made. The argumnet is a rhetorical one, not a mathematical one.

fiona-h said...

I don't think so. In fact, the wikipedia page I mentioned gives a very useful example that helps a lot:
Let's assume that there are 100 doors instead of three. Ninety-nine conceal goats; one conceals a car. All other details are the same. This time, after you make your choice, Monty opens 98 doors revealing goats, leaving only 1 still closed. In this case, wouldn't you agree that is smart to switch?

PS: And thanks a lot for not making rude remarks about my apostrophe use in my earlier comment - how embarrassing!

Mark Hanington said...

It is definately a mathematical argument, not a rhetorical one. Trust me...

Darren said...

Mr. Kite, there is no difference between actual and perceived probability. What is wrong with perception, that it cannot see actual probability?

If Monty simply hid one of the other doors at random (rather than revealing it to be a goat) after you picked a door, and said "want to change to the remaining door?", you'd be indifferent. why?
If you picked a car (P = 1/3), you should not change.
If you picked a goat (P = 2/3) there is a 50% change that you'll get the other goat if you switch, and 50% chance of getting a car. 1/3 plus 50% of 2/3 equals one half.... so you don't care if you switch or not.

By revealing the goat Monty is getting rid of that 50% uncertainty for you. In other words, there is a 2/3 chance that Monty has helped you by revealing the goat. That's why the conditional probability has changed... he hasn't *certainly* helped you, but he may have.

Simon said...

It can best be explained if you ignore the option to swap but simpy guess what is more likely to be behind the door you picked.

So first you choose a door. At this point there is definitely a 2/3 chance it hides a goat.

Whether you chose a goat or not, there is a 100% certainty that at least one goat is hidden among the other two doors. You also know that the host must open one of the other doors and reveal a goat. He will always do this regardless of whether one or both the other doors have goats.

In other words the host's revelation of a goat is a constant event that occurs regardless of what you chose. He only confirms what you already knew - that at least one of the other doors was hiding a goat.

After the host has done what was inevitable, you are then asked to restate what your door is most likely to be hiding.

The answer is exactly the same as before: 2/3 in favour of a goat.

If you did choose a goat, the other door naturally hides the prize.

To summarise:

If you originally choose a goat, the host must reveal a goat.

If you originally choose the prize, the host must reveal a goat.

Odds of having originally chosen a goat = 2/3 - unaffected by the host's constant action.

Odds of having originally chosen the prize = 1/3 - unchanged by the host's constant action.

Whatever you chose orignally, the remaining door hides the opposite.

You're more likely to have chosen a goat originally - so switch.

Everyone convinced now?


fiona-h said...

beautiful, Simon!

Uri Kalish said...

Got my own version:

Uri Kalish said...

( found you through Google blog search: http://blogsearch.google.com/ )