Comments on: Let’s Make a Deal!

By: Caveat

Caveat — Thu, 10 Apr 2008 02:52:48 +0000

Very interesting explanations of the effect and how it works.It’s a fascinating topic when discussed by people who understand it, I must say.

By: Greg Laden

Greg Laden — Wed, 09 Apr 2008 21:59:26 +0000

Paco: I was wondering if anyone was to notice that.

By: Paco

Paco — Wed, 09 Apr 2008 18:54:52 +0000

I love the continuing references to “Monte Hall” here. Because the real guy is Monty Hall, and calling him “Monte” only makes sense if you have Monte Carlo simulations on the brain :-).Is the particle in detector #1 or detector #2?

By: DTis

DTis — Wed, 09 Apr 2008 12:09:13 +0000

Sorry guys, you are mistaken.3 doors, Monty opens a goat door:Switch wins 2/3,Stay wins 1/3.Maximise by switching.3 doors, Monty opens Cadillac door:Switch loses 100%,Stay loses 100%.DON’T CARE Condition.It doesn’t matter if Monte knows or Monte guesses.Therefore, what Monte “knows” is irrelevant. Irrelevant aint equal to essential.

By: Tim Bartik

Tim Bartik — Wed, 09 Apr 2008 12:04:03 +0000

Two points:1. I think it’s easier to get intuition if you assume a million doors, with 999,999 goats and one car. You pick one door. You are almost certainly picking a door with a goat behind it (or at least, 999,999 times out of 1 million you are). Monty then opens up 999,998 doors to reveal 999,998 goats. Clearly, you should switch. One way to put it is that Monty has filtered out most of the bad choices from the remaining doors.2. The entire problem rests on the assumption that Monty is fair and not devious, and always, regardless of the contestant’s choice, opens up goat doors. If Monty is devious, and ONLY opens up goat doors if you have picked the door with a car, then you should NOT switch.

By: Armchair Dissident

Armchair Dissident — Wed, 09 Apr 2008 08:10:14 +0000

The big problem here is that the Monty Hall effect doesn’t guarantee you’ll win by switching, but it does increase the probability.People look at the last two doors, and assume that – despite the way the game is played – there is a 50-50 chance that the car is behind either door. There isn’t. There is a two-thirds chance that the car is behind the door you haven’t picked, because the door that is openned is not arbitrary.Remember that when you start there is a two in three chance you have lost. This probability doesn’t suddenly change because a known loosing door is opened. Similarly, the probability that the other closed door contains the car hasn’t magically leapt to one in three — it’s still actually one in three out of the three doors! But you now not only know the location of one of the loosing doors, but it was opened in a rigged game — it was known before the door was opened that it was a looser. It’s the fact that it was known to be a loosing door that creates the Monty Hall effect.You still have that original one-in-three probability that you picked a wrong door, because you didn’t know the location of ANY of the loosing doors when you picked yours. And you can argue that the probability of the other door containing the car is still one-in-three, but if you pick the door that was just opened, you’d be an idiot, so the remaining door has a two-in-three probability that it contains the car.You can write it out:D C D (Donkey Car Donkey)Pick 1, door 3 is opened. Switch, you winPick 2, either door 1 or 3 is opened. Switch, you loosePick 3, door 1 is opened. Switch you win.Note, again, that the Monty Hall effect works purely because Monty *knows beforehand* which door is the winner.

By: Greg Laden

Greg Laden — Wed, 09 Apr 2008 07:57:11 +0000

The essence of the Monty Hall effect is that Monty always knows where the car and the goats are, and always reveals the goat and never reveals the car.

By: DTis

DTis — Wed, 09 Apr 2008 01:32:18 +0000

First, the essence of the problem is the division of the solution universe into two subsets; subset A (your door), and subset notA (the rest of the universe). Obviously, subset notA is N-1 times more likely to contain the car than is subset A (assuming N=the total number of doors).You never add information to the system. You know that there are N-1 goat doors, of which at least N-2 are in subset notA. The beautiful Wanda can open doors in notA all day long and (unless she opens all the doors) all she is doing is demonstrating that there are at least N-2 goat doors in notA. We already know that. Therefore, the odds never change. Likewise, it is completely irrelevant whether she, or Monty, know what is behind a door. It is given, in the problem definition that Wanda’s picks are goat doors.So, your choice is actually between your one door (your original choice), or all the other doors. Always go with N-1 rather than 1.

By: Brandon

Brandon — Wed, 09 Apr 2008 01:29:23 +0000

If you pick the wrong door, do you actually get to keep the goat?

By: A Lurker

A Lurker — Tue, 08 Apr 2008 23:23:40 +0000

It is important to remember the reasoning here only works if Monty knows in advance where the car is and must reveal a goat each and every time. If Monty was just guessing when he had door number 3 opened then the odds revert back to fifty-fifty. If Monty gets to choose whether or not to reveal a goat and make the offer then he might be more likely to make the offer when your guess is true.