Do you know what the Monty Hall effect is? Let me explain.
OK, I’m Monty Hall and you are a hapless game show guest. I show you three doors and tell you that behind one door is a nice brand new car, and behind each of the other two doors is a goat.You get to pick one of the door. Say you pick Door Number 1.Now, I have my lovely assistant Gwenda (assume her name is Gwenda) open Door Number 3 to show you that there is a goat behind Door # 3. I give you the option: Are you going to stick with Door Number 1, or are you going to switch to the only remaining unopened door?Statistically, scientifically, what is the correct answer to this question? Obviously, the probability that the car was behind the door you chose has not changed, so you should not change your mind without any other information, right?BBBBZZZZTTT Nope, you would have gotten that wrong. In a sense, the probability that the car is behind the door you picked has not changed (or at least you can think of it that way if you like), but the probability that the car is behind the still closed door has gone way up. And, this means you should switch every time.Here is a place to go to actually play the Monte Hall game several times and see which strategy actually gets you more cars.Here is a post on the Evolution Blog that reports on a number of very interesting findings that link Monte Hall, the Goat and some Monkeys. Oh, and some M&M’s … Enjoy. By the way, there is yet another strategy for this particular game, with the car and the goats: Prefer goats. You’ll win way more often.
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If the game is played with a mechanism, then the usual mathematical analysis applies. However, on an actual game show, the contestant has no idea how many other players are involved. For example, if there is a test audience to gauge the popularity of the contestant, then if the perception is ‘egg-head know-it-all’ then the contestant is guaranteed to lose every time.Mathematicians are fond of abstracting real-world situations and then analyzing their abstractions, and they always naively assume their analyses map back to reality exactly. Pointing out their mistakes gets nowhere with them.
It’s funny how hard your mind struggles against believing the odds on this are. The game really does an excellent job illustrating the point.
Lummox -What?Please explain your point again. Are you suggesting that Monte Hall will play favorites? I’m not sure that really changes the point of the puzzler.
Lummox: Yes, explain yourself, please!!!I happen to think you are dead on correct, but do go on.Also, let’s now work on applying this to that other TV game show with the suit cases and the babes.
Lummox: Mathematicians are fond of abstracting real-world situations and then analyzing their abstractions, and they always naively assume their analyses map back to reality exactly. Pointing out their mistakes gets nowhere with them.FAIL.As if the reason this analysis is interesting is that we think it might improve our chances of winning on Let’s Make a Deal.
Lummox,I think you assume that mathematicians think there is a 2/3 probability of winning from the beginning of the game show. This is not the case. The Monty Hall effect only governs the last part of the game show, where only the last contestant gets to choose the prize from three doors. The probability for this problem, 2/3 chance of winning if you switch, does hold for that part of the game according to the rules of the game. For that part, it maps completely to the real world. The simulation is just a chance to test how valid that mapping is.Mathematicians map probability to such games constantly, just not how you think. In the suitcase game that Greg mentioned, the offer from the bank is determined using expected values from probability. It is always a lower estimate then the true expected value, but it is not randomly generated. Casinos and video gambling use expected values to help determine profit, and it maps rather accurately. Insurance companies also use mathematicians to determine policy costs to insure profit.Also physicists, biologists, and chemists all abstract the real-world and then analyze it. Any form of measure is an abstraction. I doubt we will ever have the ability to measure or model with 100% accuracy.I will admit that I’ve seen mathematicians, and others, who assume that because the math works for a situation, it must be true. That always seemed to me to be a lesson on why experiments should be designed to disprove a hypothesis.
..??Lummox seems to imply that these ‘TV games’ are somehow rigged; that Monty would switch the prize for an unpopular player. Or that they would ‘switch’ suitcases or otherwise rig that game somehow..??I thought we already had this controversy in the late 50’s, something about rigging the Twenty One game show.flummoxed
For someone like me, with no mathematical ability, it doesn’t make much sense the way they explain it over at NYT. I get that the odds of winning increase with fewer choices but I don’t understand how switching leads to more wins, since it seems like a 50-50 chance.This is bringing back bad memories, guys….Anyway, since I’d be just as happy with a goat as I would with a car, I win no matter which door I pick! Nyuk!
There’s 2 phases, phase one where you have 3 doors to chose from, and phase 2 where you have 2 doors to chose from.Phase 1 doesn’t count. It doesn’t even exist. One goat will be removed anyways, no matter what. So you always have 50% chances of chosing the car, even if there is *apparently* 3 doors to chose from.So at phase 2, stick to the same door, or switch, as you please, it doesn’t matter at all, except to Greg Laden, who is mistaken. 😉
Arthur,I am not mistaken. Nor am I correct. This is not my thing, I’m just reporting it.But, you are mistaken! I was thinking the same thing you were thinking but then it all came home at once and I realized the Monty Hall effect is real.Consider the following two possibilities.The condition is that you win if you end up with the queen of hearts.First, you pick a card randomly from a deck of 52 cards, and place it face down on the table.Then, I take deck of cards, and I remove one card that is the queen of hearts, if in fact this deck of cards has a queen of hearts, and I place that down on the table.Now, you get to pick the card you pulled from the deck or the card I pulled from the deck. Which one is more likely to be the queen of hearts? The one I pulled from the deck, obviously.Now, imagine this version of Lets Make a Deal:Pick a card any card, and place it down on the table. You are hoping it is the queen of hearts.Now, I’m going to take the remaining 51 cards, and if the queen of hearts is in there, I’ll lay it down on the table. If it is not, I’ll lay down some random card, but you can’t see it so you don’t know. It is possible that you already picked the queen of hears, in which case the card I put down is not the queen of hearts. But it is also possible that you did not pick that card, which means that the card I put down IS the queen of hearts.Would you stick to the card you picked, or switch? You would switch.The car/goat thing is the same.There, did I explain it better than the NYT, or what? Huh?
Arthur – You should always switch. After your first choice (say, Door 1), the chance of the car being behind doors 2 or 3 is 2/3. Monte knows where the car is, so he always shows you a goat. So 2/3rds of the time, he has told you exactly where the car is. 1/3rd of the time, you’d do better to stick with your first choice. Therefore, you increase your chances by switching.
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.
If you pick the wrong door, do you actually get to keep the goat?
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.
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.
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.
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.
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.
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?
Paco: I was wondering if anyone was to notice that.
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.