Monty Hall Problem

24 点作者 lnyan将近 3 年前

15 条评论

antris将近 3 年前

The fact that the show host knows what's behind the doors and never chooses a door with the prize behind it is crucial. The host leaks information about what's behind, as he chooses a door.I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. Then everyone who heard of the problem before piled against me like I'm a dumbass, but never really explaining why I was wrong.Yes, I'm still salty about it.

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em500将近 3 年前

The Month Hall Problem has been discussed many many times on HN and elsewhere. I've read many of the discussions (but of course not nearly all of them. My observations:1. The linked Wikipedia article is very extensive, and already covers usually all the points and insights that forum posters bring up. And a lot more.2. Few people will probably bother to read the entire (or even most of the) Wikipedia article, because it's so extensive.3. I'll copy and highlight the IMO most crucial part the article here:The most famous formulation, a column in Parade magazine in 1990, went thus:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?In this formulation the solution is ambiguous. Switching is *not necesarily* the optimal strategy (it depends on the precise strategy/behavior of the host, which is not well formulated).Under the unambigous "standard assumptions":- The host must always open a door that was not picked by the contestant.- The host must always open a door to reveal a goat and never the car.- The host must always offer the chance to switch between the originally chosen door and the remaining closed door.switching is the better strategy. But in this formulation the reason is also a lot more obvious.

nkrisc将近 3 年前

Here's how I finally understood the problem intuitively:If your initial pick is correct, then switching loses 100% of the time.If your initial pick is incorrect, then switching wins 100% of the time (because you picked wrong, and host will reveal the other incorrect door, leaving only the correct door to switch to).So switching always inverts your initial guess. Well what are the odds of picking incorrectly on your initial guess? 2/3. Switching wins 2/3 times.

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superb-owl将近 3 年前

The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.

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cupofpython将近 3 年前

I had this problem beaten to death in my probability and statistics courses in college.To make the problem much more obviously intuitive, simply imagine a thousand or a million doors instead of 3.. and after you make your pick - the host reveals all doors except one to be "failures". Your choice was very obviously 1 in a million, and because the reveal procedure that followed is conditional - if the "success" was behind ANY of the doors you did not pick then it WILL be the only door left in front of you.It is only a 50-50 IF the revealed doors were opened randomly. In this case, sometimes the "success" would be revealed by accident - and so IF you do happen to find yourself in a position of only having 2 doors left THEN it was more likely to happen if you had actually chosen the "success". So arriving at that situation at all is the conditional information that alters the probability of your original choice - but in this case from 1/3 to 1/2 instead of 2/3. (this is how the probabilities in "deal or no deal" work btw)Even knowing the solution, it is a good reminder about the power of information derived from systemic processes. the information leveraged in the processes that lead to circumstances is just as important as the circumstances themselves.The key general point that bypasses most peoples intuition the first time they hear the problem is that the revealed box (or door, etc) is not chosen at random but by a conditional process. The condition is that the host reveals a "failure". In order to pick a failure, he must leverage knowledge about where the "success" is, else he risks revealing it by mistake. Because knowledge of the "success" is embedded into the decision process, more information about the "success" is revealed to the observer which allows them to improve their position.

chrisbrandow将近 3 年前

I think the 100 door example is good, but is often not described in a way that was intuitive to me. I've added bit that I found very helpful.The entire trick of the game is that Monty opens 1 (or 98) of the remaining doors. That is because it is completely irrelevant to the odds, but has big impact on the contestants perception.It is not obvious, but Monty actually does two separate and unrelated things:First, a step that basically goes unnoticed, is to propose a new game to the contestant in which the contestant opens either 1 door or 2 (99) doors chosen by Monty, rather than the original game which was to select 1 door out of 3 (100). Obviously, the new game gives them the chance to make a choice that has 2/3 odds (99/100) versus the original game whose odds were 1/3 (1/100).His second step, which is typically the only step noticed and which very effectively misleads the contestant, is to open 1 (98) of the doors he selected.The result is that clearly the 2nd game gives the contestant a definitively better choice, but this is obscured by the fact the contestant only sees two closed doors, so they believe they are comparing 1 door to another, and thus only have 50-50 odds.

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paxys将近 3 年前

Lots of "intuitive" explanations of the problem shared here, but here's one that works for me.You pick door A. The hosts says you can either keep that door, or instead pick both doors B & C. If either one of them has the prize you get to keep it.You opt to switch. The host opens B, which turns out to be a goat, and then opens C.Regardless of what C contains, it's obvious that switching was the better option.

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ksaj将近 3 年前

An easy way to word this exact problem in a way that makes clear the reason you should always change your choice is:I have 52 playing cards. I know which one is the Ace of Spades. Point to one of the cards and leave your finger on it.Okay, without telling you whether you are right or not, I'm going to remove every card, leaving only yours and one other on the table, that is the right card if the one you picked is not, or an indifferent card if your choice was right.Do you wish to choose the card I left on the table, or do you still think your 1/52 choice was correct?Notice that I followed the exact rules of the game, but changed the wording and used a lot more "doors." There was a 1/52 chance you would pick the right card, and even after I remove all but one of the remaining cards, it is still a 1/52 chance you originally picked the right card, and a 51/52 chance that the one I left behind is the right card.Now just make it 3 cards, and use the traditional script, performing the exact same steps, and you'll understand why you should always change from your original choice.

headsupftw将近 3 年前

To me the key to understand the problem is paying attention to this detail: Monty knows where the car was so he wasn't just randomly opening A door. He always opened a door that didn't have a car behind it. It's this non-randomness that's giving you more information about the remaining door.

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k3lsi3r将近 3 年前

The host revealing the door is irrelevant, right?An equivalent offer would be if you had the chance to either: - keep your door OR - choose both of the other doors (since you know at least one of the other doors is a goat/non-prize)

paxys将近 3 年前

One thing that trips a lot of people up is that the problem is never explicit about the knowledge or strategy of the host. Would he offer you the choice to switch if you initially picked the wrong door?If I'm playing this game in real life, my intuition isn't going to be calculating the correct mathematical odds, but rather thinking "this guy is trying to screw me".So a 1/3 chance that you can be sure of is better than a 2/3 one which is based on trusting your opponent.

VikingCoder将近 3 年前

I didn't believe the result. So, I had to sit down and decide to write the source code to simulate it, and as I'm typing the code my internal monologue is saying "if you stay, 2/3 of the time you lose, but if you change, 2/3 of the time you win," and I was like, "CRAP!"It was like - my fingers knew how to type the correct solution, and once my brain saw my fingers type it, my brain finally caught up.

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dtagames将近 3 年前

What's missing from this thread is not that Monty always reveals a goat... It's that he never reveals the car! In other words, the "you can switch doors" routine is always part of the show.By removing one door which does not have the car, Monty has narrowed your best choice to one of the other two.

ortusdux将近 3 年前

My favorite Monty Hall explanation:"I have randomly picked a number between 0 and 1 million, and I want you to guess my choice.""Lets go with 408,235.""I will narrow it down for you. Lets eliminate 999,998 incorrect guesses, leaving either 174,999 or your guess of 408,235. Would you like to change your guess?"

PUSH_AX将近 3 年前

The way I understood this was that if you pick a door, the chances are you didn't pick the prize, given that you probably didn't pick the prize and the host shows you the other non prize door, it makes sense that switching will probably give you the prize more than it doesn't.