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I'm trying to understand the Monty Hall problem. When the contestant switches the door the probability of them winning a car is higher than if they continued with their first choice. Why would the probability change from 1/3 go up to 2/3 if both doors had the same chances of being opened?

Update: There are many posts about this subject in SO but none of then answer my question. Let me re define my question.

I know that there are many models out there that shows that if you change the doors you have 2/3 of probability (higher than 1/3). After Graham Kemp answer I will update a scenario that I have in my mind.

You choose door1 (1/3) and the host opens the door 2 (the goat). By the current solution the door 3 has 2/3 of probability to be the one with the car. Now let's think a little change. After the host open the door 2 he will give you a coin with numbers 1 and 3 and you should torn it. Now is not a choice, you should stay with the number you get in the coin. In this new scenario based on the previous model the door 3 has 2/3 of success probability than door 1. With clearly makes no sense. In this new scenario why door 3 is not 1/2 of probability? I am pretty sure that a model with 10M attempts will show 50% for each door. Is it correct?

IgorAlves
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  • If "both doors had the same chances of being opened" is talking about the door Monty Hall opens, then the point is that he only opens a door with a goat behind it, and that affects the information about the door he does not open (perhaps he could not, if it does not have a goat) – Henry Mar 02 '21 at 01:11
  • The main Monte Hall answer seems to be https://math.stackexchange.com/questions/96826/the-monty-hall-problem – Henry Mar 02 '21 at 01:13
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    In my experience it helps to rephrase the problem as follows. "Suppose you pick a door, and then are given the opportunity to switch from your door to whichever is best of the two remaining doors. Should you switch?" The whole theater around opening a door to reveal a goat is just window dressing. – Noah Schweber Mar 02 '21 at 01:13
  • whining is a word but it does not mean the same as winning – Will Jagy Mar 02 '21 at 01:13
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    @WillJagy To be fair, one's odds of whining in the Monty Hall problem do often increase. :P – Noah Schweber Mar 02 '21 at 01:14
  • @NoahSchweber I don't recall the details: evidently considering 1000 doors rather than 3 shows a more intuitive outcome – Will Jagy Mar 02 '21 at 01:16
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    @WillJagy I think that variant is to consider 1000 doors and have the host open 1000-2 of them, so you're again left with a choice between two doors. – Noah Schweber Mar 02 '21 at 01:18
  • @NoahSchweber how many cars and how many goats? – Will Jagy Mar 02 '21 at 01:19
  • Does this answer your question? The Monty Hall problem – Ethan Bolker Mar 02 '21 at 01:20
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    The odds of me whining about yet more discussion of the Monty Hall problem are very high. Probability theory tends to be counter-intuitive until you learn to forget your intuitions and rely on the combinatorics. – Rob Arthan Mar 02 '21 at 01:21
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    @WillJagy One car, 1000-1 goats. (At a certain point you do have to wonder where Monty is getting all these goats.) – Noah Schweber Mar 02 '21 at 01:21
  • @NoahSchweber I don't knw if they still do this: in the hillside grounds around MSRI, they used to hire a herd of goats to eat the vegetation, lower danger of fire. – Will Jagy Mar 02 '21 at 01:24
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    @RobArthan If I wanted to pick a fight with you, I would say that intuition in probability/combinatorics problems is both indispensable and (for students new to the topic) often wrong. Therefore, the goal for the new student should be to accept that their intuition may be defective and to stretch their intuition by studying the material, rather than forgetting about their intuition. However, since I don't really want to fight with you, let's pretend that I didn't just make this comment. – user2661923 Mar 02 '21 at 01:45
  • @user2661923: I am not a teacher of mathematics (other than responding on MSE) and I have a great respect for those like yourself who are on the frontline of mathematics education. I am sure your point about stretching intuition rather than forgetting is very valuable to your students. So let us definitely agree not to fight about this. – Rob Arthan Mar 02 '21 at 01:57
  • @RobArthan I have thought of a compromise. I am going to invent a time machine and go back and murder one of Monty Hall's grandfathers. Then, staying in the pertinent timeline-reality, mathSE never got infested, we never had these comments, and I never invented the time machine. – user2661923 Mar 02 '21 at 02:07
  • There is not a well defined question here. Voting to close as unclear what you are asking. – Ross Millikan Mar 02 '21 at 04:23
  • @RossMillikan, yes close it. And kill my pain. I need to sleep and tomorrow I have a lot of code to be done. Maybe SO Math is only for true false questions not to explore thoughts, like maths should be. Imagine what would be your comment in SO of Philosophy - you would spend many decades to find an answer (if possible). https://philosophy.stackexchange.com/ – IgorAlves Mar 02 '21 at 04:58

1 Answers1

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You select a door at random, knowing nothing about the placement of the car, other than that is presumably behind any particular door with equal probability (1/3).

Monty opens one of the other two doors, knowing the location of the car, he always chooses to reveal a goat.

You are asked if you want to switch to the remaining door.

  • If the car was behind your door, you cannot win if you switch. It was there with probability 1/3.

  • If the car was not behind your door, you surely win if you switch. It wasn't there with probability 2/3.

Therefore the probability that you win if you switch is 1 minus the probability that the car was behind the door which you selected. IE: 2/3

Graham Kemp
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  • This is the best answer I've read so far. Almost convinced but not 100%. I think there are something in this second sentence that doesn't smell good. I know that there are many models that proves that we should change but I want to compare with my feeling. Let's do the follow, you choose door 1 ans after the host open a door 2 (with the goat) he will give you a coin with the 2 numbers 1 and 3. He ask you to turn the coin and you will have what you get in your result. Is that meaning that the side 3 has 2/3 of probability compared with side 1 of the coin? – IgorAlves Mar 02 '21 at 03:18
  • Sure. Now an unbiased coin has probability of $1/2$ for showing 1, and likewise for showing $3$. That does not change, but, when the coin shows 1, there is a $1/3$ change of winning, else when it shows 3, there is a $2/3$ chance of winning. Therefore the total probability of winning when deciding to switch or not by coin toss is : $$\tfrac 12\cdot\tfrac 13+\tfrac 12\cdot\tfrac 23 ~=~\tfrac 12$$ – Graham Kemp Mar 02 '21 at 03:45
  • Another way of looking at the Monty Hall situation is that Monty says: "There is a 1/3 chance that the car is behind the door you chose. You may take whatever is there, ooorrr you can give me a goat and take whatever is behind both the other doors." – Graham Kemp Mar 02 '21 at 03:51
  • @zwitterion Since you are forced to choose whichever door the coin says, at that point it no longer matters what you know. So it is safe now for Monty to open the other two doors for a moment and then close them again so you get to peek behind them, even before you flip the coin. Suppose he does that, and suppose you see that the car is behind door number $3$. But due to the coin toss you have only $\frac12$ chance of winning the car. Does that mean that the car has only $\frac12$ probability of being behind door number $3$? Of course not. – David K Mar 02 '21 at 03:54
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    The coin tells you nothing about the probability of the car being behind door $3.$ What the coin actually does is to wipe out any advantage that you could have gained by understanding the probabilities of this game. – David K Mar 02 '21 at 03:56
  • @GrahamKemp Following the same logic if y have 5 doors and the host opens 3 you end up with your choice 1/5 and the other 4/5. That will give you not 66.66%(2/3) but 80%. So I would change the door. It seems that as much more doors, higher is my chance if I change the first option since it is going to zero and the other option is going to 1. But if you have infinite doors that will force me to think that my chances are almost 100%? This is not correct. – IgorAlves Mar 02 '21 at 04:48
  • No, it is correct - When you choose one from $n$ doors (one of which hides a car), and $n-2$ other doors which hold a goat are then revealed, therefore the probability that you win if you switch will be $(n-1)/n$. This will indeed trend towards a limit of $1$ as $n$ increases towards infinitude. – Graham Kemp Mar 02 '21 at 07:57