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I was thinking about the Monty Hall problem and I thought of a possible intuitive explanation:

  • You choose a door.
  • Monty gives you the option of sticking with your original choice or instead choosing both of the other two doors.
  • If you decided to switch (which now becomes an obvious choice), Monty first opens the door with the goat behind it (say, to add to the excitement), and then opens the other door.

My question then is, is this reasoning flawed? Is this even the same problem as before? Because now, choosing to switch from one door to two doors becomes quite obvious, and so does the $2/3^{rd}$ chance of winning the car on switching.

SilverSlash
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    I think this is one reasonable way of explaining the problem, yes. Never thought of it that way, but I like it! Good job! – 5xum Jun 09 '16 at 08:13
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    This is how I first figured out why the odds favor switching, and from this line of reasoning you can recognize a VERY important subtlety: "Monty knows". What is meant by that, is that the reason it is favorable to switch is because you knew monty would open a door with a goat behind it. If Monty didn't know, then in the event that you found yourself with one door open which had a goat, you would actually have 50/50 odds whether or not you switch. (!) – Justin Benfield Jun 09 '16 at 09:29
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    This is nice. The only problem is that is a bit difficult to vote an answer since the answer is: 'no, your reasoning is correct' – MiKo Jul 22 '16 at 08:56
  • Seriously appreciate this, thanks for sharing. Made it far more clear for me. :) – nhooyr Feb 12 '21 at 10:22
  • Does this answer your question? [The Monty Hall problem](https://math.stackexchange.com/questions/96826/the-monty-hall-problem) – NNOX Apps Dec 31 '21 at 05:45

5 Answers5

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I cannot find a flaw in your reasoning.

My own reasoning (if you are interested).

Someone who sticks to his original choice will win if his original choice was correct.

Probability on that: $\frac13$.

Someone who switches will win if his original choice was wrong.

Probability on that: $\frac23$.

drhab
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  • I like that... 1) *I think it's door A* 2) **GOAT BEHIND DOOR B REVEALED** 3) *Actually, I thought it was behind one of the other 2 doors all along, so let's go with door C* – Juergen Apr 15 '21 at 22:46
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I believe the best way to intuitively understand the Monty Hall problem is by playing the game with a $100$ doors, $99$ goats and one supercar.

I can choose a door, doing so will give me a probability of $1\%$ of choosing the car. The host then opens $98$ doors, showing $98$ goats. At this point I know that the door I chose either contains the supercar (with a probability of $1\%$) or more likely a goat ($99\%$).

Now I'm given the oppurtinity to switch doors, it's clear that doing so will increase my chance of getting the supercar to $99\%$.

Mathematician 42
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    This explanation never really did it for me. Why would this be equivalent? Wouldn't the equivalent situation be one where you have $99$ goats and $1$ car, then you pick one door, the host opens *one* other door, and asks me if I want to switch? – 5xum Jun 09 '16 at 08:39
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    I mean, I understand the original problem, I just never considered this a good "intuitive" explanation. – 5xum Jun 09 '16 at 08:39
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    Well actually, it doesn't matter how many doors he opens, the chance will always increase if you pick another door. I chose to open $98$ doors to get an extreme situation were it becomes clear that you should switch. If he opens one other door, switching to some other door is still better, but the increase in chance isn't as dramatic. – Mathematician 42 Jun 09 '16 at 09:05
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    I understand that, but what I am trying to say is that for me, the intuitive situation is not "he opens all other dors", but "he opens **one** other door", and then, switching being the better option is no more obvious than in the base case. – 5xum Jun 09 '16 at 09:06
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    Work backwards then, opening $98$ doors clearly is awesome, $97$ is still awesome, $\dots$ , $50$ better, but not that great, $\dots$ , $1$ still better than nothing. You can work your way down from one extreme situation that is very intuitive to the case you want. For the first so many cases, the intuition is very clear, so why wouldn't this extend down the sequence. It's intuition after all. – Mathematician 42 Jun 09 '16 at 09:12
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Yes this is the same problem. Since in the original solution Monty opens one door, and thus that one is bad, hence choosing that one too does not make a difference.

My favourite way of convincing people is to, instead of considering 3 doors, consider 1000 doors. If, after you have choosen 1 door, Monty opens all other doors except one, then obviously you will switch.

Ove Ahlman
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Here is the way I convinced myself of the truth that: "the odds of winning the car by switching the original choice is 2/3," given that the host knows the locations of the car and the two goats in advance (stated fact in the problem.)

I changed my thinking to consider that the host is not choosing a DOOR, but rather he is choosing a PRIZE...i.e. he will always pick a goat-door in every case, regardless of its number (assumption that must be inferred because we must assume he won't give the car away by opening its door.) I think much confusion is generated by the wording in the problem (and much of the debates) that the contestant chooses "Door 1" and the host chooses "Door 3."

Then when we look at the three possible arrangements of prizes and doors, and all possible door-choice and prize-choice sequences (see matrices elsewhere), the 2/3 probability becomes clearer.

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In my opinion this intution is not quite right. you say that the incident "the prize is behind door 1 OR door 2" is equivalent to switching to doors 1 OR 2 when the instructor opens, lets say, door number 1 and reveals a goat.

but when the instructor reveals a goat behind door number 1, the probability of the incident "the prize is behind door 1 OR door 2" is not the same as before, it shouldn't stay 66% because there is new information we added to the game.

so pure intution - this is not a convincing one in my opinion.

more convincing(imo) is this one: if the prize is behind the door i chose - I DONT WANT TO SWITCH, but the probability that the prize is behind the door i chose is exactly 1/3. and the probability that the prize is behind any other door is exactly 2/3. in this case, I WANT TO SWITCH. this means that in 66% of the cases you would need to switch doors in order th win, and in 33% of the cases you would need to stay with your choice in order to win.

mcr0yal
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  • @Moo yes, the first answer says almost the same as my intuion, which i think is the most "intuitive" way to see this. – mcr0yal Mar 31 '22 at 14:51