The only way I can get my head around this is to think about it as being like finding a ship in Battleships
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In university, I had my explanation of the Monty Hall problem to a friend grow very heated as my friend steadfastly refused to believe that it worked. I finally grew so frustrated that I physically drew 20 cards from a deck and told him that once I’d revealed the other “doors”, I’d pay him $20 if the alternate to his choice wasn’t the Ace of Spades.
I spread the cards, he chose, and then one by one I looked and flipped the cards, conspicuously skipping over the Ace of Spaces. Except that I flipped the next 18 cards as each one proved to not be the Ace of Spades leaving only his choice and the last alternate.
I could see fate as well as anyone, but gamely asked if he wanted to switch. He said no, and dutifully flipped over his original choice to reveal the Ace.
Shoulder’s slumped, I could only whisper “but it’s true” as I fished out my wallet.
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The postscript was that he laughed, explained that actually my illustration finally made the Monty Hall problem make sense and, because he was a scholar and a gentleman, didn’t collect on the $20.
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I would agree with you if I had to make that choice BEFORE door 3 was revealed.
Pretend you have 2 buckets. Bucket 1 is a 1 gallon container and bucket 2 is a 2 gallon container. One of them holds half a gallon of water. What are your odds of selecting the bucket with the water?
Oh… and also there is another bucket there, but it clearly contains no water.
Ok wait… is that the issue? Somehow the bucket that clearly contains no water is STILL considered in the odds??
I’m trying to figure out how to explain this without direct reference to to the numbers. Ironically, it’s using the wrong math that confuses people. When you select the first door, everything is unknown, and it is more likely that you will be wrong than right. After your selection, and another door is eliminated, your next choice is different because you’ve already selected a door. So you are not merely making a choice between two doors, you are making a choice between one door that is likely to be wrong, and one other door. All you really need to know to make the decision to switch is that your first choice was probably wrong, and there is only one other choice.
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I dunno but I imagine a lot of cocaine would be involved
Yes, if it was among your original options. That’s why your original choice has a 1/3 chance of being correct if there are three doors (with a 2/3 chance of being correct if you switch), 1/10 chance of being correct of there are 10 doors (with a 9/10 chance of being correct if you switch), a 1/100 chance of being correct if there are 100 doors (with a 99/100 chance of being correct if you switch), and so on. Make sense?
If not, I strongly recommend watching the video that accompanies the article. The mathematician does a very good job of explaining how this works in real-world terms.
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Okay, so the thing is you are wrong. Like, that’s a fact. You might as well be arguing that 2+2=5. I think I’ve given a lot of explanations of why, but if you are still arguing with facts then there is nothing to be argued. Run a simulator, or get some cards and do it with a friend.
As for your explanation with 12 possibilities, it doesn’t work because those possibilities are not equally probable. The car is equally probable to be behind door A, B or C. Given that the car is behind door A, B or C it is equally probable that you pick door A, B, or C. However, given that you picked the door the car is behind, the probability of Hall opening either door is 50%. Given that you picked a door the car is not behind the probability that Hall opens the open losing door is 100%. So A/A/B and A/A/C are only half as likely as A/B/C.
Now we’ve could make a chart with 36 possibilities - car in three places, three choices of doors, hall opening leftmost or rightmost remaining wrong door, and you switch or stay. Of course in the case where you pick a wrong door, Hall’s choice of rightmost or leftmost ends up opening the same door, but that way we end up with all scenarios equally probable. I’m not going to write it all out, because it’s the same as everything else. Stay gives 1/3, switch gives 2/3.
Imagine you don’t have a choice at all. You pick a door, and that’s your pick. Next, the host dramatically reveals a non-winning door you didn’t pick. Finally, the host opens your door to see if you won. You picked a door, so you had a 1/3 chance of winning the game, we know that.
That’s all the opening of the door is, drama. It doesn’t actually reveal any information about whether your pick was correct, any more than it would if you were not given a choice and the only point of doing it was to build tension for the audience.
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I feel like I’ve been presented with a number of “optimal strategies” by economics professors that don’t feel that optimal. Things like, “burn BOTH houses down” or “drop the baby off the skyscraper” or “let the cannibal eat your mother.” So please forgive my skepticism when a mathematician presents me with a scenario where I can only ever hope to win 50% of the available goats, and then suggests how to decrease that likelihood.
But after he reveals the door that does not contain the prize is shown, there is only 2 possible locations for the prize.
I really think you are calculating the odds of switching at the beginning of the scenario (3 possible locations for the prize) while I’m thinking about the odds after there are only 2 possible locations for the prize.
Statistics and economics are not the same thing. There is only one correct answer to this problem. The correct solution is not a spectrum of opinion.
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Think of it this way. There were only two possible locations for the prize from the very beginning. It’s either behind your single door or Monty Hall’s two doors. Would you prefer to look behind your single door, or both the other two? You’re defining these two sets. Monty Hall only exists in your mind. When his hand turns that doorknob, it’s actually your hand. You’re half way to opening both doors. Now open the other door.
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I get it. I’m suggesting why people might take these scenarios with a grain of salt.
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There are also only two possible locations for the prize if there are 100 doors, you’ve picked one, and Monty opens 98 other doors revealing no prize. Are the odds that you have picked the prize door on the first try 50/50? Should you switch? Why or why not?
Solution: Of course, the odds that you have picked the correct door on the first try are 1/100. There is a 99/100 chance that the car is behind one of the other doors. Monty knows which of the doors there is a car behind. Of the 99 other doors, he opens 98, none of which contain the prize, because he would not open a door that had a car behind it. He has thus done you a huge favor: you now know that there is a 99/100 chance that there is a car behind the single door that he didn’t open, vs the 1/100 chance that the door you initially picked has the car behind it. You’d be foolish to remain.
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My intuition, good or for ill, is to assume that both closed doors are affected equally, and that the prize should be equally likely behind either. What is the mechanism that makes that false?
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I just edited my passage to explain it better – see if that helps!
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And when there are only two remaining possible locations for the prize, the odds are 1/3 that the original choice contains the prize and 2/3 that the remaining door does.
But, it’s not subjective or hypothetical situation in which reasonable people can agree or disagree–it’s just demonstrable, objective reality. Another person posted this earlier, but it’s a pretty interesting thing to watch when you run the simulation 1k times and see the percentages fluctuate within a couple points of 66% and 33%.
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But there aren’t two possible locations; there are three. The number of unrevealed doors doesn’t matter. We aren’t re-randomizing and placing the object behind one of the remaining doors. So your odds of picking the right door start at 33% and don’t change.
Telling you that one of the remaining doors is empty doesn’t change that percentage—we knew at least one of those doors was empty when we made our choice.
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And yet you and I both know that “reasonable people” is not a viable concept. For instance, you are here arguing with someone who agrees with you.
