I have tried asking chatgpt, i have tried searching animations, I just dont get it!
Edit: I finally get it. If you choose a wrong door, then the other wrong door gets opened and if you switch you win, that can happen twice, so 2/3 of the time.
In: 296
The thing that finally made it click for me was an exaggerated example.
Suppose, instead of starting with 3 doors, we start with 100. After you pick one door, the host opens 98 doors, leaving one other unopened door. Which do you think is more likely: you correctly picked the winning door out of 100 doors, or the other door has the grand prize behind it?
Since you already understand the problem (as in what it is) I am going to modify it in a way that made it click for me
Instead of the 3 doors.
Let’s assume there are 3 *million* doors and only one of them has a prize.
You pick one of them at random. And then they get rid of all but 2 of the doors so that one has the prize and one has nothing. Just like normal
Now the idea is that you should definitely switch to the other door, right?
So Ask yourself.
Did you pick the correct door out of 3 million on your first try and then the remaining door has no prize?
OR
You picked one of the 2,999,999 wrong doors and the other door has the prize behind it?
You probably think it reasonable that you picked one of the many wrong doors. So your best course of action is to switch to the remaining door, a door remaining precisely *because* it probably has the prize.
I’ll walk you through, hopefully it helps.
First off – when you’re looking at probability, we often frame it as “The chance that I’m right,” but it’s actually more helpful to think of probabilistic events in the context of, “What’s the most likely thing to happen?”
For example, take rolling a six-sided dice. What are the odds of you rolling a 6 within six rolls? Well, an intuitive (but wrong) answer would be to say, “well, there’s a 1 in 6 chance, and I have six rolls, so I think they’re pretty good! 100%!” And that’s wrong.
If you look at it that way, you’re computing the wrong probability. You need to rephrase the question: It’s not whether you *will* roll a six. It’s, how likely is it that you *won’t* roll a six? So if you do that math – yes, you probably will roll a six at some point in those six rolls, but there’s a pretty good chance that you won’t. There’s a 33% chance that you could roll that die six times and not get a six.
So, how does that relate to the Monty Hall problem? Well, you’re looking at it as if the doors are independent of each other. There was a 33% chance that you guessed correctly the first time, and now that there’s two doors, you have a 50% chance to have it right now. But if you think about it that way – you’re thinking about it incorrectly.
It’s not about whether you guessed right the first time: It’s whether you guessed *wrong* the first time. Do it that way, and it should become clearer. What are the odds that you guessed incorrectly on your first guess? Well, two doors are wrong, one is right. So you had a 67% chance of being wrong.
Once you come to the second pass – the car and the goat can’t switch places. So the odds are that there’s a goat behind the door you already picked. Then the host removed the OTHER door that had the goat. Meaning, the only door that’s left has the car behind it.
If the car and the goat could switch places, then your intuition that it’s a 50:50 chance would be correct. But since they can’t, once the host removes one of the goats, the only way for you to lose is that the door you initially picked had the car behind it. There’s only a 33% chance of that happening, so if you always switch doors after the host reveals the goat you have a 67% chance of winning the car.
You’ll still lose sometimes, of course, but that’s the strategy that will get you a car more often.
I don’t know if this is what you are talking about but the old game show from the 70’s would have a part in it where you had to choose between three curtains, each equally likely to have the big prize behind it and the other two would have a crap prize behind it. Since equally likely the one you pick would have a 1/3 chance of having the big prize while the other two would have a 2/3 chance of having the big prize behind them. But Monty knows which has the big prize and which have crap and so he shows you what is behind the one of the two that has crap behind it. So now you are given the choice, stick with your first pick which has a 1/3 chance of having the big prize or switching to the remaining curtain which has a 2/3 chance of having the big prize behind it. You switch because you are twice as likely to win
Adding more doors makes sense for a lot of people but to me it never clicked in that way, so this is how I always understood it:
In the scenario, you have a 1/3 chance of choosing the correct door on your first guess, and a 2/3 chance of choosing an incorrect door on your first guess.
Now assume we’re in the first scenario where you chose correctly initially (a 1/3 chance). That means that if you choose to switch later on, you will switch to a wrong door and lose.
In the next scenario, assume you choose incorrectly the first time (with 2/3 probability). In this case, there will be two doors unchosen, one with a prize and one with nothing. The door with nothing must be opened by the gameshow host, and so in this scenario you switching will lead to choosing the right door.
Therefore, think about it like this: if you don’t switch, you must choose the correct door initially to win, which is a 1/3 probability. If you do switch, you must incorrectly choose on your first guess to win, which has a probability of 2/3.
The main point that made me understand the logic is that if you pick an empty door at the start (which happens 2/3 of the time), the host HAS NO CHOICE in which door they can open – they can only open the only other empty door.
That certainty of which door gets opened if your first choice was incorrect gives you the information that you need to know to switch.
There’s 3 doors. 2 don’t have a prize, and one does.
The odds you pick the door correctly on the first try is 1/3. They reveal a door that you didn’t pick and doesn’t have a prize. The odds your door is correct is still 1/3 because the reveal of that other door doesn’t change that first decision you made. That means switching doors will give you the prize 2/3 of the time.
Imagine instead there were 100 doors and 1 prize. If you oick a door at random, it’s a 1% chance to be right. We then open 98 of the doors that don’t have a prize. There’s a 99% chance the other remaining door has the prize and a 1% chance you guessed correctly on the first try. It’s not the 50/50 that people think it is.
Initially you have a 1/3 chance of picking right. Or said better, a 2/3 chance of picking wrong.
Then the host reveals a _losing_ door.
So what’s more likely? That you picked a winning door or losing door? That you picked a losing door.
So now 2 losing doors are _probably_ accounted for. The one you picked. And the one the host revealed.
So your chances are best if you switch your pick.
The way I think about/simplify the problem is to remove the concepts of changing your selection and of Monty removing the doors without prizes.
You are given 3 doors, and a choice at the beginning. You can either pick any door, or you can choose any 2 doors (choose one door not to pick). Obviously, you’d want to eliminate one door and you have a 2/3 chance at winning.
Going back to the original problem. When you select a door and don’t change your guess, that’s choosing a single door. When you select a door and change your guess, that’s choosing the door not to pick. This is because the prize isn’t reshuffled after the doors are eliminated, and Monty can only eliminate doors with prizes.
The options are:
– Prize is behind the door you picked (1/3 chance, Monty can remove either door)
– Prize isn’t behind the door you picked (2/3 chance, Monty *has* to eliminate the door without the prize, leaving the door remaining with the prize)
* There are 3 doors. You pick one.
* The odds of “the door you picked” are 1/3. The odds of “**not** the door you picked” are 2/3.
* The host opens one door, then gives you a chance to change your pick. *Nothing has changed about the odds* – it’s still either 1/3 “the door you picked”, or 2/3 “**not** the door you picked”.
* Do you want to stick with 1/3 “the door you picked”, or change to 2/3 “**not** the door you picked”?
If you still don’t get it, this may help:
* Pretend the host gives you a chance to change your choice – not to another door, but to “*both* of these other doors you didn’t pick”. You’d *obviously* change your pick, right? Because then you get 2 doors instead of 1. Which would make it a 2/3 chance to win. Which is what happens if you change your pick.
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