# monty halls door problem please

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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

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?

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?

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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 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

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.

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.