I have read through around 10 articles and webpages on this problem, and still don’t understand. I’ve run simulations and yes, switching does get you better odds, but why?
In: Mathematics
If the host picks randomly, then you will indeed not see better odds. 1/3 chance your original has prize, 1/3 the switched one has the prize, and 1/3 the one the host picked has the prize.
But the host didn’t pick randomly, they are guaranteed to always pick one without the prize. This changes the probability of all steps that proceed the host’s selection.
Here’s a way of framing the problem that allowed me to understand it. (Let’s use the 100 doors example). Let’s think about it inversely.
Instead of making it so the goal of your first choice is to pick the prize door, make it so the goal of your first choice is to pick the *wrong door.* With that in mind, you have a 99% of making the right decision (aka the wrong door).
Then the host eliminates doors until the only there are only two remaining. Since the door you first picked had a 99% chance of being the wrong door, then odds are, the remaining door has to be the prize door.
Maybe thinking about it this way might help.
10 doors.
You pick one.
There is a 1 in 10 chance you picked the right door.
Another way to say this is there is a 9 in 10 chance the winning door is NOT your door.
Someone who knows which door has the prize opens 8 of those 9 doors.
There is still a 9 in 10 chance the winning door is NOT your door.
Your door has a 1 in 10 chance of having the prize, the other a 9 in 10 chance of having the prize.
Do you switch?
You had 2/3 chances of being wrong with your random guess.
You switching doors basically means you saying “My first Guess was wrong”, which has a 2/3 chance of being right.
Okay, so you do understand that Monty Hall knows where the car is and doesn’t open a losing door ever. So look at the game from his point of view.
If the contestant picks the car the first time around, Monty Hall has his choice of which door to open. But only if the contestant chose the car the first time around before anything else happened, which makes it a 1 in 3 chance for this situation. Staying makes the contestant win.
If the contestant picks a losing door the first time around, Monty Hall is **forced** to open the other losing door. The rules force him to. This is a 2 in 3 chance. Switching makes the contestant win.
Since your odds of winning based on your first door pick alone were not very good, and the fact that Monty Hall knows where the car is, when he opens a door you should be asking yourself, “why *didn’t* he open the other door?”
Okay so the essence of switching and staying is that switching gives you the opposite of what’s behind the door you picked, and staying gets you what’s behind the door. Since 2/3 options are goats then you should always switch because switching essentially makes 2/3 options cars.
Here’s a way of explaining it that helped me see it:
Imagine you pick the left-hand door, and then Monty says “*you can either stick with the left-hand door, or you can choose to have the prize if it’s behind either the middle or right hand door*”. Obviously you would switch because you are now being offered 2/3 odds instead of 1/3 odds.
So after you’ve switched and your choice is locked in, Monty says “*OK, go ahead and see if you’ve won, but to save you time opening doors, I’m going to let you know it’s not behind the middle door*”.
When you switch around the order like this it is obvious that the decision to switch is correct. But there is no actual difference between this and the original problem – either way Monty is offering you a 2/3 choice – the only difference is *when* he tells you which door (out of the two that you didn’t originally pick) does not have the prize.
When you picked your door, you had a one in three chance of choosing the right door, so there was a two in three chance that the right door is one of the others.
Then Monty opens one door, and gives you the choice to swap.
Nothing has changed about the probabilities. You still have a one in three chance of having picked the right door, and there’s still a two in three chance that the right door is one of the other two. The only thing that changed is that now you know that, if you picked the wrong door to begin with, there’s only one option left, so, if you were wrong with your guess, that has to be the right door. So if you were wrong 2/3 of the time, that is likewise the right door 2/3 of the time.
The key to the Monty Hall Problem is that the host knows where the goats are and where the car is.
Whether you’re dealing with 3 doors or 1000 doors, as other posters have suggested you think about it, your options become choosing 1 door (your original choice) or *every other door*.
If the host *didn’t* know where the goats and car are, and the game resets if Monty chooses to open the car door, then the odds at the end are truly 50/50.
That seemingly innocuous statement changes the game entirely.
It’s not 2 doors, 50-50. It’s 2 doors, 33-67 (67 being 66.6666… rounded up). How is this possible?
Think of the three doors as two groups. Group #1 is the door you picked. Group #2 are the two doors you did not pick. Group #1 and Group #2 are NOT 50-50 because group 1 has one door and group 2 has two doors. So it’s 33-67.
When one of group #2’s doors is eliminated, and group #2 now only has one door, the odds DO NOT CHANGE. Group #1 is STILL 33% and Group #2 (despite having only one door remaining now) is STILL 67%.
So of course you wouldn’t stick with the door in group #1, it only has a 33% chance of winning. You’d switch to the door in group #2, which has a 67% chance of winning.
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