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
So let’s start off with the exaggerated problem. There’s 52 doors. The thing I find that most people, including myself, miss when presented the premise is this:
The decision the host made wasn’t baseless. He’s not supposed to open the door with the prize in it, the game wouldn’t make sense.
It’s important in probability that probability is informationally based. If you have a deck of 52 cards, your chance of pulling the ace of spades is 1/52 right? Then the next one, 1/51 because you eliminated one card, then 1/50 and so on and so forth. But if you made that deck face up, you know your probability on every card draw is either 0 or 100%.
Or take poker for example, every game has the stats of each player winning the hand written next to them. They’re using the same deck of cards, yet their probability of winning stat changes as the round goes. How is that possible? Nothing has technically changed. The dealer has the same deck as the very beginning, and the players are holding the same cards. It’s because new information is revealed with every new card the dealer reveals.
So this host, who has true knowledge over the real distribution, had to make an active choice to close almost every door. He provided you implicit information that changed your probabilities.
So let’s start at the beginning of our card example and treat the game like a deck of cards. You picked a card at random, just like picking a card at random in a deck. You have a 1/52 chance of picking the winning card and 51/52 chance of picking anything else (aka a losing card).
The host, who has perfect knowledge of this deck of cards, combs through the entire deck and eliminates everything but one other card.
Now there are two scenarios, assume you always switch.
Scenario A: you picked the right card, a MINISCULE 1/52 chance, and it didn’t matter what he eliminated. You switch your choice, you lose.
Scenario B: 51/52 odds you DID NOT pick the right card and the host was FORCED by the rules of the game to eliminate every single wrong card except the winning one. You switch your choice, you win.
The probability was never about which card it was, the probability was always about the chances you picked the right card from the very beginning.
Get a friend and actually play the game with them 100 times. 50 where you switch and 50 where you don’t. You’ll see it actually works.
There’s 3 doors: [?], [?], [?]. One of them has a car, the other two have goats.
There’s 3 universes:
Universe 1: [C], [G], [G]
Universe 2: [G], [C], [G]
Universe 3: [G], [G], [C]
You choose door 1: [X], [?], [?]. Monty, the gameshow host, now opens a door that’s not the car.
Universe 1: [X], [G], [G] where Monty opens door 2 or 3.
Universe 2: [X], [C], [G] where Monty opens door 3.
Universe 3: [X], [G], [C] where Monty opens door 2.
In 2 of the 3 universes, universes 2 and 3, the car is behind the door you didn’t choose, which is why you should switch.
Note that in a gameshow like Deal or No Deal where this “that’s not the car” contigency doesn’t exist, it doesn’t matter if you switch at the end or not.
What made it easier for me to understand is to run through all the outcomes.
To make things easier let’s say the correct door is door 1 for all the examples.
If you pick door one, you can stay and win, or switch and lose.
If you pick door two, you can stay and lose, or switch and win.
If you pick door three, you can stay and lose, or switch and win.
If you chose to switch in every example, you would’ve won whether you picked doors 2 or 3. If you had chosen to stay in every example, you would’ve had to choose door 1 off the start to win.
Your first chooce is probably wrong.
The other wrong choice gets removed, so by changing your probably wrong choice, you are picking something that is probably right.
When you initially pick, your chances of being right are 1/3. The chances that **one of the other two doors** is correct is 2/3.
After Monty opens a goat door, the 2/3 probability collapses onto the one remaining door.
What made it click for me was drawing a picture of all possible scenarios, which I’ll attempt to do using text here.
There are 3 possibilities for the 3 doors where only 1 has a car, the rest have goats. Let’s say in all 3 cases, you pick Door #1 (**BOLD)**
Case1) **Goat** Goat Car
Case2) **Goat** Car Goat
Case3) **Car** Goat Goat
So in 1/3 of the cases above, you’ve picked the door with the Car. Now, Monty Hall comes along and opens a door. As others have points out, Monty knows which door the car is behind, so he’ll never open that door. The door Monty opens will be ~~Struckthrough~~
Case1) **Goat** ~~Goat~~ Car
Case2) **Goat** Car ~~Goat~~
Case3) **Car** ~~Goat~~ Goat
So now, in 2/3 of the cases above, switching doors means you get a new car! Therefore, you’re statistically more likely than not to win a car if you switch doors
Just clicked for me so I’ll add my 2 cents in how I think of it:
Chance of picking right door=1/3
Chance of picking empty door=2/3
Host opens one empty door. Odds of you picking the other empty door BEFORE host opened other one are still 2/3, so it’s more likely you picked the other empty door instead of prize door, so switching is better.
If you pick the loser you will switch to the winner. If you pick a winner then you switch to the loser so you have to pick the loser to start with. There is a 2/3 chance of picking the loser so thats the probability.
Are you and Kevin still arguing about this? You two just need to [bone](https://youtube.com/watch?v=AD6eJlbFa2I&feature=share7)
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