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
Normally the host will ask ‘which of these doors do you want? If the one you pick has the prize, you win it’
That part is rather straightforward. You have three doors and one chance to get the prize. 33% chance.
But then the host takes a door out of the situation, one that definitely doesn’t have a prize.
Two doors remain. And you can pick one of those with a new choice of ‘stay or switch.’
It does look like the odds are now one out of two. 50%
But let’s go back to the beginning.
There’s three doors, and you already know that the host is going to do the whole remove a door and give you another choice thing.
So you think about it and realize you can actually pick two doors. And if either has a prize, you win.
Those odds are 2 out of 3. 66% And the best option.
Let’s say you want to pick doors B and C. So you say A as the first choice, then the host removes B or C if there’s no prize on said door. You then switch to the remaining door of your actual B&C choice.
There are 3 doors and 1 car. Car could be behind door A, B, or C. The ‘X’ represents the car:
| A | | B | |C |
| X | | | | |
| | | X | | |
| | | | | X|
Here you clearly see the 3 possibilities. Let’s say you choose “A.” In one out of the 3 possibilities, you selected the door where the car already is, and switching from A to B or C means you lose the car.
However, in 2/3 of the scenarios, you have chosen A where the car is actually behind B or C. If you get shown the empty door, then you will 100% choose the correct door by switching. Remember, this scenario happens 2/3 of the time, so by switching you actually have 2/3 probability.
This is, again, because in the scenario where you were already on the car, that only happens 1/3 of the time. By not switching, you keep your odds at that same 1/3 for the initial guess. By knowing an extra door and switching, your odds improve to 2/3.
The key point that is crucial to understanding this.
#The host knows which door the prize is behind.
The host’s choice is **not** random.
The host will **always** open a door that has no prize behind it. Always.
So. If you choose an empty door first time round, the host will show you the other empty door, so switching will get you the prize.
If you choose the prize door first time around, the host will show you one of the empty doors, you switch and you lose.
But how likely are you to pick the prize first time round? One in three right? Which means picking an empty door first time round is two in three likelihood. Which also means, switching gives you a 2 in 3 likelihood of winning, as the only time that doesn’t get you the prize door is if you picked the prize door first time around. Which is 1 in 3 chance.
Because they will never show you a door that has the prize after the first step it creates a bias. There are 3 places for the prize (and goats for the rest) so it can be:
Door| 123
—|—
A| PGG
B| GPG
C | GGP
Lets assume you chose door 1. Out of the three setups, you would win 2/3 if you switch. This seems obvious, but its also because the door they show you isn’t random. If you chose door 1 in setup B they would never show you door 2 and ask if you want to switch. Because they remove that outcomes it creates a bias in your selection. It rigs the game in your favor by eliminating a you lose condition.
This simple explanation is that if you switch you win by losing.
3 doors, so your odds of choosing right are 1/3, which means you have a 2/3 chance of losing.
If you pick wrong and switch you win, and your initial odds of picking wrong are 2/3.
So by switching the odds of losing become the odds of winning.
There are three doors, the prize is behind one of them. You pick one at random, your chance of having chosen correctly is 1 in 3. One is eliminated because it is a known losing door, and it *can not* be the one you picked, nor can it be the one with the prize behind it. They aren’t removing doors at random, they’re removing the door that can’t possibly win. The door you picked might be the winner, but there is a 2 in 3 chance that it wasn’t, and was only kept in the game because it was the one you randomly chose. So there is a 2 in 3 chance that the door you didn’t pick *and* wasn’t removed is the winner.
But you’re going to feel way worse if you switched and the door you chose originally was the winner, so there’s that.
Think about a game where the goal is to pick the ace of spades out of a deck of cards. The way the game is played is you get to pick on card from the face down deck, then I get to look through the whole deck face up and pick the card I want. Who is more likely to get the ace of spades?
Probability locked in when you chose the first door, the odds of it being correct are 1 in 3.
So flip that now, the chances you were incorrect are 2 in 3.
In the Monty hall problem, you are basically choosing the 1 in 3 odds over the 2 in 3 by staying at the first door.
Go further. There are 100 billion doors, you can either choose one, or choose 99,999,999,999. So either your first choice was correct or one of the subsequent 99… would be. That’s the real heart of it.
Okay… look at the problem not from the contestant’s point of view, but Monty Hall’s. First he has rules: the contestant chooses a door, Monty Hall opens a *different* door that contains *only a goat* (the Zonk prize), and offers the switch. Remember, Monty Hall knows where the car is. *He put it there*. And he’s not going to reveal the car before giving you the switching option now is he…
If the contestant chose the car door the first time around – a 1 in 3 chance – then Monty Hall has his choice of which of the two doors to open. Staying wins the contestant the car.
If the contestant chose a goat the first time – a 2 in 3 chance – then Monty Hall is **forced** to open the other goat door. It’s the only door meeting his rules above. In this case, switching to the other door wins them the car.
So when Monty Hall opens one of the two doors, ask yourself… He opened this door…. so why the heck didn’t he open *that* door?
The easiest explanation is that you are treating all the events as if they were random. But it’s not random. The host can’t open the door that has a prize behind it. That restriction gives away additional information, making the game no longer strictly random.
If you have picked the door without the prize, then you should switch. The prize is not the door you picked. It’s also not the door the host just showed you. So obviously, it’s going to be the remaining door.
If you pick the door with the prize then you obviously should not switch.
So what are the odds you picked the door with the prize? Only 1 in 3.
What are the odds you picked the door without the prize? 2 in 3.
So your best strategy is to assume that the door you have chosen is the wrong door and switch doors.
Latest Answers