I was watching Brooklyn 99 Season 4 Episode 8 around the 5 minute mark
The problem goes “There are 3 doors behind one of which is a car. You pick a door and the host, who knows where the car is, opens a different door showing nothing behind it. He asks if you want to change your answer.
Apparently the math dictates that you have better chances if you change your decision. Why? 2 doors 50/50 chance, no?
One character (Kevin) says it’s 2/3 if you switch 1/3 if you don’t. What? How? Please help.
In: Mathematics
The Monty Hall problems come down to this, is it better to choose one door, or two doors? If you decide to stay with your original pick, you are basically saying that you want to keep your one door instead of the two other ones. When you change doors, you are picking both the other ones.
A lot of people get stumped by the door reveal, but that door reveal is actually 100% inconsequential. The prize can only be behind a single door, which means that most doors will be empty. In the two doors that you did not pick, one door *must* be empty. That does not improve or worsen your odds in any way, because it is expected that the door will be empty. Furthermore, the host knows not to open a door that has the prize, so the door will be empty.
Since this step is inconsequential, you can skip it. So, Monty says pick a single door. Then once you pick a single door, he skip opening an empty door and says “okay, do you want to keep your single door, or switch to the other two doors?” Which do you choose, one door or two?
Another way to help this problem is add the amount of doors. Let’s say that there was 100 doors, and you have to pick one. You pick one, and then Monty says “do you want to switch to the other 99.” Keep in mind that with those 99 doors, at least 98 of them will be empty. If you see a bunch of empty doors, that should not be a surprise. Which do you chose?
The host never reveals a door which has the prize behind it. You already know from the start that two doors don’t have prizes, so when you pick one you already know at least one of the doors you don’t pick has no prize.
Which has a better chance of having the prize, 1 door or 2 doors? Obviously 2 doors. One of those 2 doors is always going to not have the prize but the host eliminated it by showing you. So by switching you in essence pick the 2 doors option.
If you don’t switch, your odds are always one in three of being correct.
Now break down the odds when you switch:
There is a one in three chance you picked the right door originally and are losing it by switching.
There is a two in three chance you picked the wrong door originally, and since Monty eliminated the other incorrect door you are switching to the correct one.
Simple.
Which is more likely?
You picked the one door out of three that has the prize.
Or;
You picked one of the two doors that does not have the prize.
Chances are that you picked one of the two doors that does not have the prize, it is a 2/3 chance that you chose wrong. It is a 1/3 chance that you chose the one with the prize.
Since it is more likely that your choice was wrong, it would be wise to change your choice.
I still get confused by it, but the simplest way I can explain is that, when you pick a door, the chances you picked an empty door (no prize) is 2/3. The fact that the host reveals a door doesn’t change much. You still know there is a 2/3 chance that you picked the wrong door on your first choice and a 1/3 chance that you picked the right door. So when he asks if you want to switch, he’s really showing you that there were really always only 2 options: opening the 1 of 2 incorrect doors you most likely picked or picking the 1 door left that the host knows is correct.
_____________________________________________________________
|___ **your door** (which is most likely empty) ___|
______________________________________________________________
|___**the last unopened door** (that is most likely correct)___|
hope this helped
One way to solve it is to list all the scenarios:
Assume car is behind door #1.
Car | You choose | Monty opens | You switch | Don’t switch
———|———-|———-|———-|———-
1 | 1 | either 2 or 3 | lose | win
1 | 2 | 3 | win | lose
1 | 3 | 2 | win | lose
Doors 2 and 3 are the same.
The reason is quite simple.
If your initial choice was incorrect, and you switch, you have a 100% chance of getting the car.
Since the odds of your initial choice being incorrect were 2/3rds, therefore the odds of winning if you switch is *also* 2/3rds.
Caveat to what everyone else is saying: if the problem is not explained very precisely, the true answer is undetermined as there is not enough information to compute.
In particular, it is important that
1. the host knows where the car is, and
2. the host deliberately chooses never to reveal the car
Saying that the host knows where the car is and happened to open nothing in this particular instance is not sufficient information to create the 2/3, 1/3 solution. It is important that the host deliberately avoids the prize, as that is what causes the door he didn’t choose to become special. He has special knowledge, and you can gain partial knowledge from observing his behavior if his behavior depends on his knowledge.
If, instead, the host opens doors randomly and just happened to open nothing this time, then we are not in the true Monty Hall problem, but the “Monty Fall” variant, in which the answer is 1/2, 1/2 as people intuitively suspect.
In a lot of cases, the problem is difficult not because people being given the problem are bad at math, but because the presenters are, and provide an ambiguous scenario which could either be the Monty Hall problem or the Monty Fall problem, and people assume it’s the Monty Fall problem since that’s a reasonable assumption someone would make if not given information that rules it out.
Monty always reveals the dud prize. If you did not originally choose the car, you will always get it by switching. The probability of getting the car initially is 1/3 so switching is 2/3.
It becomes easier to imagine if you think about 100 doors. You pick one. I open 98 donkey doors. Do you want to switch?
Here is a better way to look at it.
First, understand that Monty only reveals doors that do NOT have the car. This is very important.
Let’s say you are playing with 1 Million doors. You choose a number let’s say door 2. Monty opens all the doors except door 789543. Do you keep your original door or do you choose door 789543? The car is either in door 2 or door 789543. The probability of it being door 2 was 1/Million. The probability of it being door 789543 is 999999/million. This is because Montey has only opened doors that do NOT contain the car.
Now let’s say you’re playing with 10000 doors. You again choose door 2. Monty opens all the doors except door 3897. Again, the probability that door 2 was correct was 1/10000. The probability that door 3897 is correct is 9999/10000 because Monty only opens doors without the car.
Now let’s say you’re playing with 10 doors. You again choose door 2. Monty opens all the doors except door 10. Again, the probability that door 2 was correct was 1/10. The probability that door 10 is correct is 9/10 because Monty only opens doors without the car.
Now finally let’s say you’re playing with 3 doors. You again choose door 2. Monty opens all the doors except door 1. Again, the probability that door 2 was correct was 1/3. The probability that door 1 is correct is 2/3 because money only opens doors without the car.
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