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
It’s easier to explain backwards. When you pick one option out of three there is a 2/3 chance that the correct answer was behind the other doors. You picked one there were two left. The odds say that one of the ones you didn’t pick has the prize. If you could pick 2 doors instead of one you would have a 2/3 shot of winning. However if you pick 2 doors you know one has to be empty. In this scenario it is like picking 2 doors and the host tells you which one is empty and then let’s you have the last one.
Let’s do the experiment:
Take 3 cups and 1 ping pong ball. Line the cups up in a row. Place the ball under cup 2. You switch cups every time offered by the host.
First experiment you pick cup 1. The ball is under cup 2. The host shows that cup 3 is empty and asks if you want to switch to cup 2. You do and win.
Second experiment you pick cup 2. The ball was under this cup. Host shows you cup 3 is empty and asks if you want to switch to cup 1
You do and lose.
Third experiment you pick cup 3. The ball is under cup 2. Host shows you cup 1 is empty and asks if you want to switch to cup 2. You do and win.
You win 2 out of 3.
On your first choice you have a 33.3% of being right. But you also have a 66.6% of being wrong.
The opening of another door does not change the fact you only had a 33.3% of choosing the right door in the first place.
But with one door opened the full 66% chance of you being wrong “collapses” to the other closed door you didn’t choose.
Imagine there was 1,000,000,000 doors. You pick one. The host then opens up 999,999,998 doors. And asks if you want to switch. You clearly did not have a 50:50 chance of picking the right door even though there’s only two closed doors out of a billion. You had only a 1 in a billion chance of picking the correct door. And a 999,999,999:1billion odds you chose wrong.
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
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.
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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