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
https://youtu.be/kJzSzGbfc0k here is a link to a Vsauce2 video that I find explains it really well. There’s also two other paradoxes in there that I find interesting. If you don’t care about them the Monty Hall paradox explanation starts at 4:44
I completely understand how the breakdown of the chances works but it still doesn’t click for me. Even imagining 100 doors. Okay, I pick 2 and he opens all but 66. I understand from a maths perspective, because he knows which door doesn’t have the prize, it’s 99/100 (I’m not sure why not 98?). But that doesn’t invalidate your original choice. Your door may have always had the prize or it didn’t; at the time of choosing all options were equally valid. The prize doesn’t move. If we had the door with the prize the host would be just leaving an arbitrary door unopened, but because of the how the ratio now splits thats now the more likely choice? Some people have described it as saying you are picking between 1 door and the 99 other doors, but that’s not the choice that’s offered to you; you’ve just had your original choice confirmed in that its definitely not one of those 98 other doors and it’s either in yours or the door the host leaves (which they know for sure does or doesn’t have the prize)
Let’s put it this way: I’m thinking of a number. Any number. See if you can guess what number I’m thinking of.
Got something in mind?
Okay. I’ll give you a hint. It’s either the number you’re guessing, or it’s 767676767676. Do you want to change your guess?
Suppose it’s 100 doors instead of 3. Only one door has a car. You randomly pick some door. Then the host opens up 98 of the remaining doors which he knows don’t have the car, and gives you the choice to switch to the last remaining door.
What’s more likely: the 1-in-100 chance that the door you initially picked magically had the car, or the chance that the one remaining door this guy carefully did not pick has the car?
Clearly the latter. If you stay with your original doors, that’s a 1/100 chance of winning. But if you switch after this guy has done all this, you have a 99/100 chance of winning. One easy way to see this is that it *must* be one of the two doors, and your initial pick has a 1/100 chance, so the other door has whatever probability is left over which is 99/100. B
But thinking about it in terms of 100 doors makes the problem much easier to gain intuition about than 3 doors, and that intuition is – some *information* is clearly given when the guy eliminates 98 doors. The information is that the door he doesn’t eliminate is potentially extremely important, as there is a high chance the one door he magically *couldn’t* eliminate is the door that has the car.
The regular Monty Hall problem is the same way, just with 3 doors and less extreme probabilities.
The explanation that made this click for me was to imagine 100 doors. Your chance of being right is 1%, and that leaves a 99% chance that you guessed wrong. The host opens 98 doors but this does not change the chance of your initial guess being wrong.
Pretty sure the experts would say this is a weird way to think about it but the visual is pretty convincing 👍
Let’s make a small change to the game — instead of two doors being empty, one has a blue ball inside, and the other has a red ball.
You can see there are four cases:
1. You pick blue 1/3 of the time, monty reveals red. You win by switching
2. You pick red 1/3 of the time, monty reveals blue. You win by switching
3. You pick the car 1/3 of the time
1. monty reveals blue 1/2 of the time. You lose by switching
2. monty reveals red 1/2 of the time. You lose by switching
Crucially 3a + 3b put together are still a total of 1/3, and cases 1 + 2 put together add up to 2/3 probability of winning by switching.
Now, imagine the balls are invisible. You still have the exact same scenarios with the same probabilities, so 2/3 probability you’re in case 1 or 2 and win by switching, you just can’t tell them apart. Invisible balls or no balls at all are fundamentally the same thing.
Because the first door you choose has a 66% chance of being a losing door and a 33% chance of being the winning door. When the host reveals one of the losing doors. THOSE PERCENTAGES DON’T CHANGE.
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.
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?
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.
Latest Answers