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
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.
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