Question

You're a genius if you can solve this (don't cheat!)

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Btw, the probability of getting a car is NOT 50%

Answer 1

i remember this from a movie XD...
is that where you got it?

Answer 2

Blah blah blah. You switch and double your odds. Thanks, Monty! Now tell me how to carve up one sphere and turn it into two.

Answer 3

I got this from the novel "the curious incident of the dog in the nighttime" :)
Aw its too famous

Answer 4

lol

Answer 5

But I still don't understand why that's the answer :/

Answer 6

monty hall problem says you should switch :P

Answer 7

It would be great if someone could explain how this works

Answer 8

I'll try and explain. This problem has stumped a lot of bright minds. It even fooled Paul Erdos. Gimme a second....

Answer 9

So the gameshow host only takes away a door that has a goat behind it. One of the two doors MUST have the sports car, right?

So if you decide to switch, you would be switching either TO or FROM the winning door. This makes or breaks you. So what are the odds that by switching you will be switching TO the car? Well it's exactly the odds that the first door you selected was a goat: 2 in three!

Seen form the other way around, what if you decide to stay? You only win if you picked the car in the first round. What are the odds of that? One in three.

Sort of a good way to see it is to imagine a much bigger game: What if you played with five doors and four goats? What if you played with 100 doors and 99 goats? In each game you pick a door, Monty takes away all but one door you haven't picked, and you get to switch if you want. In the 100 doors version, you have a 99% chance of having picked a goat the first time. So the other door almost certainly has a car behind it.

Answer 10

https://www.youtube.com/watch?v=Zr_xWfThjJ0

Answer 11

@jh99 it's based on dependent probabilities.

If you picked the correct door the first time round, you have a 1/3 chance of doing so.
Now the host reveals one of the other doors is out of the question, leaving 2 doors for which the car has to be in. Hence the probability of picking the correct door after this at that stage is 1/2 chance.

The probability of picking the correct door to begin with and then switching to an incorrect door is 1/3 x 1/2 = 1/6 (or staying, which is still 1/2 chance)
The probability of picking the incorrect door to start and then picking the right door after (by switching) is 2/3 x 1/2 = 1/3
If you stay with the incorrect door your probability is still 1/3.

So to pick the correct door eventually, you either stay with correct (1/6) chance or you change from an incorrect pick (1/3).

If you pick incorrect eventually, you either stay with incorrect (1/3) chance or change from correct (1/6)

Either way CHANGING to get a correct answer has a higher probability (1/3) than changing to get an incorrect answer or staying with a correct answer (both 1/6).

Answer 12

For me, the best was looking at it this way:

Imagine 3 people are in the exact same situation, and all of them pick a different door. 1 will pick right and 2 will pick wrong. So now when they remove one door, they can either all stay or switch.

2 of them will switch to the right door, while 1 of them will switch to the wrong door. So 2/3 of the time you get the right answer as long as you flip.