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Monty Hall problem question?
I've recently heard of the Monty Hall problem, and i have some concerns about it, of course i trust the geniuses who have spent the time to figure it out, but i'm not one to follow blindly until i get my concerns answered.
After you know door #3 does not have the car you are left with doors # 1 and 2, door #3 may as well not exist, but what if it didn't? what if you simply had to choose between two doors, that would be a 50/50% chance to get it right, well how is it different when there are 2 possible doors and one that may as well not exist compared to when there are only 2 possible doors to ever choose from, why does the first have a 2/3 chance for door #2 while the 2nd has a 1/2 chance for door #2 when both situations are almost identical (i say almost because i'm sure there is some math that explains how they are not the same).
Thanks in advance.
I understand it, i just want to know why is it that if there the game show had only two doors and the host didn't reveal any, you simply chose between the two, you have a 50% chance of getting it right, yet when there are 2 doors, same as before, but also 1 irrelevant door, the chances suddenly become 2/3. i understand why it's 2/3, but both cases seem to be clashing against each other,
how do you explain this? that in two identical situations the chances are so different.
it's like flipping a coin, if you flip it three times and the last two times you got heads, wouldn't logic lead you to believe that you still have a 50% chance to get either heads or tails (disregarding that there is no such thing as chance outside of theory), because if the past 2 flips change the chances of flip # 3 wouldn't all flips that have ever occurred in the world effect this flip? that is, in theory, in real life it's all simply dependent on physics.
But in the Monty hall problem can't you only choose between staying or switching to door #2? you can't, and even if you could you never would switch to door #3, so it is irrelevant, it's like the coin toss example i gave, if the irrelevant door chances your chances of getting it right then every coin toss before a certain toss changes your chances of getting heads.
PS: as a future reference to people who are going to answer this question, please read the question, i'm NOT looking for a definition of the Monty Hall problem, i already understand it, what i am looking for is an answer to how the given situation fits in with it.
2 Answers
- ?Lv 71 decade agoFavourite answer
Hi
i couldn't quite follow what part you don't understand, so I'll explain the whole problem…
Since there are 2 doors with goats and 1 door with a car, you have a 2/3 chance of getting a goat on the first try and a 1/3 chance of getting a car on the first try.
Let's consider the case where you pick a goat. Since you picked a goat, the host will open the door with the other goat. The only remaining door is the one with the car, so you should change your choice to the unopened door 100% of the time (if you chose a goat).
Let's say that instead you chose a car. Then the host opens a door with a goat, and you shouldn't change your choice because the only remaining door would contain a goat, so you should switch 0% of the time.
Since you should switch 100%, 2/3 of the time and 0%, 1/3 of the time, we can multiply each of them and add them together to get how often you should actually switch:
1*(2/3) + 0*(1/3)
= 2/3 + 0
= 2/3
So you should switch 2/3 of the time.
I hope this helps!
EDIT:
That "irrelevant" door is what changes the chances. Since you could choose that door like the other two doors, that causes the chances of picking the other two doors to go down to 2/3. Every little piece of information could alter the chances of an event. I hope that answers your question.
I see what you mean now. That's what makes the question a paradox. It's different than just a 50/50 chance with only two doors, in which there is only one case, because there are two cases with the monty hall problem: one where you choose the car and one where you choose a goat, and the fact that you'll choose the goat twice as often as the car is what makes the difference.
- Charles DLv 61 decade ago
This is a well known problem. A good explanation is given at the website below.
A short answer is that you have a 2/3 chance of picking a bad door.
Monte knows which are the bad doors so he will always show you one. What he does doesn't matter.
Since at the begining you have a 2./3 change of picking you switch to have a 2/3 chance of picking a good door.
One of my students ran a simulation of this. After 1,000,000 tries is came out that you would win 2/3 of the time if you switched.
