The Monty-Hall game (from the movie 21)
The Monty-Hall game originated from a tv show called Let's Make a Deal in the 1950's.
The game basically consisted of 3 doors, behind one of which was a new car. Behind the other 2, were goats.
|—| |—| |—| | 1 | | 2 | | 3 | |—| |—| |—|
Logically, each door holds a 1/3 chance of being the car. (33.333333%)
Let's say we pick door #1.
The host, would then open up either door #2 or door #3, and show you a goat. (He knows what is behind each of the doors).
The host would then ask you if you would like to keep the original case you had selected, or if you would like to switch cases.
The probabilities for each case would look something like this:
|—| |—| |—| | 1 | | 2 | | 3 | |—| |—| |—|
1/2 1/2 0/2 (50%) (50%) (0%)
Most people, out of superstitious-ness, would keep their original case, and hope for the best out of a fifty fifty chance.
The movie 21, suggests that switching cases improves your odds, to 2/3 (66.66666%).
You may have been baffled by this for a while now, so here's the explanation.
Originally, it is true that each door held a 1/3 probability of being the one with the car.
So let's go back to our initial problem, where we picked door #1…
|—| |—| |—| | 1 | | 2 | | 3 | |—| |—| |—|
1/3 1/3 + 1/3= 2/3 (33.33%) (66.666%)
So looking at that, we have a 33.33% chance of winning the car, and 66.66% chance of getting a goat.
But then, the host opens up door let's say, door #3 (he could have opened door number 2 it doesn't matter, he opens whichever has nothing in it). Behind door #3, a goat.
So now door #3 holds a 0% chance of having the car.
And now if we re-look at our original assumption, keeping in mind that door #3 has 0% chance…
|—| |—| |—| | 1 | | 2 | | 3 | |—| |—| |—|
1/3 x + 0 = 2/3 (33.33%) (66.666%)
The probability of doors 2 and 3 combined is still 66.66%, but since door #3 doesn't have anything behind it, door number 2 must compensate for the lost 1/3.
Therefore the odds of it being door #2 have changed from 1/3 to 2/3 (because 1/3+1/3).
So by changing from your original choice of door #1 to door #2, you're in fact doubling your odds of winning.
And that's how it's done with simple math.
Here's the clip from the movie:
I get the concept, but that's just not how it works.
The probability is dependent.
In the beginning, P(getting car) = 33.3333…% But after you mess up, the probability changes.
Because I picked door number 3, it doesn't mean that door number 2 is more likely than door number 1 to have the car behind it; because, if you just switched around doors number 1 and 2, then door 1 would have the higher probability, by your thinking.
So after the first door is picked, there's a 50% probability of either door having the car behind it.
That's exactly what I said when I denied it as an article. There's three doors to start. So you have a 1/3 (.333%) chance to get the right one. One of them opens, and you see that it's wrong. Now there's only two. So you have a 1/2 (50%) chance. You can't just add percentages like that. It's not how probability works.
It's a psychology play. You pick one of three doors, the host shows you one of the three doors that loses… at which point you can assume that the host is covering up the winning door by picking another. You're both "banking" on the fact that you will probably pick wrong at first (with 33% chances) and that the host will try to keep you away from the winning door. The only reason it's not a straight 50%-50% chance is because, in numerous cases of this situation, you are least likely to pick a winning door and most likely to be shown a losing one.
Futility wrote: That's exactly what I said when I denied it as an article. You can't just add percentages like that. It's not how probability works. You were absolute right to deny it as an article… since this information is not deep enough to base an article on. However, the interaction between psychology and probability is an interesting one. It seems as if this is a straight probability case but, really… it's influenced by the revelation of "non-chances". It's an interesting study.
Okaaaay.
I was just mis-understanding the whole situation. I thought that you pick a door, and then they open it. But in reality, you pick a door, and then they show you one of the others.
So now I understand where you're coming from with the whole 'choosing the other door gives you a higher probability' thing.
But still, I don't think you can actually put a real percentage on that since it's basically reverse psychology, and everybody responds to that differently.
s3klyma wrote: But still, I don't think you can actually put a real percentage on that since it's basically reverse psychology, and everybody responds to that differently. The probabilities in this case are based upon chances of success. Thus, it doesn't matter how people respond to reverse psychology… the chances of success are the same no matter the choice.
Zephyr_Pure wrote: The probabilities in this case are based upon chances of success. Thus, it doesn't matter how people respond to reverse psychology… the chances of success are the same no matter the choice.
This is true. There is an advantage when you switch. [large explanation] Google "Monty Hall problem" [/large explanation]
Quick example; say I hold fifty cards. 49 of them are red ones, and 1 is black. I offer you one card, you can choose it, but not see what it is. Now, I drop 48 red cards on the table, in my hand holding only the card you chose and another. Is it likely that you chose correctly the first (1/50) time? No, switching increases your chances of winning.
Anyway, you can do the three door thing yourself, you can play this game with anyone. You'll see that constant switching will balance your winnings to 66.66%
s3klyma wrote: But still, I don't think you can actually put a real percentage on that since it's basically reverse psychology, and everybody responds to that differently. Zephyr_Pure wrote: The probabilities in this case are based upon chances of success. Thus, it doesn't matter how people respond to reverse psychology… the chances of success are the same no matter the choice. spyware wrote: Actually, this isn't true. There is an advantage when you switch. [large explanation] Google "Monty Hall problem" [/large explanation]
Yes, I already agreed with that. What he wrote above was that a person's response to the "reverse psychology" would affect the original percentages on the choices. It would not… That's what I meant when I said "the chances of success are the same no matter the choice". Not the same as each other, but the same no matter who is doing the choosing.
jjbutler88 wrote: Surely this isnt even a maths problem? If you get to pick 1 door out of 'n' doors, then your chances of winning are 1/n. If n = 50, you have a 1/50 chance of winning. But if n = 2, you have a 1/2 chance….
Am I missing the complex bit? Coz I did this stuff in like year 5… Yes, you're missing the complex part. A great deal of it banks on the fact that you're getting shown an incorrect choice, which eliminates that and increases the chances that the one that wasn't shown is the winning one. Can't really explain it any more than that.
Right, none of you know shit.
Look at cases:
Case 1)
You pick door with car. Door without car opens, you can either change or stay. Let's call the two options from here, 1a and 1b. so 1a) you change, and you don't get the car. so 2b) you don't change, and you DO get the car.
So for case 1) you basically have a 50/50 shot of getting the car.
Case 2)
You pick a door without the car. So then you have two options again, 2a) you change, you get a car. 2b) you don't change, you don't get a car.
right, so it's all 50/50 right?
No, Case 2) is twice as likely to occur.
Always change, 2/3 of the time you'll get a car.
I can demonstrate this mathematically but I don't think anyone posting to this thread will understand the precepts involved anyway.
Zephyr_Pure wrote: Yes, I already agreed with that. What he wrote above was that a person's response to the "reverse psychology" would affect the original percentages on the choices. It would not… That's what I meant when I said "the chances of success are the same no matter the choice". Not the same as each other, but the same no matter who is doing the choosing.
Whoops, didn't read your post right. Edited.
richohealey wrote: Right, none of you know shit.
I think only one person in this thread is disagreeing with this explanation :+.
Anyway, you know what's funny? Go to IMDB, search for the 21 forum and look at dumb people being dumb. Laughs guaranteed.
spyware wrote: So?
Why are you posting this?
Because after watching the movie I really wanted to know how, so I went to the math center at my school and a teacher explained it to me. I just thought it would be interesting to people who have seen the movie and didn't understand the reasoning behind the problem.
**Frogguy wrote:**Because after watching the movie I really wanted to know how, so I went to the math center at my school and a teacher explained it to me. I just thought it would be interesting to people who have seen the movie and didn't understand the reasoning behind the problem.
- It's not a problem
- It's a flame-bait.
It's kinda like making threads on religion. Don't.
spyware wrote: [quote]**Frogguy wrote:**Because after watching the movie I really wanted to know how, so I went to the math center at my school and a teacher explained it to me. I just thought it would be interesting to people who have seen the movie and didn't understand the reasoning behind the problem.
- It's not a problem
- It's a flame-bait.
It's kinda like making threads on religion. Don't.[/quote]
Hahaha alright thanks for the tip
heres how i believe it goes…. there are three possibilities here, assuming the contestant chooses door no. 1…
- The player picked the door hiding the car. The game host must open one of the two remaining doors randomly.
- The car is behind Door 2 and the host must open Door 3.
- The car is behind Door 3 and the host must open Door 2.
so you have 1/3 of guessing the car door, and a 2/3 chance of getting a goat door. therefore, if you chose a door with a goat (2/3 chance) and you switch, you win the car!
if you look at it another way, your trying to eliminate the goats, because once you know where the goats are, when prompted to switch, you avoid the goat doors. so you have a 2/3 chance of eliminating a goat by choosing it, and then being able to switch after the host eliminates one, therefore you have a 2/3 chance of getting the car
If your confused on this, watch this video http://www.youtube.com/watch?v=mhlc7peGlGg
MinDistortionist wrote: If your confused on this, watch this video http://www.youtube.com/watch?v=mhlc7peGlGg
Nice video, explains the problem very clearly haha