monty hall problem

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ID:6439
 
Dec 10 2005, 8:50 pm
there have been a few other math-related blog posts, but i don't think the monty hall problem has been mentioned, so i'll mention it =)

a contestant on a game show must pick one of three doors. only one of those doors has a prize behind it. after the contestant makes a selection, the host opens one door. the door that the host opens is one of the doors that the contestand *did not* select. and, since the host knows where the prize is, the host never reveals the prize door. once that door is revealed to the contestant, they can choose to stay with their original choice, or they can pick the other unopened door.

the question is: is it advantageous for the contestant to switch? does the contestant have a better chance of picking the prize door if they switch or if they don't switch?

Dec 10 2005, 8:52 pm
there are a zillion websites that explain (or try to explain) this problem. there are also some (mostly crappy) online simulations you can try.

google search
Dec 10 2005, 9:39 pm
is it advantageous for the contestant to switch? does the contestant have a better chance of picking the prize door if they switch or if they don't switch?

Nope. The idea of "a better chance" implies that it would be more likely to win by choosing Door B as opposed to Door A, when in fact each has a 50% chance of being the correct door.
Dec 10 2005, 10:01 pm
intuitively, that's how it would seem.

but things aren't always intuitive ;-)
Dec 11 2005, 4:41 am
There's a better chance if the contestant switches. The first decision was 1/3, if you switch, it's 1/2. Therefore, illogical as it may seem, you're better off switching.
Dec 11 2005, 5:25 am
They did that on Numbers(The TV show), although I never thought I would see it again, much less remember it.
It still kind of bugs me though, because it's 50/50 once the two doors are left, so sticking with the current door is just one of the choices. What if you chose the correct door in the first place, and then switch? When he removes an option, shouldn't that change the 1/3 chance that your answer is right to 1/2? I'm so confused >_>
Dec 11 2005, 7:05 am
There's a better chance if the contestant switches. The first decision was 1/3, if you switch, it's 1/2. Therefore, illogical as it may seem, you're better off switching.

Not exactly. Your odds go up to 2/3 if you switch. Why? Well, look at it this way. When you first picked, your odds of picking a junk door are 2/3. Monty then eliminates a junk choice. Here's where stacking probabilities come in because you just introduced a new action. There is a 1/2 chance the new door is junk. So if you pick it, your odds of ending with junk are 2/3*1/2 or 1/3. Stick with your old choice and there is no choice and essentially you just continue revealing the results of the original choice. Your odds remain locked at 2/3 chance of getting junk.


Neat, huh?

People seem to confuse two selections on one event with two selections on two events (which would give a 50% shot at the last stage). However, this would require that the prize be randomly replaced after the first door is revealed. It is not. That's where the confusion lies.
Dec 11 2005, 7:24 am
The first decision was 1/3, if you switch, it's 1/2.

if its 1/2 if you switch, it would have to be 1/2 if you stick, because sticking is the only other choice.

there are two possbilities, that the prize is behind the door you picked, or its not. but, there are three different situations.

suppose you pick door A:

situation 1: the prize is behind door A. it doesn't matter which door he reveals because sticking would win.

situation 2: the prize is behind door B. door C is revealed, switching would win.

situation 3: the prize is behind door C. door B is revealed, switching would win.

in two of the three cases, switching would win. in one of three cases, sticking would win.

one explanation which might make it a bit clearer is to think about the same problem, but with more doors. imagine that there are 1,000 doors. you pick one, monty eliminates a non-prize door, and you're allowed to switch to any other door, or stick with your choice. that process goes on until there are two doors left. when should you switch and when should you stick?
Jan 23 2006, 7:53 am
imagine that there are 1,000 doors. you pick one, monty eliminates a non-prize door, and you're allowed to switch to any other door, or stick with your choice. that process goes on until there are two doors left.

Or, for a quicker variation: you pick an initial door, Monty opens all but your chosen door and one other (i.e., he opens 998 of the doors), and you have to choose between your original choice or the one he left closed. Since the door you originally chose has a 1 in 1000 chance of being right, the other closed door has a 999 in 1000 chance.
Jan 23 2006, 8:57 am
So...wait. What becomes the better option then? This is what I'm seeing.

You pick one of three doors, but it doesn't matter because one bad door will be taken out (regardless of what you picked) and you have two left to pick from. Wouldn't the 1/3 guess become moot since you're being asked to pick from 2 afterwards anyway?
Jan 23 2006, 2:21 pm
It's definitely a 1/3 chance if you stay, and a 2/3 chance if you switch.

Draw up a diagram:

G = Picked a goat door.
P = Picked the prize door.

S = Switch
T = Stick

S - P
G
T - G

S - P
G
T - G

S - G
P
T - P

If you picked a goat, originally, then switching gets you a prize. If you picked a prize, switching gets you a goat.

There are two goats and one prize. So 2/3 times you pick a goat originally, and switching gets you the prize.
Mar 15 2006, 12:53 pm
But it's not about the original pick. It's about the final pick. The original pick should have nothing to do with it, and can be seen as never even happened with no change in the outcome of the final pick. Either way, when you are asked to stay or switch, either choice is 50/50. The first pick, or first 998 picks, are irrelivant.
Mar 15 2006, 12:55 pm
I was thinking about this just two days ago.

The only way I could see it being 1/3 if you stayed is if Monty did NOT take an option away if you wanted to switch.
Mar 15 2006, 1:15 pm
It is about the first few picks, Xooxer. Because they determine the initial conditions for the final pick. You have a 1/3 chance of starting the final pick with the prize behind your door, and a 2/3 chance of starting with a goat behind your door. Switching with a goat behind gets you a prize, switching with a prize gets you a goat.

So, 2/3 of the time, switching gets you a prize, 1/3 of the time, switching gets you a goat. That sounds like a 2/3 and 1/3 chance to me.

Write a program to simulate it. Run it for a few thousand trials, and look at the probability.

proc/MontyHall()
    var/list/doors[3]
    var/list/takenindices=list()
    var/a=rand(1,3)
    doors[a]="Prize"
    takenindices+=a
    for(var/i=1,i<=2,i++)
        while(a in takenindices)
            a=rand(1,3)
        doors[a]="Goat"
        takenindices+=a
    var/firstpick=doors[rand(1,3)]
    doors-="Goat"
    var/list/results=list()
    results+=firstpick //Results[1] is when you stay
    doors-=firstpick
    results+=doors[1] //Results[2] is when you switch
    return results


Run that a few times, and check how many goats you get if you stay, and how many if you switch.

*Note that I'm tired and the above program may not work. It may also be really crappily written. If you think it's bad, write your own.
Mar 15 2006, 1:18 pm
From the initial pick, sure it's 1/3. Hoever, there is no third option in the second pick. The third door does not exist, and neither does the math related to it. you have to recalculate the odds from the second pick only, as there are no longer the proper number of options. The 1/3 chance does not factor into the final pick.
Mar 15 2006, 1:31 pm
Yes, it does.

It's only 50/50 if you start with a goat or prize evenly. Think about it. If, 2/3 of the time, you start with a goat, then switching will earn you a prize 2/3 of the time.

Try the program (Or a better one) out some time. And make sure to simulate the whole process - not just the final pick. Because, otherwise, you won't have the 2/3 goat and 1/3 prize distribution for the door you're on initially.

Think about it - When you start the FINAL pick, what happens is determined by what is behind the door you picked for the FIRST pick - that is the door you are starting at. 2/3 of the time, that door has a goat behind it.
Mar 15 2006, 7:41 pm
No, you're wrong. The first pick has nothing at all to do with the second pick. If you pick door A, and he opens door C, door B has the same chance of holding a prize as the door you've picked. The statistical probablility is the SAME if you picked door B instead. It's 50/50, becuase door C does not count as a choice in the final pick! You cannot choose C, so it does not factor in at all. It's not 2 out of 3 now, it's 1 out of 2. Only two doors, only two outcomes. Either you picked it right from the start, or you didn't. There is no other choice available.

The only way you can have a 1/3 2/3 setup is if they allowed you to choose the revealed door. Otheriwse, at the moment you are given the final choice, you only have two options. There is no third door. It doesn't exist!
Mar 15 2006, 8:51 pm
It's not 2 out of 3 now, it's 1 out of 2.
The only way you can have a 1/3 2/3 setup is if they allowed you to choose the revealed door. Otheriwse, at the moment you are given the final choice, you only have two options. There is no third door. It doesn't exist!

1/3 and 2/3 don't necessarily mean that you're picking one out of three or two out of three, those are just the probabilities that you'll win or lose.

it will either rain tomorrow or it won't rain at all. there are two possible outcomes but that doesn't mean that each event has a 50% chance of happening. it could have a 1/3 chance of raining even though there are only two possible outcomes to pick from.
Mar 15 2006, 9:18 pm
Xooxer, run the simulation yourself and figure it out.

You can switch, or you can stay. You 'start' at door A. Door A was chosen when there were three doors, and two goats - the chance door A is a goat is 2/3. If Door A is a goat, then the other door has the prize - and switching gets you that prize. You have a 2/3 chance that you're sitting on a goat, so there's a 2/3 chance that you're switching to the prize.

Run the damned simulation.
Mar 16 2006, 5:02 am
When asked to switch, there are only two doors. One goat, one prize. 50/50 If you're going to factor in the probablility of the third door, which no longer even counts as a choice, why not factor in the probability that the person was chosen to play the game? Why not factor in the chance that they could have died on the way to the studio. Why not factor in the probablility that they were never born? The third door is irrelivant, just as these other outcomes are.
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