SRL Archive - Discussions & Debates

You're in a quiz. The host shows you 3 doors. Behind one of them, there's a sports car. Behind the 2 others, there's a goat. Host asks you to pick one of those doors. After you have picked one, he opens one of the 2 remaining doors and says "ahaa, there was a goat behind this door. There are still 2 doors left. The sports car is either behind the door you already chose, or it is behind this one here. You have a chance to change your mind now if you want to."

Do you change your door?

Mathematics claim that you should change the door, thus giving you a better chance to win. I made a scar script to check if this is true.

Ofcourse, even if i ran the script a billion times, it doesn't prove anything 100%, because it's all about propabilities, which cannot be proved by testing something. But, believe whatever you want to.

I ran the script 100000 times, and it seems that if you change the door, your chance is 50%. If you don't change, your chance is 33%.

DoorsAndGoat.scar: you can pick the door yourself, this is to show that my script works correctly.
DoorsAndGoat_AUTOMATED.scar: Here the whole process is automated.
[PHP]
~ Wins: 569
~ Losses: 568
~ WinRate: 0,500439753737907
[/PHP]

EDIT: Fixed a bug! Please redownload!

Lol 21….

O.k, at the beginning you have a 33% chance. And the others have 33%.

You still hold 33% and the other one is gone so the other door gets the additional %33. So, the chance of the current one you have is 33% and the other one is 66%.

So switch. And chances are good you = win.

From: ZaSz
Lol 21….

O.k, at the beginning you have a 33% chance. And the others have 33%.

You still hold 33% and the other one is gone so the other door gets the additional %33. So, the chance of the current one you have is 33% and the other one is 66%.

So switch. And chances are good you = win.

No that isn't true at all.

You have a 33% chance of picking the right door in the beginning, if you make it to round 2 the door you have and the other one have an exactly 50% of being the car on each one.

Simply math, stupid movie.

This theory is not proved with SCAR at all, it's entirely a concept not subject to any trial.

1: begin
           if Doors[2] then
             TakeAway := 3
           else
             TakeAway := 3
         end;

Doesn't look right…

Rest assured the script is fine and so is the math of variable chance is 66% as it would show with any statistic or computer simulation.

But when it boils down to actual probability of you winning it's really 50% a coin flip.

The real problem is people are applying math to something that math can't determine.

From: marpis
You're in a quiz. The host shows you 3 doors. Behind one of them, there's a sports car. Behind the 2 others, there's a goat. Host asks you to pick one of those doors. After you have picked one, he opens one of the 2 remaining doors and says "ahaa, there was a goat behind this door. There are still 2 doors left. The sports car is either behind the door you already chose, or it is behind this one here. You have a chance to change your mind now if you want to."

Do you change your door?

Mathematics claim that you should change the door, thus giving you a better chance to win. I made a scar script to check if this is true.

Ofcourse, even if i ran the script a billion times, it doesn't prove anything 100%, because it's all about propabilities, which cannot be proved by testing something. But, believe whatever you want to.

I ran the script 100000 times, and it seems that if you change the door, your chance is 50%. If you don't change, your chance is 33%.

DoorsAndGoat.scar: you can pick the door yourself, this is to show that my script works correctly.
DoorsAndGoat_AUTOMATED.scar: Here the whole process is automated.
[PHP]
~ Wins: 569
~ Losses: 568
~ WinRate: 0,500439753737907
[/PHP]

lol da owner showed me this

From: IceFire908
Rest assured the script is fine and so is the math of variable chance is 66% as it would show with any statistic or computer simulation.

But when it boils down to actual probability of you winning it's really 50% a coin flip.

The real problem is people are applying math to something that math can't determine.

It's not longer random, since they do not open the door randomly.

If they opened one of the other doors randomly and it was a car, you would have a 0% chance of winning, even if you switched. Since they involve themselves by selecting a door with a goat, it makes sense that the probabilities would be skewed.

And yes, that code I pasted is wrong. When you fix it, you win 66% of the time when you switch every time, and win 33% of the time if you refuse.

This was in the movie "21" :p

Thats why i said lol 21. In my post noidea :p

Omg sorry there are a few typos in the script, i update them now
edit: 1 typo, the one tarajunky mentioned

From: IceFire908
Rest assured the script is fine and so is the math of variable chance is 66% as it would show with any statistic or computer simulation.

But when it boils down to actual probability of you winning it's really 50% a coin flip.

The real problem is people are applying math to something that math can't determine.

Ice, look up the Monty Hall problem.

This is one of those storys that mathematicians come up with. Heard it quite a few times lately, still don't get the mathematic explanation though :(.

It doesn't need to be explained with mathematics…
If you narrow a choice down to 2 doors, one being a goat and one being a car, then you have a 50% chance of getting the car no matter what. It is not more logical to switch doors, because either way, you have a 50% chance.

From: Runaway Cop
It doesn't need to be explained with mathematics…
If you narrow a choice down to 2 doors, one being a goat and one being a car, then you have a 50% chance of getting the car no matter what. It is not more logical to switch doors, because either way, you have a 50% chance.

The common misconception is that the Monty Hall problem holds true when it really isn't.

From a computer/statistical/mathematical trial/theory the switch gives you an advantage, but in actualization there is absolutely no difference.

The probability if you flipping heads is 50% the probability of you getting heads twice is 25%, but the second time you flip the coin it's still a 50% chance of heads. Probability is a flawed concept in so many ways.

From: Runaway Cop
It doesn't need to be explained with mathematics…
If you narrow a choice down to 2 doors, one being a goat and one being a car, then you have a 50% chance of getting the car no matter what. It is not more logical to switch doors, because either way, you have a 50% chance.

You'd think so, but that's not true. Read up on the Monty Hall problem.

Oooooh. That is kind of confusing.

but pretty much it's saying… You have a 66% chance of picking the car if you swap every time because you have a 66% chance of picking a goat. and only a 33% chance if you don't swap because you have a 1 in 3 chance of picking the car the first time…

damn. if the monty hall problem was ice cream, it would be a deep, rich flavor :)

From: Runaway Cop
Oooooh. That is kind of confusing.

but pretty much it's saying… You have a 66% chance of picking the car if you swap every time because you have a 66% chance of picking a goat. and only a 33% chance if you don't swap because you have a 1 in 3 chance of picking the car the first time…

damn. if the monty hall problem was ice cream, it would be a deep, rich flavor :)

Thats what I explained in my first post…. :/

From: ZaSz
Thats what I explained in my first post…. :/

All it did was confuse me at first :p
I think you have it worded differently than the way I worded it.

I don't get it. Why would the second door have a 66% chance then? I could see someone forgetting to change the value of the first door and saying the first one has a 33% chance and the second one has a 50% chance but that's apparently not the case.

From: Widget
I don't get it. Why would the second door have a 66% chance then? I could see someone forgetting to change the value of the first door and saying the first one has a 33% chance and the second one has a 50% chance but that's apparently not the case.

1. You have a 66% chance of picking a door with a goat behind it, since there are 2 goats and 3 doors.

2. If you do pick one of the doors with a goat behind it, the door with the other goat will be opened.

3. Therefore, if you switch doors when you pick a goat, (66% chance) then you will get the car.

The only time you don't get the car when you are switching doors is when you pick it on your first try (33% chance). So, all in all, you have a 66% chance of picking the car if you swap doors every time.

How about this?

Take 1 million doors. Choose one and leave it closed. Then open 999,998 doors with goats in them, leaving one additional door closed.

Is there really a 50/50 chance that you chose the right door? Every time you repeated this experiment you would end up with 2 doors, and a "50/50" chance, even though you know the real odds of choosing the right door from the beginning are 1 in a million.

you gotta look at it from the beginning.

there is a 33% chance of picking the car.
so you probably picked a goat (66%)
if you eliminate a goat, and assume you picked a goat at the start (as you probabilistically did) you should change.

edit: and Dr. tarajunky just did a great job elaborating.

When the host has to eliminate door(s) such that a switch means you get the opposite of what you originally picked, then the chance of winning after a switch is 1-ChanceBeforeEliminations. So because you were probably wrong the first time, you will probably be right after the switch.


Edit: or look at the picture
http://upload.wikimedia.org/wikipedia/commons/d/de/Monty_tree_door1.svg

I think it's confusing to people because it is so counter-intuitive.

No one thinks of it as: Since there's a 66% chance of picking a goat, there's a 66% chance of getting the car if you switch doors.

From: Brain
you gotta look at it from the beginning.

there is a 33% chance of picking the car.
so you probably picked a goat (66%)
if you eliminate a goat, and assume you picked a goat at the start (as you probabilistically did) you should change.

edit: and Dr. tarajunky just did a great job elaborating.

Ahh ok. Now it makes sense (I guess). I didn't realize that you never actually opened the door (though I blame the first post in the thread for that :p)

1) You got this from 21 (it was on tonight). 2) Doesn't this f'ing make sense without a proof?

From: R0b0t1
1) You got this from 21 (it was on tonight). 2) Doesn't this f'ing make sense without a proof?

I saw the movie like a year ago (it was great btw) and I couldn't see the sense in it before now :)

I may be late with this reply, but I read this in a book:

"Let the doors be called X,Y and Z.

Let Cx be the event that the car is behind door X and so on.

Let Hx be the event that the host opens door X and so on.

Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula:

[INDENT]P(Hz ^ Cy) + P(Hy ^ Cz)[/INDENT]

[INDENT]= P(Cy) x P(Hz|Cy) + P(Cz) x P(Hy|Cz)[/INDENT]

[INDENT]= (1/3 x 1) + (1/3 x 1) = 2/3[/INDENT]"

Which basically explains that you should always switch doors because you havea 2/3 chance of getting the car.

Well, the idea of the problem is that if you don't switch doors, you're not actually taking advantage of the fact that the announcer told you that one of the doors has a goat behind it, so you keep the 33% chance. If you switch doors, knowing that one of the the other doors has a goat behind it, it would give you the extra 33 percent chance.

Lol i did not know you could do somthing like that in scar. I think thats really cool:)

And a bit off topic i lvoe the movie 21 :D