Variable Change v1

HOW TO PLAY:

All you have to do is follow the on screen instructions. You have two chances to input your choice. The first is your initial decision of which door has the Sports Car in it. The second is your last choice to either stay with what you originally selected, or switch to the other door remaining.

Basics:

You are playing on a game show. You have 3 doors. Behind one of them is a Sports Car, behind the other two are Goats. The odds are stacked against you to select the Sports Car, so you get two chances to select it. The host will also remove one of the Goats after you have made your first decision. Then after he removes one of the Goats from play, he will ask you if you would like to keep your current decision, or if you would like to change it to the other door. Their are nine ways to win. Only three of them include sticking with your first answer throughout both selections. You make a selection by moving your Terran Civilian to the 2x2 terrain closest to the Door.

STRATEGY:

The idea of the game is to pick the Sports Car. The hard part is that the odds are against you. The odds that you will pick a Goat rather then the Sports Car are 2/3. You can counter act this by using Variable Change. This can be done by instead of initially going to choose the Goat. He can’t show you your answer, or the correct answer. This will force him to reveal and remove the other Goat. Then when you get your second chance to change your choice or stick with it, you just have to change to the next available answer. The chance are greater that you will win more often using this process then if you select what you believe to be the Sports Car and you keep your answer.

ODDS:

The beginning odds are 1/3 (33.33%) that you will initially pick the Sports Car. The odds are 2/3 (66.66%) that you will pick one of the two Goats. If you select at Goat for your first choice, when your second choice comes around and the host removes the other Goat, your chances double to 2/3 (66.66%) if you change your decision to the other available door. Now as these are not 100%, you are not guaranteed to win every single game. All this method does is increase your odds of winning.

Credits:

-Testers:
Clan DyNe

-Original Idea From:
The Movie “21”

-Concept Help By:
My Math Teacher

-Completely Triggered By:
Brontobyte@USEast

HELP:

This Wiki explains this theory more in depth.

Post has been edited 4 time(s), last time on May 18 2008, 7:04 pm by Brontobyte.

lol, i remember that.

Technically, if we deal with just probability:

Your first choice is 1/3 (33.3%). Once the goat is displayed, you now have TWO choices (50.0%). Your odds do not increase to 2/3, they increase to 1/2. However, either way, the odds are better to change doors once the goat is shown then to stay. That wiki makes the assumption thats odds change due to past expirience...

Post has been edited 1 time(s), last time on Apr 26 2008, 4:49 am by A_of-s_t.

False.

http://en.wikipedia.org/wiki/Monty_Hall_problem
(Ah, I see that Brontobyte has linked to it as well)

The wiki doesn't make the assumption that odds are changed due to past experience. It's ironic because you are making that assumption.

If you select a door, and out of the remaining two doors the host shows you which one contains a goat, switching the door you chose will give you a 2/3 probability of picking the door with the car.

There's three cases, one of which (where you initially chose the door with the car) results in a loss. The other two remaining cases result in a win. That's 2/3.

Post has been edited 5 time(s), last time on Apr 26 2008, 5:35 am by devilesk.

There are only 3 choices. Sports Car, Goat1, Goat2. The idea is to try to pick the goat you first chance. You have a better chance of picking that your first try then you do if you try to pick the car. Then because the host cannot reveal your answer or the sports car, if you have one of the goats he will have to select the other one and remove it. Then the odds of you winning increase to 2/3 (66.66%) if you switch your door. Its just probability, its not an exact science. You could do this ever single time and still loose. Its highly unlikely but it could happen. I have included a trigger for Vision of P8, (Owner of burrowed units) so you just have to "enable" it and then you could actually see what happens. (But that wouldn't be fun)

Ya, I knew about that Monty Hall problem
I have to say that the first time I heard about it, I couldn't believe you can increase of that much your chances by changing of decision, but after, I got how it worked

Its a pretty simple concept if you actually think about it. A lot of criteria has to be met for this to work, as in you have to know that the host actually knows where the two goats are and where the car is, and he also has to give you another chance to change your original decision. I doubt that any game show would ever do this because of the drastic odd increase one has in playing to choose the goat, then switching to the car. I'm not 100% sure that my map works, with every possibility, so if you encounter problems, errors, ideas, feel free to pm me.

SEN should make a game for this, you play for 20 minerals, and you choose between 6 doors, and if you pick the one door that holds the 100 minerals, you gain 80 minerals! and there will be one that will have a random item that is on the 100 or less item thing, and the other 4 doors [unless the SEN mods disable the item] are 0 through -60 minerals, so the odds are really , really, against you

I'm confused with this Monty Hall problem...

You randomly pick a door, there's a 1/3 chance you are correct. If you are given another chance to pick (not told if you are right or wrong), how does your odds increase? There's still a 1/3 chance that the one you switched over to is the car. The only way this would work would be if the host told you if you were wrong with your first choice.

Correct. As stated before, the host must know where the two goats are and where the car is.

Break Down:

First choice - 1/3 change of selecting car 2/3 change of selecting one of two goats
One of the goats is removed
Second choice (switched answer to remaining answer) 2/3 selecting car

The odds increase because if you (odds 2/3) select a goat for your first choice, the host can't remove your answer or the car, so he is forced to remove the second goat (the one you didn't pick). Then when you change your answer you have a better change of picking the car. Try my map. You will win (with this method) 2/3 of the time.

(NOTE: Odds are not exact)

Maybe I misunderstood the game show, I didn't know that a goat would be removed. I'll play it after I install SC on this computer though. Let me just do a visual representation because I'm semi confused still:

Doors:
1 2 3

Doors 1 and 2 have goats.
The person picks door 1
The host then removes door 2 (reveals a goat) which narrows it down to doors 1 and 3
You should switch your choice now to the other door because you most likely selected a goat with your first choice and if the host removed the other, the car is likely the other choice.

Is that right?

I didn't actually get it the first time I watched 21. glad I do now.

Btw, more you have doors, more you get chances of winning when you change of decision

the principle is very simple, its just explained very craptacularly here(not saying i could do better, but nevertheless, its still very craptacular).

explanation of the game:

basically, you get to choose a door two times. the first chosen door will result in the opening of a door with a goat behind it. you will then be asked to keep the door, or switch your choice to the other door. the second door will be opened and you will win w/e is there.

the principle:

because there are two goats out of the three doors and because one goat will ALWAYS be eliminated, you are more likely to have chosen a goat as your first choice, by 'inverting'(switching) your decision, you'd have chosen the correct door.

misconceptions:

someone [incorrectly] stated that when you get to the second door, your odds are 1:2. HOWEVER, the error is that you can't start calculating at that point because you will be leaving out a crucial factor, whether or not your first door had a car or a goat. THEREFORE, you must start at the beginning of the problem.


the diagram in the wiki helps alot.

Post has been edited 1 time(s), last time on Apr 30 2008, 5:22 am by Sauceover.

please delete

Which means your odds are never 1:2. Saying your odds are 1:2 at any point is just plain wrong. Justifying it by saying that one needs to ignore the basic premise of the problem is completely meaningless.

Post has been edited 3 time(s), last time on Apr 30 2008, 1:32 am by devilesk.

about the "1:2", its kinda obvious that i'm paraphrasing someone else's notion/conceptualization...if not in the first three words, then at least in the 'rebuttal'

You stated that you thought the explanation was terribly explained. Well, I'm saying that yours isn't that good either, specifically the first sentence.

You state that the odds are 1:2, but then disprove it with the rest of the explanation. Either the post is just to remain completely incoherent, or the first sentence must be worded differently.

It is obvious that you're paraphrasing what someone said, but it's not obvious that you are disagreeing with that person's statement.

Just so I seem to be justifying myself even though I really don't care:

Let n be the amount of total choices. The probability of picking the right choice is: 1/n

One choice is eliminated:

1/ (n-1)


Here is another:

You are told to either switch or stay, that means you have two choices. Thus, 1/2.

Faultly logic ftw.