Variable Change (Topic)

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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.

the whole practice of calculating probability is to do it at the 'highest level'/first step where things become inconsistent across different scenarios*('point of deviation' - my own term for this activity, do not use in public).(this may or may not be how it is in proper mathematical sense, but its always been the case in my math classes and it works perfectly fine in this scenario). if you don't follow this practice, then you end up calculating the sub-scenarios of a particular scenario(ala, only a PART of ALL scenarios).

so, lets break this down:
1. player picks a door
2. host opens a door to reveal a goat
3. player picks between remaining doors

the fault is that most people will think that 'point of deviation' is at st.3, whereas it is ACTUALLY at st.1. this is because the activity of picking a door is RANDOM.

when you start your calculations on st.3, it is INDEED 1:2 odds, BUT like i said above, you're only calculating a PORTION of ALL possible scenarios, making your calculations invalid.

*maybe a better phrasing is: the FIRST 'step' where randomness comes 'into play' in a given case study.

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Quote from

A_of-s_t

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.

That's completely false.

For example, let's say I have a bag of 100 black marbles and 1 white marble.

I blindly pick two marbles out of the bag, one in my right hand, one in my left hand.

According to your logic I will have a 1/2 chance of picking a white marble, because I only have two choices.

I think the problem occurs because there are two identical goats, which causes people to take the two different cases with the goats and think they are one case. If the problem was worded with let's say a goat and something else, I think it would be less counterintuitive, because in their minds they would be forced to make a distinction and recognize the three cases.

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ᴄʜᴇᴇsᴇ ɪᴛ!

Quote from

A_of-s_t

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.

You are correct, but you're not including past experience.

3 doors. You choose 1, so you have a 33% chance of getting a car. Now, the host opens up one of the OTHER two doors to show a goat. Your chosen door still has a 33% chance of getting a car. The only thing that changes is the fact that now you only have 2 choices, one of which is set at a 33% success rate. So the other door HAS to be 67%.

Also, I'd rather have a goat than a car. The goat increases in value, while the car decreases. Plus, I'd start a business by killing kudzu.

"Parliamentary inquiry, Mr. Chairman - do we have to call the Gentleman a gentleman if he's not one?"

Quote from

rockz

Quote from

A_of-s_t

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.

You are correct, but you're not including past experience.

3 doors. You choose 1, so you have a 33% chance of getting a car. Now, the host opens up one of the OTHER two doors to show a goat. Your chosen door still has a 33% chance of getting a car. The only thing that changes is the fact that now you only have 2 choices, one of which is set at a 33% success rate. So the other door HAS to be 67%.

Also, I'd rather have a goat than a car. The goat increases in value, while the car decreases. Plus, I'd start a business by killing kudzu.

A B C, let A and B represent the goats and C represent the car
Door Chosen | Remaining Doors
A | BC
B | AC
C | AB
Due to the nature of the problem, one A or B can be eliminated from whats in the remaining doors
After this is done your choices are

Door Chosen | Remaining Doors
A | C
B | C
C | A or B

It's clear that only 1/3 of the time the door you have chosen has the car and 2/3 of the time switching gives you the car.

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Thats pretty much how the game works. The first choices (Goat1, Goat2, Car) are randomly placed in 3 locations. Where you (devilesk) have C | A or b its A.

This means that if you pick:

Goat 1 the host will always remove Goat 2
Goat 2 the host will always remove Goat 1
Car the host will always remove Goat 1

I didn't want to have the first choice of the car to be random because it wouldn't matter. I went over this configuration many times and felt that it was correct. If your a disbeliever, the reason that the removal choices are set is because not matter what, their location will always be random at the start of the game. This means that if you pick the Car as your first choice, the host will remove Goat1 and because its randomly placed behind one of the other 2 doors.

None.

So then the second choice (where the host picks one of the remaining Goats) is not random. The way you stated it is simple for triggering as I would imagine but its not a random choice. The choice I speak of is the one where you first select the Car.

Car -> Goat 1
Goat 1 -> Goat 2
Goat 2 -> Goat 1

If you pick the Car the host should randomly select one of the two wrong goats. Goat 1 or 2. He should be able to do this because if you select the Car, then the other two possible answers are wrong. The host should pick between one of those two answers.
EX:

Car -> Goat 1
50/50 Chance only 1 switch randomized.
Car -> Goat 2

Yes I understand when you said that it doesn't matter, their position will be randomized at the start of each game. Their is still only 6 different combinations of where they can be placed. This means at least 2 out of every 7 games has a chance of having the same position of burrowed units that has already been played.

Post has been edited 30 time(s), last time on Jun 12 2008, 11:41 am by Impersonation.

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No, it is random. The positions of the units, when the game starts, is random. Their are 6 different positions for the units.

Car - Goat 1 - Goat 2 = 1/6
Car - Goat 2 - Goat 1 = 1/6
Goat 1 - Car - Goat 2 = 1/6
Goat 2 - Car - Goat 1 = 1/6
Goat 1 - Goat 2 - Car = 1/6
Goat 2 - Goat 1 - Car = 1/6

Now that you understand this part, we will move on to the next part. When you select a unit, in this case we will use a car. A Death Counter is set to a value. This value will correspond to the next set of triggers in which the host will remove one of the two Goats, granted you didn't select it as your first choice. The conditions of the trigger are specific to each possible scenario. As you mentioned, if you pick the Goat 1, the host will pick Goat 2. If you pick Goat 2, the host will pick Goat 1. If you pick the Car, the host will pick Goat 1. Now you have to remember that I use two different units (Goat 1 and Goat 2) because it will dramatically reduce the number of triggers and the thought that needs to be set in place for the triggers to work correctly. You said that it is not random. If you select the Car as your first choice, the host will always select Goat 1 and remove it. You are correct. You have to think of the Goats, not as Goat 1 or Goat 2, but as just a Goat. They both are the same thing in the real game. All I did was make them two separate units. Goat 1 is just Goat. Goat 2 is just Goat. The position of the Goats in the start of the game are randomized (as mentioned numerous times in this post and throughout the thread) each time. You can't see (unless you switch the vision trigger to "active") the burrowed units, so you have no idea what you picked and where Goat 1, Goat 2, and the Car are located.

All in all, it is random, eight ways to Sunday.

If you have any more questions on this specific area, just message me and I will explain it to you in even more, plain English. (as I already did)

I plan on adding an "Odds" meter into the game eventually. I want the Ore to represent the Whole number and I want the Gas to represent the Decimal. EX: Ore = 66 Gas = 66 -> 66.66% Chance of wining. I'm not sure how to set it up. I want it to change based on your current and secondary choices. I also want it to be something like a system, I don't want to have a lot of triggers for ever single outcome/possibility.

-Thanks!

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