Car or Goat game

[The Subhuman]

this has kept waccoe arguing for a while, and we've had some clever people disagreeing..

Let's say you're on a 1970 cheesy American game show.

3 doors, behind one is a car, the other two have goats.
You pick a door. Host builds tension... Opens one of the other two doors. It's a goat. He then offers you the chance to switch doors...

Do you stick with your original choice, or do you switch?

[oneshortplanck]

stick,
there is no mathematical advantage to switching.

if i were on deal or no deal, i would just pick boxes in ascending numerical order. maybe if number 1 was a hottie, then i would go in descending order so i could squeeze her bum if i win / lose. thats the only advantage you can get from the box sequence.

[Dr.Leeds]

I would take the goat out from behind the open door and place it between the other two doors. On the assumption the goat is more likely to approach the door that smells of the other goat I would make my decision then.

[oneshortplanck]

oohhhhhh, i like

[oneshortplanck]

(the idea, not the goat handling)

[gessa]

(the idea, not the goat handling)

If you say so mate.

[MeeTa]

The Monty Hall problem.

There are three equal possibilities. For example, if you originally pick A:

1) It’s in A (sticking wins)
2) It’s in B, so C is revealed (switching wins)
3) It’s in C, so B is revealed (switching wins)

So switching is more likely to win.

[gessa]

Think I'd go with the 8ft wide door.

[The Subhuman]

interesting...

[YorkshireSquare]

As MeeTa says it's the Monty Hall problem.

Most people would assume that there is a 1/3 chance of picking the correct door, which there is when you first pick. This changes when the host opens one of the doors. Remember this is not random, it's is opened based on the information that there is not a car behind that door.

So technically when you picked your door you picked it at a 1/3 chance, switching gives you a 2/3 chance of winning the car.

[Martyn]

If you switch you have a 2 thirds chance of winning, if you stick you have a one third chance of winning

[The Subhuman]

As MeeTa says it's the Monty Hall problem.

Most people would assume that there is a 1/3 chance of picking the correct door, which there is when you first pick. This changes when the host opens one of the doors. Remember this is not random, it's is opened based on the information that there is not a car behind that door.

So technically when you picked your door you picked it at a 1/3 chance, switching gives you a 2/3 chance of winning the car.

Has someone tested this and proven it's not 50/50, because I'm not seeing it.

[MeeTa]

Has someone tested this and proven it's not 50/50, because I'm not seeing it.

Its not 50/50, this is simple probability math.

[theleedsmango]

Has someone tested this and proven it's not 50/50, because I'm not seeing it.

Easy enough to prove yourself if you test it out. I may suggest using something easier to handle than two goats and a car though.

We can all agree when the options are [goat][goat][car] that the chance of picking a car are 1/3.

If you don't switch, the odds of you picking the car will be 1/3.

What switching does is change the object of the game. Now you want to pick the goat in the first round. The host will eliminate the other goat, so that when you switch you get a car.

The chance of picking a goat are 2/3. Therefore there is a 2/3 chance you will win the car if you switch. 1/3 if you don't.

[The Subhuman]

Its not 50/50, this is simple probability math.

gonna need to test it for myself...I don't understand how the initial selection comes into play

[YorkshireSquare]

gonna need to test it for myself...I don't understand how the initial selection comes into play

Initially when you pick there is 1 car and 2 goats, a chance of 1/3 of picking the car.

You make a choice and the host removes one that he knows is a goat from the remaining two. So the reamining door which you did not pick has a 2/3 chance of being the car, your initial choice still has a 1/3 chance.

[Martyn]

The key is that the host's role takes away the apparent randomness.

If you initially picked a goat (0.67 chance) he is forced to open the door behind which is the second goat.

If you initially picked the car (0.33 chance) he has a free choice of door to open.

The easy way to look at it is that you have one chance in three that your first choice was the car, and it remains one chance in three after the host opens a door, therefore you should switch which gives you two chances in three

[The Subhuman]

The key is that the host's role takes away the apparent randomness.

If you initially picked a goat (0.67 chance) he is forced to open the door behind which is the second goat.

If you initially picked the car (0.33 chance) he has a free choice of door to open.

The easy way to look at it is that you have one chance in three that your first choice was the car, and it remains one chance in three after the host opens a door, therefore you should switch which gives you two chances in three

but have you tested it?

[YorkshireSquare]

but have you tested it?

[MeeTa]

As the video shields posted shows, its quite simple. How this has kept Waccoe arguing for a while i dont understand..