Would you switch doors?

Would you switch doors?

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en.wikipedia.org/wiki/Monty_Hall_problem
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well?

yes because statistics

you have to if you chose 1 or 2 not now
arab would be set good

1/3 x 1/2
or 1/3 x (unspoken)1/2

Either way im picking 1 because door 2 is the goats ass

Switching gives you a 2/3 chance of winning, not 1/2

en.wikipedia.org/wiki/Monty_Hall_problem

Yes

The answer is Yes.

YES

That's all there is to it.

A 2/3 increase from an unspoken number is still an unspoken number. This isn't a hard concept.

Changing doors does nothing.

No, because I actually want to win the goat.

Can't tell if retarded or not

/thread

Always switch. Initially there was a 2/3 chance of the price being behind one of the doors you didn't pick. This remains the same even if one door is shown to have a goat behind it.

Oh, yes do elaborate.

No.
This user is right.

Explanation: OP didn't actually describe the problem accurately.
>Pick a door
>Oh, BTW, there's an ass behind door 3.

The whole point of Monty Hall is that you're shown what's behind one of the two doors you didn't pick.

>Pick a door
>I pick Door 1
>Oh, btw there's a goat behind door 3, wanna change?
>Obs Yes b/c statistics, door 2.

nah goats r radical duderino

I'd skull fuck that goat

Come on guys this isn't hard

whoever came up with this riddle is a real dick

1-1/3=2/3, 2/3/2=1/3
1-1/2=1/2.
Your odds go up, like 16.6% switching.
Just do it, faggot!

You didn't specify a context so who knows? If it's the typical Monty Hall problem then yes, of course.

here faggots: prior to monty opening a door with a goat, there are two goats and a car and you don't know where they are. you are initially more likely to have picked a goat than the car because its 2/3 and the car is 1/3. thus, when monty opens a door to reveal a goat, it is in your favor to switch because your chances of having initially chosen a car are less than the goat. prior to monty's information, it is a 2/3 chance you picked a goat simply because there are more goats than cars. if you initially picked the car, which is a 1/3 rather than 2/3 chance, you lose if you switch. three doors, but only two options. it isn't goat, car, house. it is more goats than cars.

Context : doors contain
* object of all lust and desire
* your mom
* a goat

/ easy choice for the oedipal crowd

Well the text on the top talks about switching doors so I figured it was obvious

Monty hall problem has a few 'assumptions' that that are identified before even playing:

1) There are 2 goats and 1 car
2) You must choose a door
3) The host will then ask if you would like to switch your choice, but only after showing one of the two goats 100% of the time before being asked to swap.

With these set of assumptions the following logic can be proposed.

The actual odds of you choosing a car on the first pick are 1/3.

This means that you have an initial 2/3 chance of picking a goat on your first pick.

So let's play this out:
1) You choose a door
2) The host shows you a goat behind 1 of 2 remaining doors. (This doesn't change your initial 2/3 chance of picking a goat)
3) You are then asked to switch. Which you should take if you're not retarded.

On average you have a 66% (2/3) chance of winning a car if you swap because you're more likely to initially pick a goat over the prize. This can only work because under the rules it states he always shows 1 of the 2 goats 100% of the time.