Intuition or reasoning?

Question

Imagine you are on a game show and the idea is to win a car. There are 3 doors and behind 2 of them, there is a goat and behind the last 1- the car.
You choose a door. The host (Who knows what is behind the doors) opens a different door with a goat behind.
Now there are 2 doors left and you have 2 options:

1. Stick with your door

2. Swap doors

Should you stick, swap or does it not matter? (10 points for the correct answer and best reason) have fun! :-)

Answer 1

There is a good explanation of this problem in "The Curious Incident of the Dog in the Night". Basically as summarised by Lacy B.

This isn't a proof but another way of thinking about it is if you had the same problem but with 100 doors. You make your choice then the host opens 98 doors to show you 98 goats... Your dliemma is now to stick with your original choice or to switch to the other remaining closed door. You would be very daft not to switch now.

Answer 2

You should swap and here's why:

The probability you initially pick a goat is 2/3, whereas the probability you immediately pick the car is 1/3.

Now the host will always show you a goat!

If you stick with your initial choice your chance of getting the car is 1/3 * 1/2 = 1/6

If you swap, your chance of getting the car is:
2/3 * 1/2 = 2/6

Hence twice the chance of winning when you swap!

Answer 3

there is a blindingly obvious error in Lacy B's and others argument where they claim that by switching you double the chances

Lacy B says:

player picks door 1; host shows goat behind 2; wins if switch
player picks door 2; host shows goat behind 1; wins if switch
player picks door 3; host shows either goat; wins if stays
.
The error is in the last line. The "host shows either goat" actually means that there are 2 possible outcomes behind this third outcome. In other words the correct table should have been:

player picks door 1; host shows goat behind 2; wins if switch
player picks door 2; host shows goat behind 1; wins if switch
player picks door 3; host shows goat behind 1; wins if stays
player picks door 3; host shows goat behind 2; wins if stays

It now should be obvious that the chances have always been 50/50 and the choice really does not matter. All this is in line with what alrivera_1 says and I believe she is correct.

Answer 4

you swap doors.

we did this problem in math club it was a real intuition shocker.

basically the first time you before you know what's behind one of the doors, there's a 1 in 3 chance of getting it right

everyone always thinks that the odds will increase to just 1 in 2 but that's incorrect. you actually double your chances (2 in 3) of winning just by switching. here's why:
we can set up a table that shows the results of the pickin'. lets assume that the car is behind door 3.

player picks door 1; host shows goat behind 2; wins if switch
player picks door 2; host shows goat behind 1; wins if switch
player picks door 3; host shows either goat; wins if stays

there's now a 2 in 3 chance that the player will win if he/she switches

Answer 5

Suppose the doors are marked A, B and C.

I thought that car is behind B.

The host opens door A or C and reveals a goat.

Thus it is clear that the car is behind the door B as the host knew where the car was and if he opened the door B then I would have got the car.

So I would stick to my choice

If my choice was initially incorrect then the host must have opened the door B.

Answer 6

Ok, let's model the problem:

Three doors, A, B, C

Prizes: a car, goat, goat

A priori probability for the goat = 2/3
A priori probability for the car = 1/3

You choose a door...and he opens a door with a goat.

Now, the problem has transformed...raising the probability for the car to 0.50 (2 doors, one has the car)
And the probability of the goat has decreased to 0.50 (from 2/3)

There is not additional information concerning the two doors...the fact that he showed you a goat does not add additional information to assist you in your selection.

P(Car/Given Goat) = 0.50
P(Goat/Give Goat) = 0.50

It's your call, keeping the door or swapping doors, the probability is the same.

Answer 7

take a risk, because u get all or nothing thru sheer goodluck or a blunder.
If u use intution u can get something and loose nothing ( for example taking a Gre where u can fail, but u still have many more chances to take the exam again).
On the other hand if u use ur probability theories, you could result in.
1. winning the car,
2. being a loser
3. waste others time
4. be percieved as others by a big fool for wasting time energy and resources.

Answer 8

stick with the door you originally chose - unless of course it is the one the host has now opened - if that is the case - chose another and just go for it - the chance of you winnning is of course 50/50 (unless of course you are on the gameshow 321 where no matter what one you chose even if it was obvious - you would still end up with the muppet and i dont mean the host i mean the animated dustbin)

Answer 9

It doesnt matter whether you stick or swap as the chance of winning is the same in each case (50%).

I would probably stick as in such situations I stick with my 'gut feeling'.

Answer 10

hi
the probability of u r winning increases to 2/3 if u swap doors.this is a well known concept in this area of research.