There are 3 doors, and you have a car behind one of the doors, and nothing behind the other 2.You choose a door(say door 1), and a person who is aware of what is behind all three doors, opens one of doors 2 or 3 showing an empty door.

Now you are given a choice of switching, from door 1 to the other unopened door. Would you choose to switch, and why?

## 11 comments:

You switch.

Reason: After the omniscient person opens one of the other doors which surely doesn't have the car, there are 3 cases to consider.

One: You have picked the door with the car. One third chance of this happening. You get screwed when you swtich.

Two: You have picked the door without a car . Two thrids choice of this happening. You win if you switch.

So, you swtich cuz two thirds is better than one thirds.

Terrific, more details( confusing) here

monty hall

The probability of winning is more if you change the door.

P(WIN|change_door)

= P(WIN|wrong_door & change_door)*P(wrong_door) + P(WIN|right_door & change_door)*P(right_door)

= 1*2/3 + 0*1/3 = 2/3

Similarly,

P(WIN|no_change_door) = 1*1/3 + 0*2/3 = 1/3

Hence, more chances of winning if u change.

Intuitively, the prior probability of choosing the wrong door is greater, hence probability of win after change is more.

in the calculation .. why shud one consider the already eliminated door ?

i think the chances are equal 1/2 .

as per anastasia's calculations ..

it shud be 1*1/2 + 0*1/2 and 1*1/2 which makes it equal .

this is just my view .

correction ..

*athanasia

good one :) got it now :) the previous one was a mistake :) i was thinking on "probablity of winning " so had ditched one of the doors. its actually the probablity of switching to win :)

the devil also got it right :) ; I have to say that the first time that I heard this problem, I was of the opinion that sticking to your choice should be fine, as there are now only 2 sample points, and one of them contains the winner :(

Monty Hall problem here too!

I just mentioned it on my blog.

Nice coincidence!

Richie, "Curious Incident......" is awesome indeed.

I am in love with that whole feeling that the book emanates.

Been reading anything else?

Yup :).

Other books:

* No 1 ladies detective agency ( a detective in Botswana) - a very light and nice read, after I got through the initial pages.

* Disgrace - J.M.Coetzee

* Focault's pendulum - Umberto Eco, gave up on this, got too difficult, to follow the plot :)

got a question.

Am I interested in the car?

Let me assmume for the moment, albeit reluctantly that im interested in the car.

Anasthasia is right. with one wrong door eliminated, the probability is 1/2 on either side. so it doesnt matter.

even if the probablity is 2/3 vis a vis 1/3, it only means that outta 3 attempts, 2 r won by switiching and 1 is won by not switching. we r not sure which 3 attempts are being considered, which one will be won and not swithcing etc. there may be 3 random experiments all of which are won by not switching. that may be made up by the next 6 experiments. So, to make a call, over what range of random experiments the observed probability holds and the meta distribution of probability. By meta distribution, I refer to this. Are we sure that the probability of the probability of winning by not switching is 1/3? if not, what is the kind of distribution this meta probability follows - poisson or binomial or normal....?

Summary - When you are gambling, with ur knowledge of probability, you are not a lot better than an illiterate. So, trust in God.

Interesting, the theory being I guess that in probability you generate mathematical models, that explain ( or try to ), on the chance (supposedly or not) occurences of events. If the mathematical model works for you (use it, like casinos and insurance companies do).

This is like asking what is the proof of the Bernoulli's trials (which is what I inferred you meant by meta probability ) . There is none. You just assume that nature doesn't have memory for this experiment. People could argue quoting the Law of Averages, but you need to pick the model that works.

(proving that it works :), is a different beast altogether, I guess.)

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