tregor: i was talking bt the first problem, hari..

Posted by: tregor October 7, 2010

i was talking bt the first problem, hariyo...why did you have to use ith children, i mean can you elaborate a lil on that???

Let's say the car is behind the door Ci and the host reveals the door Rj.

For the player, probabilities p(c1), p(c2) and p(c3) are 1/3 each.

Then, considering the car is in door i (Ci), the host does not have the choice to open the door i (Ri) anymore.

For the host, therefore, p (R1\C1)=0

p (R2\C1)=1/2

p(R3\C1) =1/2

and so on for c2 and c3

Now, If the player initially has chosen door 1, and the third door is revealed (it doesn't matter what assumption you make),

Probability that the car is in door 1 using conditional probability,

p(C1\R3) = P(R3\C1)*P(C1)/P(R3)

=(1/2)* (1/3 )/ ( P(C1R3)+P(C2R3)+P(C3R3) ) {the denominator is equal to P(3)}

=(1/6)/(1/6+1/3+0)

=1/3

Since the third door was revealed, P(c3)=0

that makes P(C2\R3) =1-1/3 =2/3

Sure enough 2/3 is better than 1/3, so you should always switch.

Last edited: 07-Oct-10 12:53 AM