Consider 3 doors A,B,C and each of them can have a car worth Rs 3 lac with 0.5 probability. Now, you chose door A and the show host opened one of the door which contained a car. Now, he has given you 3 offers

Offer 1: You can stick to door A

Offer 2: You can change your door

Offer 3: Take 1,25,000 and leave the game

Which offer would you chose? What is the minimum amount the show host must offer you so that you can take his offer and leave the show?

Assumption: Show host knows everything behind the doors

Offer 1: You can stick to door A

Offer 2: You can change your door

Offer 3: Take 1,25,000 and leave the game

Which offer would you chose? What is the minimum amount the show host must offer you so that you can take his offer and leave the show?

Assumption: Show host knows everything behind the doors

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ReplyDeleteClarification: This 0.5 probability was the probability that you attributed for a car behind each of the door.

ReplyDelete1st method: Sample Space One

Total Sample space ={111,110,101,100,011,010,001,000}

1 meaning car behind the door and 0 otherwise

101 would mean car behind 1st and 3rd only

Clearly the show host has opened a car from the 2nd or the 3rd door.

So, 100 and 000 won't be possible.

Of the remaining, Probability you will get a car if you stick to door A = 3/6 =0.5

Probability you get a car if you change = 2/6 = 0.3333

Why 2/6? because only 111 and 011 would be favorable cases

Thus, always better to stick to A or leave with more than 1,50,000

2nd method: Bayes theorem

ReplyDeleteLet X be the door which host chooses and there is a car.

Let Y be the other door

P(host chooses X) = P(host chooses X/car in Y)*P(car in Y) + P(host chooses X/no car in Y)*P(no car in Y)

= 0.5*0.5 + 1*0.5 = 0.75

Since if both X and Y contains car, he could have chosen anyone

P(car is in door A/ host chooses X)

= P(host chooses X/ car is in door A)*P(car is in door A)/P(host chooses X)

Now, P(host chooses X) is independent of car in door A since host can only chose between X and Y.

P(car is in door A/ host chooses X) = P(car is in door A) = 0.5

Now,

P(car in door Y/host chooses X) = P(host chooses X/car in door Y)*P(car in door Y)/P(host chooses X)

= 0.5(car in both X,Y. Host could choose anyone)* 0.5/ 0.75

= 0.3333

Thus, always better to choose door A