Consider the following 9 door version of the Monty Hall problem. There are 9 doo

Question

Consider the following 9 door version of the Monty Hall problem. There are 9 doors, behind one of which there is a car (which you want), and behind the rest of which there are goats (which you don't want). Initially, all possibilities are equally likely for where the car is. You choose a door. Monty Hall then opens 4 goat doors, and offers you the option of switching to any of the remaining 4 doors. Assume that Monty Hall knows which door has the car, will always open 4 goat doors and offer the option of switching, and that Monty chooses with equal probabilities from all his choices of which goat doors to open. Should you switch? What is your probability of success if you switch to one of the remaining 4 doors?

Explanation / Answer

as the door you choose initially has probabilty of getting the car =1/9

and probabilty that it is in rest of 8 doors =8/9

once he opened 4 doors, where car is not there, remaining 4 doors hold total probabilty of 8/9.

hence probabilty that one of them hold the car=(8/9)/4 =2/9

hence once you switched your probabilty of winning will get doubled to 2/9

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