Variation of Monty Hall Problem

Question

You're in a game show. There are 100 doors. A Car is there behind one of those doors.

1. You pick a door. The host opens 98 other doors where the car is absent. What is the probability that the car is behind the door you picked and what is the probability that the car is behind the door the remaining door that is not yet opened(i.e., the door which you did not choose and the host also didn't open it).

2. You pick a door. The host opens 98 other doors where the car is absent. But the Host never reveals to you whether the car is present in any of the door he opened. So, you are thinking that car might be behind any of those 100 doors. What is the probability that the car is behind the door you picked and what is the probability that the car is behind the door the remaining door that is not yet opened(i.e., the door which you did not choose and the host also didn't open it).

3. You pick door number 29. The host opens 98 other doors where the car is absent. 92nd door is left unopened. What is the probability that the car is behind the door 29 and what is the probability that the car is behind the door 92?

4. You pick door number 29. The host opens 98 other doors where the car is absent. 92nd door is left unopened. But the Host never reveals to you whether the car is present in any of the door he opened. So, you are thinking that car might be behind any of those 100 doors. What is the probability that the car is behind the door 29 and what is the probability that the car is behind the door 92?

Comments

Balaji in 2nd and 4th question you mean that even after opening doors i don't know that those 98 doors are not having any car so i can choose any of the 100 doors?

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The contestant doesn't know whether the car is there behind the door which is opened or not. But host opens only those doors where the car is absent. So contestant assumes that the chance of a car behind the door which he has initially chosen is 1/100.

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probability of car behind the door you picked initially = $\frac{1}{100}$

probability of car behind the door which is closed which you didn't picked initially = $\frac{99}{100}$

for all 4 points

Answer 1

1/100 in any door if all doors are unopened.

But if 98 doors are opened than probability in your guessed door is 1/100 and for the other unopened door is 99/100